II

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I I. Types of Mixtures

Ch. 14 – Mixtures & Solutions

A. Definitions

Mixture = Variable combination of 2 or more pure substances

Homogeneous = uniform composition throughout Heterogeneous = variable composition

Heterogeneous Homogeneous

A. Definitions

Solution – Solution – homogeneous mixture

Solvent Solvent – dissolves the solute

Solute Solute – substance being dissolved

B. Mixtures

Gases can also mix with liquids Gases are usually dissolved in

water Examples are carbonated drinks

• Homogeneous mixtures (solutions)

• Contain sugar, flavorings and carbon dioxide dissolved in water

B. Mixtures

Solution• homogeneous• very small particles

• no Tyndall effect • particles don’t settle• Ex: rubbing alcohol

Tyndall Effect

B. Mixtures Colloid

• heterogeneous• medium-sized particles

Tyndall effect :the scattering of light by colloidal particles.• particles don’t settle• Ex: milk

B. Mixtures

Suspension• Heterogeneous• large particles usually

> 1000nm• particles settle• Tyndall effect• Ex: fresh-squeezed

lemonade

B. Mixtures

Examples:

• mayonnaise

• muddy water

• fog

• saltwater

• Italian salad dressing

colloid

suspension

colloid

true solution

suspension

II

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I II. Factors Affecting Solvation(p. 489 – 496)

Ch. 14 –Solutions

A. Solvation

Solvation – Solvation – the process of dissolving

solute particles are separated and pulled into solution

solute particles are surrounded by solvent particles

Solution Process

•For a solute to be dissolved in a solvent, the attractive forces between the solute and solvent particles must be great enough to overcome the attractive forces within the pure solvent & pure solute. •The solute & the solvent molecules in a solution are expanded compared to their position within the pure substances.

B. Solvation DissociationDissociation

• Separation/Solvation of an ionic solid into aqueous ions For ionic solids, the lattice energy describes the attractive forces

between the solute molecules (i.e. ions) For an ionic solid to dissolve in water, the water-solute attractive

forces has to be strong enough to overcome the lattice energy

NaCl(s) Na+(aq) + Cl–(aq)

B. Factors Affecting Solvation

Molecules are constantly in motion according to…• Kinetic Theory

When particles collide, energy is transferred

B. Factors Affecting Solvation

Solubility = max. amount of a solute that will dissolve in a solvent @ a specific T

Smaller pieces of a substance dissolve faster b/c of larger surface area

Stirring or shaking speeds dissolving b/c particles are moving faster and colliding more

Heating speeds dissolving of solidsNot all substances dissolve

C. Solubility Water is universal solvent

b/c of its polarity If something can dissolve

in something else, it is said to be soluble or miscible

If it cannot dissolve, it is said to be insoluble or immiscible

“Like dissolves like”

C. Solubility

NONPOLAR

NONPOLAR

POLAR

POLAR

““Like Dissolves Like”Like Dissolves Like”

Saturated Solutions•A solution that can contain the maximum amount of solute at a given temperature (if the pressure is constant).•Solution said to be at a dynamic equilibrium•Any point on the line

•Ex: At 90 ° C 40 g of NaCl (s) in 100 g of H2O represent a saturated solution

Solubility curve• Any point on a line

represents a saturated solution.

• In a saturated solution, the solvent contains the maximum amount of solute.

• Example• At 90oC, 40 g of NaCl(s) in

100g H2O(l) represent a saturated solution.

Unsaturated Solution•A solution that can contain less than the maximum amount of solute at a given temperature (if the pressure is constant).•It is any value under the solid line on the solubility graph

Solubility curve• Any point below a line

represents an unsaturated solution.

• In an unsaturated solution, the solvent contains less than the maximum amount of solute.

• Example• At 90oC, 30 g of NaCl(s)

in 100g H2O(l) represent an unsaturated solution. 10 g of NaCl(s) have to be added to make the solution saturated.

Supersaturated Solutions•A solution that can contain greater than the maximum amount of solute at a given temperature (if the pressure is constant). •A supersaturated solution is very unstable & the amount of solute in excess can precipitate or crystallize.•It is any value above the solid line on the solubility graph

Solubility curve• Any point above a line

represents a supersaturated solution.

• In a supersaturated solution, the solvent contains more than the maximum amount of solute. A supersaturated solution is very unstable and the amount in excess can precipitate or crystallize.

• Example• At 90oC, 50 g of NaCl(s) in

100g H2O(l) represent a supersaturated solution. Eventually, 10 g of NaCl(s) will precipitate.

Solubility curve• Any point above a line

represents a supersaturated solution.

• In a supersaturated solution, the solvent contains more than the maximum amount of solute. A supersaturated solution is very unstable and the amount in excess can precipitate or crystallize.

• Example• At 90oC, 50 g of NaCl(s) in

100g H2O(l) represent a supersaturated solution. Eventually, 10 g of NaCl(s) will precipitate.

Solubility

SATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

concentration

SolubilitySolubility curve

Any solution can be made saturated, unsaturated, or supersaturated by changing the temperature.Any solution can be made Saturated, Unsaturated,

or Supersaturated by changing the Temperature.

C. Solubility

Solubility CurvesSolubility Curves• 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).• With larger mass (LDF)• EX: nitrogen narcosis,

the “bends,” soda

StrongElectrolyte

Non-Electrolyte

solute exists asions only

- +

salt

- +

sugar

solute exists asmolecules

only

- +

acetic acid

WeakElectrolyte

solute exists asions and

molecules DISSOCIATION IONIZATION

•Although H2O is a poor conductor of electricity, dissolved ions in an aqueous solution can conduct electricity. •Ionic aqueous solutions are known as electrolytes.

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I II. Solution Concentration

(p. 480 – 486)

Ch. 16 – 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

B. Percent Solutions Percent By Volume (%(v/v))

• Concentration of a solution when both solute and

solvent are liquids often expressed as percent by

volume

100solution ofvolume solute ofvolume (%(v/v))Volume by Percent

total combined volume

substance being dissolved

B. Percent SolutionsFind the percent by volume of ethanol

(C2H6O) in a 250 mL solution containing 85 mL ethanol.

85 mL ethanol250 mL solution

= 34% ethanol (v/v)

Solute = 85 mL ethanolSolution = 250 mL

% (v/v) = x 100

C. MolarityConcentration of a solution most often

used by chemists

solution of literssolute of moles(M)Molarity

total combined volume

substance being dissolved

C. Molarity

2M HCl

LmolM

nsol' L 1HCl mol 2HCl 2M

What does this mean?

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

LITERSOF

SOLUTION

Molarity(mol/L)

D. Molarity Calculations

D. Molarity CalculationsHow many moles of NaCl are

required to make 0.500L of 0.25M NaCl?

0.500 L sol’n 0.25 mol NaCl

1 L sol’n

= 0.013 mol NaClL 1mol0.25 0.25M

D. Molarity CalculationsHow many grams of NaCl are

required to make 0.500L of 0.25M NaCl?

0.500 L sol’n 0.25 mol NaCl

1 L sol’n

= 7.3 g NaCl

58.44 g NaCl

1 mol NaCl

L 1mol0.25 0.25M

D. Molarity Calculations

Find the molarity of a 250 mL solution containing 10.0 g of NaF.

10.0 g NaF 1 mol NaF41.99 g NaF

= 0.24 mol NaF

LmolM 0.24 mol NaF

0.25 L= 0.95 M NaF

2211 VMVM

E. Dilution

Preparation of a desired solution by adding water to a concentrate

Moles of solute remain the same

E. 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

F. Molality

solvent ofkg solute of moles(m)molality

mass of solvent only

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

G. Molality Calculations

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 waterkgmolm

= .79 mol MgCl2

.79 mol MgCl2

G. Molality Calculations

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 1mol1.54 1.54m

H. 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

H. 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!” or AA – add acid!

II

III

I IV. Colligative Properties of Solutions

(p. 498 – 504)

Ch. 14 – Mixtures & Solutions

A. Definition

Colligative PropertyColligative Property• property that depends on the

concentration of solute particles, not their identity

• Examples: vapor pressure, freezing point, boiling point

B. Types

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

Vapor Pressure Lowering• lower number of solvent particles at the surface of the

solution; therefore, this lowers the tendency for the solvent particles to escape into the vapor phase.

B. Types

Applications• salting icy roads• making ice cream• antifreeze

• cars (-64°C to 136°C)

C. Calculations

T: change in temperature (°C)i: Van’t Hoff Factor (VHF), the number of

particles into which the solute dissociatesm: molality (m)K: constant based on the solvent (°C·kg/mol)

or (°C/m)

T = i · m · K

C. Calculations T

• Change in temperature• Not actual freezing point or boiling point• Change from FP or BP of pure solvent

• Freezing Point (FP) TF i is always subtracted from FP of pure

solvent• Boiling Point (BP)

TB i is always added to BP of pure solvent

C. Calculations ii – VHF – VHF

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

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

C. Calculations

ii – VHF – VHF

• Examples

• CaCl2

• Ethanol C2H5OH

• Al2(SO4)3

• Methane CH4

•i =

• 3

• 1

• 5

• 1

C. Calculations

KK – molal constant – molal constant •KKFF – molal freezing point constant

• Changes for every solvent • 1.86 °C·kg/mol (or °C/m) for water

•KKBB – molal boiling point constant• Changes for every solvent • 0.512 °C·kg/mol (or °C/m) for water

C. Calculations: Recap!

T : subtract from F.P. : subtract from F.P. add to B.P.add to B.P. ii – VHF : covalent = 1 – VHF : covalent = 1

ionic > 2ionic > 2K : K : KKF waterF water = = 1.86 °C·kg/mol

KKB water B water = = 0.512 °C·kg/mol

T = i · m · K

At what temperature will a solution that is composed of 0.730 moles of glucose in 225 g of water boil?

C. Calculations

m = 3.24mKB = 0.512°C/mTB = i · m · KB

WORK:m = 0.730 mol ÷ 0.225 kg

GIVEN:b.p. = ?

TB = ?i = 1

TB = (1)(3.24m)(0.512°C/m)

TB = 1.66°C

b.p. = 100.00°C + 1.66°C

b.p. = 101.66°C

100 + Tb

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

NaCl containing 28 g NaCl in 100. mL water.

i = 2m = 4.8m

KF = 1.86°C/m

TF = i · m · KF

WORK:m = 0.48mol ÷ 0.100kg

GIVEN:f.p. = ?

TF = ? TF = (2)(4.8m)(1.86°C/m)

TF = 18°C

f.p. = 0.00°C – 18°C

f.p. = -18°C

0 – TF

D. Osmotic Pressure

Osmosis: The flow of solvent into a solution through a semipermeable membrane

Semipermeable Membrane: membrane that allows solvent to pass through but not solute

D. Osmotic Pressure

Net transfer of solvent molecules into thesolution until the hydrostatic pressureequalizes the solvent flowin both directions

Because the liquid level for the solution is higher, there is greater hydrostatic pressure on the solution than on the pure solvent

Osmotic Pressure: The excess hydrostatic pressure on the

solution compared to the pure solvent

D. Osmotic Pressure

Osmotic Pressure:

Minimum Pressurerequired to stop flowof solvent into the solution

D. Osmotic Pressure

D. Osmotic Pressure

Osmosis at Equilibrium

= i M R Twhere:π = osmotic pressure (atm)osmotic pressure (atm)i = VHFVHFM = Molarity (moles/L)R = Gas Law ConstantT = Temperature (Kelvin)

E. Osmotic Pressure Calculations

0.08206 L atm/mol K

E. Osmotic Pressure Calculations

Calculate the osmotic pressure (in torr) at 25oC of aqueous solution containing 1.0g/L of a protein with a molar mass of 9.0 x 104 g/mol.

i = 1M = 1.11 x 10-5 MR = 0.08206 L atm/mol K

T = 25oC = 298 K

WORK:M = 1.0 g prot.

GIVEN: = ?

1.11 x 10-5 M

= (1)(1.11x10-5)(.08206)(298)

= 2.714 x 10-4 atm

= 0.21 torr

1 mol prot. 1 L sol’n 9.0 x 104 g

=

If the external pressure is larger than the osmotic pressure, reverse osmosis occurs

One application is desalination of seawater

F. Reverse Osmosis

F. Reverse Osmosis

•Net flow of solventfrom the solution to the solvent