II

III

I I. The Nature of Solutions

(p. 401 - 410, 425 - 433)

Ch. 13 & 14 - Solutions

A. Definitions

Solution - Solution - homogeneous mixture

Solvent Solvent - present in greater amount

Solute Solute - substance being dissolved

A. Definitions

Solute Solute - KMnO4 Solvent Solvent - H2O

B. Solvation

Solvation – Solvation – the process of dissolving

solute particles are separated and pulled into solution

solute particles are surrounded by solvent particles

B. Solvation

StrongElectrolyte

Non-Electrolyte

solute exists asions only

- +

salt

- +

sugar

solute exists asmolecules

only

- +

acetic acid

WeakElectrolyte

solute exists asions and

molecules DISSOCIATION IONIZATION

View animation online.

B. Solvation

DissociationDissociation• separation of an

ionic solid into aqueous ions

NaCl(s) Na+(aq) + Cl–(aq)

B. Solvation

IonizationIonization• breaking apart of

some polar molecules into aqueous ions

HNO3(aq) + H2O(l) H3O+(aq) + NO3–(aq)

B. Solvation

Molecular Molecular SolvationSolvation• molecules

stay intact

C6H12O6(s) C6H12O6(aq)

B. Solvation

NONPOLAR

NONPOLAR

POLAR

POLAR

““Like Dissolves Like”Like Dissolves Like”

B. Solvation

Soap/DetergentSoap/Detergent• polar “head” with long nonpolar “tail”• dissolves nonpolar grease in polar water

C. Solubility

SATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

concentration

C. Solubility

SolubilitySolubility• maximum grams of solute that will

dissolve in 100 g of solvent at a given temperature

• varies with temp• based on a saturated soln

C. Solubility

Solubility CurveSolubility Curve• shows the

dependence of solubility on temperature

C. Solubility

Solids are more soluble at...Solids are more soluble at...• high temperatures.

Gases are more soluble at...Gases are more soluble at...• low temperatures &• high pressures

(Henry’s Law).• EX: nitrogen narcosis,

the “bends,” soda

II

III

I II. Concentration

(p. 412 - 418)

Ch. 13 & 14 - Solutions

A. Concentration

The amount of solute in a solution.

Describing Concentration• % by mass - medicated creams• % by volume - rubbing alcohol• ppm, ppb - water contaminants• molarity - used by chemists• molality - used by chemists

A. Concentration

SAWS Water Quality Report - June 2000

B. Molality

solvent ofkg solute of moles(m)molality

mass of solvent only

1 kg water = 1 L waterkg 1mol0.25 0.25m

B. Molality

Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.

75 g MgCl2 1 mol MgCl2

95.21 g MgCl2

= 3.2m MgCl2

0.25 kg water

kgmolm

B. Molality

How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?

0.500 kg water 1.54 mol NaCl

1 kg water

= 45.0 g NaCl

58.44 g NaCl

1 mol NaCl

kg 1mol1.5 1.5m

2211 VMVM

C. Dilution

Preparation of a desired solution by adding water to a concentrate.

Moles of solute remain the same.

C. Dilution

What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:M1 = 15.8MV1 = ?

M2 = 6.0MV2 = 250 mL

WORK:M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

D. Preparing Solutions 500 mL of 1.54M NaCl

500 mLwater

45.0 gNaCl

• mass 45.0 g of NaCl• add water until total

volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water

500 mLmark

500 mLvolumetric

flask

1.54m NaCl in 0.500 kg of water

D. Preparing Solutions

Copyright © 1995-1996 NT Curriculum Project, UW-Madison(above: “Filling the volumetric flask”)

D. Preparing Solutions

Copyright © 1995-1996 NT Curriculum Project, UW-Madison(above: “Using your hand as a stopper”)

D. Preparing Solutions

250 mL of 6.0M HNO3 by dilution• measure 95 mL

of 15.8M HNO3

95 mL of15.8M HNO3

water for

safety

250 mL mark

• combine with water until total volume is 250 mL

• Safety: “Do as you oughtta, add the acid to the watta!”

Solution Preparation Lab Turn in one paper per team. Complete the following steps:

A) Show the necessary calculations.B) Write out directions for preparing the solution.C) Prepare the solution.

For each of the following solutions:1) 100.0 mL of 0.50M NaCl2) 0.25m NaCl in 100.0 mL of water3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

II

III

I III. Colligative Properties

(p. 436 - 446)

Ch. 13 & 14 - Solutions

A. Definition

Colligative PropertyColligative Property• property that depends on the

concentration of solute particles, not their identity

B. Types

Freezing Point DepressionFreezing Point Depression (tf)• f.p. of a solution is lower than f.p. of

the pure solvent

Boiling Point ElevationBoiling Point Elevation (tb)• b.p. of a solution is higher than b.p. of

the pure solvent

B. Types

View Flash animation.

Freezing Point Depression

B. Types

Solute particles weaken IMF in the solvent.

Boiling Point Elevation

B. Types

Applications• salting icy roads• making ice cream• antifreeze

• cars (-64°C to 136°C)• fish & insects

C. Calculations

t: change in temperature (°C)k: constant based on the solvent (°C·kg/mol)m: molality (m)n:# of particles

t = k · m · n

C. Calculations # of Particles# of Particles

• Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle

• Electrolytes (ionic)• dissociate into ions when dissolved• 2 or more particles

C. Calculations At what temperature will a solution that is composed

of 0.73 moles of glucose in 225 g of phenol boil?

m = 3.2mn = 1tb = kb · m · n

WORK:m = 0.73mol ÷ 0.225kg

GIVEN:b.p. = ?tb = ?kb = 3.60°C·kg/mol

tb = (3.60°C·kg/mol)(3.2m)(1)

tb = 12°C

b.p. = 181.8°C + 12°C

b.p. = 194°C

C. Calculations Find the freezing point of a saturated solution of

NaCl containing 28 g NaCl in 100. mL water.

m = 4.8mn = 2tf = kf · m · n

WORK:m = 0.48mol ÷ 0.100kg

GIVEN:f.p. = ?tf = ?

kf = 1.86°C·kg/mol

tf = (1.86°C·kg/mol)(4.8m)(2)

tf = 18°C

f.p. = 0.00°C - 18°C

f.p. = -18°C