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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 12. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 12 SOLUTIONS. SOLUTION. - A homogeneous mixture of two or more substances Solvent

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Principles of chemistry ii chem 1212 chapter 12

PRINCIPLES OF CHEMISTRY II CHEM 1212CHAPTER 12

DR. AUGUSTINE OFORI AGYEMAN

Assistant professor of chemistry

Department of natural sciences

Clayton state university


Principles of chemistry ii chem 1212 chapter 12

CHAPTER 12

SOLUTIONS


Principles of chemistry ii chem 1212 chapter 12

SOLUTION

- A homogeneous mixture of two or more substances

Solvent

- The substance present in the greatest quantity

Solute

- The other substance(s) dissolved in the solvent


Principles of chemistry ii chem 1212 chapter 12

SOLUTION

- Solutions can exist in any of the physical states

Solid Solution

- dental fillings, metal alloys (steel), polymers

Liquid Solution

- sugar in water, salt in water, wine, vinegar

Gas Solution

- air (O2, Ar, etc. in N2),

- NOx, SO2, CO2 in the atmosphere


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

- The amount of solute dissolved in a given quantity of

solvent or solution

Molarity (M: molar)

- The number of moles of solute per liter of solution

- A solution of 1.00 M (read as 1.00 molar)

contains 1.00 mole of solute per liter of solution


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

  • Calculate the molarity of a solution made by dissolving 2.56 g

  • of NaCl in enough water to make 2.00 L of solution

  • - Calculate moles of NaCl using grams and molar mass

  • Convert volume of solution to liters

  • - Calculate molarity using moles and liters


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

After dissolving 1.56 g of NaOH in a certain volume of water,

the resulting solution had a concentration of 1.60 M. Calculate the

volume of the resulting NaOH solution

- Convert grams NaOH to moles using molar mass

- Calculate volume (L) using moles and molarity


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Mole Fraction (χ)

- Fraction of moles of a component of solution

The sum of mole fractions of all components = 1


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Given that the total moles of an aqueous solution of NaCl and

other solutes is 1.75 mol. Calculate the mole fraction of NaCl

if the solution contains 4.56 g NaCl.


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Percent Concentration

- Percent by mass [mass-mass percent, %(m/m)]

mass of solution = mass of solute + mass of solvent


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

A sugar solution is made by dissolving 5.8 g of sugar in

82.5 g of water. Calculate the percent by mass concentration

of sugar.


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Percent Concentration

- Percent by volume [volume-volume percent, %(v/v)]

volume of solution ≠ volume of solvent + volume of solute

- Due differences in bond lengths and angles


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Calculate the volume percent of solute if 345 mL of ethyl

alcohol is dissolved in enough water to produce 1257 mL

of solution


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

Percent Concentration

- Mass-volume percent [%(m/v)]

- Units are specified because they do not cancel


Principles of chemistry ii chem 1212 chapter 12

CONCENTRATION OF SOLUTIONS

The concentration of a solution of NaCl is 0.92 %(m/v)

used to dissolve drugs for intravenous use. What is the

amount, in grams, of NaCl needed to prepare 41.50 mL

of the solution?

g solute = [%(m/v)] x [volume of solution (mL)]/[100 %]

= [(0.92 % g/mL) x (41.50 mL)]/(100 %)

= 0.38 g


Principles of chemistry ii chem 1212 chapter 12

PARTS PER MILLION (PPM)

Percent can be defined as parts per hundred

1 ppm ≈ 1 µg/mL or 1 mg/L


Principles of chemistry ii chem 1212 chapter 12

PARTS PER MILLION (PPM)

If 0.250 L of aqueous solution with a density of

1.00 g/mL contains 13.7 μg of pesticide, express

the concentration of pesticide in ppm

ppm = µg/mL

0.250 L = 250 mL

Density = 1.00 g/mL

Implies mass solution = 250 g


Principles of chemistry ii chem 1212 chapter 12

PARTS PER BILLION (PPB)

1 ppb ≈ 1 ng/mL or 1 µg/L


Principles of chemistry ii chem 1212 chapter 12

PARTS PER MILLION (PPB)

If 0.250 L of aqueous solution with a density of

1.00 g/mL contains 13.7 μg of pesticide, express

the concentration of pesticide in ppb

ppm = µg/L

Volume of solution = 0.250 L

Density = 1.00 g/mL

Implies mass solution = 250 g


Principles of chemistry ii chem 1212 chapter 12

MOLALITY (m)

Moles of solute per kg of solvent

Unit: m or molal


Principles of chemistry ii chem 1212 chapter 12

MOLALITY (m)

What is the molality of a solution that contains

2.50 g NaCl in 100.0 g water?

- Calculate moles NaCl

- Convert g water to kg water

- Divide to get molality


Principles of chemistry ii chem 1212 chapter 12

CONVERTING CONCENTRATION UNITS

Calculate the molality of a 6.75 %(m/m)

solution of ethanol (C2H5OH) in water

Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water


Principles of chemistry ii chem 1212 chapter 12

CONVERTING CONCENTRATION UNITS

Calculate the mole fraction of a 6.75 %(m/m)

solution of ethanol (C2H5OH) in water

Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water


Principles of chemistry ii chem 1212 chapter 12

CONVERTING CONCENTRATION UNITS

Practice Question

Given that the mole fraction of ammonia (NH3) in water is 0.088

Calculate the molality of the ammonia solution


Principles of chemistry ii chem 1212 chapter 12

CONVERTING CONCENTRATION UNITS

- Molarity is temperature dependent

(changes with change in temperature)

- Volume increases with increase in temperature

hence molarity decreases

On the other hand

- Molality

- Mass percent

- Mole fraction

are temperature independent


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY

- A measure of how much of a solute can be dissolved in a solvent

- Grams of solute per 100 mL of solvent

- Units: grams/100 mL

Three factors that affect solubility

- Temperature

- Pressure

- Polarity


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY

Unsaturated Solution

- More solute can still be dissolved at a given temperature

- Concentration of the solute is less than the solubility

Saturated Solution

- No more solute can be dissolved at a given temperature

- Concentration of the solute is equal to the solubility

- Dynamic equilibrium is reached

Supersaturated Solution

- Too much solute has temporarily been dissolved

- Concentration of solute is temporarily greater than the solubility

- Unstable condition


Principles of chemistry ii chem 1212 chapter 12

DISSOLUTION

The process of dissolving (known as dissolution) is

contributed by factors such as

- Enthalpy change due to solute-solvent interactions

and

- Change in disorder


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

- Change in enthalpy arises mainly from changes in

intermolecular attractions

Three types of intermolecular attractions are involved

- Solute-solute

- Solvent-solvent

- Solute-solvent


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

- The relative strengths of these interactions determine

the formation of a solution by two substances

- Substances with similar properties (strong solute-solvent interactions) tend to form solutions

- Like dissolves like


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

- Solvent molecules move apart to accommodate solute molecules

- Energy is required to separate solvent molecules attracting each other (ΔH1)

- Energy is also required to separate solute molecules (ΔH2)

- Energy is released when solvent and solute molecules come together due to attractive forces between them (ΔH3)


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

Enthalpy of Solution

- The overall enthalpy change that accompanies the dissolution

of one mole of a solution

ΔHsoln = ΔH1 + ΔH2 + ΔH3


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

Endothermic Heat of Solution

- Energy released by solute-solvent interactions is less than the

energy absorbed by separating the solvent and solute molecules

ΔHsoln is positive

ΔH3 < (ΔH1 + ΔH2)

Example

Ammonium nitrate in water


Principles of chemistry ii chem 1212 chapter 12

ENTHALPY OF SOLUTION

Endothermic Heat of Solution

Separated

solute

Separated

solvent

+

∆H2

Enthalpy

Solute

+

Separated solvent

∆H3

∆H1

Solution

∆Hsoln

Solute + Solvent


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

Exothermic Heat of Solution

- Energy released by solute-solvent interactions is greater than the

energy absorbed by separating the solvent and solute molecules

ΔHsoln is negative

ΔH3 > (ΔH1 + ΔH2)

Example

Sulfuric acid in water

NaOH in water


Principles of chemistry ii chem 1212 chapter 12

ENTHALPY OF SOLUTION

Exothermic Heat of Solution

Separated

solute

Separated

solvent

+

∆H2

Enthalpy

Solute

+

Separated solvent

∆H3

∆H1

Solute + Solvent

∆Hsoln

Solution


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

Generally

- Substances with similar intermolecular forces and hence similar

properties have strong solute-solvent interactions

- Such substances tend to form solutions

- Like dissolves like

Example

CH3OH readily dissolves in H2O (hydrogen bonding in both)

CCl4 readily dissolves in C7H16 (London forces in both)


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

- Increase in disorder on mixing is another contributing factor

in the dissolution process

- Increase in disorder tends to occur spontaneously in processes

- The main driving force in the formation of solutions

Consider NH4NO3 in H2O (used in cold packs)

- Enthalpy change on mixing is positive (+28 kJ/mol)

- NH4NO3 dissolves to form solution due to increase in disorder


Principles of chemistry ii chem 1212 chapter 12

SOLUTE-SOLVENT INTERACTIONS

Spontaneous Process

- Takes place with no apparent cause

Nonspontaneous Process

- Requires something to be applied in order for it to occur

(usually in the form of energy)


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF IONIC COMPOUNDS

- Strong electrostatic attractions between oppositely charged ions

hold ionic solids together

- For soluble ionic compounds the enthalpy of attraction between

solvent molecules and ions must be comparable to the crystal

lattice enthalpy in the solid

Example

- NaCl solution contains Na+ and Cl- ions

- Each ion is surrounded by water molecules

- Good conductor of electricity


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF IONIC COMPOUNDS

  • Solvation Process (Hydration)

  • - Ions in aqueous solution are surrounded by the H2O molecules

  • The O atom in each H2O molecule has partial negative charge

  • and attract cations

  • - The H atoms have partial positive charge

  • and attract anions

  • - Cations and anions are prevented from recombining

  • - About 4 to 10 water molecules surround each cation


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF IONIC COMPOUNDS

- There is an increase in disorder of the solute as it separates into ions

- There is an increase or decrease in disorder of the solvent

depending on the solute

- Solubilities are difficult to predict due to these several

contributing factors


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF MOLECULAR COMPOUNDS

- Most molecular compounds do not form ions in solution

- The molecules disperse throughout the solution

Example

- Sucrose in water solution contains neutral sucrose molecules

- Each molecule is surrounded by water molecules

- Poor conductor of electricity

- A few molecular compounds form ions in aqueous solution

- HCl dissociates into H+(aq) and Cl-(aq)

- HNO3 dissociates into H+(aq) and NO3-(aq)


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF MOLECULAR COMPOUNDS

Consider mixing hydrocarbons such as C6H14 and C7H16

- London dispersion forces dominate within the molecules

- Attraction between C6H14 and C7H16 molecules are also

due to London dispersion forces

- These two substances mix because the attractions are

close in energy

- Increase in disorder is the controlling factor


Principles of chemistry ii chem 1212 chapter 12

SOLUBILITY OF MOLECULAR COMPOUNDS

Consider mixing water and a hydrocarbon

- Strong hydrogen bonding dominates

intermolecular attractions between water molecules

- London dispersion forces dominate

intermolecular attractions between hydrocarbon molecules

- Attraction between water and hydrocarbon molecules

are due to weak London dispersion forces

- Increase in disorder is not sufficient to overcome the unfavorable

enthalpy change hence very low solubility results


Principles of chemistry ii chem 1212 chapter 12

EFFECT OF PRESSURE ON SOLUBILITY

- Solubilities of gases in liquids are sensitive to pressure changes

- Increase in pressure increases solubility of gases

- An increase in pressure of a saturated solution results

in dissolving more gas molecules

- Solubilities of liquids and solids change very little with pressure

due to very little change in volume


Principles of chemistry ii chem 1212 chapter 12

EFFECT OF PRESSURE ON SOLUBILITY

Henry’s Law

- The solubility of a gas is directly proportional to its

partial pressure at any given temperature

C = kP

C = concentration of the gaseous solute

k = proportionality constant (units depend on units of C)

P = partial pressure of gaseous solute


Principles of chemistry ii chem 1212 chapter 12

EFFECT OF TEMPERATURE ON SOLUBILITY

- Effect of temperature depends on the sign of the enthalpy change

- Solubility increases with increasing temperature when the

enthalpy change is positive (+∆H, endothermic process)

- Solubility decreases with increasing temperature when the

enthalpy change is negative (-∆H, exothermic process)

Generally

- The more positive the ∆H the greater the change in solubility

with temperature


Principles of chemistry ii chem 1212 chapter 12

EFFECT OF TEMPERATURE ON SOLUBILITY

- Solubility of most ionic solids increase with increase in

temperature

- Solubility of most gases decrease with increase in

temperature

- Enthalpy of solution of most gases in water is negative

- There is little or no attraction between gas molecules

but there are attractions between solvent and solute

molecules

- Hence the negative enthalpy change


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

- The physical properties of a solution differ from those of the

pure solvent

- The physical properties of a solvent change when a solute is

added to a solvent

- Four physical properties change based on the amount of

solute added but not the solute’s chemical identity


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

- These are known as the Colligative Properties

- Vapor-pressure lowering

- Boiling-point elevation

- Freezing-point depression

- Osmotic pressure


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Vapor-Pressure Depression

- When a nonvolatile solute (low tendency to vaporize) is

added to a solvent, the vapor pressure of the resulting

solution is lowered below that of the pure solvent at the

same temperature


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Vapor-Pressure Depression

Raoult’s Law

- The partial pressure of a substance in equilibrium with a solution

is equal to the product of its mole fraction in the solution

and the vapor pressure of the pure substance

Psolv = partial pressure exerted by solvent’s vapor above a solution

Xsolv = mole fraction of the solvent in the solution

Posolv = vapor pressure of the pure solvent


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Vapor-Pressure Depression


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Vapor-Pressure Depression

- The vapor pressure depression (∆P) is proportional to the

mole fraction (concentration) of solute

- Raoult’s law only applies to dilute solutions


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Vapor-Pressure Depression

At 25 oC, the vapor pressure of pure benzene is 93.9 torr.

What is the solute concentration in a benzene solution

that has a vapor pressure of 92.1 torr?

∆P = 93.9 torr – 92.1 torr = 1.8 torr


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Boiling-Point Elevation

- When a nonvolatile solute (low tendency to vaporize) is

added to a solvent, the boiling point of the resulting solution

is raised above that of the pure solvent

- Since vapor pressure is lowered, a higher temperature is

needed to raise the depressed vapor pressure to atmospheric

pressure ( a condition required for boiling)


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Boiling-Point Elevation

ΔTb = mkb

ΔTb = increase in boiling point

kb = molal boiling-point elevation constant

(depends only on solvent)

m = molality (molal concentration)

- Applies to dilute solutions only


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Boiling-Point Elevation

What is the boiling point of a 0.21-molal aqueous solution

of sodium chloride at 1 atm (kb of water = 0.512 oC/m)?

ΔTb = mkb = (0.21)(0.512 oC/m) = 0.11 oC

Boiling point of pure water = 100 oC

Tb = 100 oC + 0.11 oC = 100.11 oC


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Freezing-Point Depression

- When a nonvolatile solute (low tendency to vaporize) is

added to a solvent, the freezing point of the resulting

solution is lowered below that of the pure solvent

- The triple-point temperature of a solution decreases with increasing concentration of solute

(due to decrease in vapor pressure)

- The solid-liquid equilibrium line moves to lower temperatures


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Freezing-Point Depression

ΔTf = mkf

ΔTf = decrease in freezing point

kf = molal freezing-point-depression constant

(depends only on solvent)

m = molality (molal concentration)

- Applies to dilute solutions only


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Freezing-Point Depression

What is the freezing point of a 0.21-molal aqueous solution

of sodium chloride at 1 atm (kf of water = 1.86 oC/m)?

ΔTf = mkf = (0.21)(1.86 oC/m) = 0.39 oC

Freezing point of pure water = 0 oC

Tf = 0 oC – 0.39 oC = – 0.39 oC


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Osmosis

- The passage of a solvent through a semipermeable membrane

that separates a solution of lower solute concentration from a

solution of higher solute concentration

- Flow of solvent is in both directions so it is actually a net flow

Semipermeable Membrane

- Allows certain types of molecules to pass through but prohibits

other types of molecules (usually based on size)


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Osmotic Pressure

- Pressure required to prevent osmosis by pure solvent

- Pressure difference needed for no net transfer of solvent

- Very sensitive and useful for measuring molar mass of large molecules

- Aqueous solutions with higher osmotic pressure take up more water than aqueous solutions with lower osmotic pressure


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Osmotic Pressure

πV = nRT

(similar to the ideal gas law)

π = osmotic pressure

V = volume of solution

n = number of moles of solute

R = ideal-gas constant

T = absolute temperature

M = molarity of solution


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Electrolyte Solutions

Electrolytes dissociate into ions in solution

NaCl(aq) → Na+(aq) + Cl-(aq)

1 mole of NaCl in solution produces 2 moles of ions

AlCl3(aq) → Al3+(aq) + 3Cl-(aq)

1 mole of AlCl3 in solution produces 4 moles of ions

For example

- The osmotic pressure of NaCl is twice that of glucose

- Glucose does not dissociate in solution


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Electrolyte Solutions

van‘t Hoff Factor (i)

i = number of particles produced from the dissociation

of one formula unit of solute (for dilute solutions)

- The number of particles present determines the measured

colligative property


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Electrolyte Solutions

Taking van‘t Hoff Factor (i) into account

ΔTb = imkb

ΔTf = imkf


Principles of chemistry ii chem 1212 chapter 12

COLLIGATIVE PROPERTIES

Nonideal Solutions

- The value of i tends to be smaller than expected at greater

solution concentrations (> 0.01 m)

- Some ions cluster in solution and behave as a single unit as a

result of strong electrostatic attractions


Principles of chemistry ii chem 1212 chapter 12

MIXTURES OF VOLATILE SUBSTANCES

Consider a solution of two volatile substances A and B

- The vapor phase in equilibrium with the solution contains all

the volatile components

- According to Raoult’s law

and


Principles of chemistry ii chem 1212 chapter 12

MIXTURES OF VOLATILE SUBSTANCES

- Total vapor pressure is the sum of the partial pressures of the components

PT = PA + PB

- The vapor in equilibrium with the mixture is richer in the more volatile component

- This is applied in fractional distillation to separate volatile substances


Principles of chemistry ii chem 1212 chapter 12

MIXTURES OF VOLATILE SUBSTANCES

Ideal Solution

- Obeys Raoult’s law throughout the entire range of composition

Considering the mixture of substances A and B

- Solution is ideal when A-B attractions are closer to

A-A and B-B attractions

- Nearly true for similar liquids

Benzene and toleune

Hexane and heptane

- Strictly for very dilute solutions


Principles of chemistry ii chem 1212 chapter 12

MIXTURES OF VOLATILE SUBSTANCES

Positive Deviation from Raoult’s Law

- Occurs when the A-B attractions are weaker than the average

A-A and B-B attractions

- The observed vapor pressure is greater than expected

Pressure

0

0.5

1

Mole fraction

- Straight lines (dotted) in ideal situation become curves


Principles of chemistry ii chem 1212 chapter 12

MIXTURES OF VOLATILE SUBSTANCES

Negative Deviation from Raoult’s Law

- Occurs when the A-B attractions are stronger than the average

A-A and B-B attractions

- The observed vapor pressure is less than expected

Pressure

0

0.5

1

Mole fraction

- Straight lines (dotted) in ideal situation become curves


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