chapter 4 type of chemical reactions and solution stoichiometric water, nature of aqueous solutions,...

Chapter 4Type of Chemical Reactions and

Solution Stoichiometric

Water, Nature of aqueous solutions, types of electrolytes, dilution.

Types of chemical reactions: precipitation, acid-base and oxidation reactions.

Stoichiometry of reactions and balancing the chemical equations.

The water molecule is polar.

A space-filling model of the water molecule.

Water is a dissolving media or solvent

Aqueous Solutions

Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt assisting

in the dissolving process.

Cl-

Na+

Cl-

Na+

H2O

Some Properties of Water

Water is “bent” or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration occurs when salts dissolve in

water.

Figure 4.3: (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of "like dissolving like."

Solvent

retains its phase (if different from the solute)

is present in greater amount (if the same phase as the solute)

Solute

dissolves in water (or other “solvent”)

changes phase (if different from the solvent)

is present in lesser amount (if the same phase as the solvent)

General Rule for dissolution

Like dissolve likewise

Polar dissolve polar (water dissolves in ethanol)

Non-polar dissolve nonpolar (Alkane H/C dissolves in fat)

Figure 4.5: When solid NaCl dissolves, the Na+ and Cl- ions are randomly dispersed in the water.

Example Example

Electrolytes

Strong - conduct current efficiently

NaCl, HNO3

Weak - conduct only a small current

vinegar, tap water

Non - no current flows

pure water (non-ionic or de-ionized), sugar solution

Figure 4.4: Electrical conductivity of aqueous solutions.

Acids

Strong acids - dissociate completely to produce H+ in solution

hydrochloric and sulfuric acid

HCl , H2SO4

Weak acids - dissociate to a slight extent to give H+ in solution

acetic and formic acid

CH3COOH, CH2O

Bases

Strong bases - react completely with water to give OH ions.

sodium hydroxide

Weak bases - react only slightly with water to give OH ions.

ammonia

Figure 4.6: HCl (aq) is completely ionized.

Figure 4.7: An aqueous solution of sodium hydroxide.

Figure 4.8: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized such as 1 in 100 molecules.

Molarity

Molarity (M) = moles of solute per volume of solution in liters:

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

Calculate the molarity (M) of the solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

Find the concentration of each type of ion in 0.26 M Al(NO3)3 and 0.15 M CaCl2

Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2.

Common Terms of Solution Concentration

Stock - routinely used solutions prepared in concentrated form.

Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl)

Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl): (MV)initial=(MV)Final

Figure 4.10: Steps involved in the preparation of a standard aqueous solution.

Figure 4.12: Dilution Procedure (a) A measuring pipette is used to transfer 28.7mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The resulting solution is 1.00 M acetic acid.

Diluted from 17.4 M to 1.00 M

Moles of solute after dilution = moles of solute before dilution

Prepare 2.00 L of each of the following solution:

a) 0.250 M NaOH from solid NaOH

b) 0.250 M NaOH from 1.00 M NaOH stock solution

M1V1 = M2V2

Before dilution after dilution

2.00 L × NaOHmol

NaOHg00.40

L

NaOHmol250.0 = 20.0 g NaOH

Types of Solution Reactions

Precipitation reactionsAgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Acid-base reactionsNaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Oxidation-reduction (redox) reactionsFe2O3(s) + Al(s) Fe(l) + Al2O3(s)

Simple Rules for the Solubility of Salts in Water

1. Most nitrate (NO3) salts are soluble.

2. Most alkali (group 1A) salts and NH4+ are soluble.

3. Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg2

2+)

4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)

5. Most OH salts are only slightly soluble (NaOH, KOH are soluble, and Ba(OH)2, Ca(OH)2 are marginally soluble)

6. Most S2, CO32, CrO4

2, PO43 salts are only slightly soluble.

Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

Describing Reactions in SolutionPrecipitation

1. Molecular equation (reactants and products as compounds)

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

2. Complete ionic equation (all strong electrolytes shown as ions)

Ag+(aq) + NO3 (aq) + Na+ (aq) + Cl(aq)

AgCl(s) + Na+ (aq) + NO3 (aq)

Describing Reactions in Solution (continued)

3. Net ionic equation (show only components that actually reacts)

Ag+(aq) + Cl(aq) AgCl(s)

Na+ and NO3 are spectator ions.

Stoichiometry of Precipitation Reactions

How to calculate quantities of reactants and products involved in a chemical reaction?

1. Convert all quantities to moles.

2. Use the balancing coefficients to relate the moles of reactants &products in a balanced equation.

3. Determine the solid product (precipitate) and spectator ions from the rules of solubility.

4. Calculate the limiting reactant and theoretical yield.

5. Convert moles to mass if required.

Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

Hint: Na+ and NO3- are spectator ions, AgCl is the only product.

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.

Hint: Na+ and NO3- are spectator ions, PbSO4 is the only product.

Acid-Base Reactions

An acid produces H+ ions in water

A base produces OH- ions in water

An acid is a proton donor

A base is a proton acceptor

Arrhenius’s concept

Bronsted & Lowrys’ concept

Performing Calculations for Acid-Base Reactions

1. List initial species and predict reaction.

2. Write balanced net ionic reaction.

3. Calculate moles of reactants.

4. Determine limiting reactant.

5. Calculate moles of required reactant/product.

6. Convert to grams or volume, as required.Remember: n H+ = n OH- (MV) H+ = (MV) OH-

Neutralization Reaction

acid + base salt + water

HCl (aq) + NaOH (aq) NaCl (aq) + H2O

H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O

H+ + OH- H2O

4.3

What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH?

In certain experiment, 28.0 ml of 0.0250 M HNO3 and 53.0 ml of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH-

ions in excess after reaction goes to completion?

Neutralization Reaction of Strong Acid and Strong Base

M1V1f1 = M2V2f2

Where: M & V are molarities & volumes of respective acid & base

And f = number of H+ or OH- per formula unit of acid & base

For example in acids: in HCl, f = 1 while in H2SO4, f = 2 and in H3PO4, f = 3

Similarly in bases:In NaOH, f = 1, Ca(OH)2, f = 2 and so on

Key Titration Terms

Titrant - solution of known concentration used in titration

Analyte - substance being analyzed

Equivalence point - enough titrant added to react exactly with the analyte

Endpoint - the indicator changes color so you can tell the equivalence point has been reached.

movie

In one experiment, 1.3009 g sample of potassium hydrogen phthalate (KHC8H4O4 often abbreviated KHP) was titrated against NaOH solution. Exactly 41.20 mL of NaOH solution was required for complete titration of 1.3009 g of KHP. Calculate the concentration of the sodium hydroxide solution.

KHC8H4O4 K+ (aq) + HC8H4O4- (aq)

NaOH Na+ (aq) + OH- (aq)

OH- (aq) + HC8H4O4- (aq) H2O (l) + HC8H4O4

2- (aq)

Practice Example

How many moles are in 18.2 g of CO2?

Practice Example

Consider the reaction

N2 + 3H2 2NH3

How many moles of H2 are needed to completely react 56 g of N2?

Practice Example

How many grams are in 0.0150 mole of caffeine C8H10N4O2

Practice Example

A solution containing Ni2+ is prepared by dissolving 1.485 g of pure nickel in nitric acid and diluting to 1.00 L. A 10.00 mL aliquot is then diluted to 500.0 mL. What is the molarity of the final solution?(Atomic weight: Ni = 58.70).

Practice Example

Calculate the number of molecules of vitamin A, C20H30O in 1.5 mg of this compound.

Practice Example

What is the mass percent of hydrogen in acetic acid HC2H3O2

Oxidation-Reduction Reactions(electron transfer reactions)

2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

Reducing agent: Oxidizes it self

Oxidizing agent: Reduces it self

Redox Reactions• Many practical or everyday examples of

redox reactions:– Corrosion of iron (rust formation)– Forest fire– Charcoal grill– Natural gas burning– Batteries

– Production of Al metal from Al2O3 (alumina)

– Metabolic processes

combustion

Rules for Assigning Oxidation States

1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic element = charge

3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)

4. H = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)

Zn is oxidizedZn Zn2+ + 2e-

Cu2+ is reducedCu2+ + 2e- Cu

Zn is the reducing agent

Cu2+ is the oxidizing agent

4.4

Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?

Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)

Cu Cu2+ + 2e-

Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent

NaIO3

Na = +1 O = -2

3x(-2) + 1 + ? = 0

I = +5

IF7

F = -1

7x(-1) + ? = 0

I = +7

K2Cr2O7

O = -2 K = +1

7x(-2) + 2x(+1) + 2x(?) = 0

Cr = +6

Oxidation numbers of all the elements in the following ?

4.4

Balancing by Half-Reaction Method

1. Write separate reduction and oxidation half reactions.

2. For each half-reaction:

Balance other elements first (except H, O)

Balance O using H2O

Balance H using H+

Balance charges using electrons

Balancing by Half-Reaction Method (continued)

3. If necessary, multiply by integer to equalize number of electrons on both sides.

4. Must cancel electrons on both sides

5. Add both half-reactions

6. Check that number of elements and charges are balanced on both sides.

7. Cancel same number of H+ from both sides

8. This will be your balanced redox reaction in acidic medium.

Half-Reaction Method - Balancing in Base

1. Start balancing as in acidic medium.

2. Then add OH ions equal to H+ ions present on both sides of the reaction.

3. Form water by combining H+ and OH.

4. Cancel same number of H2O from both sides

5. Check number of elements and charges are balanced on both sides of the reaction..

Balancing Redox Equations

•Example: Balance the following redox reaction:

•Cr2O72- + Fe2+ Cr3+ + Fe3+ (acidic soln)

1) Break into half reactions:

Cr2O72- Cr3+

Fe2+ Fe3+


2) Balance each half reaction:

Cr2O72- Cr3+

Cr2O72- 2 Cr3+

Cr2O72- 2 Cr3+ + 7 H2O

Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O


2) Balance each half reaction (cont)

Fe2+ Fe3+

Fe2+ Fe3+ + 1 e-

Balancing Redox Reactions

3) Multiply by integer so e- lost = e- gained

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

Fe2+ Fe3+ + 1 e- x 6


3) Multiply by integer so e- lost = e- gained

6 Fe2+ 6 Fe3+ + 6 e-

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

4) Add both half reactions

Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O


Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

5) Check the equation

2 Cr 2 Cr

7 O 7 O

6 Fe 6 Fe

14 H 14 H

+24 + 24


• Procedure for Basic Solutions:– Divide the equation into 2 incomplete half

reactions• one for oxidation

• one for reduction

Balancing Redox Reactions– Balance each half-reaction:

• balance elements except H and O

• balance O atoms by adding H2O

• balance H atoms by adding H+

• add 1 OH- to both sides for every H+ added

• combine H+ and OH- on same side to make H2O

• cancel the same # of H2O from each side

• balance charge by adding e- to side with greater overall + charge

different

Balancing Redox Equations– Multiply each half reaction by an integer so that

• # e- lost = # e- gained

– Add the half reactions together.• Simply where possible by canceling species

appearing on both sides of equation

– Check the equation• # of atoms

• total charge on each side


Example: Balance the following redox reaction.

NH3 + ClO- Cl2 + N2H4 (basic soln)

NH3 N2H4

ClO- Cl2

1) Break into half reactions:


NH3 N2H4


2 NH3 N2H4

2 NH3 N2H4 + 2 H+

2 NH3 + 2 OH- N2H4 + 2 H2O

+ 2 OH- + 2 OH-

2 H2O

2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-


2 ClO- Cl2


2 ClO- Cl2 + 2 H2O2 ClO- + 4 H+ Cl2 + 2 H2O

+ 4 OH- + 4 OH-

2 ClO- + 4 H2O Cl2 + 2 H2O + 4 OH-

2 ClO- + 2 H2O Cl2 + 4 OH-

2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-

ClO- Cl2


3) Multiply by integer so # e- lost = # e- gained

2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-

2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-

4) Add both half reactions

2 NH3 + 2 OH- + 2ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-

Balancing Redox Reactions5) Cancel out common species

2 NH3 + 2 OH- + 2 ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-

2

2 NH3 + 2 ClO- N2H4 + Cl2 + 2 OH-

6) Check final equation:

2 N 2 N

6 H 6 H

2 Cl 2 Cl

2 O 2 O

-2 -2

Practice ExampleIn the following the oxidizing agent is:

5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 8H2O + 5O2

a. MnO4-

b. H2O2

c. H+

d. Mn2+

e. O2

Practice ExampleDetermine the coefficient of Sn in acidic solution

Sn + HNO3 SnO2 + NO2 + H2O

1

Practice ExampleThe sum of the coefficients when they are

whole numbers in basic solution:

Bi(OH)3 + SnO22- Bi + SnO3

2-

13

http://www.chemistrycoach.com/balancing_redox_in_acid.htm#BalancingRedoxEquationsinAcidicorBasicMedium



http://www.chemistrycoach.com/tutorials-5.htm#Oxidation-Reduction

http://www.chemistrycoach.com/tutorials-5.htm#Oxidation-Reduction

chapter 4 type of chemical reactions and solution stoichiometric water, nature of aqueous solutions,...

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