chemistry: atoms first julia burdge & jason overby chapter 13 physical properties of solutions...

Chemistry: Atoms FirstJulia Burdge & Jason Overby

Chapter 13

Physical Properties of Solutions

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

Physical Properties of SolutionsPhysical Properties of Solutions13

13.1 Types of Solutions13.2 A Molecular View of the Solution Process

The Importance of Intermolecular ForcesEnergy and Entropy in Solution Formation

13.3 Concentration UnitsMolalityPercent by MassComparison of Concentration Units

13.4 Factors that Affect SolubilityTemperaturePressure

13.5 Colligative PropertiesVapor-Pressure LoweringBoiling-Point ElevationFreezing-Point DepressionOsmotic PressureElectrolyte Solutions

13.6 Calculations Using Colligative Properties13.7 Colloids

Types of SolutionsTypes of Solutions

A solution is a homogeneous mixture of two or more substances.

A solution consists of a solvent and one or more solutes.

13.1

Solutions can be classified by the amount of solute dissolved.

An unsaturated solution is one that contains less solute than the solvent has the capacity to dissolve at a specific temperature.


Solutions can be classified by the amount of solute dissolved.

A saturated solution is one that contains the maximum amount of solute that will dissolve in a solvent at a specific temperature.


Supersaturated solutions are generally unstable.


A Molecular View of the Solution ProcessA Molecular View of the Solution Process

Solvation occurs when solute molecules are separated from one another and surrounded by solvent molecules.

Solvation depends on three types of interactions:

1) Solute-solute interactions

2) Solvent-solvent interactions

3) Solute-solvent interactions

13.2

A Molecular View of the Solution ProcessA Molecular View of the Solution Process

Solute Solvent

SolventSeparated solute

Separated solute Separated solvent

Solution

Step 2

ΔH2 > 0

Step 1

ΔH1 > 0

Step 3

ΔH3 < 0

ΔHsoln > 0

ΔHsoln = ΔH1 + ΔH2 + ΔH3

“Like dissolves like”

Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other.

The Importance of Intermolecular ForcesThe Importance of Intermolecular Forces

Toluene, C7H8 Octane, C8H18

Both non-polar liquids,solution forms when mixed

Two liquids are said to be miscible if they are completely soluble in each other in all proportions.




Polar and non-polar liquids,solution does not form when mixed

Water, H2O Octane, C8H18




Ethanol, C2H6O

Both polar liquids, solution forms when mixed

Water, H2O

Worked Example 13.1

Strategy Consider the structure of each solute to determine whether or not it is polar. For molecular solutes, start with a Lewis structure and apply the VSEPR theory. We expect polar solutes, including ionic compounds, to be more soluble in water. Nonpolar solutes will be more soluble in benzene.

Determine for each solute whether the solubility will be greater in water, which is

polar, or in benzene (C6H6), which is nonpolar: (a) Br2, (b) sodium iodide (NaI),

(c) carbon tetrachloride, and (d) formaldehyde (CH2O).

Solution (a) Bromine is a homonuclear diatomic molecule and is nonpolar. Bromine is more soluble in benzene.

(b) Sodium iodide is ionic and more soluble in water.

Worked Example 13.1 (cont.)

Solution (c) Carbon tetrachloride has the following Lewis structure:

With four electron domains around the central atom, we expect a tetrahedral arrangement. A symmetrical arrangement of identical bonds results in a nonpolar molecule. Carbon tetrachloride is more soluble in benzene.

(d) Formaldehyde has the following Lewis structure:

Crossed arrows represent individual bond dipoles. This molecule is polar and can form hydrogen bonds in water. Formaldehyde is more soluble in water.

C

Cl

Cl

Cl

Cl

C

O

H H

Think About It Remember that molecular formula alone is not sufficient to determine the shape or polarity of a polyatomic molecule. It must be determined by starting with a correct Lewis structure and applying VSEPR theory.

Concentration UnitsConcentration Units

The amount of solute relative to the volume of a solution or to the amount of solvent in a solution is called concentration.

Molarity:

Mole fraction:

13.3

moles of solutemolarity = =

liters of solutionM

A

moles of Amole fraction of component A = =

sum of moles of all components

Concentration UnitsConcentration Units

Molality (m) is the number of moles of solute dissolved in 1 kg (1000 g) solvent:

Percent by Mass:

moles of solute

molality = = mass of solvent in kg

m

mass of solutepercent by mass = 100 %

mass of solute + mass of solvent

Worked Example 13.2

Strategy Use the molar mass of glucose to determine the number of moles of glucose in a liter of solution. Use the density (in g/L) to calculate the mass of a liter of solution. Subtract the mass of glucose from the mass of solution to determine the mass of water. The molar mass of glucose is 180.2 g/mol.

A solution is made by dissolving 170.1 g of glucose (C6H12O6) in enough water to

make a liter of solution. The density of the solution is 1.062 g/mL. Express the concentration in (a) molality, (b) percent by mass, and (c) parts per million.

Solution (a)

1 liter of solution ×

1062 g – 170.1 g = 892 g water = 0.892 kg water

170.1 g180.2 g/mol = 0.9440 mol glucose per liter of solution

1062 gL

= 1062 g

0.9440 mol glucose0.892 kg water

= 1.06 m


Solution (b)

(c)

170.1 g1062 g solution × 100% = 16.02% glucose by mass

170.1 g1062 g solution × 1,000,000 = 1.602×105 ppm glucose

Think About It Pay careful attention to units in problems such as this. Most require conversions between grams and kilograms and/or liters and milliliters.

Worked Example 13.3

Strategy (a) Use density to determine the total mass of a liter of solution, and use percent by mass to determine the mass of isopropyl alcohol in a liter of solution. Convert the mass of isopropyl alcohol to moles, and divide moles by liters of solution to get molarity.

(b) Subtract the mass of C3H7OH from the mass of solution to get the mass of water. Divide moles of C3H7OH by the mass of water (in kg) to get molality.

The mass of a liter of rubbing alcohol is 790 g, and the molar mass of isopropyl alcohol is 60.09 g/mol.

“Rubbing alcohol” is a mixture of isopropyl alcohol (C3H7OH) and water that is

70 percent isopropyl alcohol by mass (density = 0.79 g/mL at 20°C). Express the concentration of rubbing alcohol in (a) molarity and (b) molality.


Solution (a)

(b) 790 g solution – 553 g C3H7OH = 237 g water = 0.237 kg water

Rubbing alcohol is 9.2 M and 39 m in isopropyl alcohol.

790 g solution

L solution

×

Think About It Note the large difference between molarity and molality in this case. Molarity and molality are the same (or similar) only for very dilute aqueous solutions.

70 g C3H7OH100 g

solution

=553 g C3H7OH

L solution

553 g C3H7OHL solution

×1 mol

60.09 g C3H7OH=

9.20 mol C3H7OHL solution

= 9.2 M

9.20 mol C3H7OH0.237 kg water

= 39 m

Factors That Affect SolubilityFactors That Affect Solubility

Temperature affects the solubility of most substances.

13.4

Pressure greatly influences the solubility of a gas.

Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.

c molar concentration (mol/L)

P pressure (atm)

k proportionality constant called Henry’s law constant

Factors That Affect SolubilityFactors That Affect Solubility

c = kP

Worked Example 13.4

Strategy Use c = kP and the given Henry’s law constant to solve for the molar concentration (mol/L) of CO2 at 25°C and the two pressures given.

Calculate the concentration of carbon dioxide in a soft drink that was bottled

under a partial pressure of 5.0 atm CO2 at 25°C (a) before the bottle is opened and

(b) after the soda has gone “flat” at 25°C . The Henry’s law constant for CO2 in

water at this temperature is 3.1×10-2 mol/L∙atm. Assume that the partial pressure

of CO2 in air is 0.0003 atm and that Henry’s law constant for the soft drink is the

same as that for water.

Solution (a) c = (3.1×10-2 mol/L∙atm)(5.0 atm) = 1.6×10-1 mol/L

(b) c = (3.1×10-2 mol/L∙atm)(0.0003 atm) = 9×10-6 mol/L

Think About It With a pressure approximately 15,000 smaller in part (b) than in part (a), we expect the concentration of CO2 to be approximately 15,000 times smaller–and it is.

Colligative PropertiesColligative Properties

Colligative properties are properties that depend on the number of solute particles in solution.

Colligative properties do not depend on the nature of the solute particles.

The colligative properties are:vapor-pressure lowering

boiling-point elevation

freezing-point depression

osmotic pressure

13.5

Raoult’s law states that the partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution.

P1 partial pressure ofsolvent over solution

P° vapor pressure of pure solvent

χ1 mole fraction of solvent

ΔP vapor pressure lowering

χ2 mole fraction of solute


P P1 1 1 P P 2 1

Worked Example 13.5

Strategy Convert the masses of glucose and water to moles, determine the mole fraction of water, and use P1 = χ1P°1 to find the vapor pressure over the solution. The molar masses of glucose and water are 180.2 and 18.02 g/mol, respectively.

Calculate the vapor pressure of water over a solution made by dissolving 225 g of glucose in 575 g of water at 35°C. (At 35°C, P° = 42.2 mmHg.)

Solution

P = χwaterP° = 0.962 × 42.4 mmHg = 40.6 mmHg

The vapor pressure of water over the solution is 40.6 mmHg.

H2O

225 g glucose180.2 g/mol = 1.25 mol glucose

575 g water18.02 g/mol = 31.9 mol waterand

31.9 mol water1.25 mol glucose + 31.9 mol

water

= 0.962

H2OH2O

χwater =

Think About It This problem can also be solved using Equation 13.5 to calculate the vapor-pressure lowering, ΔP.

If both components of a solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures.


o oT A A B BP P P

oA A AP P o

B B BP P

CH3

Benzene Toluene

An ideal solution obeys Raoult’s law.


o oT A A B BP P P

CH3

Benzene Toluene

Solutions boil at a higher temperature than the pure solvent.

ΔTb boiling point elevation

Kb boiling point elevation constant (°C/m)

m molality


ob b bT T T

b bT K m

Solutions freeze at a lower temperature than the pure solvent.

ΔTf freezing point depression

Kf freezing point depression constant

(°C/m)

m molality


off fT T T

ffT K m

Worked Example 13.6

Strategy Convert grams of ethylene glycol to moles, and divide by the mass of water in kilograms to get molal concentration. Use molal concentrations and ΔTb = Kbm and ΔTf = Kfm, respectively. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. Kf and Kb for water are 1.86°C/m and 0.52°C/m, respectively.

Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is

water soluble and fairly nonvolatile (b.p. 197°C). Calculate (a) the freezing point and (b) the boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water.

Solution

(a) ΔTf = Kfm = (1.86°C/m)(5.32 m) = 9.89°CThe freezing point of the solution is (0 – 9.89)°C = – 9.89°C

(b) ΔTb = Kbm = (0.52°C/m)(5.32 m) = 2.8°CThe boiling point of the solution is (100.0 + 2.8)°C = 102.8°C

685 g C2H6O2

62.07 g/mol= 11.04 mol C2H6O2

11.04 mol C2H6O2

2.075 kg water= 5.32 m C2H6O2and

Think About It Because it both lowers the freezing point and raises the boiling point, antifreeze is useful at both temperature extremes.

Osmosis is the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated one.


Osmotic pressure () of a solution is the pressure required to stop osmosis.

Osmotic pressure (atm)

M molarity (moles/L)

R gas constant (0.08206 L∙atm/mol∙K)

T absolute temperature (kelvins)


MRT

iMRT

Electrolytes undergo dissociation when dissolved in water.

The van’t Hoff factor (i) accounts for this effect.

actual number of particles in solution after dissociationnumber of formulas units initially dissolved in solution

i


b bT iK m

ffT iK m

The van’t Hoff factor (i) is 1 for all nonelectrolytes:

For strong electrolytes i should be equal to the number of ions:

1 particle dissolved, i = 1


NaCl(s) Na+(aq) + Cl–(aq)H2O

C12H22O11(s) C12H22O11(aq)H2O

Na2SO4(s) 2Na+(aq) + SO42–(aq)H2O

2 particles dissolved, i = 2

3 particles dissolved, i = 3

The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.

An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.


ion pair

The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.

An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.


Concentration has an effect on experimentally measured van’t Hoff factors (i).


Worked Example 13.7

Strategy Use osmotic pressure to calculate the molar concentration of KI, and divide by the nominal concentration of 0.01000 M. R = 0.08206 L∙atm/K∙mol, and T = 298 K.

The osmotic pressure of a 0.0100 M potassium iodide (KI) solution at 25°C is 0.465 atm. Determine the experiment van’t Hoff factor for KI at this concentration.

Solution Solving π = MRT for M,

M = =

i =

The experimental van’t Hoff factor for KI at this concentration is 1.90.

πRT = 0.0190 M

0.465 atm(0.08206 L∙atm/K∙mol)(298 K)

0.0190 M0.0100 M

= 1.90

Think About It The calculated van’t Hoff factor for KI is 2. The experimental van’t Hoff factor must be less than or equal to the calculated value.

Worked Example 13.8

Strategy Use ΔTf = Kfm to determine the molal concentration of the solution. Use the density of ethanol to determine the mass of the solvent. The molal concentration of quinine multiplied by the mass of ethanol (in kg) gives moles of quinine. The mass of quinine (in grams) divided by moles of quinine gives the molar mass. Kf for ethanol is 1.99°C/m.

Quinine was the first drug widely used to treat malaria, and it remains the treatment of choice for severe cases. A solution prepared by dissolving 10.0 g of quinine in 50.0 mL of ethanol has a freezing point 1.55°C below that of pure ethanol. Determine the molar mass of quinine. (The density of ethanol is 0.789 g/mL.) Assume that quinine is a nonelectrolyte.

Solution mass of ethanol = 50.0 mL × 0.789 g/mL = 39.5 g or 3.95×10-2 kg

Solving ΔTf = Kfm for molal concentration

m = =ΔTf Kf

= 0.779 m1.55°C

1.99°C/m


Solution The solution is 0.779 m in quinine (i.e., 0.779 mol of quinine/kg ethanol solvent.)

molar mass of quinine = = 325 g/mol

(3.95×10-2 kg ethanol)0.779 mol quinine

kg ethanol

10.0 g quinine0.00308 mol

quinine

= 0.00308 mol quinine

Think About It Check the result using the molecular formula of quinine: C20H24N2O2 (324.4 g/mol). Multistep problems such as this one require careful tracking of units at each step.

Worked Example 13.9

Strategy Use π = MRT to calculate the molarity of the solution. Because the solution volume is 1 L, the molarity is equal to the number of moles of hemoglobin. Dividing the given mass of hemoglobin by the number of moles gives the molar mass. R = 0.08206 L∙atm/K∙mol, T = 298 K, and π = 14.3 mmHg/(760 mmHg/atm) = 1.88×10-2 atm.

A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is measured and found to be 14.3 mmHg at 25°C. Calculate the molar mass of hemoglobin. (Assume that there is no change in volume when the hemoglobin is added to water.)


Solution Rearranging π = MRT to solve for molarity we get,

M = =

Thus, the solution contains 7.69×10-4 moles of hemoglobin.

molar mass of hemoglobin =

πRT = 7.69×10-4 M

1.88×10-2 atm(0.08206 L∙atm/K∙mol)(298 K)

50.0 g7.69×10-4 mol

= 6.50×104 g/mol

Think About It Biological molecules can have very high molar masses.

Percent dissociation is the percentage of dissolved molecules (or formula units, in the case of an ionic compound) that separate into ions in a solution.

Strong electrolytes should have complete, or 100%, dissociation, however, experimentally determined van’t Hoff factors indicate that this is not the case.

Percent dissociation of a strong electrolyte is more complete at lower concentration.

Percent ionization of weak electrolytes is also dependent on concentration.

Calculations Using Colligative PropertiesCalculations Using Colligative Properties13.6

Worked Example 13.10

Strategy Use the osmotic pressure and π = MRT to determine the molar concentration of the particles in solution. Compare the concentration of particles to the nominal concentration (0.100 M) to determine what percentage of the original HF molecules are ionized. R = 0.08206 L∙atm/K∙mol, and T = 298 K.

A solution that is 0.100 M in hydrofluoric acid (HF) has an osmotic pressure of 2.64 atm at 25°C. Calculate the percent ionization of HF at this concentration.

Solution Rearranging π = MRT to solve for molarity,

M = =π

RT = 0.108 M2.64 atm

(0.08206 L∙atm/K∙mol)(298 K)


Solution The concentration of dissolved particles is 0.108 M. Consider the ionization of HF:

HF(aq) H⇌ +(aq) + F-(aq)

According to this equation, if x HF molecules ionize, we get x H+ ions and x F- ions. Thus, the total concentration of particles in solution will be the original concentration of HF minus x, which gives the concentration of intact HF molecules, plus 2x, which is the concentration of ions (H+ and F-):

(0.100 – x) + 2x = 0.100 + x

Therefore, 0.108 = 0.100 + x and x = 0.008. Because we earlier defined x as the amount of HF ionized, the percent ionization is given by

percent ionization =

At this concentration HF is 8 percent ionized.

0.008 M0.100 M

×100% =8%

Think About It For weak acids, the lower the concentration, the greater the percent ionization. A 0.010 M solution of HF has an osmotic pressure of 0.30 atm, corresponding to 23 percent ionization. A 0.0010 M solution of HF has an osmotic pressure of 3.8×10-2 atm, corresponding to 56 percent ionization.

A colloid is a dispersion of particles of one substance throughout another substance. Colloid particles are much larger than the normal solute molecules.

Categories of colloids:

aerosols

foams

emulsions

sols

gels

ColloidsColloids13.7

Examples of colloids

ColloidsColloids

Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing).

Hydrophilic groups on thesurface of a large moleculestabilize the molecule inwater.

ColloidsColloids

Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing).

Negative ions are adsorbed onto the surface of hydrophobic colloids.

The repulsion between like charges prevents aggregation of the articles.

ColloidsColloids

Hydrophobic colloids can be stabilized by the presence of hydrophilic groups on their surface.

ColloidsColloids

Emulsification is the process of stabilizing a colloid that would otherwise not stay dispersed.

ColloidsColloids

Key ConceptsKey Concepts13

Types of SolutionsA Molecular View of the Solution Process

The Importance of Intermolecular ForcesEnergy and Entropy in Solution Formation

Concentration UnitsMolalityPercent by MassComparison of Concentration Units

Factors that Affect SolubilityTemperaturePressure

Colligative PropertiesVapor-Pressure LoweringBoiling-Point ElevationFreezing-Point DepressionOsmotic PressureElectrolyte Solutions

Calculations Using Colligative PropertiesColloids

chemistry: atoms first julia burdge & jason overby chapter 13 physical properties of solutions...

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