Solutions

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. CovalentCovalent 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 concentration4 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

100xsolutionmasssolutemass

100xsolutionvolumesolutemass

100xsolutionvolumesolutevolume

% 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 ExampleIf 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.

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 / Solvent1. Nature of Solute / Solvent. - Like dissolves like (IMF)

2. Temperature -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 -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 Pc = k P

c = solubility of the gas (M)k = Henry’s Law ConstantP = 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 numbernumber of solute particles in the solution, NOT on the identityidentity of 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 Pwater = (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

Ideal Mixture Behaviour Intermediate-boiling Systems, including Raoult’s Law

Behaviour

Ideal Phase Behaviour: P-xy Diagrams

Non-Ideal Vapour-Liquid Equilibria Systems having a minimum boiling azeotrope:

We also observe systems with a maximum boiling azeotrope.

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 Express by Boiling point Elevation point Elevation equationequation

Boiling Point ElevationTb = (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 Water 100.00100.00 +0.5121 +0.5121 0.0 0.0 1.861.86BenzeneBenzene 80.10 80.10 +2.53+2.53 5.50 5.50 4.904.90CamphorCamphor 207 207 +5.611 +5.611 179.75 179.75 39.739.7Chloroform Chloroform 61.70 61.70 +3.63+3.63 - 63.5 - 63.5 4.70 4.70 (CH(CH33Cl)Cl)

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 Water 100.00100.00 +0.5121 +0.5121 0.0 0.0 1.861.86BenzeneBenzene 80.10 80.10 +2.53+2.53 5.50 5.50 4.904.90CamphorCamphor 207 207 +5.611 +5.611 179.75 179.75 39.739.7Chloroform Chloroform 61.70 61.70 +3.63+3.63 - 63.5 - 63.5 4.70 4.70 (CH(CH33Cl)Cl)

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 Factorm = 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.