Solutions

M. Patenaude

Advanced Chemistry 30S

GPHS Science Dept

Definitions

A solution is a homogeneous mixture

A solute is dissolved in a solvent.

– solute is the substance being dissolved

– solvent is the liquid in which the solute is dissolved

– an aqueous solution has water as solvent

A saturated solution is one where the concentration is

at a maximum - no more solute is able to dissolve.

– A saturated solution represents an equilibrium: the rate of

dissolving is equal to the rate of crystallization. The salt

continues to dissolve, but crystallizes at the same rate so

that there “appears” to be nothing happening.

Dissolution of Solid Solute

What are the driving forces which cause solutes to

dissolve to form solutions?

1. Covalent solutes dissolve by H-bonding to water or by LDF

2. Ionic solutes dissolve by dissociation into their ions.

Solution and Concentration

4 ways of expressing concentration

–Molarity(M): moles solute / Liter solution

–Mass percent: (mass solute / mass of solution) * 100

–Molality* (m) - moles solute / Kg solvent

–Mole Fraction( A) - moles solute / total moles solution

* Note that molality is the only concentration unit in which denominator contains only solvent information rather than

solution.

% (w/w) =

% (w/v) =

% (v/v) =

% Concentration

100xsolutionmass

solutemass

100xsolutionvolume

solutemass

100xsolutionvolume

solutevolume

% Concentration: % Mass Example

3.5 g of CoCl2 is

dissolved in 100mL

solution.

Assuming the

density of the

solution is 1.0 g/mL,

what is concentration

of the solution in %

mass?

%m = 3.5 g CoCl2

100g H2O

= 3.5% (m/m)

Concentration: Molarity Example

If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of

solution, what is the molarity of KMnO4?

Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L .

As is almost always the case, the first step is to

convert the mass of material to moles.

0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4

158.0 g KMnO4

Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M

0.250 L solution

When a solution is diluted,

solvent is added to lower its

concentration.

The amount of solute remains

constant before and after the

dilution:

moles BEFORE = moles AFTER

C1V1 = C2V2

Suppose you have 0.500

M sucrose stock solution.

How do you prepare 250

mL of 0.348 M sucrose

solution ?

Concentration

0.500 M

Sucrose

250 mL of 0.348 M sucrose

Dilution

A bottle of 0.500 M standard

sucrose stock solution is in the lab.

Give precise instructions to your

assistant on how to use the stock

solution to prepare 250.0 mL of a

0.348 M sucrose solution.

3 Stages of Solution Process

Separation of Solute

– must overcome IMF or ion-ion attractions in solute

– requires energy, ENDOTHERMIC ( + H)

Separation of Solvent

– must overcome IMF of solvent particles

– requires energy, ENDOTHERMIC (+ H)

Interaction of Solute & Solvent

– attractive bonds form between solute particles and solvent

particles

– “Solvation” or “Hydration” (where water = solvent)

– releases energy, EXOTHERMIC (- H)

Dissolution at the molecular level?

Consider the dissolution of NaOH in H2O

Factors Affecting Solubility

1. Nature of Solute / Solvent. - Like dissolves like (IMF)

2. Temperature -

i) Solids/Liquids- Solubility increases with Temperature

Increase K.E. increases motion and collision between solute / solvent.

ii) gas - Solubility decreases with Temperature

Increase K.E. result in gas escaping to atmosphere.

3. Pressure Factor -

i) Solids/Liquids - Very little effect

Solids and Liquids are already close together, extra pressure will not

increase solubility.

ii) gas - Solubility increases with Pressure.

Increase pressure squeezes gas solute into solvent.

Solubilities of Solids vs Temperature

Solubilities of

several ionic solid

as a function of

temperature. MOST

salts have greater

solubility in hot

water.

A few salts have

negative heat of

solution,

(exothermic

process) and they

become less

soluble with

increasing

temperature.

Temperature & the Solubility of GasesThe solubility of gases DECREASES at higher temperatures

Henry’s LawThe effect of partial pressure on solubility of gases

At pressure of few atmosphere or less, solubility of gas solute

follows Henry Law which states that the amount of solute gas

dissolved in solution is directly proportional to the amount of

pressure above the solution.

c = k P

c = solubility of the gas (M)

k = Henry’s Law Constant

P = partial pressure of gas

Henry’s Law Constants (25°C), k

N2 8.42 •10-7 M/mmHg

O2 1.66 •10-6 M/mmHg

CO2 4.48•10-5 M/mmHg

Henry’s Law & Soft Drinks

Soft drinks contain “carbonated

water” – water with dissolved

carbon dioxide gas.

The drinks are bottled with a CO2

pressure greater than 1 atm.

When the bottle is opened, the

pressure of CO2 decreases and

the solubility of CO2 also

decreases, according to Henry’s

Law.

Therefore, bubbles of CO2

escape from solution.

Henry’s Law Application

The solubility of pure N2 (g) at 25oC and 1.00 atm

pressure is 6.8 x 10-4 mol/L. What is the solubility of

N2 under atmospheric conditions if the partial

pressure of N2 is 0.78 atm?

Step 1: Use the first set of data to find “k” for N2 at 25 C

Step 2: Use this constant to find the solubility

(concentration) when P is 0.78 atm:

44 16.8 10

6.8 101.00

c x Mk x M atm

P atm

4 1 4(6.8 10 )(0.78 ) 5.3 10c kP x M atm atm x M

Colligative Properties

Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure

Colligative Properties are properties of a liquid that change when a solute is added.

The magnitude of the change depends on the numberof solute particles in the solution, NOT on the identityof the solute particles.

Vapor Pressure Lowering for a Solution

The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent.

The black curve represents the pure liquid and the blue curve represents the solution.

Notice the changes in the freezing & boiling points.

Vapor Pressure Lowering

The presence of a non-volatile solute means that

fewer solvent particles are at the solution’s surface,

so less solvent evaporates!

Application of Vapor Pressure Lowering

Describe what is happening in the pictures below.

Use the concept of vapor pressure lowering to

explain this phenomenon.

Raoult’s LawDescribes vapor pressure lowering mathematically.

The lowering of the vapour pressure when a non-volatile solute is dissolved in a volatile solvent (A) can be described by Raoult’s Law:

PA = AP°A

PA = vapour pressure of solvent A above the solution

cA = mole fraction of the solvent A in the solution

P°A = vapour pressure of pure solvent A

only the solvent (A) contributes to

the vapour pressure of the solution

What is the vapor pressure of water above a sucrose

(MW=342.3 g/mol) solution prepared by dissolving

158.0 g of sucrose in 641.6 g of water at 25 ºC?

The vapor pressure of pure water at 25 ºC is 23.76

mmHg.

mol sucrose = (158.0 g)/(342.3 g/mol) = 0.462 mol

mol water = (641.6 g)/(18 g/mol) = 35.6 mol

Xwater =

mol water

(mol water)+(mol sucrose)=

35.6

35.6+0.462= 0.987

Psol’n = Xwater P water = (0.987)(23.76 mm Hg)

= 23.5 mm Hg

Mixtures of Volatile LiquidsBoth liquids evaporate & contribute to the vapor pressure

Raoult’s Law: Mixing Two Volatile Liquids

Since BOTH liquids are volatile and contribute to the

vapour, the total vapor pressure can be represented

using Dalton’s Law:

PT = PA + PB

The vapor pressure from each component follows

Raoult’s Law:

PT = AP°A + BP°B

Also, A + B = 1 (since there are 2 components)

Benzene and Toluene

Consider a two solvent (volatile) system

– The vapor pressure from each component follows

Raoult's Law.

– Benzene - Toluene mixture:

• Recall that with only two components, Bz + Tol = 1

• Benzene: when Bz = 1, PBz = P°Bz = 384 torr &

when Bz = 0 , PBz = 0

• Toluene: when Tol = 1, PTol = P°Tol = 133 torr &

when Tol = 0, PBz = 0

384 torr

133 torr

X Benzene

X Toluene

0 1

1 0

P (Total)

P (Benzene)

P (Toluene)

133 torr

384 torr

Normal Boiling ProcessExtension of vapor pressure concept:Normal Boiling Point: BP of Substance @ 1atm

When solute is added, BP > Normal BP

Boiling point is elevated when solute inhibits solvent from escaping.

Elevation of B. pt.

Express by Boiling

point Elevation

equation

Boiling Point Elevation

Tb = (Tb -Tb°) = i ·m ·kb

Where, Tb = BP. Elevation

Tb = BP of solvent in solution

Tb° = BP of pure solvent

m = molality , kb = BP Constant

Some Boiling Point Elevation and Freezing Point Depression Constants

Normal bp (°C) Kb Normal fp (°C) Kf

Solvent pure solvent (°C/m) pure solvent (°C/m)

Water 100.00 +0.5121 0.0 1.86

Benzene 80.10 +2.53 5.50 4.90

Camphor 207 +5.611 179.75 39.7

Chloroform 61.70 +3.63 - 63.5 4.70(CH3Cl)

When solution freezes the solid form is almost

always pure.

Solute particles does not fit into the crystal lattice

of the solvent because of the differences in size.

The solute essentially remains in solution and

blocks other solvent from fitting into the crystal

lattice during the freezing process.

Freezing Point Depression Normal Freezing Point: FP of Substance @ 1atm

When solute is added, FP < Normal FP

FP is depressed when solute inhibits solvent from crystallizing.

Freezing Point Depression

Phase Diagram and the lowering of the

freezing point.

Tf = i ·m ·kf

Where, Tf = FP depression

i = van’t Hoff Factor

m = molality , kf = FP Constant

Generally freezing point depression is

used to determine the molar mass of an unknown substance.

Derive an equation to find molar mass from the equation above.

Osmotic pressure Osmosis is the spontaneous movement of water across a semi-

permeable membrane from an area of low solute concentration to an area of high solute concentration

Osmotic Pressure - The Pressure that must be applied to stop osmosis

= i CRT

where P = osmotic pressure

i = van’t Hoff factor

C = molarity

R = ideal gas constant

T = Kelvin temperature

Osmosis and Blood Cells(a) A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same.

(b) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die.

(c) In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst.