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Chaptr 4 HW Packet—Answer Key

Multiple Choice—1 point; Calculations—three points unless otherwise specified—1 for formula, 1 for substitution, 1 for calculation.

1. An unknown substance dissolves readily in water but not in benzene (a nonpolar solvent). Molecules of what type are present in the substance?

a) nonpolar b) none of these c) neither polar nor nonpolar d) polar e) either polar or nonpolar

Remember that “it’s all about interparticle forces.” In order for a solute to be able to mix in a liquid, the particles of the liquid have to be able to exert forces of attraction on the particles of the solute, breaking the interparticle forces of the solute, and then, to keep the particles of solute from reconnecting, the particles of solvent have to form bonds with the solvent particles that are more stable than the bonds between the solvent particles themselves. If a solute can dissolve in water, which is polar, the solute must therefore either be polar or ionic, because these are the types of substances with which water can develop forces of attraction. Benzene, which is non-‐polar, will not be able to develop forces of attraction with polar or ionic substances, and therefore, cannot dissolve these types of substances. In this list of answers, the substance must therefore be polar. Answer D (1 point/1)

2. The interaction between solute particles and water molecules, which tends to cause a salt to fall apart in water, is called

a) dispersion b) coagulation c) hydration d) polarization e) conductivity

The action of polar water molecules being able to develop forces of attraction with particles of ionic solids, and then being able to surround individual ions with a number of water particles that effectively separates the solute particles and keeps them mixed with the water particles is called hydration. Answer C (1 point/2)

3. Consider two organic molecules, ethanol and benzene. One dissolves in water and the other does not. Why?

a) They have different molar masses. b) One is ionic, the other is not.

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c) One is an electrolyte, the other is not. d) Ethanol contains a polar O–H bond, and benzene does not. e) Two of these are correct.

Again, in order for a solute to be able to mix in a liquid, the particles of the liquid have to be able to exert forces of attraction on the particles of the solute, breaking the interparticle forces of the solute, and then, to keep the particles of solute from reconnecting, the particles of solvent have to form bonds with the solvent particles that are more stable than the bonds between the solvent particles themselves. If a solute can dissolve in water, which is polar, the solute must therefore either be polar or ionic, because these are the types of substances with which water can develop forces of attraction. Because of its –OH group, ethanol is polar and can therefore dissolve in water. Benzene, which is entirely hydrocarbon, is non-‐polar and will not be able to dissolve in water. (a) Although the substances have different molar masses, and this could potentially play a role in solubility, in this case it does not. (b) Both substances are covalent so neither substance is ionic. (c) Because neither substance can dissociate into ions neither substance is an electrolyte. (d) Ethanol does contain a polar –OH bond while benzene does not and this is why, as explained above, ethanol can dissolve while benzene cannot. (e) only one answer suffices. Answer D (1 point/3)

4. T F Polar molecules have an unequal distribution of charge within the molecule. The reason that some molecules are polar is because there is an unequal distribution of charge within the molecule. Remember that this is due to differences in electronegativity between the atoms forming covalent bonds. Answer True (1 point/4)

5. Which of the following is a strong acid?

a) HF b) HClO c) HBrO d) KOH e) HClO4 You should have memorized the six “strong” acids—those that can completely dissociate in water—memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Therefore, (a) HF (hydrofluoric acid) ), as stated above, is not one of our six strong acids and so is a weak acid. (b) HClO (hypochlorous acid) is not HClO4—as stated above, it is not one of our six strong acids and so is a weak acid.. (c) HBrO (hypobromous acid) is not HBr— as stated above, it is not one of our six strong acids and so is a weak acid.. (d) KOH



(potassium hydroxide) is a strong base. (e) HClO4 is one of our strong acids. Answer E (1 point/5) 6. All of the following are weak acids except

a) HCNO b) HBr c) HF d) HNO2 e) HCN You should have memorized the six “strong” acids—those that can completely dissociate in water—memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Therefore, (a) The CNO-‐ ion is called the fulminate ion. It is highly unstable and explosive. Its corresponding acid is called fulminic acid—this follows our convention of “ate” polyatomic ions adding the “ic” ending to the name of the acid. In any case, it is a weak acid. (b) HBr (hydrobromic acid) is one of our strong acids, so for this problem, it is the exception and the correct answer. (c) HF (hydrofluoric acid), as stated above, is not one of our six strong acids and so is a weak acid. (d) HNO2 (nitrous acid), as stated above, is not one of our strong acids and so is a weak acid. (e) HCN (hydrocyanic acid), as stated above, is not one of our strong acids and so is a weak acid. Answer B (1 point/6)

7. Which of the following is not a strong base?

a) Ca(OH)2 b) KOH c) LiOH d) Sr(OH)2 e) NH3 Strong bases will either (1) directly give up OH-‐ to a solution when they dissociate, or (2) consist of a soluble metal oxide, which, when it dissociates, will form 2 OH-‐ ions as the dissociated oxygen will pull a hydrogen ion away from a water molecule (thus creating two hydroxide ions). Weak bases are those substances that can remove a hydrogen ion from a water molecule leaving a single OH-‐ behind, but they can only accomplish this weakly. (a) Ca(OH)2 (calcium hydroxide) will give up OH-‐ ions directly to solution, and so is a strong base. (b) KOH (potassium hydroxide) will give up OH-‐ ions directly to solution, and so is a strong base. (c) LiOH (lithium hydroxide) will give up OH-‐ ions directly to solution, and so is a strong base. (d) Sr(OH)2 (strontium hydroxide) will give up OH-‐ ions directly to solution, and so is a strong base. (e) NH3 can only weakly remove a hydrogen ion from a water molecule, and so is a weak acid. This is the exception in the list. Answer E (1 point/7) 8. Which of the following is paired incorrectly?

a) H2SO4 – strong acid b) HNO3 – weak acid



c) Ba(OH)2 – strong base d) HCl – strong acid e) NH3 – weak base You should have memorized the six “strong” acids—those that can completely dissociate in water—memorizing these six means that anything else that is not one of these is a weak acid. They are HCl, HBr, HI, H2SO4, HNO3, HClO4. Strong bases will either (1) directly give up OH-‐ to a solution when they dissociate, or (2) consist of a soluble metal oxide, which, when it dissociates, will form 2 OH-‐ ions as the dissociated oxygen will pull a hydrogen ion away from a water molecule (thus creating two hydroxide ions). Weak bases are those substances that can remove a hydrogen ion from a water molecule leaving a single OH-‐ behind, but they can only accomplish this weakly. Therefore, (a) H2SO4 is a strong acid and so is paired correctly. (b) HNO3 is a strong acid and so is not paired correctly. (c) Ba(OH)2 will give up OH-‐ ions directly to solution, and so is a strong base, and so, is paired correctly. (d) HCl is a strong acid and so is paired correctly. (e) NH3 can only weakly remove a hydrogen ion from a water molecule, and so is a weak acid. Answer B (1 point/8)

9. The man who discovered the essential nature of acids through solution conductivity studies is

a) Priestly b) Boyle c) Einstein d) Mendeleev e) Arrhenius Arrhenius discovered that the essential nature of acids was in the dissociation of the acid molecule into a hydrogen ion (which we now know is “donated” to a water molecule to form the hydronium ion) and the anion corresponding to whatever the specific acid is. This dissociation was discovered in the larger context of discovering that when ionic substances in general dissociate completely, they easily conduct electricity. So, acids become one type of a larger class of compounds called electrolytes. Acids (or ionic substances) that completely dissociate in water are strong electrolytes. Acids (or ionic substances) that only incompletely dissociate in water are called weak electrolytes. Answer E (1 point/9)

10. A solid acid HX is mixed with water. Two possible solutions can be obtained. Which of the following is true?

I. II.



a) In case I, HX is acting like a weak acid, and in case II, HX is acting like a strong acid. b) In case I, HX is acting like a strong acid, and in case II, HX is acting like a weak acid. c) In both cases, HX is acting like a strong acid. d) In both cases, HX is acting like a weak acid. e) HX is not soluble in water. These diagrams test your true understanding of the concept of “complete” vs “partial” dissociation. In the first diagram, all of the HX particles have completely dissociated to form only individual ions, H+(which we know in fact are not individual protons because they will immediately form a covalent bond with a water molecule to form a hydronium ion) and X-‐. This is what a strong acid does—completely dissociates. A weak acid only partially dissociates, and so, while there will be some H+ ions (which again will immediately bond with a water molecule to form a hydronium ion) and an equal amount of X-‐ ions, some of the HX particles will remain intact. So, diagram I depicts a strong acid while diagram II depicts a weak acid. Answer B (1 point/10) 11. T F An acid is a substance that produces OH– ions in water. An acid is a substance that produces H3O+ ions in water. Answer False (1 point/11)



12. A 17.0-‐g sample of HF is dissolved in water to give 2.0 × 102 mL of solution. The concentration of the solution is: (See “Stoichiometry Skills” above) You are asked to find the concentration so you immediately write down. M = mol

L Upon inspection you realize that you do not have moles—because you know mol = g

mm you immediately then substitute g

mm for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. Plug in the given values and solve. 2 SF. Answer 4.2 M (3 points/14)

M = mol

L =g

mm

L=

17.0g(1.01+19.00)g /mol

.20 L= 4.2M

13. 1.00 mL of a 3.95 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. What is the concentration of solution B? (See “Dilution Skills” above) In this problem, there is a sequence of dilutions and you should take them one at a time, the answer to the first becoming the given information for the next. In the first dilution you are given V1 (1.00 mL=.001 L) and M1 (3.95 x 10-‐4M) and told that it is diluted with 9.00 mL of petroleum ether to form solution A. Be careful—9.00 mL IS NOT V2. V2 is the total final volume of the solution, and this is 1.00 mL + 9.00 mL which is 10.0 mL (.0100L). Therefore, you can plug these three values into the dilution relationship and find the molarity of solution A.

A : M1V1 = M2V2 → M2 =

M1V1

V2

=(3.95x10−4 mol

L )(0.001L).0100L

= 3.95x10−5 M



This molarity then becomes M1 of the next dilution. You are given V1 of this next dilution—2.00 mL (.002 L). You are also told that this is mixed with 8.00 mL (.008 L) of petroleum ether to give solution B. Be careful—again, 8.00 L IS NOT V2. V2 is the total final volume of the solution, and this is 2.00 mL + 8.00 mL which is 10.0 mL (.0100L). Therefore, again, you can plug these three values into the dilution relationship and find the molarity of solution B. (6 points/20)

B : M1V1 = M2V2 → M2 =

M1V1

V2

=(3.95x10−5 mol

L )(0.002L).0100L

= 7.90x10−6 M

14. 1.00 mL of a 3.60 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid are 5.00 mL of solution B? (Molar mass for oleic acid = 282 g/mol) (Note: this problem is different from the previous problem in two ways. First, you are not asked to find the concentration of the solution B, you are asked to find the number of grams of oleic acid in a certain amount of solution B. But, to do this, you still have to go through the complete process of finding the concentration of solution B before you can calculate the number of grams. Second, recognize that the molarity of the given solution is slightly different than what it was in question 13. You cannot just use the value of the concentration of solution B in this question—again, you have to go through the whole process of calculating it again with this different given concentration.) (See “Dilution Skills” and “Stoichiometry Skills” above) In this problem, there is a sequence of dilutions and you should take them one at a time, the answer to the first becoming the given information for the next. In the first dilution you are given V1 (1.00 mL=.001 L) and M1 (3.60 x 10-‐4M) and told that it is diluted with 9.00 mL of petroleum ether to form solution A. Be careful—9.00 mL IS NOT V2. V2 is the total final volume of the solution, and this is 1.00 mL + 9.00 mL which is 10.0 mL (.0100L). Therefore, you can plug these three values into the dilution relationship and find the molarity of solution A.

A : M1V1 = M2V2 → M2 =

M1V1

V2

=(3.60x10−4 mol

L )(0.001L).0100L

= 3.60x10−5 M

This molarity then becomes M1 of the next dilution. You are given V1 of this next dilution—2.00 mL (.002 L). You are also told that this is mixed with 8.00 mL (.008 L) of petroleum ether to give solution B. Be careful—again, 8.00 L IS NOT V2. V2 is the total final volume of the solution, and this is 2.00 mL + 8.00 mL which is 10.0 mL (.0100L). Therefore, again, you can plug these three values into the dilution relationship and find the molarity of solution B.

B : M1V1 = M2V2 → M2 =

M1V1

V2

=(3.60x10−5 mol

L )(0.002L).0100L

= 7.20x10−6 M



Finally, you are asked for grams of oleic acid, so immediately write down g = (mol)(mm) . You are given the molar mass so you do not have to calculate this or include it within the rest of your calculation—you simply use it. However, upon inspecting this formula you do not have mol, but because you know mol = g

mm you immediately then substitute

gmm for moles. The calculation should be simple enough

that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been calculated molarity of solution B and you have been given a volume (5.00 mL = .00500 L) so plug these values in and solve. (9 points/29)

g = (mol)(mm) = ( M )(L)(mm) = (7.20x10−6 mol

L )(.00500 L)(282 g / mol) = 1.02x10−5 g

15. How many grams of NaCl are contained in 350. mL of a 0.334 M solution of sodium chloride? (See “Stoichiometry Skills” above) You are asked for grams of NaCl, so immediately write down g = (mol)(mm) . As you inspect this equation you know you can find the mm of NaCl easily—just include its calculation in your final calculation. You are not given the number of moles, but because you know mol = ( M )(L) you immediately then substitute ( M )(L) for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given the molarity (.334 M) and volume (350 mL=.350L). Plug in the values and solve. (3 points/32)

g = (mol)(mm) = ( M )(L)(mm) = (.334 mol

L )(.350 L)((22.99+ 35.45) g / mol) = 6.83g

16. Which of the following aqueous solutions contains the greatest number of ions?

a) 400.0 mL of 0.10 M NaCl b) 300.0 mL of 0.10 M CaCl2 c) 200.0 mL of 0.10 M FeCl3 d) 200.0 mL of 0.10 M KBr e) 800.0 mL of 0.10 M sucrose Remember that for every one particle of an ionic substance that dissociates completely, you will obtain a number of particles that is equal to the number of individual particles present within the original particle. If 1 particle of NaCl dissociates, you will obtain 2 particles, one Na+ and one Cl-‐. If 1 particle of CaCl2 dissociates, you will obtain 3 particles, one particle of Ca2+ and two particles of Cl-‐. If 1 particle of FeCl3 dissociates, you will obtain 4 particles, one particle of Fe3+ and three particles of Cl-‐. If 1 particle of KBr dissociates, you will obtain 2 particles, one K+ and one Br-‐. Sucrose is not an ionic compound so this answer can be eliminated. Therefore,



If you have a certain number of moles of NaCl particles, when it all dissociates you will have twice the number of moles of ions in solution. If you have a certain number of moles of CaCl2 particles, when it all dissociates you will have three times the number of moles of ions in solution. If you have a certain number of moles of FeCl3 particles, when it all dissociates you will have four the number of moles of ions in solution. If you have a certain number of moles of KBr particles, when it all dissociates you will have twice the number of moles of ions in solution.

You know a formula to find the number of moles of particles: mol = ( M )(L) . You have been given the molarity and volume of each substance—simply find the number of moles for each and multiply times the factor you have determined for each substance:

ions for NaCl = (2)(mol NaCl) = (2)( M )(L) = (2)(.10 molL )(.400L) = .08 mol ions

ions for CaCl2 = (3)(mol CaCl2 ) = (3)( M )(L) = (3)(.10 molL )(.300L) = .09 mol ions

ions for FeCl3 = (4)(mol NaCl) = (4)( M )(L) = (4)(.10 molL )(.200L) = .08 mol ions

ions for NaCl = (2)(mol NaCl) = (2)( M )(L) = (2)(.10 molL )(.200L) = .04 mol ions

Answer B (1 point/33)

17. What mass of calcium chloride, CaCl2, is needed to prepare 3.650 L of a 1.75 M solution? (See “Stoichiometry Skills” above) You are asked for grams of NaCl, so immediately write down g = (mol)(mm) . As you inspect this equation you know you can find the mm of CaCl2 easily—just include its calculation in your final calculation. . You are not given the number of moles, but because you know mol = ( M )(L) you immediately then substitute ( M )(L) for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given the molarity (1.75 M) and volume (3.650 L). Plug in the values and solve. (3 points/36)

g = (mol)(mm) = ( M )(L)(mm) = (1.75 mol

L )(3.650 L)((40.08 + (2)(35.45))g / mol) = 709 g

18. A 30.1-‐g sample of SrCl2 is dissolved in 112.5 mL of solution. Calculate the molarity of this solution. (See “Stoichiometry Skills” above) You are asked to find molarity. Immediately write

down M = mol

L. You are given volume (112.5 mL=.1125 L), but not moles. but

because you know mol = gmm you immediately then substitute

gmm for moles. The

calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have grams and molar mass can be calculated in the problem. Plug in the values and solve:



M = mol

L=

gmm

L=

30.1g((87.62+(2)(35.45))g /mol )

.1125 L= 1.69 M

(3 points/39)

19. What mass of solute is contained in 256 mL of a 0.820 M ammonium chloride solution? (See “Stoichiometry Skills” above) You are asked to find grams. Immediately write down g = (mol)(mm) . You realize that you haven’t been given moles, but because you know mol = ( M )(L) you immediately then substitute ( M )(L) for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given liters (.256 L) and molarity (.820 M) so, plug in values and solve. (3 points/42)

g = (mol)(mm) = ( M )(L)(mm) = (.820 mol

L )(.256L)((14.01+ 4.04+ 35.45)g / mol) = 11.2g

20. A 57.17-‐g sample of Ba(OH)2 is dissolved in enough water to make 1.800 liters of solution. How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)2?

(See “Dilution Skills” and “Stoichiometry Skills” above) As this is a dilution problem, immediately write down the dilution relationship. You are asked to find V1 (“How many mL of solution must be diluted”), so isolate this value. Upon inspecting this formula you realize that you have been given M2 (.100 M) and V2 (1.000 L), but have not been given M1—but you have been given the information to calculate it. You

know that M1 =

molLso write this down separately. Inspecting this formula you do

not have mol but know that mol = g

mmso substitute this in for moles.

You have been given grams and can easily find mm as part of the calculation. Once you find the value for M1 plug this value into the original dilution relationship and solve for V1. Remember to turn L into mL. (6 points/48)

M1V1 = M2V2 → V1 =M2V2

M1

= (.100M )(1.000L).1854M

= .539L = 539 mL

M1 =mol

L=

gmm

L=

57.17 g(137.3+32.00+2.02)

1.800L= .1854M

21. What volume of 18 M sulfuric acid must be used to prepare 1.80 L of 0.215 M H2SO4? (See “Dilution Skills” above) As this is a dilution problem, immediately write down the dilution relationship. You have been asked to find “what volume . . . must be used,” which indicates that you need to find V1, so isolate this variable. Inspecting the



formula, find that you have been given M1 (18 M), V2 (1.80 L), and M2 (.215 M). Simply plug the values in and solve. (3 points/51)

M1V1 = M2V2 → V1 =

M2V2

M1

= (.215M )(1.80L)18M

= .022 L = 22mL

22. How many grams of NaOH are contained in 5.0 × 102 mL of a 0.77 M sodium hydroxide solution? (See “Stoichiometry Skills” above) You are asked for number of grams so you should immediately write out g = (mol)(mm) and let this lead you through the rest of the problem. On inspecting this formula you know that you can easily find mm and show this within the final calculation. You realize that you haven’t been given moles, but because you know mol = ( M )(L) you immediately then substitute ( M )(L) for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations. You have been given M (.77M) and volume (5.0 x 102 mL=.50 L). Plug these values in and solve. (3 points/54)

g = (mol)(mm) = ( M )(L)(mm) = (.77 M )(.50L)(22.99+16.00+1.01) = 15g

23. An analytical procedure requires a solution of chloride ions. How many grams of CaCl2 must be dissolved to make 2.15 L of 0.0520 M Cl–? (See “Stoichiometry Skills” above) You are asked for number of grams so you should immediately write out g = (mol)(mm) and let this lead you through the rest of the problem. On inspecting this formula you know that you can easily find mm and show this within the final calculation. . You realize that you haven’t been given moles, but because you know mol = ( M )(L) you immediately then substitute ( M )(L) for moles. The calculation should be simple enough that you can put all of this into one line of calculation rather than separating into multiple calculations.

Note—you have one additional complicating factor at this point. You have been asked to find the number of g of CaCl2 that must be dissolved to give you .0520 M chloride ion (Cl-‐), NOT .0520 CaCl2 M solution. Every formula unit of CaCl2, when it dissociates, will give 2 particles of Cl-‐. Therefore, you would need only ½ the number of moles of CaCl2 to give .0520 M Cl-‐. Add this into your factor for moles in the equation.

You have been given M (.0520 M) and volume (2.15 L). Plug these values in and solve. (3 points/57)

g = (mol)(mm) = (.5)( M )(L)(mm) = (.5)(.0520 M )(2.15 L)(40.08+ (2)(35.45)) = 6.20g 24. T F The concentration of a salt water solution that sits in an open beaker decreases over time. False—as water evaporates and the number of water particles relative to the number of salt particles decreases, the concentration will increase. Answer False (1



point/52) 25. You have two solutions of chemical A. To determine which has the highest concentration of A in molarity, what is the minimum number of the following you must know?

I. the mass in grams of A in each solution

II. the molar mass of A

III. the volume of water added to each solution

IV. the total volume of the solution

a) 0 b) 1 c) 2 d) 3 e) You must know all of them. Understand that while we are seeking which solution has the highest concentration, we don’t need to know the actual concentration.

Remember, concentration is moles per liter: M = mol

L. From this you can see that to

determine which solution had the greatest concentration you would need to know the number of moles and number of L. Note, for molarity it is total liters of solution, not liters of water added—so, IV is a definite, while III is not necessary.

Moles is not given as a choice or we would choose that. Normally, to find moles, you would also need to have the number of grams and the molar mass. However, as the molar mass would be the same for each solute (chemical A for both), just knowing the number of grams for each, given a certain volume, would be enough. So, I is a definite while II is not necessary. Therefore, 2 of the 4 values would need to be known. Answer C (1 point/58)

26. Diabetics often need injections of insulin to help maintain the proper blood glucose levels in their bodies. How many moles of insulin are needed to make up 45 mL of 0.0059 M insulin solution? (See “Stoichiometry Skills” above) You are asked to find moles. You have two

formulas that can help you do this: mol = g

mmor mol = ( M )(L)

You should be able to quickly decide that in this case you need the latter formula because you are given M and L. You immediately write this formula down, plug in the values and solve. (1 point/59)

mol = ( M )(L) = (.0059 mol

L )(.045L) = 2.7x10−4 mol



27. You have two solutions of sodium chloride. One is a 2.00 M solution, the other is a 4.00 M solution. You have much more of the 4.00 M solution and you add the solutions together. Which of the following could be the concentration of the final solution?

a) 2.50 M b) 3.00 M c) 3.70 M d) 6.00 M e) 7.50 M First, reason that if you have two solutions of different known concentrations and you mix them together, you would be combining the number of moles present in both solutions and combining the volumes of both solutions and the result could never be less than 2.00 or more than 4.00 mole per liter. This eliminates d and e. Then, if you were to mix equal amounts of solution together it would be easy to see that you should get a solution that is exactly halfway between—that is, 3.0 M. However, since you have much more of the 4.00 M solution rather than equal amounts, the only valid choice here would be 3.70 M. Answer C (1 point/60)

28. You have equal masses of different solutes dissolved in equal volumes of solution. Which of the solutes would make the solution having the highest molar concentration?

a) NaOH b) KCl c) KOH d) LiOH e) all the same

As mol = g

mm, saying that you have the same mass of different substances would

mean that the less the molar mass, the greater the number of moles and the greater the concentration. Therefore, the substance that has the lowest molar mass would have the greatest concentration. The chloride ion has a mm of 35.45 g/mol while the hydroxide ion has a molar mass of 16.00 + 1.01 = 17.01 g/mol. Because there is both a KOH and a KCl, the KOH would have the lesser molar mass between these two. The others are also both hydroxides, so the cation with the least molar mass will give the substance with the least molar mass—this would be Li. So, LiOH would give the least molar mass, and so, the greatest concentration. Answer D. (1 point/61) 29. Which of the following do you need to know to be able to calculate the molarity of a salt solution?

I. the mass of salt added

II. the molar mass of the salt

III . the volume of water added



IV. the total volume of the solution

a) I, III b) I, II, III c) II, III d) I, II, IV e) You need all of the information.

Remember, concentration is moles per liter: M = mol

L. From this you can see that to

determine molarity you would need to know the number of moles and number of L. Note, for molarity it is total liters of solution, not liters of water added—so, IV is a definite, while III is not necessary.

Moles is not given as a choice or we would choose that. To find moles, you know

mol = g

mm, so you can see that you also need grams (mass of the salt) and the molar

mass of the salt. Therefore, you need I, II, and IV. Answer D (1 point/62)

30. A 230.0-‐mL sample of a 0.275 M solution is left on a hot plate overnight; the following morning the solution is 1.50 M. What volume of solvent has evaporated from the 0.275 M solution? Although it may not initially look like it, this problem is basically a dilution problem—just in reverse. But is still works the same. (See “Dilution Skills” above)

To find the volume lost from the solution you will need to find the final volume and subtract this from the initial volume. (see below)

You are given the initial molarity (M1 = .275 M), the initial volume (230.0 mL = .2300 L) and the final concentration (M2 = 1.50 M). Write down the dilution relationship, isolate final volume (V2), plug the values in and solve. (4 points/66)

Vevap =Vinitial −Vfinal =V1 −V2 = .2300L− .0422L = .1878 L = 187.8 mL

M1V1 = M2V2 → V2 =M1V1

M2

= (.275M )(.230L)1.50M

= .0422 L

31. For the reaction 4FeCl2(aq) + 3O2(g) → 2Fe2O3(s) + 4Cl2(g), what volume of a 0.890 M solution of FeCl2 is required to react completely with 8.71 × 1021 molecules of O2? You have been asked to find volume (liters) of a solution of FeCl2 so you immediately

write down LFeCl2

=molFeCl2

M FeCl2

. (You do not have to use subscripts but it will make like

much easier later in the calculation because you will need to say that a certain number of moles of FeCl2 will be equivalent to a certain number of moles of O2, and the two values for moles can easily get confused if they are not labeled.) This formula



tells you that in order to find the volume (number of liters) of the .890M solution, you need to know the number of moles of FeCl2 that must be reacted with 8.71 × 1021 molecules of O2. According to the mole ratio of FeCl2 to O2 stated to us by the balanced chemical equation, for every 3 mol of O2, 4 mol of FeCl2 are required. Therefore, we can substitute

43( ) molO2( ) in for

molFeCl2

in our calculation (see below).

You have not been given the number of moles of oxygen, but you have been given the number of molecules, and you always know that you can turn a number of molecules into a number of moles by dividing by the number of particles in 1 mole, which is Avogadro’s number. Substitute

molecules O2AN in for molO2 (see below). Then, substitute in

the given values and AN and solve: (3 points/69)

LFeCl2=

molFeCl2

M FeCl2

=43( ) molO2( )

M FeCl2

=43( ) molecules O2

AN( )M FeCl2

=43( ) (8.71x1021 molecules)

(6.022x1023 molecules/mol )

.890molL

= .0216L = 21.6mL

32. Phosphoric acid, H3PO4, is a triprotic acid. What is the total number of moles of H+ available for reaction in 3.50 L of 0.400 M H3PO4? You are asked for mol H+ and you have been given L and M, and you know that for a triprotic acid like H3PO4, for every mol of acid, 3 mol of H+ will dissociate. So, you immediately write down mol H + = (3)( M )(L) . You already have all of the information so you plug in the values and solve. (3 points/72)

mol H + = (3)( M )(L) = (3)(.400 mol

L )(3.50L) = 4.20L

33. The following reactions:

Pb2+ + 2I– → PbI2

2Ce4+ + 2I– → I2 + 2Ce3+

HOAc + NH3 → NH4+ + OAc–

are examples of

a) acid-‐base reactions b) unbalanced reactions c) precipitation, acid-‐base, and redox reactions, respectively d) redox, acid-‐base, and precipitation reactions, respectively e) precipitation, redox, and acid-‐base reactions, respectively In the first reaction you can see that all is happening is that lead 2+ ions and iodine 1-‐ ions are combining to form the solid compound lead (II) iodide. The atoms are not changing oxidation states in the reaction, they are merely combining to form an ionic solid, which is the essential feature of a precipitation reaction (note—oxidation and reduction do not take place during precipitation reactions).



In the second reaction we see cerium being reduced from 4+ to 3+ and iodine being oxidized from 1-‐ to 0, so this is an oxidation reduction reaction. These are also called redox reactions. Beside seeing charges change from the reactant side of the equation to the product side, the quickest way to recognize a redox reaction is by seeing an elemental atom be changed into an atom in a compound (or vice versa) because you know that the oxidation state of an elemental substance is 0, while that of an atom in a compound is not 0. In the third reaction we see a molecule with an available hydrogen at the beginning of its formula (meaning that it is an acid—not one of the six strong acids, so you also know that it is a weak acid) and the weak base, NH3. The reaction shows the donation of the hydrogen by the weak acid to the weak base to form the conjugate acid, NH4+ and the conjugate base, OAc-‐. Therefore, the reactions are: precipitation, redox, acid-‐base, respectively. Answer E (1 point/73)

34. The following reactions

2K(s) + Br2(l) → 2KBr(s)

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)

are examples of

a) precipitation reactions b) redox, precipitation, and acid-‐base, respectively c) precipitation (two) and acid-‐base reactions, respectively d) redox reactions e) none of these In the first reaction we see that two elemental substances (oxidation state 0) are changed to atoms that are part of a compound (oxidation states “not zero”). You don’t even need to know what the non-‐zero oxidation states are—you know that they have changed from 0 to something else so it must involve oxidation and reduction. As two elements are combining to form a compound this would also be a synthesis reaction. In the second reaction you can see that this is a double replacement reaction, and that the solid compound silver chloride is a precipitate. This is therefore, a precipitation reaction. Note that as consistent with a precipitation reaction, no elements change oxidation states—a precipitation reaction does not involve oxidation and reduction. In the third reaction we can see that a strong acid (HCl) is interacting with a strong base (KOH) resulting in neutralization. This is an acid base reaction.



Therefore, the reactions are: redox, precipitation, and acid-‐base respectively. Answer B (1/74)

35. The following reactions

ZnBr2(aq) + 2AgNO3(aq) → Zn(NO3)2(aq) + 2AgBr(s)

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

are examples of

a) oxidation-‐reduction reactions b) acid-‐base reactions c) precipitation reactions d) A and C e) none of these In both circumstances you can see that both reactants are ionic compounds and that the cations “switch places” with each other with respect to their corresponding anions. They are therefore both double replacement reactions. As you can also see, both reactions result in the formation of a solid compound which precipitates out, which means that both are precipitation reactions—all double replacement reactions are precipitation reactions. Answer C (1/75)

36. All of the following reactions

2Al(s) + 3Br2(l) → 2AlBr3(s)

2Ag2O(s) → 4Ag(s) + O2(g)

CH4(l) + 2O2(g) → CO2(g) + 2H2O(g)

can be classified as

a) oxidation-‐reduction reactions b) combustion reactions c) precipitation reactions d) A and B e) A and C In the first reaction we see that two elemental substances (oxidation state 0) are changed to atoms that are part of a compound (oxidation states “not zero”). You don’t even need to know what the non-‐zero oxidation states are—you know that they have changed from 0 to something else so it must involve oxidation and reduction. As two elements are combining to form a compound this would also be a synthesis reaction. In the second reaction we see the opposite occurring—we see atoms that are part of an ionic compound (oxidation states “not zero”) being changed into elemental atoms (oxidation state 0). As with the first reaction, this involves oxidation and reduction.



However, in contrast to the first reaction, as a compound is being broken into its component elements this is a decomposition reaction. In the third reaction, we have a carbon compound being reacted with oxygen to form carbon dioxide and water. This is the hallmark of a combustion reaction. However, as the other two are not combustion reactions, we cannot say that all the reactions are combustion reactions. But, we also know that all combustion reactions are oxidation reduction reactions—especially as they all involve the change of elemental oxygen (oxidation state 0) to oxygen in a compound (oxidation state -‐2). Therefore, all of the reactions could be considered to be oxidation reduction reactions. (Note, you also could not say A and B because not all of the reactions are combustion). Answer A (1/76)

37. You have exposed electrodes of a light bulb in a solution of H2SO4 such that the light bulb is on. You add a dilute solution and the bulb grows dim. Which of the following could be in the solution?

a) Ba(OH)2 b) NaNO3 c) K2SO4 d) Cu(NO3)2 e) none of these Because sulfuric acid is a strong acid, the first hydrogen will dissociate completely. This means that sulfuric acid is a strong electrolyte and will allow electric current to flow strongly—the light bulb is on. If we add a solution and the bulb grows progressively dimmer, this must mean that something is “cancelling out” the ionic (charged) particles in the solution. The only solution capable of accomplishing this in an acid solution is a basic solution—one that provides OH-‐ ions. Therefore, Ba(OH)2 is the logical choice. Answer A (1/77)

38. Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct?

a) Both NaCl and CuS precipitate from solution. b) No reaction will occur. c) CuS will precipitate from solution. d) NaCl will precipitate from solution. e) A gas is released. The answer to this question is based on knowledge of solubility rules. We are told that the initial compounds are both soluble, but when mixed together, we might expect something to precipitate out. Except for group one (IA) metals, when combined with other metals sulfides are highly insoluble. Therefore, we would expect copper sulfide to precipitate out. Answer C (1/78) 39. Aqueous solutions of potassium sulfate and ammonium nitrate are mixed together. Which statement is correct?



a) NH4SO4 will precipitate from solution. b) KNO3 will precipitate from solution. c) No reaction will occur. d) Both KNO3 and NH4SO4 precipitate from solution. e) A gas is released. The answer to this question is based on knowledge of solubility rules. We are told that the initial compounds are both soluble, but when mixed together, we might expect something to precipitate out. However, ammonium, nitrate, and potassium compounds are almost always soluble, and sulfates are often soluble, so, according to the solubility rules, there is no combination of ions in the mix given that should precipitate out—no reaction should occur. Answer C (1 point/79) 40. Which of the following salts is insoluble in water?

a) Na2S b) K2CO3 c) Pb(NO3)2 d) CaCl2 e) All of these are soluble in water. (a) Most sulfides are insoluble—however, those of 1A metal cations are soluble—Na2S is soluble. (b) Most carbonates are insoluble—however, those of 1A metal cations are soluble—K2CO3 is soluble. (c) Most nitrates, including lead nitrate, are soluble. (d) Most chlorides, including those of 2A metal cations, are soluble. Answer E (1 point/80)

41. How many of the following salts are expected to be insoluble in water?

sodium sulfide barium nitrate

ammonium sulfate potassium phosphate

a) none b) 1 c) 2 d) 3 e) 4 Most sulfides are insoluble—however, those of 1A metal cations are soluble—Na2S is soluble. Most nitrates, including barium nitrate, are soluble. Most ammonium compound, including ammonium sulfate are soluble. Most phosphates are insoluble. However, those of 1A metal cations are soluble. None of these compounds should be insoluble. Answer A (1 point/81)

42. When NH3(aq) is added to Cu2+(aq), a precipitate initially forms. Its formula is:

a) Cu(NH)3 b) Cu(NO3)2



c) Cu(OH)2 d) Cu(NH3)22+ e) CuO

This is a tricky question. Recognize that NH3 is a weak base and its mechanism of action is that it pulls a hydrogen ion from H2O to form OH-‐ ions. This means that Cu2+ ions that are in solution now are exposed to OH-‐ ions. Most hydroxides of transition metal cations are insoluble, including Cu(OH)2. Therefore, this will precipitate out initially. (However, as the reaction proceeds, Cu2+ions will secondarily form complex ions with ammonia molecules—Cu(NH3)22+ ions—and these are soluble—so after a short time, the precipitate disappears as it is converted to this complex ion.) Answer C (1 point/82)

43. Which of the following ions is most likely to form an insoluble sulfate?

a) K+ b) Ca2+ c) S2– d) Li+ e) Cl– According to solubility rules— most sulfates are soluble, except for those of Ba, Pb, Hg and Ca—so, Ca2+ is most likely to form an insoluble sulfate. You should be able to eliminate the anions S2-‐ and Cl-‐ immediately because these are not going to form a sulfate. Answer B (1 point/83) 44. Which of the following compounds is soluble in water?

a) Ni(OH)2 b) K3PO4 c) BaSO4 d) CoCO3 e) PbCl2 According to solubility rules— most hydroxides are insoluble except for a few in column 1A and 2A, so (a) will precipitate; most phosphates are insoluble, except for those of the column 1A—so, (b) will actually dissolve; most sulfates are soluble, except for those of Ba, Pb, Hg and Ca so, (c) will precipitate; most carbonates are insoluble, except for those of the column 1A, so (d) will precipitate; most chlorides are soluble, except for Ag, Pb, and Hg, so (e) will precipitate. Answer B (1 point/84)

45. Which pair of ions would not be expected to form a precipitate when dilute solutions of each are mixed?

a) Al3+, S2– b) Pb2+, Cl– c) Ba2+, PO43– d) Pb2+, OH– e) Mg2+, SO42–



According to solubility rules—sulfide precipitates out almost everything so (a) will precipitate; most chlorides are soluble, except for Ag, Pb, and Hg, so (b) will precipitate; most phosphates are insoluble, except for those of the column 1A, so (c) will precipitate; most hydroxides are insoluble except for a few in column 1A and 2A, so (d) will precipitate; most sulfates are soluble, especially those of columns 1A and 2A, so, (e) will not precipitate. Answer E (1 point/85)

46. A solution contains the ions Ag+, Pb2+, and Ni2+. Dilute solutions of NaCl, Na2SO4, and Na2S are available to separate the positive ions from each other. In order to effect separation, the solutions should be added in which order?

a) Na2SO4, NaCl, Na2S b) Na2SO4, Na2S, NaCl c) Na2S, NaCl, Na2SO4 d) NaCl, Na2S, Na2SO4 e) NaCl, Na2SO4, Na2S

Most sulfate salts are soluble, except for those of Ba, Pb, Hg, and Ca. So, add NaSO4 first and only PbSO4 will precipitate out. Filter this out of the solution, leaving Ag and Ni ions in solution. Next, most chloride salts are soluble except for those of Ag, Pb and Hg. There is no more Pb, so add NaCl and AgCl will precipitate out. Filter this out of the solution leaving only Ni ions in solution. Then, add Na2S. This will precipitate almost anything, including the nickel. Answer A (1 point/86)

47. Consider an aqueous solution of calcium nitrate added to an aqueous solution of sodium phosphate. What is the formula of the solid formed in the reaction?

a) Ca(PO4)2 b) CaPO4 c) Ca3(PO4)2 d) Ca3(PO3)2 e) none of these Make sure you know how to write the correct formulas for all compounds involved. If necessary to help visualize write out the formulas for the compounds, or even the reactant side of a chemical equation so you can “see” the correct chemical formulas:

Ca(NO3)2 (aq) and Na3PO4 (aq) In this case, both substances are individually soluble. However, added together your options for possible precipitates are:

Ca3(PO4 )2 and NaNO3 From your solubility rules you know that all nitrates are solube—especially one combined with column 1A metals. You also know that phosphates (except for those combined with metals in column 1A) are insoluble. So, Ca3(PO4 )2 would be the correct answer. Answer C (1 point/87)

48. T F The filtrate is the solid formed when two solutions are mixed.



False—a solid formed when two solutions are mixed is called the precipitate. The filtrate is liquid that passes through a filter when a liquid mixture containing solid particles is passed over a filter. (1 point/88)

Given the following information for questions 49-‐51:

Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. Note—these test questions use the term “molecular equation” to indicate the “formula equation” we talked about in lecture.

49. The balanced molecular equation contains which one of the following terms?

a) AgCl (s) b) 2AgCl (s) c) 2Ba(NO3)2 (aq) d) BaNO3 (aq) e) 3AgCl (aq) Write out the complete balanced formula equation (molecular equation). Remember that this is all of the substances involved, all written in compound form, with states of matter included. You are given the names of all of the compounds so it should be easy to write this equation out and balance it (include states of matter).

BaCl2 (aq) + 2AgNO3(aq) → 2AgCl(s) + Ba(NO3)2 From the balanced equation you should quickly be able to see that 2AgCl is the correct choice). Answer B (1 point/89)

50. The balanced complete ionic equation contains which of the following terms?

a) 2Ba2+(aq) b) Cl–(aq) c) 2Ag+(aq) d) NO3– (aq) e) AgCl(aq) Remember, this is all substances involved. For dissolved ions, the specific ions are written out, including the “aq” state of matter. For precipitated substances, the compound formula is written out including the “s” state of matter. From the formula (molecular) equation you should quickly be able to write out the complete ionic equation, inspect it, and select the correct answer. Answer C (1 point/90)

Ba2+ (aq) + 2Cl− (aq)+ 2Ag+ (aq)+ 2 NO3− (aq) → 2AgCl(s)+ Ba2+ (aq)+ 2NO3

− (aq)

51. The net ionic equation contains which of the following terms?

a) Ag+(aq) b) Ba2+(aq)



c) NO3– (aq) d) H+ (aq) e) AgCl(aq) Remember that the net ionic equation includes only those ions/substances involved in the “action.” In this case, silver and chloride ions combined to form solid silver chloride. Therefore, these are the only substances listed in the equation. This should appear in reduced form. You should be able to quickly write out these formulas and select the correct answer from the list. Answer A (1 point/91)

Ag+ (aq)+ Cl− (aq)→ AgCl(s)

52. In writing the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of KOH and Mg(NO3)2 are mixed, which of the following would not be written as ionic species?

a) KOH b) Mg(NO3)2 c) Mg(OH)2 d) KNO3 e) All of the above would be written as ionic species. You should be able to create a complete ionic equation from the description of a reaction without having to write down the complete formula (molecular) equation. You know that on the reactant side, both substances will be in solution. Therefore, all four ions making up these substances will be written as an ionic species on the reactant side.

K+ (aq) + OH − (aq) + Mg 2+ (aq) + 2 NO3

− (aq)→ (At this point, you wouldn’t know it, but after balancing, the K+ and OH-‐ ions would have coefficients of 2) Then, because you know your solubility rules, you would know that Mg(OH)2 will precipitate out, but the other ions would remain in solution. Add these substances on the product side, and balance the equation.

2K + (aq) + 2OH − (aq) + Mg 2+ (aq) + 2 NO3− (aq)→ Mg(OH )2 (s) + 2K + (aq)+ 2 NO3

− (aq) Then, select the correct answer—Mg(OH)2 would not be written as an ionic species. Answer C (1 point/92)

53. The net ionic equation for the reaction of calcium bromide and sodium phosphate contains which of the following species?

a) 2Br–(aq) b) PO43–(aq) c) 2Ca3(PO4)2(s) d) 6NaBr(aq) e) 3Ca2+(aq)

You should be able to create a net ionic equation from the description of an equation. You understand that from the description, before a reaction occurs, Ca2+, Br-‐, Na+, and



PO43-‐ ions will be in the solution. From solubility rules, you know that compounds with Na+ ions are almost always soluble; compounds with Br-‐ ions are usually soluble; compounds with Ca2+ are soluble with some things but not others; phosphates are usually insoluble (although sodium phosphate is soluble). Mixing calcium bromide and sodium phosphate, the two possible compounds you could have precipitate out would be calcium phosphate and sodium bromide. Again, because of knowledge of solubility rules, sodium bromide would not precipitate but calcium phosphate would. Therefore, Ca2+ and PO43-‐ ions would be the reactants and calcium phosphate would be the solid that precipitates. So, writing the net ionic equation down we get:

3Ca2+ (aq)+ 2PO43− (aq)→ Ca3(PO4 )2(s)

From this you can then pick the correct answer from the list. The only answer that matches what is in the equation is 3Ca2+ . Answer E (1 point/93)

54. When sodium chloride and lead(II) nitrate react in an aqueous solution, which of the following terms will be present in the balanced molecular equation?

a) PbCl(s) b) Pb2Cl(s) c) NaNO3(aq) d) 2NaNO3(aq) e) 2PbCl2(s) Write out the complete balanced formula equation (molecular equation). Remember that this is all of the substances involved, all written in compound form, with states of matter included. You are given the names of the reactants so it should be easy to write this equation out and balance it (include states of matter).

2NaCl (aq) + Pb(NO3)2(aq) → PbCl2(s) + 2Na(NO3) From the balanced equation you should quickly be able to see that 2NaNO3 is the correct choice. Answer D (1 point/94)

55. Consider an aqueous solution of calcium nitrate added to an aqueous solution of sodium phosphate. Write and balance the equation for this reaction to answer the following question. What is the sum of the coefficients when the molecular equation is balanced in standard form?

a) 4 b) 5 c) 7 d) 11 e) 12 Write out the complete balanced formula equation (molecular equation). Remember that this is all of the substances involved, all written in compound form, with states of matter included. You are given the names of the reactants so it should be easy to write this equation out and balance it (include states of matter). From the balanced equation you can quickly determine the sum of the coefficients: 3+2+1+6=12 Answer



E (1 point/95)

3Ca(NO3)2 (aq) + 2Na3PO4(aq) → Ca3(PO4 )2(s) + 6NaNO3(aq)

56. When solutions of phosphoric acid and iron(III) nitrate react, which of the following terms will be present in the balanced molecular equation?

a) HNO3(aq) b) 3HNO3(aq) c) 2FePO4(s) d) 3FePO4(s) e) 2HNO3(aq) Write out the complete balanced formula equation (molecular equation). Remember that this is all of the substances involved, all written in compound form, with states of matter included. You are given the names of the reactants so it should be easy to write this equation out and balance it (include states of matter).

H3PO4 (aq) + Fe(NO3)3(aq) → FePO4(s) + 3HNO3 (aq) From the balanced equation you should quickly be able to see that 3HNO3 is the correct choice. Answer B (1 point/96)

57. When solutions of cobalt(II) chloride and carbonic acid react, which of the following terms will be present in the net ionic equation?

a) CoCO3(s) b) H+(aq) c) 2CoCO3(s) d) 2Cl–(aq) e) two of these

You should be able to create a net ionic equation from the description of an equation. You understand that from the description, before a reaction occurs, Co2+, Cl-‐, H+, and CO32-‐ ions will be in the solution. From solubility rules, you know that compounds with transition metal cations may or may not be soluble, but are often insoluble; compounds with Cl-‐ ions are usually soluble; compounds with H+ are acids and would be considered soluble; carbonates are usually insoluble (although carbonic acid is soluble). Mixing cobalt chloride and carbonic acid, the two possible compounds you could have precipitate out would be cobalt carbonate and hydrochloric acid. Again, because of knowledge of solubility rules, hydrochloric acid would not precipitate but cobalt carbonate would. Therefore, Co2+ and CO32-‐ ions would be the reactants and cobalt carbonate would be the solid that precipitates. So, writing the net ionic equation down we get:

Co2+ (aq)+ CO32− (aq)→ CoCO3(s)

From this you can then pick the correct answer from the list. The only answer that matches what is in the equation is CoCO3. Answer A (1 point/97)

58. When solutions of strontium chloride and sodium carbonate react, which of the following is a spectator ion?



a) strontium ion b) chloride ion c) sodium ion d) carbonate ion e) two of these Writing out the complete ionic equation will reveal which ions participate in some type of a reaction versus any that don’t participate—those ions that do not participate will remain in solution and are a called spectator ions. You should be able to create a complete ionic equation from the description of a reaction without having to write down the complete formula (molecular) equation. You know that on the reactant side, both substances will be in solution. Therefore, all four ions making up these substances will be written as an ionic species on the reactant side.

Sr 2+ (aq) + 2Cl− (aq) + 2Na+ (aq) + CO32− (aq)→

Then, because you know your solubility rules, you would know that SrCO3 will precipitate out, but the other ions would remain in solution. Add these substances on the product side, and balance the equation.

Sr 2+ (aq) + 2Cl− (aq) + 2Na+ (aq) + CO32− (aq)→ SrCO3 (s)+ 2Na+ (aq)+ 2Cl− (aq)

From the complete ionic equation you can see that both the sodium and chloride ions will remain in solution and so are spectator ions. Answer E (1 point/98)

59. The net ionic equation for the reaction of aluminum sulfate and sodium hydroxide contains which of the following species?

a) 3Al3+(aq) b) OH–(aq) c) 3OH–(aq) d) 2Al3+(aq) e) 2Al(OH)3(s)

You should be able to create a net ionic equation from the description of an equation. You understand that from the description, before a reaction occurs, Al3+, SO42-‐, Na+, and OH-‐ ions will be in the solution. From solubility rules, you know that compounds with aluminum (with a 3+ charge) will only likely be soluble with column 7A anions; compounds with SO42-‐ ions are often soluble; compounds with Na+ are always soluble; hydroxides are often insoluble. Mixing aluminum sulfate and sodium hydroxide, the two possible compounds you could have precipitate out would be aluminum hydroxide and sodium sulfate. Again, because of knowledge of solubility rules, sodium sulfate would not precipitate but aluminum hydroxide would. Therefore, Al3+ and OH-‐ ions would be the reactants and aluminum would be the solid that precipitates. So, writing the net ionic equation down we get:

Al3+ (aq)+ 3OH − (aq)→ Al(OH )3(s) From this you can then pick the correct answer from the list. The only answer that matches what is in the equation is 3OH-‐. Answer C (1 point/99)



Precipitation Reaction Skills As tedious as it is, for any precipitation reaction for which you are going to have to provide information about reactants or products (including precipitates), or do stoichiometry calculations, it is probably best to write out the complete ionic equation to make sure you understand all of its pieces and have them in front of your eyes. You should be able to tell what the specific ionic species are on the reactant side by inspecting the names/formulas of the stated reactants. You should then be able, using solubility rules, to predict what will precipitate (note, we don’t always have to have a precipitate—sometimes, in mixing ionic compounds together nothing will precipitate out—that is, there is no reaction) and what will remain in solution. There are a variety of possible problems requiring a variety of skills. Limiting Reagent Type Problems (see problems 60, 61, 71) You may be given information to be able to find the amounts (moles) of reactants (usually given M and L). After writing out the complete ionic equation, inspecting the equation you can more easily visualize what ions will precipitate out and what the concentrations and volumes of the reactant ions that give rise to this precipitate are. Determine moles or grams of a precipitate (see problem 61, 71) One reason for determining a limiting reactant is to determine how many moles or grams of a precipitate are formed. As with any limiting reactant problem, you would simply multiply the moles of each reactant times the mole ratio of the precipitate to the reactant to calculate the number of moles of precipitate each reactant would produce.

moles precipitate = (moles reactant)( mol from equation of precipitate

mol from equation of reactant ) Whichever reactant produces less precipitate is the limiting reactant. This number of moles can then be multiplied by the molar mass of the precipitate to get the number of grams. Amount of excess of non-‐limiting precipitating ion (especially see problem 61) As an extension of the above process, if there is a limiting reactant in a precipitation reaction, not only will the soluble ions remain in the solution after the precipitation, any excess precipitating ion that was not consumed will also remain in solution. You may have to determine how much excess ion remains, or, at the very least determine which ion still remains (question 60 is an example of this). NEVER ASSUME THAT THE ONLY IONS REMAINING AFTER



A PRECIPITATION ARE JUST THOSE IONS THAT WON’T CREATE A PRECIPITATE!!!!!! If you have to determine how much excess remains, you already had to calculate the number of moles of that ion you started with. Next, calculate the number of moles of that ion that were consumed by the limiting reagent. It will usually be the same as the number of moles of the precipitate, but it might not be depending on the mole ratios in the chemical equation. (For example, maybe 2 moles of the excess ion are precipitated for every mole of limiting ion). Finally, subtract the number of moles of excess ion that were consumed from the beginning number of moles, and this will give you the number of moles remaining. If you need grams of excess precipitating ion, multiply the above number of moles time the molar mass. Grams or moles precipitated by a precipitating reagent.

Precipitation reactions designed to determine an amount of substance (moles/grams/mass percent) in a mixture

60. Consider the reaction between 15.0 mL of a 1.00 M aqueous solution of AgNO3 and 10.0 mL of a 1.00 M aqueous solution of K2CrO4. When these react, a precipitate is observed. What is present in solution after the reaction is complete? Note: the solid is not considered to be in solution.

a) Ag+, NO3–, K+, CrO42–, water b) Ag+, NO3–, K+, water c) K+, CrO42–, water d) NO3–, K+, CrO42–, water e) water (See Precipitation Reaction Skills) Write out the complete ionic equation:

2Ag+ (aq) + 2NO3− (aq) + 2K + (aq) + CrO4

2− (aq)→ Ag2CrO4 (s)+ 2K + (aq)+ 2NO3− (aq)

Even though it provides stoichiometry information, this question SEEMS to only ask for you to identify the substances remaining in solution after the precipitate is formed. But, the stoichiometry information is given for a reason—either Ag+ or CrO42-‐ will also be left in solution because there will not be enough of one or the other to precipitate out all of the corresponding ion—as with almost all precipitation reactions, this problem involves aspects of limiting reagent problems. Therefore, in addition to K+ and NO3-‐, one of the precipitating ions will also be left in solution. Unfortunately, both are options in the answer list so we will have to figure out which one. Fortunately, the problem does not ask for how much excess of this ion there will be. We could perform calculations to help us out but the information given is actually reasonable enough that we won’t need to (and this is a multiple choice question so we should be able to do it without a calculator)



To find out which precipitating ion will be in excess we first have to find out which will be the limiting reactant—which will give the least amount of Ag2CrO4 precipitate. We would proceed like any other limiting reactant problem, and test how many moles of precipitate we get with the specific amount of each reactant. We can see from the given information that although we haven’t been given the number of grams or moles of each reactant, we have been given the information to find out the number of moles—M and L.

mol AgNO3 = ( M )(L) = (1.00M )(.0150L) = .0150molmol K2CrO4 = ( M )(L) = (1.00M )(.0100L) = .0100mol

These values are easy to determine without a calculator. Then, we could write out calculations to determine the actual number of moles of Ag2CrO4 each reactant would provide. But, we can actually reason it out without much of a problem. We can see that for every mole of AgNO3, we would get half a mole of Ag2CrO4. Therefore, as we would be starting with .0150 mol of AgNO3, we would get .0075 mol of Ag2CrO4. We can see that for every mole of K2CrO4, we would get one mole of Ag2CrO4. Therefore, as we would be starting with .0100 mol of AgNO3, we would get .0100 mol of Ag2CrO4. Therefore, AgNO3 would be limiting and we would have excess K2CrO4. So, some CrO42-‐ ion would also be left in the solution at the end of the process because there would not have been enough Ag+ to precipitate it all out. Answer D (1 point/100)

61. You mix 265.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.55 M potassium iodide. The lead(II) iodide is insoluble. Which of the following is false?

a) The final concentration of Pb2+ ions is 0.151 M. b) You form 107 g of lead(II) iodide. c) The final concentration of K+ is 0.823 M. d) The final concentration of NO3– is 0.823 M. e) All are true. (See Precipitation Reaction Skills) Write out the complete ionic equation:

Pb2+ (aq) + 2NO3− (aq) + 2K + (aq) + 2I − (aq)→ PbI2 (s)+ 2K + (aq)+ 2NO3

− (aq)

You are given a problem that contains stoichiometry information—in particular, you are given (M) and (L) for each reactant and it would be wise to assume that you are going to need to determine the number of moles for each and then determine a liiting reagent in order to be able to select the correct answer from the list. In particular, you can see by looking at the equation that after the precipitate is formed, in addition to potassium and nitrate remaining in solution, one of the precipitating ions, Pb2+ or



I-‐ will be in excess and will remain in solution as well. In scanning the list of answers, the type of information required to answer this question will be somewhat varied so you will likely have to simply take each answer one at a time.

(a) Right away, you are going to have to calculate the number of moles of each reactant solution, determine which will be a limiting reactant, and then determine if lead is in excess compared to iodide, and if so, what is its remaining concentration. To determine whether lead or iodide will be limiting, you will need to determine moles of each and test how much PbI2 would be formed:

mol Pb(NO3)2 = ( M )(L) = (1.20 molL )(.2650L) = .318mol

mol KI = ( M )(L) = (1.55 molL )(.3000L) = .465mol

mol PbI2( for PbNO3)2 ) = (.318 mol PbNO3)( 1 mol PbI21 mol Pb( NO3 )2

) = .318 mol

mol PbI2( for KI ) = (.465 mol KI )(1 mol PbI22 mol KI ) = .233 mol

The amount of KI given will provide less PbI2 so it is limiting. Then you can see that if .233 mol of PbI2 is formed, as 1 mol of Pb(NO3) will be consumed for every mole of PbI2 formed, we can find the amount of excess Pb(NO3)2.

excess Pb(NO3)2 = beginning mol Pb(NO3)2 − consumed mol Pb(NO3)2 = .318mol − .233= .085 mol Then, you can also find the total volume of final solution, so you can determine the concentration of lead ion remaining in the solution.

total volume = .3000L+ .2650L = .5650L

[Pb2+ ]= molL = .085mol

.5650 L = .151 M

So, (a) is true. (b) You already have determined the number of moles of lead iodide formed in determining how much excess lead ion you have. You determined that .233 mol of lead iodide would be formed for the amount of KI given. To find grams just write down g = (mol)(mm) , plug in the values and solve:

g = (mol)(mm) = (.233mol)(461.01 g / mol) = 107 g So, (b) is true. (c) You have already determined the number of moles of KI in part (a). As KI dissociates into 1 particle each of K+ and I-‐, the number of moles of KI on the reactant side will also be the number of moles of K+ on the product side. Therefore, the number of moles of K+ in solution after the precipitation will be .465 mol. You also have already calculated the total final volume, .5650 L, so write down the molarity formula, plug the values in and solve.

M = molL = .465 mol

.5650 L = .823 M So, (c) is true. (d) You have also already determined the number of moles of Pb(NO3)2 in part (a).



As Pb(NO3)2 dissociates into 1 particle of Pb2+ and 2 particles of NO3-‐, the number of moles of Pb(NO3)2 on the reactant side will equal twice the number of moles of NO3-‐ on the product side. Therefore, the number of moles of NO3-‐ in solution after the precipitation will be (2)(.318 mol)=.636 mol. Again, you already have the total final volume, so write down the molarity formula, plug the values in and solve.

M = molL = .636 mol

.5650 L = 1.13 M So, (d) is false. Answer D (4 points/104)

62. If all of the chloride in a 4.776-‐g sample of an unknown metal chloride is precipitated as AgCl with 70.90 mL of 0.2010 M AgNO3, what is the percentage of chloride in the sample?

(See Precipitation Reaction Skills) Write out the complete ionic equation (just use an X+ for the unknown metal)

X+ (aq) + Cl− (aq) + Ag+ (aq) + NO3

− (aq)→ AgCl (s)+ X + (aq)+NO3− (aq)

You are asked to determine the percentage of chloride in the sample and you know this will be mass of chloride divided by the mass of the sample times 100 (see below). Inspecting this formula, you have the sample mass but need to find the mass of chloride in the sample. You know that g=(mol Cl-‐)(mm Cl-‐), so write this down (you don’t have to label the “mol” and “mm” with “Cl-‐“ but it will help keep things straight later in the calculation). You know the mm of Cl-‐, but how are you going to find the number of moles of Cl-‐? You have been told that all of the Cl-‐ has been precipitated by .07090 L of .2010 M AgNO3. As Ag+ has a 1+ charge, every Ag+ ion used will precipitate one Cl-‐, so the number of moles of AgNO3 used to precipitate, will also equal the number of moles of Cl-‐, so you can substitute this in for mol Cl-‐ in your calculation. Finally, as we have already stated, we know that mol =(M)(L), and we have M and volume of AgNO3 used to precipitate the Cl-‐—just substitute this in for mol AgNO3, plug in the values and solve. Once you have the grams of Cl-‐ you can plug this back into the percent Cl-‐ equation and solve for this. (6 points/110) 63. A mixture of BaCl2 and NaCl is analyzed by precipitating all the barium as BaSO4. After addition of an excess of Na2SO4 to a 3.743-‐g sample of the mixture, the mass of precipitate collected is 2.014 g. What is the mass percentage of barium chloride in the mixture? (See Precipitation Reaction Skills) You are asked to determine the percentage of



barium chloride in the sample and you know this will be mass of barium chloride divided by the mass of the sample times 100 (see below). Inspecting this formula, you have the sample mass but need to find the mass of barium chloride in the sample. You know that g=(mol BaCl2)(mm BaCl2), so write this down (you don’t have to label the “mol” and “mm” with “BaCl2“ but it will help keep things straight later in the calculation). Inspecting this formula, you can find the mm of BaCl2, but how are you going to find the number of moles of BaCl2? You have been told that all of the barium has been precipitated by sodium sulfate, forming barium sulfate. Therefore, however, many moles of barium sulfate there are, will equal the number of moles of barium chloride you had to start with. Substitute mol BaSO4 in for mol BaCl2. Then, to find the number of mol of BaSO4, you know mol equal g/mm so substitute this in for mol BaSO4, because you have been given the number of grams of the precipitate. Plug in values and solve to get grams of BaCl2. Then plug this value into the original formula to get the percentage of BaCl2 in the mixture.

64. A 3.00-‐g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.35 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)

(See Precipitation Reaction Skills) You are asked to determine the percentage of tin in the sample and you know this will be mass of Sn divided by the mass of the sample times 100 (see below). Inspecting this formula, you have the sample mass but need to find the mass of tin in the sample. Because you are wise and have read through the whole problem, you realize that the precipitation you are performing actually precipitates out the lead in the sample, so you reason that the grams of Sn in the sample will be the sample mass minus the grams of lead—modify this initial formula appropriately (see below).

You know that g Pb=(mol Pb)(mm Pb), so write this down (you don’t have to label the “mol” and “mm” with “Pb“ but it will help keep things straight later in the calculation). Inspecting this formula, you can find the mm of Pb, but how are you going to find the number of moles of Pb? First, you have been told that the alloy is “dissolved” by nitric acid. This is not dissolved in the traditional sense in that Pb and Sn atoms are now



simply mixed with H+ and NO3 ions. There is actually a chemical change that occurs in which Pb is oxidized to 2+ and Sn is oxidized to 4+, leaving the following ions in solution: Pb2+, Sn4+, NO3-‐. This information is not necessary to solve this problem as long as you make the connection that in some way, the first reaction allowed the lead to be present in the following reaction as Pb2+ so it could be precipitated out.

Then, the addition of the sulfuric acid provided the sulfate ion which caused PbSO4 to precipitate out, whereas SnSO4 is soluble and will not precipitate out. Therefore, however, many moles of PbSO4 there are, will equal the number of moles of you had to start with. Substitute mol PbSO4 in for mol Pb. Then, to find the number of mol of PbSO4, you know mol equal g/mm so substitute this in for mol PbSO4, because you have been given the number of grams of the precipitate. Plug in values and solve to get grams of Pb. Finally, plug this value back into the original formula to solve for percentage of Pb in the alloy.

65. A mixture contained no fluorine compound except methyl fluoroacetate, FCH2COOCH3 (molar mass = 92.07 g/mol). When chemically treated, all the fluorine was converted to CaF2 (molar mass = 78.08 g/mol). The mass of CaF2 obtained was 35.3 g. Find the mass of methyl fluoroacetate in the original mixture. This problem is more straightforward than it looks. There is not really a chemical equation here that you can write down so don’t waste time trying to think of one, just start with what you are asked to find. You are asked to find the mass of methyl fluoroacetate and you know a formula to find grams: g=(mol MFA)(mm MFA). Write this down. You have been given the mm, but how are you going to find the number of moles? Because the final product, CaF2 has twice the number of F atoms than the original substance, you should be able to reason that the number of moles of methyl fluoroacetate should be twice the number of moles of CaF2. Substitute (2)(mol CaF2) in for (mol MFA).

g = (mol MFA)(mm MFA) = (2)(mol CaF2 )(mm MFA) =

(2) g CaF2mm CaF2( )(mm MFA) = (2) 35.3mol

78.08g /mol( )(92.07g / mol) = 83.2g

66. A 1.57-‐g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.47 g. Calculate the molar mass of M.



Write out the complete ionic equation:

M2+ (aq) + 2 Cl− (aq) + 2Ag+ (aq) + 2NO3

− (aq)→ 2AgCl (s) + M 2+ (aq) + 2NO3− (aq)

You know that mm M =g M/mol M, so write this down (see below). You do not have grams of M, but you can write a formula that will help you. Because you are going to be precipitating AgCl it looks like you will be able to find grams and moles of Cl-‐. You know that the number of grams of M will equal the number of grams of sample minus the number of grams of Cl-‐, so substitute this into the original formula. How do you find grams of Cl-‐. You have been told that the mass of precipitate is 3.47g. You know that all of the Cl-‐ in the precipitate must have originally come from the MCl2. This means you can find the grams of Cl-‐ in the original sample by multiplying the mass fraction of Cl-‐ in AgCl by the precipitate mass. Because you know the molar masses of Cl and Ag, the mass fraction would be (mm Cl)/(mm Cl+mm Ag) (see below). Plug the values in and calculate the grams of Cl-‐. You can then plug this into the original equation. Likewise, you don’t know mol of M. However, you can reason because of the formula MCl2, that the number of moles of M will be ½ the number of moles of Cl-‐. Substitute this in the original formula as well. Then, because you have just calculated the number of grams of Cl-‐, and you know that mol=g/mm, you can calculate number of moles of Cl-‐. Then also plug this value back into the original formula, and calculate the molar mass of M.

mm M = g Mmol M

= g sample− g Cl−

(.5)(mol Cl− )= 1.57g − .8581g

(.5)(.02421mol)= 58.5

g Cl− = (mass fraction of Cl− in AgCl)( precipitate mass) = (35.45g)(107.9+ 35.45)

⎛⎝⎜

⎞⎠⎟

(3.47g) = .8581g Cl−

mol Cl− = gmm = .8581g

35.45 = .02421 mol

67. You have 135.8 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 2.50 M solution of AgNO3(aq). Calculate the concentration of Na+ after the two solutions are mixed together. (See Precipitation Reaction Skills) Write out the complete ionic equation:

2Na+ (aq)+CrO42− (aq)+ 2Ag+ (aq)+ 2NO3

− (aq)→ Ag2CrO4(s)+ 2Na+ (aq)+ 2NO3− (aq)

Don’t make this question any harder than it needs to be. Normally, when we are given the M and L of both reactants this is a clue that we can find the moles of both reactants and determine which reactant is a limiting reagent. However, this is only necessary when we want to determine information about the precipitate. However, here we are asked how much Na+ will remain in the solution. For an ion that will not precipitate under any circumstances, it does not matter which reactant is limiting—however many moles of Na+ we start with, that will be the same number of moles of Na+ that will remain after the precipitation has taken place—none of the Na+ will be



consumed by the precipitation. So, all we have to do is determine the number of moles of Na+ that are present at the start of the reaction, and divide it by the total number of L present at the end of the reaction to get the final concentration of Na+. In this case, the number of moles of Na+ will be twice the number of moles of Na2CrO4 because for every mole of this compound, 2 moles of Na+ will dissociate.

mol Na+ = (2)(mol Na2CrO4 ) = (2)( M )(L) = (2)(2.50 molL )(.1358L) = .679 mol Na+

M = molL = .679 mol

.1358L+.125L = 2.60 M

68. You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 1.52 M solution of AgNO3(aq). Calculate the concentration of CrO42– after the two solutions are mixed together. (See Precipitation Reaction Skills) Write out the complete ionic equation:



Normally, when we are given the M and L of both reactants this is a clue that we can find the moles of both reactants and determine which reactant is a limiting reagent. As we can see from the complete ionic equation, if there is any CrO42-‐ left in the solution, it will be because CrO42-‐ was present in excess and was not all consumed by the precipitation. So, we will need to test both reactants to see how much Ag2CrO4 (s) each would provide in the precipitation. We find that AgNO3 is limiting, meaning that only .095 mol of Ag2CrO4 is formed. This means that of the .1875 mol of CrO42-‐ originally present (from the Na2CrO4), .1875 mol minus .095 mol remains—show this calculation as well (see below). As we want the concentration of this ion we need to divide by the total volume, which is .0750L + .125 L = .200L

mol Na2CrO4 = ( M )(L) = (2.50 molL )(.0750L) = .1875 mol

mol AgNO3 = ( M )(L) = (1.52 molL )(.125L) = .190 mol

mol Ag2CrO4( for Na2CrO4 ) = (.1875 mol Na2CrO4 )( 1 mol AgCrO41 mol Na2CrO4

) = .1875 mol

mol Ag2CrO4( for AgNO3) = (.190 mol)(1 mol Ag2CrO42 mol AgNO3

) = .095 mol (limiting)

excess mol CrO42− = beginning mol CrO4

2− − consumed mol CrO42− = .1875mol − .095 mol = .0925 mol

[CrO42− ]= mol

L= .0925 mol

(.0750L + .125 L)= .463 M

69. You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 2.23 M solution of AgNO3(aq). Calculate the concentration of Ag+ after the two solutions are mixed together.

(See Precipitation Reaction Skills) Write out the complete ionic equation:



Normally, when we are given the M and L of both reactants this is a clue that we can find the moles of both reactants and determine which reactant is a limiting reagent. As we can see from the complete ionic equation, if there is any Ag+ left in the solution, it will be because was Ag+ present in excess and was not all consumed by the



precipitation. So, we will need to test both reactants to see how much Ag2CrO4 (s) each would provide in the precipitation. We find that AgNO3 is limiting, meaning that only .1394 mol of Ag2CrO4 is formed. This means that ALL of the Ag+ is consumed in the reaction so no Ag+ will be left in solution.

mol Na2CrO4 = ( M )(L) = (2.50 molL )(.0750L) = .1875 mol


mol Ag2CrO4( for Na2CrO4 ) = (.1875 mol Na2CrO4 )( 1 mol AgCrO41 mol Na2CrO4

) = .1875 mol

mol Ag2CrO4( for AgNO3) = (.2278mol)(1 mol Ag2CrO42 mol AgNO3

) = .1394mol (limiting)

AgNO3 is limiting so no Ag+ remains [Ag+ ]= 0.00 M

70. You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). You also have 125 mL of a 1.74 M solution of AgNO3(aq). Calculate the concentration of NO3– after the two solutions are mixed together. (See Precipitation Reaction Skills) Write out the complete ionic equation:



Don’t make this question any harder than it needs to be. Normally, when we are given the M and L of both reactants this is a clue that we can find the moles of both reactants and determine which reactant is a limiting reagent. However, this is only necessary when we want to determine information about the precipitate. However, here we are asked how much nitrate (NO3-‐) will be in the solution. For an ion that will not precipitate under any circumstances, it does not matter which reactant is limiting—however many moles of NO3-‐ we start with, that will be the same number of moles of NO3-‐ that will remain after the precipitation has taken place—none of the NO3 will be consumed by the precipitation. So, all we have to do is determine the number of moles of NO3-‐ are present at the start of the reaction, and divide it by the total number of L present at the end of the reaction to get the final concentration of NO3.

mol NO3 = mol AgNO3 = ( M )(L) = (1.74 molL )(.125L) = .2175 mol NO3

M = molL = .2175 mol

.0750 L+.125L = 1.09 M

71. You mix 55 mL of 1.00 M silver nitrate with 25 mL of 0.81 M sodium chloride. What mass of silver chloride should you form? (See Precipitation Reaction Skills-‐Limiting Reagent Type Problem) Write out the complete ionic equation:

Ag+ (aq)+ NO3− (aq)+ Na+ (aq)+Cl− (aq)→ AgCl(s)+ Na+ (aq)+ NO3

− (aq) This is a straightforward limiting reagent type of precipitation reaction problem. You are given the molarities and volumes of the reactants, which means you can find the moles of the reactants, which is the major clue that you will have to test the number of moles of AgCl formed by each to determine the limiting reactant. Once you determine the number of moles of AgCl provided using the limiting reactant, multiply



by the molar mass of silver chloride to get the number of grams.


mol NaCl = ( M )(L) = (.81 molL )(.025L) = .02025 mol

mol AgCl( for AgNO3) = (.0550 mol AgNO3)( 1 mol AgCl1 mol AgNO3

) = .0550 mol

mol AgCl( for NaCl) = (.02025 mol)( 1 mol AgCl1 mol AgNO3

) = .02025 mol (limiting)

g AgCl = (mol)(mm) = (.02025 mol)(107.87 + 35.45) = 2.9g

72. When solutions of formic acid and sodium hydroxide react, which of the following are NOT present in the complete ionic equation?

a) hydrogen ion b) formate ion c) sodium ion d) hydroxide ion e) water Even though this is a multiple choice question it is probably best to write out some form of equation so that you can visualize it. Formic acid is an organic acid. We will often abbreviate these rather than writing out the specific formula. You might abbreviate this one as HFor. As this is not one of our strong acids, it is a weak acid—as such, it does not ionize to a significant degree. Therefore, as with all substances that do not ionize completely in water, for purposes of writing ionic equations we treat it as a molecular substance—we write its complete formula—not dissociated into ions. On the other hand, as NaOH ionizes completely we write these as ions in the equation. On the product side, the hydroxide will have neutralized all of the acidic hydrogen forming water. Therefore, only the formate ion (For-‐) will be left of the formic acid molecule. Only Na+ will remain from the NaOH. Because it is a complete ionic equation, we keep all of these pieces in the equation. Inspecting the equation we see that what is not present is the hydrogen ion. Answer A

HFor (aq) + Na+ (aq) + OH − (aq)→ H2O (l) + For − (aq)+ Na+ (aq) Note: the net ionic equation would be: HFor (aq) + OH − (aq)→ H2O (l) + For − (aq)

73. When solutions of carbonic acid and sodium hydroxide react, which of the following are NOT present in the net ionic equation?

I. hydrogen ion

II. carbonate ion

III. sodium ion

IV. hydroxide ion

a) I and II b) I, II, and III



c) I and IV d) I and III e) II and III Even though this is a multiple choice question it is probably best to write out some form of equation so that you can visualize it. If a net ionic equation is not immediately obvious, it may be useful to write out the complete ionic equation and then cancel substances that appear on both sides of the equation to get the net ionic equation. As carbonic acid is not one of our strong acids, it is a weak acid—as such, it does not ionize to a significant degree. Therefore, as with all substances that do not ionize completely in water, for purposes of writing ionic equations we treat it as a molecular substance—we write its complete formula—not dissociated into ions. On the other hand, as NaOH ionizes completely we write these as ions in the equation. On the product side, the hydroxide will have neutralized all of the acidic hydrogen forming water. As carbonic acid contains 2 hydrogens, however, two water molecules will form for each carbonic acid. (This will require 2 Na+ and 2 OH-‐ on the reactant side). As all of the acidic hydrogen is neutralized, only the carbonate ion will be left on the product side. As all of the OH-‐ is neutralized, only Na+ will be left on the product side. The complete ionic equation is:

H2CO3 (aq) + Na+ (aq) + OH − (aq)→ 2H2O (l) + Na+ (aq) + CO3− (aq)

To create the net ionic equation we would cancel out everything that appears on both sides of the equation, in this case, Na+.

H2CO3 (aq) + OH − (aq)→ 2H2O (l) + CO3− (aq)

Inspecting the equation, hydrogen and sodium ions are not present in the net ionic equation. Answer D

74. When solutions of carbonic acid and magnesium hydroxide react, which of the following are NOT present in the net ionic equation?

I. hydrogen ion

II. carbonate ion

III. magnesium ion

IV. hydroxide ion

a) I and II b) I, II, and III c) I and IV d) I and III e) II and III

Even though this is a multiple choice question it is probably best to write out some form of equation so that you can visualize it. If a net ionic equation is not immediately obvious, it may be useful to write out the complete ionic equation and then cancel substances that appear on both sides of the equation to get the net ionic equation. As carbonic acid is not one of our strong acids, it is a weak acid—as such, it



does not ionize to a significant degree. Therefore, as with all substances that do not ionize completely in water, for purposes of writing ionic equations we treat it as a molecular substance—we write its complete formula—not dissociated into ions.

However, Mg(OH)2 is also considered to be insoluble, and as such, we need to write the entire formula on the reactant side as well.

Then, something interesting happens. As a few H+ ions and a few OH-‐ ions interact to form water, more H+ and OH-‐ ions dissociate, and then form water, and then a few more H+ and OH-‐ ions dissociate, and this slow dissociation process continues until all of the H+ and OH-‐ ions are consumed, leaving water and just Mg2+ and CO32-‐ ions.

But wait! MgCO3 is also insoluble. Therefore, there are actually no ions left in solution. All three equations, the formula (molecular) equation, the complete ionic equation, and the net ionic equation all reduce to the same equation. In essence, because of the way we write these equations, all of the ions are “part of the action.”

H2CO3 (aq) + Mg(OH )2 (aq)→ H2O (l) + MgCO3 (s) Therefore, NONE of the listed ions are actually present in the official net ionic equation. However, I, II, III, IV is not a listed answer—the list is wrong. None of these ions are present in the net ionic equation. (Note, the official answer key says only I and IV are not present suggesting the Mg2+ and CO32-‐ are present in the final solution—however, in reality these would precipitate out as MgCO3.)

75. When solutions of acetic acid and copper(II) hydroxide react, which of the following are spectator ions?

a) hydrogen ion b) acetate ion c) copper(II) ion d) hydroxide ion e) none of these When asked about spectator ions the best equation to visualize these will be the complete ionic equation. As acetic acid is not one of our strong acids, it is a weak acid—as such, it does not ionize to a significant degree. Therefore, as with all substances that do not ionize completely in water, for purposes of writing ionic equations we treat it as a molecular substance—we write its complete formula—not dissociated into ions. We will abbreviate it as HAc. However, Cu(OH)2 is also considered to be insoluble, and as such, we need to write the entire formula on the reactant side as well.

Then, something interesting happens. As a few H+ ions and a few OH-‐ ions interact to form water, more H+ and OH-‐ ions dissociate, and then form water, and then a few more H+ and OH-‐ ions dissociate, and this slow dissociation process continues until all of the H+ and OH-‐ ions are consumed, leaving water and just Cu2+ and acetate (C2H3O2-‐) ions. Copper acetate is soluble so on the product side these two ions would be in solution—the complete ionic equation would look like this:



2HAc (aq) + Cu(OH )2 (s)→ 2H2O (l) + Cu2− (aq) + 2Ac− Because of the way we write this equation, all of the ions present are changing form, and no isolated ions are present on both sides of the equation. This is also the form of the net ionic equation. There are no spectator ions. Answer E

76. In the balanced molecular equation for the neutralization of sodium hydroxide with sulfuric acid, the products are:

a) NaSO4 + H2O b) NaSO3 + 2H2O c) 2NaSO4 + H2O d) Na2S + 2H2O e) Na2SO4 + 2H2O In writing down the reactants we have:

H2SO4 + NaOH → We know that both the acid and base are strong, and so, dissociate completely. Because there are two hydrogens in the acid (it is diprotic), each will be neutralized by an OH-‐ so we will require two OH-‐ ions and get two water molecules as a product. Also, the sulfate ion has a 2-‐ charge, so two Na+ ions will be required—combining the sulfate and two sodium ions we get Na2SO4. So, the complete molecular equation is

H2SO4 (aq) + 2NaOH (s)→ 2H2O (l) + Na2SO4 (aq)

77. A 0.307-‐g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 mL of 0.106 M NaOH. Calculate the molar mass of the acid. A triprotic acid is an acid that has the ability to lose up to three H+ ions. You are aware that one such acid is H3PO4 or phosphoric acid. In contrast to hydroxide bases, which dissociate all of their hydroxides at once (for example, Ca(OH)2, for which one particle will dissociate into one particles of Ca2+ and two particles of OH-‐ all at once), acids with more than one H+ to dissociate only dissociate one hydrogen until most particles have dissociated the first hydrogen, and then will begin dissociate the second hydrogen, and so on. In a titration, where you are neutralizing an amount of acid with a base, if the acid is monoprotic, the point at which enough base has been added to neutralize all of the acid is called the equivalence point. If the acid is diprotic, the point at which base has neutralized the first hydrogen of all of the particles is called the first equivalence point. At this point, all of the particles still have the second hydrogen left, and addition of further hydroxide will now begin to neutralize this second hydrogen. The point at which the second hydrogen is completely neutralized would be the second neutralization point. And so on, for a triprotic acid. After the (final) equivalence point is reached, addition of further base just adds base to the solution. So, if a triprotic acid is titrated to the third equivalence point, this means enough base has been added to neutralize all three hydrogens. A formula (molecular) equation that would represent this is (letting “A” be the anion of the acid):

H3 A(aq) + 3NaOH (s)→ 3H2O (l) + Na3 A (aq)



You can see that it would require three OH-‐ ions to neutralize the 3 H+ ions, and because the anion of the acid molecule would have a 3-‐ charge, there would need to be three sodium ions to balance this on the product side. Having said all this, you have been asked to find the molar mass of the acid and you

know: mm H3 A =

g H3 Amol H3 A

, so write this down. You have already been given g (.307 g),

all you need to do is find moles of H3A In a titration of an acid, the way we are going to find the unknown number of moles acid, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of base used to titrate the acid. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of H3A, we need 3 moles of NaOH. Therefore, you can reason that the number of moles of NaOH required to titrate the acid would be 3 times the number of moles of H3A, or, 1 mole of H3A is equivalent to (1/3)(mol NaOH), and you could substitute this into your calculation (see below). Now, how do you find the number of moles of NaOH used to titrate? You have been given M (.106 M) and L (.0352 L) and you know mol = (M)(L), so substitute this in for mol NaOH, plug in the values and solve:

mm H3 A =

g H3 Amol H3 A

=g H3 A

( 13)(mol NaOH )

=g H3 A

( 13)( M NaOH )(L NaOH )

= .307 g( 1

3)(.106 molL )(.0352L)

= 247 g / mol

78. An unknown diprotic acid requires 26.66 mL of 0.117 M NaOH to completely neutralize a 0.845-‐g sample. Calculate the approximate molar mass of the acid.

A diprotic acid is an acid that has the ability to lose up to two H+ ions. You are aware that one such acid is H2SO4 or sulfuric acid. In contrast to hydroxide bases, which dissociate all of their hydroxides at once (for example, Ca(OH)2, for which one particle will dissociate into one particles of Ca2+ and two particles of OH-‐ all at once), acids with more than one H+ to dissociate only dissociate one hydrogen until most particles have dissociated the first hydrogen, and then will begin dissociate the second hydrogen, and so on. In a titration, where you are neutralizing an amount of acid with a base, if the acid is monoprotic, the point at which enough base has been added to neutralize all of the acid is called the equivalence point. If the acid is diprotic, the point at which base has neutralized the first hydrogen of all of the particles is called the first equivalence point. At this point, all of the particles still have the second hydrogen left, and addition of further hydroxide will now begin to neutralize this second hydrogen. The point at which the second hydrogen is completely neutralized would be the second neutralization point. And so on, for a triprotic acid. After the (final) equivalence point is reached, addition of further base just adds base to the solution.



So, if a diprotic acid is titrated to the second equivalence point (that is, until it is completely neutralized), this means enough base has been added to neutralize both hydrogens. A formula (molecular) equation that would represent this is (letting “A” be the anion of the acid):

H2 A(aq) + 2NaOH (s)→ 2H2O (l) + Na2 A (aq) You can see that it would require two OH-‐ ions to neutralize the 2 H+ ions, and because the anion of the acid molecule would have a 2-‐ charge, there would need to be two sodium ions to balance this on the product side. Having said all this, you have been asked to find the molar mass of the acid and you

know: mm H2 A =

g H2 Amol H2 A

, so write this down. You have already been given g (.845 g),

all you need to do is find moles of H2A In the titration of an acid, the way we are going to find the unknown number of moles of the acid, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of base used to titrate the acid. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of H2A, we need 2 moles of NaOH. Therefore, you can reason that the number of moles of NaOH required to titrate the acid would be 2 times the number of moles of H2A, or, 1 mole of H2A is equivalent to (1/2)(mol NaOH), and you could substitute this into your calculation (see below). Now, how do you find the number of moles of NaOH used to titrate? You have been given M (.117 M) and L (.02666 L) and you know mol = (M)(L), so substitute this in for mol NaOH, plug in the values and solve:

mm H2 A =

g H2 Amol H2 A

=g H2 A

( 12 )(mol NaOH )

=g H2 A

( 12 )( M NaOH )(L NaOH )

= .845 g( 1

2 )(.117 molL )(.02666L)

= 542 g / mol

79. You have separate solutions of HCl and H2SO4 with the same concentrations in terms of molarity. You wish to neutralize a solution of NaOH. Which acid solution would require more volume (in mL) to neutralize the base?

a) The HCl solution. b) The H2SO4 solution. c) You need to know the acid concentrations to answer this question. d) You need to know the volume and concentration of the NaOH solution to answer this question. e) C and D The key to this question is recognizing that one of the acids is monoprotic and will only supply one mole of H+ ions per mole while the other is diprotic and will provide two moles of H+ atoms per mole. Because both acids are stated to be the same concentration, they will have the same number of acid particles for whatever that



concentration is, but no matter what that concentration is, H2SO4 will always supply twice the number of H+ particles and so will always neutralize the NaOH with half the volume, regardless of what the NaOH concentration is. Therefore, the neutralization will always require a greater volume of HCl. Answer A

80. What mass of NaOH is required to react exactly with 25.0 mL of 3.0 M H2SO4? Write down a balance chemical equation for the neutralization, remembering that 2 OH-‐ will be required to completely neutralize the 2 H+ ions of the acid:

H2SO4(aq) + 2NaOH (s)→ 2H2O (l) + Na2SO4 (aq) You are asked for mass and you know that g=(mol)(mm), and you can find molar mass of NaOH easily. Write this down. How are you going to find mol NaOH?

You know that in the neutralization of a base (for which a titration is an application), the way we are going to find the unknown number of moles of base, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of acid neutralized. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of H2SO4, we need 2 moles of NaOH. Therefore, you can reason that the number of moles of NaOH required to neutralize the acid would be 2 times the number of moles of H2SO4, and you could substitute this into your calculation (see below). Now, how do you find the number of moles of H2SO4 used to neutralize? You have been given M (3.00 M) and L (.0250 L) and you know mol = (M)(L), so substitute this in for mol H2SO4, plug in the values and solve:

g NaOH = (mol NaOH )(mm NaOH ) = (2)(mol H2SO4 )(mm NaOH ) = (2)( M H2SO4 )(L H2SO4 )(mmNaOH ) =(2)(3.00 mol

L )(.0250 L)((22.99 + 16.00 + 1.01)g / mol) = 6.00 g NaOH

81. With what volume of 5.00 M HF will 3.95 g of calcium hydroxide react completely, according to the following reaction?

2 2 22HF + Ca(OH) CaF + 2H O→

Inspecting the balanced chemical equation you can see that in this case, the base supplies 2 OH-‐ ions while the acid only supplies 1 H+. Therefore, two moles of acid molecules will be required to completely neutralize one mole of base molecules. You are asked for volume of acid and you know that L HF will equal (mol HF)/(M HF). (If you forget this you can get it from M=(mol)/(L), right?). Write this down. You have been given M HF, but how will you find the number of moles of HF? You know that in the neutralization of an acid (for which a titration is an application), the way we are going to find the unknown number of moles of acid, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of base neutralized. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of Ca(OH)2, we need 2 moles of HF. Therefore, you can reason that the number of moles of HF required to neutralize the



Ca(OH)2 would be 2 times the number of moles of Ca(OH)2, and you could substitute this into your calculation (see below). The minor twist in this problem is that you have not been given moles of Ca(OH)2, but grams, but by now it should be intuitive that to find moles, mol = g/mm. Substitute this into the equation, plug in the values and solve.

L HF = mol HF

M HF=

(2)(mol Ca(OH )2 )M HF

=(2)( g Ca(OH )2

mm Ca(OH )2)

M HF=

(2) 3.95 g Ca OH( )2(40.08+32.00+2.02)g /mol

5.00 molL

= .0213 L = 21.3 mL

82. Sulfamic acid, HSO3NH2 (molar mass = 97.1 g/mol), is a strong monoprotic acid that can be used to standardize a strong base:

3 2 2 2 2HSO NH ( ) + KOH( ) KSO NH ( ) + H O( )aq aq aq l→

A 0.177-‐g sample of HSO3NH2 required 19.4 mL of an aqueous solution of KOH for a complete reaction. What is the molarity of the KOH solution? Inspecting the balanced chemical equation you can see that in this case, the base supplies 1 OH-‐ ions while the acid supplies 1 H+. Therefore, one moles of acid molecules will be required to completely neutralize one mole of base molecules. As you are aware, before an analytical titration can take place, it is desirable to actually determine the precise concentration of the titrating substance—this process is called standardization—and takes place in sort of a reverse titration where the concentration of the substance being titrated is actually known, and you are trying to determine the precise concentration of the titrant. Regardless, the principles of the neutralization reaction taking place are exactly the same as they are in any other neutralization. In this situation, you are asked to find the molarity of the KOH solution. You know M KOH=(mol KOH)/(L KOH) so write this down. You have been given that .0194 L of the base is required to complete the standardization—how are you going to find the number of moles of base? You know that in the neutralization of a base (for which standardization is an application), the way we are going to find the unknown number of moles of base, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of acid neutralized. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of KOH, we need 1 mole of HSulf (lets abbreviate sulfamic acid this way). You could then substitute this into your calculation (see below). You have not been given moles of sulfamic acid, but you have been given number of grams and the mm, and know that mol=(g)/(mm). Substitute this into your calculation for (mol HSulf). Then, plug in the values and determine the M of KOH.



M KOH = mol KOH

L KOH= mol HSulf

L KOH=

( g HSulfmm HSulf )

L KOH=

.177 g HSulf97.1g /mol

.0194 L= .0940 M

83. A student weighs out 0.681 g of KHP (molar mass = 204.22 g/mol) and titrates to the equivalence point with 36.78 mL of a stock NaOH solution. What is the concentration of the stock NaOH solution? KHP is an acid with one acidic proton. Write a balanced chemical equation for the this neutralization/titration, recognizing that there is only 1 H+ that needs to be neutralized, so only 1 OH-‐ is required and 1 H2O is formed:

KHP (s) + NaOH (aq)→ KP− (aq) + H2O (l) You are asked for the molarity of the base and you know molarity equals mol/L. Write this down. You have been given L NaOH (.03678 L), but how will you find the number of moles of NaOH? You know that in the neutralization of an base (for which a titration is an application), the way we are going to find the unknown number of moles of base, is equate it through the stoichiometry of the balanced chemical equation to the number of moles of acid neutralized. You have already written down the balanced equation, and can see that to completely neutralize 1 mole of KHP, we need 1 moles of NaOH. Therefore, you could substitute this into your calculation for mol NaOH(see below). You have not been given moles of KHP, but grams and mm, and you know that to find moles, mol = g/mm. Substitute this into the equation, plug in the values and solve.

M NaOH = mol NaOH

L NaOH= mol KHP

L NaOH=

( g KHPmm KHP )

L NaOH=

( .681 g204.22 g /mol ).03678 L

= .0907 M

84. T F A chemical that changes color at the endpoint of a reaction is called a colorimeter. False—a colorimeter is an electronic device that measures the amount of light that passes through a solution at a particular wavelength, and is used to determine concentration of a solution based on how much light passes through—the less light of a certain wavelength that passes through the solution being evaluated, the greater the concentration. A chemical that changes color at the endpoint of a reaction is called an indicator. Answer False

85. In which of the following does nitrogen have an oxidation state of +4?

a) HNO3 b) NO2 c) N2O d) NH4Cl e) NaNO2



We have stated that it will be important for you to be able to determine oxidation states of elements in order to determine the numbers of electrons shifted in either ionic or covalent oxidation reduction reactions. In particular, the oxidation state of nitrogen will be one of the most common, because nitrogen is a powerful oxidizing agent. In a positive oxidation state, nitrogen loves to take electrons. As an oxidizing agent it will usually be seen in the nitrate (NO3-‐) or nitrite (NO2-‐) form.

NO3− nitrate NO2

− nitrite+5 −2 +3 −2

Because O usually has a -‐2 state, with nitrate this results in 6 negative charges but as the overall charge is -‐1, the nitrogen has a +5 charge; with nitrite, the 2 O atoms result in 4 negative charges but as the overall charge is -‐1, the nitrogen has a +3 state. Choice (a) includes the nitrate ion and choice (e) includes the nitrite ion, so neither of these has nitrogen in the +4 state. NO2, the nitrogen dioxide molecule, (not the nitrite polyatomic ion) is also a powerful oxidizer—it is a product combustion when fuels are burned incompletely in air and is very toxic to respiratory tissues. As a gas it has a brownish color and is a prominent air pollutant.

NO2 nitrogen dioxide+4 −2

The 2 O atoms result in 4 negative charges. As the overall charge of the molecule is 0, the nitrogen has a +4 state. Choice (b) is this molecule, and so is the correct answer. N2O, the dinitrogen monoxide molecule (also called nitrous oxide, laughing gas) is also a powerful oxidizer and is used as a rocket propellant. In contrast to nitrogen dioxide, it is not toxic and is used as an anesthetic in dental surgery.

N2O dinitrogen monoxide+1 −2

The single O atom results in 2 negative charge. As the overall charge of the molecule is 0, the nitrogen must have a +1 state. Choice (c) is this molecule, and so is incorrect. Nitrogen is also a component of the positive polyatomic ammonium ion (NH4+). In contrast to the other forms of nitrogen listed above, the ammonium ion is a reducing agent. Remember that hydrogens are stated to have a +1 state. As the overall charge of the ammonium ion is +1, this means that nitrogen must have a -‐3 state in the ammonium ion. Because of this, this ion likes to provide electrons.



NH4+ dinitrogen monoxide

−3 +1

Choice (d) is this form and so, is incorrect. Answer B

86. The oxidation state of iodine in IO3– is:

a) 0 b) +3 c) –3 d) +5 e) –5 We have stated that it will be important for you to be able to determine oxidation states of elements in order to determine the numbers of electrons shifted in either ionic or covalent oxidation reduction reactions. The oxidation state of iodine will be a common state to be determined. In a positive oxidation state, iodine loves to take electrons and so is an oxidizing agent. As an oxidizing agent it will usually be seen in the iodate (IO3-‐) form.

IO3− iodate

+5 −2

Because O usually has a -‐2 state, with iodate this results in 6 negative charges but as the overall charge is -‐1, the iodine atom has a +5 charge. Recognize that this will be the same for the –ate forms of the other halogen polyatomic ions—BrO3-‐ and ClO3-‐ (see question 87). Recognize that for the per-‐…-‐ate forms of the halogen polyatomic ions, ClO4-‐, BrO4-‐ and IO4-‐, the oxidation state of the halogen atom will be +7 Answer D

87. The oxidation state of chlorine in ClO3– is:

a) 0 b) +5 c) -‐5 d) +7 e) -‐7 We have stated that it will be important for you to be able to determine oxidation states of elements in order to determine the numbers of electrons shifted in either ionic or covalent oxidation reduction reactions. The oxidation state of chlorine will be a common state to be determined. In a positive oxidation state, chlorine loves to take electrons and so is an oxidizing agent. As an oxidizing agent it will usually be seen in the chlorate (ClO3-‐) form.

ClO3− chlorate

+5 −2



Because O usually has a -‐2 state, with chlorate this results in 6 negative charges but as the overall charge is -‐1, the chlorine atom has a +5 charge. Recognize that this will be the same for the –ate forms of the other halogen polyatomic ions—BrO3-‐ and IO3-‐ (see question 86). Recognize that for the per-‐…-‐ate forms of the halogen polyatomic ions, ClO4-‐, BrO4-‐ and IO4-‐, the oxidation state of the halogen atom will be +7 Answer B

88. Which of the following statements is not true?

a) When a metal reacts with a nonmetal, an ionic compound is formed. b) A metal-‐nonmetal reaction can always be assumed to be an oxidation-‐reduction reaction. c) Two nonmetals can undergo an oxidation-‐reduction reaction. d) When two nonmetals react, the compound formed is ionic. e) A metal-‐nonmetal reaction involves electron transfer. (a) True—one of your earliest understandings is that ionic compounds are formed between a metal cation and a non-‐metal anion. (b) True—whenever a metal and non-‐metal interact chemically, there will always be a shift of electrons. If the metal and non-‐metal are in an elemental state, the metal will give up electrons to the non-‐metal—that is, the metal is oxidized and the non-‐metal is reduced. If the metal and non-‐metal are in a compound state, the non-‐metal will “give electrons back” to the metal, and the metal will “take electrons back” from the non-‐metal—that is, the metal cation will be reduced while the non-‐metal anion will be oxidized. (c) True—we have often stated that although the shift in electrons is less obvious in covalent interactions, nevertheless, a shift of electrons does occur, and so, reactions between two non-‐metals will in some way, involve oxidation and reduction. (d) False—agains, one of your earliest understandings regarding reactions of two non-‐metals is that when two non-‐metals react, the compound formed is covalent—not ionic. (e) True—this is the same thing as saying that a metal-‐nonmetal reaction is an oxidation-‐reduction reaction (see answer b). Answer D

89. In the reaction 2Ca(s) + O2(g) → 2CaO(s), which species is oxidized?

a) O2 b) O2– c) Ca d) Ca2+ e) none of these You have been given a balanced chemical equation. You immediately realize that it is between an elemental metal (Ca) and an elemental non-‐metal (O2), and your brain immediately registers that the oxidation states of these are 0—you write these states down.



2Ca s( ) + O2 g( )→ 2CaO s( )0 0 +2 −2

You also immediately know that this is a synthesis reaction, and that Ca will lose electrons (become oxidized) to form a cation and oxygen will gain these electrons (become reduced) to form an anion, and that these will form the ionic solid, CaO. You then immediately write down the oxidation states of Ca and O in this compound (+2 and -‐2 respectively). This further helps you to visualize that Ca is going from a 0 to a +2 state and so is the substance being oxidized. Answer C

90. In the reaction 2Cs(s) + Cl2(g) → 2CsCl(s), Cl2 is

a) the reducing agent b) the oxidizing agent c) oxidized d) the electron donor e) two of these You have been given a balanced chemical equation. You immediately realize that it is between an elemental metal (Cs) and an elemental non-‐metal (Cl2), and your brain immediately registers that the oxidation states of these are 0—you write these states down.

2Cs s( ) + Cl2 g( )→ 2CsCl s( )0 0 +1 −1

You also immediately know that this is a synthesis reaction, and that Cs will lose an electron (becomes oxidized) to form a cation and chlorine will gain this electrons(becomes reduced) to form an anion, and that these will form the ionic solid, CsCl. You then immediately write down the oxidation states of Cs and Cl in this compound (+1 and -‐1 respectively). This further helps you to visualize that Cl is going from a 0 to a -‐1 state and so is the substance being reduced. If Cl is being reduced, this means it is the oxidizing agent—the agent that is cause the oxidation or loss of electrons from Cs (choice b). Therefore it is also NOT the reducing agent, it is NOT being oxidized, it is NOT donating electrons (it is taking them), and so there is only one true answer. Answer B

91. In the reaction N2(g) + 3H2(g) → 2NH3(g), N2 is

a) oxidized b) reduced c) the electron donor d) the reducing agent e) two of these You have been given a balanced chemical equation. You immediately realize that it is between two non-‐metal elemental substances, and so, each has an oxidation state of 0—this is a synthesis reaction. On the product side, in a covalent compound, hydrogen almost always has an oxidation state of +1—therefore, nitrogen must have a state of -‐3. You immediately write all of these oxidation states down so you can



visualize what is happening with reference to oxidation (loss of electrons) and reduction (gaining of electrons).

N2 g( ) + 3H2 g( )→ 2NH3 g( )0 0 −3 +1

You are asked to determine what is happening to N2—as it is going from 0 to -‐3 it is being reduced (reduction is gain in electrons). Therefore, it is NOT being oxidized (choice a), it IS being reduced (choice b), if it is gaining electrons it CANNOT be an electron donor (choice c), it CANNOT be the reducing agent, and CANNOT be two of these. Answer B

92. In the reaction P4(s) + 10Cl2(g) → 4PCl5(s), the reducing agent is

a) chlorine b) PCl5 c) phosphorus d) Cl– e) none of these You have been given a balanced chemical equation. You immediately realize that it is between two non-‐metal elemental substances, and so, each has an oxidation state of 0—this is a synthesis reaction. On the product side, in a covalent compound, neither of the elements in the compound fall under one of our rules. Therefore, the electrons involved the chemical bonds are stated to belong to the most electronegative atom—in this case, the chlorine atoms. Phosphorus has five valence electrons to share, and so, it shares a single electron with each of 5 Cl atoms. Therefore, each Cl atom will be said to have one more electron than normal, and so, an oxidation state of -‐1—that is, it becomes reduced. The phosphorus will be stated to have lost 5 electrons, and so, will have a charge of +5—that is, it will become oxidized.

You immediately write all of these oxidation states down so you can visualize what is happening with reference to oxidation (loss of electrons) and reduction (gaining of electrons).

P4 s( ) + 10Cl2 g( )→ 4PCl5 s( )0 0 +5 −1

You are asked identify the reducing agent—this will be the substance that “forces” its electrons on the substance being reduced. You have already determined that Cl is becoming reduced from 0 to -‐1, so, P must be the substance that is “forcing” its electrons on Cl—that is, P is the reducing agent. Answer C

93. In the reaction C(s) + O2(g) → CO2(g) carbon is __________.

a) the reducing agent b) the electron acceptor c) reduced d) the oxidizing agent e) more than one of these You have been given a balanced chemical equation. You immediately realize that it is



between two non-‐metal elemental substances, and so, each has an oxidation state of 0—this is a synthesis reaction. On the product side, in a covalent compound, oxygen almost always has an oxidation state of -‐2—therefore, carbon must have a state of +4. You immediately write all of these oxidation states down so you can visualize what is happening with reference to oxidation (loss of electrons) and reduction (gaining of electrons).

C s( ) + O2 g( )→ CO2 g( )0 0 +4 −2

You are asked to determine what is happening to C—as it is going from 0 to +4 it is being oxidized (oxidation is loss of electrons). Therefore, as it is being oxidized, it IS a reducing agent because it is “forcing” its electrons on O (choice a), it is NOT an electron acceptor because it is giving up electrons (choice b), it is NOT reduced because it is oxidized (choice c), it is NOT the oxidizing agent because it is oxidized (choice d), and so, it is NOT more than one of these (choice e). Answer A

94. Which of the following reactions does not involve oxidation-‐reduction?

a) CH4 + 3O2 → 2H2O + CO2 b) Zn + 2HCl → ZnCl2 + H2 c) 2Na + 2H2O → 2NaOH + H2 d) MnO2 + 4HCl → Cl2 + 2H2O + MnCl2 e) All are oxidation-‐reduction reactions. (a) You immediately see that this is a combustion reaction—you have a carbon compound reacting with oxygen to form water and carbon dioxide. All combustion reactions are oxidation reduction reactions. A further clue that this is a redox reaction is that an elemental substance on the reactant side (oxygen) will be a component of both compounds on the product side. This would involve oxygen going from a 0 state to a non-‐zero state. You don’t have to, but you could write out the oxidation states to confirm this:

CH4 + 3O2 → 2H2O + CO2−4 +1 0 +1 −2 +4 −2

(b) You immediately see that this is a single replacement reaction—elemental Zn will “switch” places with H in HCl to form ZnCl2 and elemental H2 will be formed. You should intuitively know that a single replacement reaction is an oxidation reduction reaction. Further clues are that you have elemental substances on one side of the equation (Zn on the reactant side, H2 on the product side) being changed into components of compounds on the other side. You don’t have to, but you could write out the oxidation states to confirm this:

Zn + 2HCl → ZnCl2 + H20 +1 −1 +2 −1 0

(c) You immediately see that this is a single replacement reaction—elemental Na will “switch” places with H in H2O to form NaOH and elemental H2 will be formed. You should intuitively know that a single replacement reaction is an oxidation reduction reaction. Further clues are that you have elemental substances on one side of the



equation (Na on the reactant side, H2 on the product side) being changed into components of compounds on the other side. You don’t have to, but you could write out the oxidation states to confirm this:

2Na + 2H2O → 2NaOH + H20 +1 −2 +1 −2 +1 0

(d) This reaction does not readily fit into a pattern that you can definitively intuitively know is a redox reaction (for example, a synthesis or decomposition reaction)—however, you should be able to quickly see that you have both elemental Cl and Cl as components of compounds, and this can only occur if oxidation and reduction is occurring. A further clue is that we commonly use manganese ions as oxidizing agents—in this case you should be able to see fairly quickly that manganese is being reduced from +4 in MnO2 to +2 in MnCl2. You don’t have to, but you could write out the oxidation states to confirm this:

MnO2 + 4HCl → Cl2 + 2H2O + MnCl2+4 −2 +1 −1 0 +1 −2 +2 −1

(e) Therefore, all of these reactions are oxidation reduction reactions. Answer E

95. Which of the following are oxidation-‐reduction reactions?

I. PCl3 + Cl2 → PCl5

II. Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

III. CO2 + 2LiOH → Li2CO3 + H2O

IV. FeCl2 + 2NaOH → Fe(OH)2 + 2NaCl

a) III b) IV c) I and II d) I, II, and III e) I, II, III, and IV (I) This reaction does not readily fit into a pattern that you can definitively intuitively know is a redox reaction (for example, a synthesis or decomposition reaction)—however, you should be able to quickly see that you have both elemental Cl and Cl as components of compounds, and this can only occur if oxidation and reduction is occurring. You don’t have to, but you could write out the oxidation states to confirm this:

PCl3 + Cl2 → PCl5+3 −1 0 +5 −5

(II) You immediately see that this is a single replacement reaction—elemental Cu will “switch” places with Ag in AgNO3 to form Cu(NO3)2 and elemental Ag will be formed. You should intuitively know that a single replacement reaction is an oxidation



reduction reaction. Further clues are that you have elemental substances on one side of the equation (Cu on the reactant side, Ag on the product side) being changed into components of compounds on the other side. You don’t have to, but you could write out the oxidation states to confirm this:

Cu + 2AgNO3 → Cu NO3( )2 + 2Ag

0 +1 +5 −2 +2 +5 −2 0

(III) This reaction does not readily fit into a pattern that you can definitively intuitively know is a redox reaction (for example, a synthesis or decomposition reaction). (Actually this is a well disguised acid base reaction. In aqueous solutions CO2 forms carbonic acid ( H2O + CO2 → H2CO3 ) and this then will react with the LiOH to neutralize it.) In any case, you will have to write out the oxidation states to determine whether or not it is an oxidation reduction reaction. The fact that there are no elemental substances involved would be a clue that this might not be a redox reaction. In writing out the oxidation states we see that no oxidation reduction is taking place.

CO2 + 2LiOH → Li2CO3 + H2O4+ −2 +1 −2 +1 +1 +4 −2 +1 −2

(IV) This reaction is obviously a double replacement reaction—which also means that it is a precipitation reaction. You should intuitively know that double replacement reactions do NOT involve oxidation reduction. You don’t have to, but you could write out the oxidation states to confirm this:

FeCl2 + 2NaOH → Fe OH( )2 + 2NaCl

+2 −1 +1 −2 +1 +2 −2 +1 +1 −1

Therefore, only reactions I and II represent redox reactions. Answer C

96. Which of the following statements is(are) true? Oxidation and reduction

a) cannot occur independently of each other b) accompany all chemical changes c) describe the loss and gain of electron(s), respectively d) result in a change in the oxidation states of the species involved e) A, C, and D (a) True—If something is taking electrons, it is an absolute that something must be supplying electrons—oxidation and reduction cannot occur independently of each other. (b) False—we have demonstrated many reactions in which oxidation-‐reduction does not occur. (c) True—remember, oxidation is loss (OIL) and reduction is gain (RIG). (d) True, if electrons are gained or lost, the oxidation state of the species does change. (e) A, C and D are true, so this is the answer. Answer E

97. In the reaction Zn + H2SO4 → ZnSO4 + H2, which, if any, element is oxidized?

a) zinc b) hydrogen c) sulfur



d) oxygen e) none of these You immediately see that this is a single replacement reaction—elemental Zn will “switch” places with H in H2SO4 to form ZnSO4 and elemental H2 will be formed. You should intuitively know that a single replacement reaction is an oxidation reduction reaction. Further clues are that you have elemental substances on one side of the equation (Zn on the reactant side, H on the product side) being changed into components of compounds on the other side. You don’t have to, but you could write out the oxidation states to confirm this. More importantly, it helps you visualize that the substance being oxidized (losing electrons) is Zn, as it goes from 0 to +2.

Zn + H2SO4 → ZnSO4 + H20 +1+6 −2 +2 +6 −2 0

Answer A

98. In the following reaction, which species is oxidized?

8NaI + 5H2SO4 → 4I2 + H2S + 4Na2SO4 + 4H2O

a) sodium b) iodine c) sulfur d) hydrogen e) oxygen In this reaction you should immediately see that it will be best to write out the oxidation states to determine which substance will lose electrons.

8NaI + 5H2SO4 → 4I2 + H2S + 4Na2SO4 + 4H2O+1 −1 +1 +6 −2 0 +1 −2 +1 +6 −2 +2 −2

Clues that could help in this case—H will have an unchanging state of +1, O will have an unchanging state of -‐2, and Na will have an unchanging state of +1. This means that I and S will be changing states—focus on these. You can see that I will go from -‐1 to 0 and so is oxidized. S goes from a +6 state to a -‐2 state and so is reduced. Answer B

99. How many of the following are oxidation-‐reduction reactions?

NaOH + HCl → NaCl + H2O

Cu + 2AgNO3 → 2Ag + Cu(NO3)2

Mg(OH)2 → MgO + H2O

N2 + 3H2 → 2NH3

a) 0 b) 1 c) 2 d) 3



e) 4 The first reaction is a neutralization between a strong acid and a strong base. This may or may not be a redox reaction, depending on the acid. You will have to write out oxidation states to be sure. Doing so, you see that nothing changes oxidation states.

NaOH + HCl → NaCl + H2O+1−2 +1 +1 −1 +1 −1 +1 −2

For the second reaction, you immediately see that this is a single replacement reaction—elemental Cu will “switch” places with Ag in AgNO3 to form Cu(NO3)2 and elemental Ag will be formed. You should intuitively know that a single replacement reaction is an oxidation reduction reaction. Further clues are that you have elemental substances on each side of the equation (Cu on the reactant side, Ag on the product side) being changed into components of compounds on the other side. You don’t have to, but you could write out the oxidation states to confirm this:

Cu + 2AgNO3 → Cu NO3( )2 + 2Ag

0 +1 +5 −2 +2 +5 −2 0

For the third reaction, this reaction does not readily fit into a pattern that you can definitively intuitively know is a redox reaction. You will probably have to write out the oxidation states to determine this. Doing so, you see that no substances change oxidation states so no oxidation reduction.

Mg OH( )2→ MgO + H2O

+2 −2 +1 +2 −2 +1 −2

For the fourth reaction you immediately realize that it is between two non-‐metal elemental substances, and so, each has an oxidation state of 0—this is a synthesis reaction. On the product side, in a covalent compound, hydrogen almost always has an oxidation state of +1—therefore, nitrogen must have a state of -‐3. You don’t have to, but you could write out the oxidation states to confirm that oxidation reduction is occurring:

N2 g( ) + 3H2 g( )→ 2NH3 g( )0 0 −3 +1

Therefore, the second and fourth reactions are oxidation reduction reactions, so two of the four reactions are redox. Answer C

100. In the reaction shown below, what species is oxidized?

2NaI + Br2 → 2NaBr + I2

a) Na+ b) I– c) Br2 d) Br– e) I2



In this reaction, depending on your level of comfort, you can fairly easily determine which substance will give up electrons without writing down oxidations states. However, if it will take more than just a few seconds to reason it out, you should write out the oxidation states to determine which substance will lose electrons. Quick inspection shows that Na will have a +1 state on both sides of the equation and so is a spectator. I is initially at a -‐1 state and goes to a 0 state and so is oxidized, and you can stop at this point. However, you can also see that Br is going from a 0 state to a -‐1 state. In any case, I-‐ is oxidized. If you need to write down states it would look like this:

2NaI + Br2 → 2NaBr + I2+1−1 0 +1 −1 0

Answer B

101. T F Oxidation is the gain of electrons. False—oxidation is loss—OIL RIG Answer False

102. T F A reducing agent is an electron donor. True—a reducing agent becomes oxidized, and so, loses electrons—we think of this as if the reducing agent “forces” its electrons on the substance being reduced. Therefore, a reducing agent is an electron donor. Answer True 103. Balance the following oxidation-‐reduction reaction using the oxidation number method:

Fe3+ + I– → Fe2+ + I2

In the balanced equation, the coefficient of Fe2+ is

a) 1 b) 2 c) 3 d) 4 e) none of these

Write out the equation and add oxidation states. Once you do this you should notice iron is becoming reduced (changing from +3 to +2), and I is becoming oxidized (changing from -‐1 to 0)

Fe3+ + I – → Fe2+ + I2+3 −1 +2 0

Before you add in your tie lines, however, notice that on the product side there are two iodine atoms as opposed to 1 on the reactant side. In order to make the electrons balance correctly it would be better right from the start to say that two iodine atoms will be transferring electrons as you know you will need at least two anyway. So, add tie lines by adding a two as a coefficient to I-‐, and saying that the 2 iodine atoms will “lose two electrons” (one each). Then add the tie line to the iron atoms that are



transferring and say “gain 1 electron.”

This means you will have to multiply the iron ions times 2 to balance the electrons—as two I-‐ ions each lose an electron for a total loss of 2 electrons, two Fe3+ ions will each have to take 1 electron—a total gain of two electrons—two give two Fe2+ ions.

2Fe3+ + 2I – → 2Fe2+ + I2+3 −1 +2 0

No further balancing is required, and you can now see that the coefficient of Fe2+ must be 2. Answer B

104. Balance the following oxidation-‐reduction reaction using the oxidation number method:

Al + Br2 → Al3+ + Br–

In the balanced equation, the coefficient of Br– is

a) 2 b) 3 c) 4 d) 6 e) none of these Write out the equation and add oxidation states. Once you do this you should notice aluminum is becoming oxidized (changing from 0 to +3), and Br is becoming reduced (changing from 0 to -‐1).

Al + Br2 → Al3+ + Br −

0 0 +3 −1

Before you add in your tie lines, however, notice that on the reactant side there are two bromine atoms as opposed to one on the product side. In order to make the electrons balance correctly it would be better right from the start to say that two bromine atoms will be transferring electrons as you know you will need at least two anyway. So, add tie lines by adding a two as a coefficient to Br-‐, and saying that the 2 bromine atoms will “gain two e-‐” (one each). Then add the tie line to the Al atoms that are transferring and say “lose 3 e-‐.”



This means you will have to multiply the aluminum atoms times 2 to balance the electrons and the bromine atoms times 3. As two Al atoms each lose three electron for a total loss of 6 electrons, 6 bromine atoms (3 Br2) will each have to take one electron—a total gain of six electrons—to give 6 Br-‐ ions.

2Al + 3Br2 → 2Al3+ + 6Br −

0 0 +3 −1

No further balancing is required, and you can now see that the coefficient of Br-‐ must be 6. Answer D

105. Consider the following unbalanced oxidation-‐reduction reaction:

Fe + Br2 → Fe3+ + Br–

In the balanced equation, the number of electrons transferred is

a) 3 b) 2 c) 6 d) 4 e) none of these

Write out the equation and add oxidation states. Once you do this you should notice iron is becoming oxidized (changing from 0 to +3), and Br is becoming reduced (changing from 0 to -‐1).

Fe + Br2 → Fe3+ + Br −

0 0 3+ −1

Before you add in your tie lines, however, notice that on the reactant side there are two bromine atoms as opposed to one on the product side. In order to make the electrons balance correctly it would be better right from the start to say that two bromine atoms will be transferring electrons as you know you will need at least two anyway. So, add tie lines by adding a two as a coefficient to Br-‐, and saying that the 2 bromine atoms will “gain two e-‐” (one each). Then add the tie line to the Fe atoms that are transferring and say “lose 3 e-‐.”

This means you will have to multiply the iron atoms times 2 to balance the electrons and the bromine atoms times 3. As two Fe atoms each lose three electron for a total loss of 6 electrons, 6 bromine atoms (3 Br2) will each have to take one electron—a total gain of six electrons—to give 6 Br-‐ ions.



2Fe + 3Br2 → 2Fe3+ + 6Br −

0 0 3+ −1

No further balancing is required, and you can now see that the coefficient of Br-‐ must be 6. Answer C

106. The MnO4– is often used to analyze for the Fe2+ content of an aqueous solution via the reaction

MnO4–(aq) + Fe2+(aq) + H+(aq) → Fe3+(aq) + Mn2+(aq) + H2O(l)

What is the ratio of Fe2+ : MnO4– in the balanced equation?

a) 1 : 1 b) 2 : 1 c) 3 : 1 d) 4 : 1 e) 5 : 1

Write out the equation and add oxidation states. Once you do this you should notice that Mn is becoming reduced (changing from +7 to +2) and Fe is becoming oxidized (changing from +2 to +3).

MnO4– aq( ) + Fe2+ aq( ) + H + aq( ) → Fe3+ aq( ) + Mn2+ aq( ) + H2O l( )

+7 −2 +2 +1 +3 +2 +1 −2

There are no adjustments that need to be made related to the number of Mn and Fe atoms on either side—there is 1 Mn on both sides and 1 Fe on both sides. So you can add tie lines showing Mn gaining 5 electrons, and showing Fe losing 1 electron.

From this, you can see that you will have to multiply Fe ions times 5 in order to balance the electrons. Therefore, the ratio of Fe2+:MnO4-‐ will be 5:1 in the balanced equation. Answer E

107. Given the reaction:

2MnO4– + 5H2O2 + 6H+ → 2Mn2+ + 8H2O + 5O2

determine the number of electrons involved in this reaction.

a) 10 b) 8 c) 6 d) 4



e) 2 This is the titration reaction we performed in our lab. To find the number of electrons involved, first, write out the oxidation states. Remember, that for hydrogen peroxide (H2O2), for the peroxide ion, O is at a -‐1 state rather then -‐2.

2MnO4– + 5H2O2 + 6H + → 2Mn2+ + 8H2O + 5O2

+7 −2 +1 −1 +1 +2 +1 −2 0

From this, we could reason that as each Mn is going from +7 to +2, it will gain 5 electrons. However, there are 2 Mn atoms so overall the Mn atoms will gain 10 electrons. Likewise, each oxygen atom is losing 1 electron. As there are 10 O atoms doing this, overall, O atoms will lose 10 electrons. This means that the number of electrons involved is 10. Answer A

108. A molecule with an unequal charge distribution is said to be a __________ molecule. You are highly aware that a molecule with an unequal charge distribution is called a polar molecule.

109. Soluble ionic compounds containing the hydroxide ion are called strong __________. Soluble ionic compounds containing the OH-‐ increase the amount of OH-‐ ions in solution are called strong bases.

110. A __________ is a substance dissolved in a liquid to make a solution. A substance that is dissolved in a liquid to make a solution is called a solute.

111. A __________ electrolyte dissociates to a great extent in an aqueous solution. If an ionic substance dissociates to a great extent in an aqueous solution it will conduct electricity strongly and is called strong electrolyte.

112. Molarity is defined as __________ of solute per volume of solution in ___________. Molarity is defined as moles of solute per volume of solution in liters.

Use the following to answer questions 113-‐114:

Selecting from the following reagents, indicate which reagents would be mixed to give the compounds described.

CuSO4(aq) Fe2(CO3)3(s) NH3(aq)

CuCO3(s) FeCl3(aq) Na2SO4(aq)

Cr(OH)3(s) H2SO4(aq)

113. Cu(OH)2(s) This is a fairly difficult question. In order for a solid precipitate of Cu(OH)2 to be formed, you would need an aqueous solution with Cu in it. The only item in the list that could accomplish this would be CuSO4. This would also require that a solution containing OH-‐ ions be added to the mix. However, no items containing OH-‐ are present in the list that are soluble—Cr(OH)3 is insoluble and would not provide OH-‐ ions to a solution. Remember that there are also substances that do not have OH-‐



ions in them that can provide OH-‐ ions by pulling a H+ ion from water. NH3 is in the list, and, it is soluble. So, if we add NH3 to the mix, it will cause OH-‐ ions to be released and this will cause Cu(OH)2 to precipitate out. Answer CuSO4 and NH3 114. FeCl3(aq) + Na2SO4(aq) I am not sure what the point of this question is. Both of the requested compounds are already in the list, so to get these two compounds, just add these compounds from the list. Answer FeCl3 and Na2SO4

Use the following to answer questions 115-‐119:

Write balanced equations for each of the processes, choosing from the following substances as reactants:

BaCl2 O2 H2SO4 HNO3

C2H5OH H2O Ca(OH)2 K

Na2CrO4 KOH Pb(NO3)2

115. Precipitation of BaSO4 from solution To get precipitation of BaSO4, this would require a soluble substance containing Ba—this would be BaCl2. This would also require a soluble substance containing SO4. From the list, this would be H2SO4. The resulting balanced chemical equation would be: BaCl2 (aq) + H2SO4 (aq)→ BaSO4(s) + 2HCl 116. Neutralization of sulfuric acid To get neutralization of sulfuric acid, one of the substances must be the sulfuric acid. To neutralize this we need a base. We have two choices, KOH and Ca(OH)2. (Note, C2H5OH is not a base—this is the organic alcohol, ethanol.) We should probably not use Ca(OH)2 because it will cause precipitation of CaSO4. So, using the KOH, the neutralization will be: H2SO4 (aq) + 2KOH (aq)→ 2H2O + K2SO4 117. Combustion reaction For a combustion reaction we will traditionally use an organic compound and oxygen. This means we need to select C2H5OH and O2. We know that the products are CO2 and H2O. We use our standard approach of doubling the number of moles of the organic molecule to begin the balancing process. We then determine the number of CO2 and H2O molecules that must result.

118. Dissolution of calcium hydroxide with another reagent



119. Formation of hydrogen gas

120. Balance the following equation: C3H5(NO3)3 → N2 + CO2 + H2O + O2 Write out the equation and add oxidation states. You’ll immediately notice that not everything is a whole number so it would be best to start out by multiplying everything by 3 in order to get whole numbers.

C3H5 NO3( )3→ N2 + CO2 + H2O + O2

− 23 +1 +5 −2 0 +4 −2 +1 −2 0

121. Balance the following equation: KI + HNO3 → KNO3 + NO + I2 + H2O

Write out the equation and add oxidation states. Once you do this you should notice that I is becoming oxidized (changing from -‐1 to 0) and nitrogen is becoming reduced (changing from +5 to +2).

KI + HNO3 → KNO3 + NO + I2 + H2O+1 −1 +1 +5 −2 +1+5 −2 +2 −2 0 +1 −2

Before you add in your tie lines, however, notice that on the product side there are two iodine atoms as opposed to 1 on the reactant side. In order to make the electrons balance correctly it would be better right from the start to say that two iodine atoms will be transferring electrons as you know you will need at least two anyway. So, add tie lines by adding a two as a coefficient to KI, and saying that the 2 iodine atoms will “lose two electrons” (one each). Then add the tie line to the nitrogen atoms that are transferring and say “gain 3 electrons.”

This means you will have to multiply the I containing compounds by 3 and the involved N containing compounds by 2.

Then notice that you will have to balance K atoms by adding a coefficient of 6 to the potassium nitrate, which in turn will mean you have to balance nitrogen atoms on the reactant side by changing the coefficient of nitric acid to 8.



Finally, you have 24 O atoms on the reactant side and 21 on the product side—you can balance O atoms by adding a coefficient of 4 to water, which will also balance the number of H atoms.

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