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Honors text: Chapter 15 Unit 08

NOTES: Solutions & Solubility

Solution:

The ___________________ is the medium in which the _______________________ are dissolved. (the solvent is usually the most abundant substance).

Example: saltwater (solution)

The process of dissolution is favored by:1)

2)

Liquids Dissolving in Liquids: Liquids that are soluble in one another (”mix”) are MISCIBLE. Polar liquids are generally soluble in other polar liquids. Nonpolar liquids are soluble in other nonpolar liquids.

Factors affecting the rate of dissolution:1)

2)

3)

SATURATION:A solid solute dissolves in a solvent until the solution is SATURATED (equilibrium exists between dissolved & undissolved solute)

Unsaturated solution – is able to dissolve more solute Saturated solution – has dissolved the maximum amount of solute Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid

crystals generally form when this solution is cooled.

Solubility:

Factors affecting solubility: The nature of the solute and solvent : different substances have different solubilities

____________________________: many solids substances become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps.

____________________________: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.


Notes: Concentration of Solutions

Concentration refers to amount of solute dissolved in a solution.

MOLARITY (M) =

• Molarity is the # moles solute per liter of solution.• The symbol [ ] is used to denote molarity of a substance

Dilution:When you dilute a solution, you can use this equation:

Example: Describe how you would prepare 2.50 L of 0.665 Na2SO4 solution starting with:a) solid Na2SO4 b) 5.00 M Na2SO4 solution

MASS PERCENT =

Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?

MOLALITY=

Molality = m = the number of moles of solute per kilogram of solvent.

Example: What is the molaloty of a solution that contains 12.8 g C6H12O6 in 187.5 g of water?

Example: How many grams of H2O must be used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C12H22O11?

NORMALITY =

Example: Find the normality of a 3.0 M solution of H2SO4?


NOTES: Colligative Properties of Solutions

Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

1) lowering vapor pressure2) raising boiling point3) lowering freezing point4) generating an osmotic pressure

Boiling Point Elevation

a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.

Tb = Kb mwhere:

Kb values are constants; see table 15-4, p. 472

Kb for water =

Example: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution?

Example: How many grams of glucose, C6H12O6, would need to be dissolved in 535.5 g of water to produce a solution that boils at 101.5°C?

Honors text: Chapter 15 Unit 08Freezing/Melting Point Depression

the freezing pt of a solution is always lower than that of the pure solvent.

Tf = Kf mwhere:

Kf values are constants; see table 15-5, p. 474

Kf for water =

Ex: Calculate the freezing pt of a 2.50 m glucose solution.

Ex: When 15.0 g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine K f for formic acid.

Ex: An antifreeze solution is prepared containing 50.0 cm3 of ethylene glycol, C2H6O2, (d = 1.12 g/cm3), in 50.0 g water. Calculate the freezing point of this 50-50 mixture. Would this antifreeze protect a car in Chicago on a day when the temperature gets as low as –10° F?

Honors text: Chapter 15 Unit 08Electrolytes and Colligative Properties

• Colligative properties depend on the # of particles present in solution.

• Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal concentration

• For example, the freezing pt of water is lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m.

• However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute.

• The relationships are given by the following equation:

Tf = Kf m nand

Tb = Kb m n

where: Tf = freezing pt depression and Tb =boiling point elevation

Kf = the molal freezing pt constantKb = the molal boiling pt constantm = the molality of the soluten = the # of particles formed from the dissociation of each formula unit of the solute.

Examples: What is the freezing pt of:

a) a 1.15 m sodium chloride solution?

b) a 1.15 m calcium chloride solution?

c) a 1.15 calcium phosphate solution?

Honors text: Chapter 15 Unit 08Determining Molecular Weights by Freezing Point Depression

Ex: A 1.20 g sample of an unknown molecular compound is dissolved in 50.0 g of benzene. The solution freezes at 4.92°C. Determine the molecular weight of the compound. The freezing pt of pure benzene is 5.48°C, and the Kf for benzene is 5.12°C/m.

Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0 g of water. The resulting solution froze at –5.58°C. What is the molecular weight of the compound?


SOLUBILITY CURVES FOR A NUMBER OF WATER SOLUBLE INORGANIC SUBSTANCES


Worksheet 1: SOLUBILITY CURVES1. Most substances on this graph show increased solubility as temperature increases. What are the

exceptions?

2. Each curve shows how the solubility for that substance changes as _________________ changes.

3. The solubilities of substances whose curve shows greater (steeper) slopes are (more/less) affected by temperature changes than those that have more gradual slopes.

4. If 50cm3 of water saturated with potassium chlorate (KClO3) at 23ºC is slowly evaporated to dryness, how many grams of the dry salt will be recovered?

5. What is the smallest volume of water required to dissolve completely 23g of NH4Cl at 40ºC?

6. A saturated solution of NaNO3 in 100cm3 of water at 40ºC is heated to 50 C. What is the rate of increase in solubility in grams per degree (in other words, for every degree of temp. increase, how many more grams dissolve)?

7. Which salt has solubility values that are least affected by changes in temperature?

8. If 30g of KCl is dissolved in 100cm3 of water at 45ºC, how many additional grams of KCl would be needed to make the solution saturated at 80ºC?

9. At what temperature do potassium chlorate (KClO3) and potassium chloride (KCl) have the same solubility in water?

10. At 50ºC, 100cm3 of water is saturated with cerium sulfate( Ce2(SO4)3 ). How many grams of cerium sulfate must be added to saturate the solution at 0ºC?

11. At 50ºC, 100cm3 of water is saturated with potassium nitrate (KNO3). How many grams of potassium nitrate will precipitate when the solution is cooled to 40ºC?

12. How many grams of sodium chloride (NaCl) are required to saturate 500cm3 of water at 100ºC?

13. Which compound is least soluble in water at 12ºC?

14. At 80ºC, 100cm3 of water is saturated with KCl. How many grams of KCl will precipitate when the solution is cooled to 45ºC?

15. A saturated solution of which compound contains 130g of solute per 100cm3 of water at 70ºC?

16. How many grams of NaNO3 are required to saturate 200cm3 of water at 10ºC?

17. Which saturated solution of a chloride (~Cl) has the greatest percentage by mass of solute at 60ºC?

18. A saturated solution of KNO3 was made with 300cm3 of water at 70ºC. How much KNO3 could be recovered by evaporating the solution to dryness?

19. 500cm3 of water are used to make a saturated solution of KCl at 10ºC. How many more grams of KCl could be dissolved if the temperature were raised to 80ºC?

20. How many grams of NH4Cl will dissolve in a 200,000 liter tub of water at 70ºC?


Worksheet 2: Calculating Concentration (2 pgs)

Molarity (M)

Molarity =moles of soluteliter of solution

Solve the problems below.1. What is the molarity of a solution in which 58 g of NaCl are dissolved in 1.0 L of solution?

2. What is the molarity of a solution in which 10.0 g of AgNO3 dissolved in 500. mL of solution?

3. How many grams of KNO3 should be used to prepare 2.00 L of a 0.500 M solution?

MOLARITY BY DILUTION

Acids are usually acquired from chemical supply houses in concentrated from. These acids are diluted to the desired concentrated by adding water. Since moles of acid before dilution = moles of acid after dilution, and moles of acid = M x V then, M1 x V1 = M2 x V2.

Solve the following problems.1. How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution?

2. To how much water should 100. mL of 18 M sulfuric acid be added to prepare a 1.5 M solution?


MOLALITY (m)

molality =moles of solutekg of solvent

Solve the problems below.1. What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg or water?

2. What is the molality of a solution in which 25 g of NaCl is dissolved in 2.0 Kg of water?

NORMALITY (N)

Normality = molarity x total positive oxidation number of soluteExample: What is the normality of 3.0 M of H2SO4?Solution: Since the total positive oxidation number of H2SO4 is +2 (since 2 H+)

N = +2(3.0M) = 6.0

Solve the problems below.1. What is the normality of a 2.0 M NaOH solution?

2. What is the normality of a 2.0 M H3PO4 solution?

3. A solution of H2SO4 is 3.0 N. What is the molarity?


Worksheet 3: Colligative PropertiesEquations & constants:molality(m)=mol solute/kg solventMolarity(M)=mol solute/L sol’n

ΔTb = Kbm ΔTf = Kfm

Kb for water = 0.52 °C/mKf for water = 1.86°C/m

1) What are the new freezing point and boiling point of a 3.77 m glucose, C6H12O6, solution?

2) A solution is made by dissolving 546 g of glucose, C6H12O6, in 884.5 g of water.

a) What is the molality of this solution?

b) What is the new freezing point of this solution?

c) What is the new boiling point of this solution?

3) What mass of sucrose, C12H22O11, should be dissolved in 883.5 g of water to produce a solution that freezes at –4.15°C? to produce a solution that boils at 102.6C?

4) What mass of water should be used to dissolve 345.5 g of glucose, C6H12O6, to produce a solution that boils at 102.7°C? to produce a solution that freezes at –3.15C?

5) When 33.2 g of napthalene, C10H8, are dissolved in 683.7 g of nitrobenzene, the freezing point of the solution is 3.05°C. The freezing point of pure nitrobenzene is 5.7°C. Calculate Kf for nitrobenzene.

6) When 71.2 g of anthracene, C14H10, are dissolved in 417.2 g of benzene, the boiling point of the solution is 82.53°C. The boiling point of pure benzene is 80.1°C. Calculate Kb for benzene.

7) What is the new freezing point when 59.0 g of NaCl are dissolved into 2.00 kg of water?

8) What is the boiling point of a 4.2 m solution of MgCl2?

9) What is the boiling point of a 4.2 m solution of AlCl3?

10) What is the boiling point of a 4.2 m solution of Ca3(PO4)2?


Worksheet 4: Chapter 15 Study Guide (2 pgs)

Chapter 15: Solutions Solute vs. solvent Factors affecting rate of dissolution Factors affecting solubility (temperature & pressure; how they affect solubility of a solid in liquid vs. a gas in liquid) Concentration of solutions: molarity, molality, normality, mass % Colligative Properties: boiling point elevation (Tb = Kbm);

freezing point depression (Tf = Kfm) Determining molecular weight by freezing point depression or boiling point elevation Dissociation of electrolytes & colligative properties (Tb = Kbmn, or Tf = Kfmn)

Ch. 15 Practice ProblemsEquations & Constants: (this is what you will be given on the test)

Tb = Kbm (or Tb =Kbmn) Kb for water = 0.52 °C/mTf = Kfm (or Tf = Kfmn) Kf for water = 1.86 °C/m

1. What is the molarity of a 2.05 L solution that is made by dissolving 27.5 g of sodium chloride in water?

2. What is the normality of a 6.0 M solution of phosphoric acid?

3. What is the molality of a solution that is made by dissolving 98.7 g of calcium chloride in 755 g of water?

4. How many grams of Kl are needed to prepare 325 mL of a 1.33 M solution?

5. How many grams of NaCl are needed to prepare a 4.15 m solution, using 176 g of water?

6. Ethyl iodide (C2H5I) boils at 72.5 C and has a density of 1.933 g/mL. a. A solution prepared by dissolving 0.300 mol of a nonelectrolyte in 750.0 mL of ethyl iodide boils at 73.5 C.

What is the boiling point constant (Kb) for ethyl iodide?

b. Another solution is prepared by dissolving 12.5 g of an unknown nonelectrolyte in 100.0 mL of ethyl iodide. The resulting solution boils at 74.9 C. What is the molar mass (molecular weight) of the compound?

Honors text: Chapter 15 Unit 097. What is the freezing point of a solution of 74.2 g of sucrose (C12H22O11) in 883 g of water?

8. What is the boiling point of a solution of 9.75 g of sodium chloride, an electrolyte, in 115 g of water?

9. How many grams of sodium chloride, an electrolyte, should be dissolved in 700.0 g of water to produce a solution that freezes at –3.75°C?

10. A solution was made by dissolving 4.32 g of an unknown covalent solute in 108.5 g of acetone. The solution boiled at 58.75°C. The boiling point of pure acetone is 55.95°C and Kb= 1.71 °C/m. Calculate the molar mass of the solute.

11. Describe how you would prepare 225.0 mL of a 0.775 M solution of ammonium nitrite starting with a. solid ammonium nitrite

b. 2.25 M ammonium nitrite solution

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