Lecture 15. Solutions and Colligative Properties

GENERAL CHEMISTRY

Solutions and Their

Physical Properties LECTURE

15

Solutions and Their Physical Properties

CONTENTS

15-1 Types of Solutions: Some

Terminology

15-2 Solution Concentration

15-3 Intermolecular Forces and

the Solution Process

15-4 Solubility of Gases

15-5 Vapor Pressures of

Solutions

15-6 Osmotic Pressure

15-7 Freezing-Point Depression

and Boiling-Point Elevation

of Nonelectrolyte Solutions

15-8 Solutions of Electrolytes

15-1 Types of Solution: Some

Terminology

Solutions are homogeneous mixtures and are uniform throughout.

Solvent

Determines the state of matter in which the solution exists.

Is the largest component.

Solutes

Other solution components said to be dissolved in the solution.

15-2 Solution Concentration

Mass Percent (msolute/msolution)

Volume Percent (vsolute/vsolution)

Mass/Volume percent (msolute/vsolution)

Example: Isotonic saline is prepared by dissolving 0.9 g of NaCl in 100 mL of water. The solution is said to be:

0.9% NaCl (mass/volume)

- quantity of solute in a given amount of solvent (or solution)

Mole Fraction and Mole Percent

i = Amount of component i (in moles)

Total amount of all components (in moles)

1 + 2 + 3 + n = 1

Mole % i = i 100%

1 = n1

n1 + n2 + n3 +... =

n1 ntotal

Molarity and Molality

Molarity (M) = Amount of solute (in moles)

Volume of solution (in liters)

Molality (m) = Amount of solute (in moles)

Mass of solvent (in kilograms)

SAMPLE PROBLEM 15.1 Calculating Molality

SOLUTION:

What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 (MW = 110.98 amu) in 271 g of water?

molality =

= 0.288 mol CaCl2

271 g H2O

0.288 mol CaCl2

kg

103 g x

= 1.06 m CaCl2

32.0 g CaCl2 mol CaCl2

110.98 g CaCl2 x

moles of CaCl2 kg of water

molality =

SAMPLE PROBLEM 15.2 Calculating Molality

What mass of BaCl2 (MW = 208.23 amu) is needed to prepare 250 mL of 0.680 m BaCl2 aqueous solution? (Assume the density of the solution is 1.00 g/mL)

Mole Fraction

A sample of rubbing alcohol contains 142 g of isopropyl alcohol, C3H7OH (MW =60.09 amu) and 58.0 g of water (MW =18.02 amu). What are the mole fractions of isopropyl alcohol and water in rubbing alcohol?

mol isopropyl alcohol = 142 g x mol

60.09 g = 2.36 mol C3H7OH

mol water = 58.0 g x mol

18.02 g = 3.22 mol H2O

2.36 mol C3H7OH

2.36 mol C3H7OH + 3.22 mol H2O

3.22 mol H2O

2.36 mol C3H7OH + 3.22 mol H2O

C3H7OH =

SOLUTION:

= 0.423

H2O == 0.577

SAMPLE PROBLEM 15.3

Expressing a Solution Concentration in Various Units

SOLUTION:

An ethanol-water solution is prepared by dissolving 10.00 mL of ethanol, C2H5OH (d =0.789 g/mL) in a sufficient volume of water to produce 100.0 mL solution with a density of 0.982 g/mL. What is the concentration of the solution, expressed as (a) volume percent; (b) mass percent , (c) mass/volume percent

(a) Volume percent ethanol

= 10.00% x 100 % 10.00 mL ethanol

100.0 mL soln volume percent ethanol =

SAMPLE PROBLEM 15.4

An ethanol water solution is prepared by dissolving 10.00 mL of ethanol, C2H5OH (d =0.789 g/mL) in a sufficient volume of water to produce 100.0 mL solution with a density of 0.982 g/mL.

(b) Mass percent ethanol

= 7.89 g 10.0 mL ethanol x 0.789 g ethanol

1 mL ethanol mass ethanol =

= 98.2 g 100.0 mL solution x 0.982 g ethanol

1 mL ethanol mass solution =

= 8.03 % x 100 % 0.789 g ethanol

0.982 g solution mass % ethanol =

SOLUTION:


SAMPLE PROBLEM 15.4

An ethanol water solution is prepared by dissolving 10.00 mL of ethanol, C2H5OH (d =0.789 g/mL) in a sufficient volume of water to produce 100.0 mL solution with a density of 0.982 g/mL.

(c) Mass/volume percent ethanol

= 7.89 g 10.0 mL ethanol x 0.789 g ethanol

1 mL ethanol mass ethanol =

= 7.89 % x 100 % 0.789 g ethanol

100 mL solution mass/volume percent =

SOLUTION:


SAMPLE PROBLEM 15.4

Converting Concentration Terms

SOLUTION:

Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Molar mass of H2O2 is 34.02 amu. Calculate its:

(a) Molality (b) Mole fraction of H2O2 (c) Molarity

(a) g of H2O = 100.0 g solution - 30.0 g H2O2 = 70.0 g H2O

molality =

70.0 g H2O x kg H2O 103 g

= 12.6 m H2O2

30.0 g H2O2 x 34.02 g H2O2

mol H2O2 moles of H2O2 = = 0.882 mol H2O2

0.882 mol H2O2

SAMPLE PROBLEM 15.5


0.882 mol H2O2

70.0 g H2O x mol H2O

18.02 g H2O = 3.88 mol H2O

0.882 mol H2O2 + 3.88 mol H2O = 0.185 of H2O2

(b) Mole fraction of H2O2

Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Molar mass of H2O2 is 34.02 amu. Calculate the:

30.0 g H2O2 x 34.02 g H2O2


moles of H2O =

SAMPLE PROBLEM 15.5


100.0 g solution x mL

1.11 g = 90.1 mL solution

0.882 mol H2O2

90.1 mL solution x L

103 mL

= 9.79 M H2O2

(c) Molarity

Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Molar mass of H2O2 is 34.02 amu. Calculate the:

30.0 g H2O2 x 34.02 g H2O2


molarity =

SAMPLE PROBLEM 15.5

Pure solvent separated solvent molecules DHsolvent > 0

DHsoln = DHsolute + DHsolvent + DHmix

Pure solute separated solute molecules DHsolute > 0

separated solute solution DHmix < 0

and solvent molecules

Pure solute + pure solvent solution DHsoln

(a)

(b)

(c)

15-3 Intermolecular Forces and the

Solution Process

Enthalpy diagram for solution formation

Magnitude of Hsolute, Hsolvent, and Hmix depends on intermolecular forces.

Examples

Predict the DHsoln of the following sets of solutions

Acetone and CO2 Methanol and water

Hexane and pentane

Concept Check

Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how H plays a role.

The Energy Terms for Various Types of

Solutes and Solvents

DH1 DH2 DH3 DHsoln Outcome

Polar solute, polar

solvent

Large Large Large,

negative

Small Solution

forms

Nonpolar solute, polar

solvent

Small Large Small Large,

positive

No solution

forms

Nonpolar solute,

nonpolar solvent

Small Small Small Small Solution

forms

Polar solute, nonpolar

solvent

Large Small Small Large,

positive

No solution

forms

In General

Processes that require large amounts of energy tend not to occur.

Overall, remember that like dissolves like.

(c) Ethanol. Diethyl ether can interact through dipole-dipole and dispersion forces. Ethanol can provide both while water would like to H bond.

(b) Water. Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol.

Predicting Relative Solubilities of Substances

SOLUTION:

Predict which solvent will dissolve more of the given solute:

(a) Sodium chloride in methanol (CH3OH) or in 1-propanol (CH3CH2CH2OH)

(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)

or in water. (c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)

(a) Methanol. NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and 1-propanol. However, 1-propanol is subject to the dispersion forces to a greater extent.

SAMPLE PROBLEM 15.6

15-4 Solution Formation and

Equilibrium

Formation of a saturated solution

solute (undissolved) solute (dissolved)

Equilibrium in a saturated solution

Supersaturated

Recrystallization of

KNO3

Aqueous solubility of several

salts as a function of temperature

Unsaturated

Sodium acetate crystallizing from a

supersaturated solution.

Effect of Temperature on Solubility

kJ/mol8.48Br Li LiBr-+

s aqaq

aqaq-

4

+

s4MnO K kJ/mol 6.43KMnO

Upon dissolution, will the solution feel cold or warm?

What will happen to the solubility of LiBr if the solution is warmed?

Upon dissolution, will the solution feel cold or warm?

What will happen to the solubility of KMnO4 if the solution is warmed?

Warm

Cold

Solubility decreases

Solubility increases

15-5 Solubility of Gases

Effect of temperature on the solubilities of gases

Effect of Temperature

Most gases are less soluble in water as temperature increases.

Effect of Pressure

William Henry found that the solubility of a gas increases with increasing pressure.

C = kPgas

Effect of pressure on the solubility of a gas

15-6 Vapor Pressures of Solutions

Raoults Law

Dissolved solute lowers vapor pressure of solvent.

The effect of a solute on the vapor pressure of a solution.

15-6 Vapor Pressures of Solutions

solvent Psolvent = solvent P

Psolvent = partial pressure exerted by solvent vapor above the ideal solution

solvent = mole fraction of solvent in the solution Psolvent = vapor pressure of solvent at a given temperature.

Using Raoults Law to Find the Vapor Pressure Lowering

Calculate the vapor pressure at 25 C of a solution containing 165 g of the nonvolatile solute, glucose, C6H12O6 (MW=180 amu) in 685 g H2O. The vapor pressure of water at 25 C is 23.8 mmHg

SOLUTION:

165 g C6H12O6 x mol C6H12O6

180 g C6H12O6 = 0.917 mol C3H8O3

685 g H2O x mol H2O

18.02 g H2O = 38.01 mol H2O

P = 38.01 mol H2O

38.01 mol H2O + 0.917 mol C3H8O3

23.8 mmHg x = 23.24 mmHg

Psolvent = Xsolvent x Po

solvent

SAMPLE PROBLEM 15.7

For liquid-liquid solutions, where both

components are volatile:

Ptotal = A PA+ B PB

Calculating the vapor pressure of a solution containing two liquids

A solution is prepared by mixing 5.81 g acetone (MW =58.1 amu) and 11.9 g chloroform (Mw= 119.4 amu). At 35 C, the vapor pressure of pure acetone and pure chloroform are 345 torr and 293 torr, respectively. Assuming it is an ideal solution, what is the vapor pressure exerted by the solution at 35 C?

SAMPLE PROBLEM 15.8

Answer: PTotal = 319 torr

Vapor-pressure lowering by a nonvolatile solute


and Boiling Point Elevation of

Solutions

Vapor pressure is lowered when a solute is present.

This results in boiling point elevation.

Freezing point is also effected and is lowered.

Tf = Kf m Tb = Kb m

Tf = Tf, solvent Tf, solution


and Boiling Point Elevation of

Nonelectrolyte Solutions

Tb = Tb, solution Tb, solvent

m = molality Kf and Kb= constants that depends on the solvent

Practical Applications

Determining the Boiling Point Elevation and Freezing Point Depression of a Solution

You add 1.00 kg of ethylene glycol (Mw of C2H6O2 = 62.07 amu) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution?

SOLUTION:

1.00 x 103 g C2H6O2 x mol C2H6O2

62.07 g C2H6O2 = 16.1 mol C2H6O2

DTbp = 0.512oC/m

16.1 mol C2H6O2

4.450 kg H2O = 3.62 m C2H6O2

3.62 m x = 1.85oC bp = 100.00oC + 1.85oC = 101.85oC

fp = 0.00oC - 6.73oC = -6.73oC DTfp = 1.86oC/m 3.62 m = 6.73oC x

Tf = Kf m Tb = Kb m

SAMPLE PROBLEM 15.9

A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?

Determining the Boiling Point Elevation and Freezing Point Depression of a Solution

m =

5.58 C

1.86 C/m = 3.0 m

Tf

Kf m =

Tf = Kf m mol solute

kg solvent

=

3.0 m X 0.2 kg Mol of compound = = 0.60 mol

Molar mass = 37.0 g

0.600 mol = 61.7 g/mol

SOLUTION:

SAMPLE PROBLEM 15.10

15-8 Osmotic Pressure

Osmotic Pressure

V = nRT

= RT n V

= MRT

= Osmotic pressure M = Molarity of solution

R = Ideal gas law constant (0.0821 L-atm/mol-K)

T = Temperature in K

Determining Molar Mass from Osmotic Pressure

Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studying a variety associated with a fatal disease first finds its molar mass. She dissolves 21.5 mg of the protein in water at 5.0oC to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin?

SOLUTION:

M =

RT =

3.61 torr x atm

760 torr

(0.0821 Latm/molK)(273.15 K + 5.0) = 2.08 x 10-4 M

2.08 x 10-4 mol

L x 1.50 mL x

103 mL

L = 3.12 x 10-7 mol

21.5 mg x g

103 mg 1

3.12 x 10-7 mol = 6.89 x 104 g/mol x


Colligative Properties

Vapor Pressure Lowering

Psolvent = Xsolvent x Po

solvent

Ptotal = XA Po

A + XB Po

B

Boiling Point Elevation and Freezing Point Depression

DTb = Kbm DTf = Kf m

Osmotic Pressure

M R T

15-9 Solutions of Electrolytes

Svante Arrhenius

Nobel Prize 1903.

Ions form when electrolytes dissolve in solution.

Explained anomalous colligative properties.

Tf = -Kf m = -1.86C m-1 0.0100 m = -0.0186C

Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)

Freezing point depression for NaCl is -0.0361C.

Vant Hoff

Tf = i Kf m

i = = = 1.98 measured Tf

Tb = i Kb m

expected Tf

0.0361C

0.0186C

= i M RT

i = 1 (nonelectrolyte solute) i = # of ions in solution (electrolyte solute in ideal solution)

Depicting a Solution to Find Its Colligative Properties

A 0.952-g sample of magnesium chloride is dissolved in 100. g of water.

(b) Assuming the solution is ideal, what is its freezing point (at 1 atm)?

(a) Which scene depicts the solution best?

SOLUTION:

(a) MgCl2 (s) Mg (aq) + 2Cl- (aq)

The correct depiction must be A with a ratio of 2 Cl-/ 1 Mg2+.


(b) molality (m) = 0.0100 mol MgCl2

100. g x 103 g

1 kg = 0.100 m MgCl2

Assuming this is an IDEAL solution, the vant Hoff factor, i, should be 3.

DTf = i (Kf m) = 3(1.86oC/m x 0.100 m) = 0.558oC

Tf = 0.000oC - 0.558oC = - 0.558oC

SOLUTION:

A 0.952-g sample of magnesium chloride is dissolved in 100. g of water.

(b) Assuming the solution is ideal, what is its freezing point (at 1 atm)?

DTf = i (Kf m)

Depicting a Solution to Find Its Colligative Properties


Additional Exercises

1. Which has a higher concentration of lithium cations, 0.05 M Lithium phosphate or 0.07 M lithium sulfate? (2 pts)

2. Identify which of the following vitamins are water soluble and fat soluble. (2 pts)

3. Calculate the molality and the molarity of an aqueous solution that is 10.0% (m/m) glucose, C6H12O6. The density of the solution is 1.04 g/mL. 1 mol C6H12O6 = 180 g (8pts)

1. Which has a higher concentration of lithium cations, 0.05 M lithium phosphate or 0.07 M lithium sulfate?

* 0.05 M Li3PO4 (2 pts)

Li3PO4 (s) 3Li+ (aq) + PO4

3- (aq)

0.05 M

Li2SO4 (s) 2Li+ (aq) + SO4

2- (aq)

0.07M

Answer

3(0.05M) 0.05M

2(0.07M) 0.07M

2. Identify which of the following vitamins are water soluble and fat soluble. (2 pts)

Water soluble Vitamin C

Fat Soluble- Vitamin A and D

3. Calculate the molality and the molarity of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. (1 mol C6H12O6 = 180 g)

.molalityin ion concentrat theis This

OHC 617.0

OHC g 180

OHC mol 1

OH kg 1

OH g 1000

OH g 90.0

OHC g 0.10

OH kg

OHC mol

6126

6126

6126

2

2

2

6126

2

6126

m

m

3. Calculate the molality and the molarity of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. (1 mol C6H12O6 = 180 g)

126

6126

61266126

6126

HC 578.0

OHC g 180

OHC mol 1

L 1

mL 1000

nsol' mL

nsol' g 04.1

n sol' g 100.0

OHC g 0.10

nsol' L

OHC mol

M

M

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Lecture 15. Solutions and Colligative Properties