CHAPTER 14CHAPTER 14

SolutionsSolutions

The Dissolution ProcessThe Dissolution Process

Solutions are homogeneous mixtures of two or more substances Dissolving medium is called the solventsolvent Dissolved species are called the solutesolute

Upon dissolution, solute molecules are surrounded (solvated) by solvent molecules

Solvent molecules have to rearrange around solute molecules

NaCl(s) Na+(aq) + Cl–

(aq)

H2O

Dissolution of Ionic SaltsDissolution of Ionic Salts Forces to overcome (energy loss):

Cation-anion electrostatic attraction

Attractive forces between solvent molecules

Energy gain: Solvent-cation and solvent-anion

electrostatic attractions Increased disorder

If energy gain exceeds energy loss, the compound is soluble

Colligative Properties of Colligative Properties of SolutionsSolutions

When solute is introduced into the volume of solvent, the solvent properties change

Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution

We will look at three types of colligative properties of solutions:

Vapor pressure lowering

Freezing point depression

Boiling point elevation

Raoult’s LawRaoult’s Law Rearrangement of solvent molecules and new

solvent-solute interactions affect the vapor pressure of the solvent – it becomes lower

Raoult’s Law: The vapor pressure (P ) of a solvent in an

ideal solution is directly proportional to the mole fraction (X ) of the solvent in the solution

Psolvent = Xsolvent P 0solvent

(P 0 – vapor pressure of the pure solvent)

This is true only for non-volatile non-ionizing solutes at low solute concentrations

Mole FractionMole Fraction For the mixture of components A, B, C, …

... C moles# B moles# A moles# A of moles#

A X

... C moles# B moles# A moles# B of moles#

B X

... C moles# B moles# A moles# C of moles#

C X

1 iX

Mole Fraction: ExampleMole Fraction: Example

What are the mole fractions of glucose (C6H12O6) and water in a 10.0% glucose solution?

Mole FractionMole Fraction

If a solution contains only one type of solute A

components all of moles# A of moles#

AX

solvent of moles# A of moles# A of moles#

AX

solvent of moles# A of moles# solvent of moles#

solvent X

1solventA XX

Raoult’s LawRaoult’s Law P 0 – the vapor pressure of the pure

solvent

Psolvent – the vapor pressure of the solvent over the solution of compound A:

Psolvent = Xsolvent P 0

solvent

Psolvent = (1–XA) P 0

solvent

The vapor pressure lowering:

Psolvent = P 0

solvent – Psolvent

Psolvent = XA P 0

solvent

Psolvent = (1–XA) P 0solvent

Raoult’s LawRaoult’s Law

Raoult’s Law: ExampleRaoult’s Law: Example

Determine the vapor pressure lowering for the 10.0% glucose solution?

MolalityMolality

In Chapter 3 we introduced two important concentration units:

100%solution of masssolute of mass

= mass by %

liters in solution of volumesolute of moles#

= Molarity

Now we introduce another unit - molalitymolality

MolalityMolality

Molality (mm) – the number of moles of solute per kilogram of solvent

kg in solvent of masssolute of moles#

m

Boiling point elevation and freezing point depression are calculated based on this unit of concentration

Calculate the molality of a 10.0% aqueous solution of glucose, C6H12O6

Molality: Example 1Molality: Example 1

10 g of NaOH was dissolved in 250 mL of water. What is the molality of the obtained solution?

Molality: Example 2Molality: Example 2

Boiling Point ElevationBoiling Point Elevation According to Raoult’s law, addition of a solute

lowers the vapor pressure of the solvent:

Boiling Point ElevationBoiling Point Elevation

The boiling temperature elevation is determined by the number of moles of solute dissolved in the solution:

Tb = Kbm

Tb – the change in boiling point

Kb – boiling point elevation constant (depends only on the nature of solvent)

m – molality of the solution

Freezing Point Freezing Point DepressionDepression

Addition of a nonvolatile solute to a solution lowers the freezing point of the solution relative to the pure solvent

Tf = Kfm

Tf – the change in freezing point

Kf – freezing point depression constant (depends only on the nature of solvent)

m – molality of the solution

Calculate the freezing and boiling points of a solution that contains 8.50 g of benzoic acid (C6H5COOH) in 75.0 g of benzene, C6H6

Example 1Example 1

3.75 g of a nonvolatile compound was dissolved in 108.7 g of acetone. The solution boiled at 56.58°C. The boiling point of pure acetone is 55.95°C and Kb = 1.71°C/m. Calculate the molecular weight of the compound.

Example 2Example 2

Example 2 (continued)Example 2 (continued)

Assignments & Assignments & RemindersReminders

Read Sections 14-8, 14-9, 14-11, 14-12 & 14-13

Homework #8 is due by 12/07 @ 9:00 p.m.

Review Sessions: Sunday – 5:15-8:00 p.m. in 100 Held Wednesday – 5:15-8:00 p.m. in 100 Held

Final Exam:Final Exam:

Friday – 3:00-5:00 p.m. Friday – 3:00-5:00 p.m.

in 100 Heldin 100 Held