11.1 Solution Composition

11.2 The Energies of Solution Formation

11.3 Factors Affecting Solubility

11.4 The Vapor Pressures of Solutions

11.5 Boiling-Point Elevation and Freezing-Point Depression

11.6 Osmotic Pressure

11.7 Colligative Properties of Electrolyte Solutions

11.8 Colloids

Solution Properties

Various Types of Solutions

ExampleState of Solution

State of Solute

State of Solvent

Air, natural gas Gas Gas Gas

Vodka, antifreeze Liquid Liquid Liquid

Brass Solid Solid Solid

Carbonated water (soda) Liquid Gas Liquid

Seawater, sugar solution Liquid Solid Liquid

Hydrogen in platinum Solid Gas Solid

Solution Composition

AA

moles of soluteMolarity ( ) =

liters of solution

mass of soluteMass (weight) percent = 100%

mass of solution

molesMole fraction ( ) =

total moles of solution

moles of soluteMolality ( ) =

kilogram of s

M

molvent

Molarity

moles of soluteMolarity ( ) =

liters of solution M

Exercise #1

You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

8.00 M

Exercise #2

You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

0.200 L

Exercise #3

Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity.

10.0 M NaOH

5.37 M KCl

Mass Percent

mass of soluteMass (weight) percent = 100%

mass of solution

Exercise #4

What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?

6.6%

Mole Fraction

AA

molesMole fraction ( ) =

total moles of solution

Exercise #5

A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4.

(Assume water has a density of 1.00 g/mL.)

0.0145

Molality

moles of soluteMolality ( ) =

kilogram of solvent m

Exercise #6

A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

0.816 m

Formation of a Liquid Solution

1. Separating the solute into its individual components (expanding the solute).

2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).

3. Allowing the solute and solvent to interact to form the solution.

Steps in the Dissolving Process

Steps in the Dissolving Process

• Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.

• Step 3 usually releases energy.

• Steps 1 and 2 are endothermic, and step 3 is often exothermic.

Enthalpy (Heat) of Solution

• Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps:

ΔHsoln = ΔH1 + ΔH2 + ΔH3

• ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released).

Enthalpy (Heat) of Solution

Concept Check

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

The Energy Terms for Various Types of Solutes and Solvents

H1 H2 H3 Hsoln 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

• One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed.

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

• Overall, remember that “like dissolves like”.

• Structural Effects: Polarity – “like dissolves like”

• Pressure Effects: Henry’s law – for solubility of gases

• Temperature Effects: Affecting aqueous solutions

Factors Affecting Solubility

Pressure Effects

• Henry’s law: c = kPc = concentration of dissolved gas

k = constant

P = partial pressure of gas solute above the solution

• Amount of gas dissolved in a solution is directly proportional to the partial pressure of gas above the solution.

A Gaseous Solute

Temperature Effects (for Aqueous Solutions)

• Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature.

• Predicting temperature dependence of solubility is very difficult.

• Solubility of a gas in solvent typically decreases with increasing temperature.

The Solubilities of Several Solids as a Function of Temperature

The Solubilities of Several Gases in Water

Ideal Solution: One that obeys Raoult’s Law

Ideal Solutions Consisting of Two Volatile Liquids

• Two volatile Liquids form ideal solution if: – they are structurally very similar, and

– molecular interactions between nonidentical molecules were relatively similar to identical molecules.

• The vapor of each liquid obeys Raoult’s Law:

PA = XAPoA; PB = XBPo

B

PT = PA + PB = XAPoA + XBPo

B

(X : mole fraction; Po : vapor pressure of pure liquid)

Summary of the Behavior of Various Types of Solutions of Two Volatile Liquids

Interactive Forces Between Solute (A) and

Solvent (B) ParticlesHsoln

T for Solution

Formation

Deviation from

Raoult’s Law

Example

A A, B B A B Zero ZeroNone (ideal

solution)

Benzene-toluene

A A, B B < A BNegative

(exothermic)Positive Negative

Acetone-water

A A, B B > A BPositive

(endothermic)Negative Positive

Ethanol-hexane

Vapor Pressure for a Solution of Two Volatile Liquids

Laboratory Fractional Distillation Apparatus

Fractional Distillation Towers in Oil Refinaries

Refined Crude Oil Mixtures

Concept Check

For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation?

a) Hexane (C6H14) and chloroform (CHCl3)

b) Ethyl alcohol (C2H5OH) and water

c) Hexane (C6H14) and octane (C8H18)

Exercise #7

• A solution of benzene (C6H6) and toluene (C7H8) contains 50.0% benzene by mass. The vapor pressures of benzene and pure toluene at 25oC are 94.2 torr and 28.4 torr, respectively. Assuming ideal behavior, calculate the following:

(a) The mole fractions of benzene and toluene;

(b) The vapor pressure of each component in the mixture, and the total vapor pressure above the solution.

(c) The composition of the vapor in mole percent.

Exercise #8

• A solution composed of 24.3 g acetone (CH3COCH3) and 39.5 g of carbon disilfide (CS2) has a measured vapor pressure of 645 torr at 35oC.

(a) Is the solution ideal or nonideal?

(b) If not, does it deviate positively or negatively from Raoult’s law?

(c) What can you say about the relative strength of carbon disulfide-acetone interactions compared to the acetone-acetone and carbon disulfide-carbon disulfide interaction?

(Vapor pressures at 35oC of pure acetone and pure carbon disulfide are 332 torr and 515 torr, respectively.)

An Aqueous Solution and Pure Water in a Closed Environment

Liquid/Vapor Equilibrium

Vapor Pressure Lowering: Addition of a Solute

Vapor Pressures of Solutions of Nonvolatile Solutes

• Nonvolatile solute lowers the vapor pressure of solvent.

• Raoult’s Law:

Psoln = vapor pressure of solution

solv = mole fraction of solvent

= vapor pressure of pure solvent

soln solv solv = P P

solvP

Lowering of solvent vapor pressure Freezing-point depression Boiling-point elevation Osmotic pressure

Colligative properties depend only on the number, not on the identity, of the solute particles in an ideal solution.

Colligative Properties

Lowering of Solvent Vapor Pressure

• The presence of nonvolatile solute particles lowers the number of solvent molecules in the vapor that is in equilibrium with the solution.

• The solvent vapor pressure is lowered;• Assuming ideal behavior, the lowering of vapor

pressure is proportional to the mole fraction of solute:

P = Xsolute.Posolvent (for nonelectrolytes)

= iXsolute.Posolvent (for electrolytes)

(i is the van’t Hoff’s factor, which approximately relates to the number of ions per formula unit of the compound)

Changes in Boiling Point and Freezing Point of Water

• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.

• ΔT = Kfmsolute (for nonelectrolytes)

= iKfmsolute (for electrolytes)

ΔT = freezing-point depression

Kf = freezing-point depression constant

msolute = molality of solute

Freezing-Point Depression

Freezing Point Depression: Solid/Liquid Equilibrium

Freezing Point Depression: Addition of a Solute

Freezing Point Depression: Solid/Solution Equilibrium

• Nonvolatile solute elevates the boiling point of the solvent.

• ΔT = Kbmsolute

ΔT = boiling-point elevation

Kb = boiling-point elevation constant

msolute = molality of solute

Boiling-Point Elevation

Boiling Point Elevation: Liquid/Vapor Equilibrium

Boiling Point Elevation: Addition of a Solute

Boiling Point Elevation: Solution/Vapor Equilibrium

Exercise #9

• What mass of ethylene glycol (C2H6O2), in grams, must be added to 1.50 kg of water to produce a solution that boils at 105oC?

(Boiling point elevation constant for water is Kb = 0.512oC/m)

At what temperature will the solution freeze?

(Freezing point depression constant for water is Kf = 1.86oC/m)

• Osmosis – flow of solvent into the solution through a semipermeable membrane.

• = MRT

= osmotic pressure (atm)

M = molarity of the solution

R = gas law constant

T = temperature (Kelvin)

Osmotic Pressure

Osmosis

• The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as:

van’t Hoff Factor, i

moles of particles in solution =

moles of solute dissolvedi

Modified Equations for the Colligative Properties of Electrolytes

= T imK

= iMRT

Ion Pairing• At a given instant a small percentage of the

sodium and chloride ions are paired and thus count as a single particle.

• Ion pairing is most important in concentrated solutions.

• As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.

• Ion pairing occurs to some extent in all electrolyte solutions.

• Ion pairing is most important for highly charged ions.

Ion Pairing

Exercise #10

A solution was prepared by dissolving 25.00 g glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution.

100.35 °C

Exercise #11

You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

72.8% sucrose and 27.2% sodium chloride;

mole fraction of the sucrose is 0.313

Exercise #12

A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of – 0.246°C. Will the cell explode/expand, shrivel, or do nothing?

Exercise #13

When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.

111 g/mol

• The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur).

NaCl i = 2 KNO3 i = 2

Na3PO4 i = 4

Examples

Colloidal Mixtures

• A suspension of tiny particles in some medium.

• Tyndall effect – scattering of light by particles.

• Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.

Scattering of Light by Colloid Particles

Tyndall Effect of Colloidal Mixture

Tyndall Effect of Morning Mist

Types of Colloids

Micelle – A Colloidal Suspension

Micelle in Soap Bubbles

• Destruction of a colloid.

• Usually accomplished either by heating or by adding an electrolyte.

Coagulation