Properties of solutions

COMPOSITION of SOLUTIONS• Solute – the substance that is being dissolved

• Solvent – the substance doing the dissolving

• If both are liquids, the one that is present in higher proportions is the solvent.

MOLARITYM = moles of solute

Liters of solutionEx. Calculate the molarity of a solution of glucose, if 250.0

g are dissolved in 350.0 ml of solution.

MASS %% by mass = mass of solute x 100

mass of solution

Ex. A solution of vinegar is 5.0% by mass acetic acid. Calculate the mass of acetic acid dissolved in 5.0 L water if the solution has a 1.08 g/ml.

MOLE FRACTIONRecall from the gas laws chapter.

Compares the number of moles of one part of the solution to the total number of moles in the solution.

X = na /nt

Practice problem:A solution was made by adding 5.84 g of H2CO to 100.0 g of

water. The final volume of solution was 104.0 ml.Calculate Molarity

Molality

% by mass

Mole fraction

Energy of Solution Formation3 steps to solution formation:1.Break the solid into individual components

Expand the solute Endothermic

2.Overcome IMF in solvent to make room for the solute Expand the solvent endothermic

3.Solute/Solvent interaction Endo or exo-thermic

Heat of solution ΔHsolnOverall endothermic = + ΔHsoln

Feels cold to the touch

Overall exothermic = - ΔHsoln

Feels warm to the touch

FACTORS affecting SOLUBILITY1. Structural effects

“Likes dissolve likes” Polar substances are more soluble in polar solvents. Hydrophillic – water loving

Ionic and polar covalent compounds

Hydrophobic – water fearing Non-polar substances

Which of the following would be miscible (mutually soluble)?C6H6 H20 MgCl2 CH3OH I2

2. PressureHas little effect on solid or liquid solubility.Higher pressure increases the solubility of a gas.

Henry’s lawP = kC

P = partial pressure of the gas above the solutionK = constant for a particular solutionC = concentration of dissolved gas

Henry’s law (cont.)

The amount of dissolved gas is directly proportional to the pressure of the gas above the solution.

This only applies to solutions where the gas does not react or dissociate in the solvent.

Ex. The solubility of oxygen is 2.2x10-4 M at 0°C and 0.10 atm. Calculate the solubility at 0°C and 0.35 atm.

3. Temperature effectsFOR SOLIDS, dissolving always occurs more

rapidly at higher temperatures BUT

The amount of solute able to be dissolved may increase or decrease with the increased temperature.

Solubility (the total amount of solute that may be dissolved at a certain temperature) must be determined experimentally.

FOR GASES, solubility decreases with increasing temperature.

Higher kinetic energy of the gas causes higher Pvap of the dissolved gas, more gas molecules escape the surface to the solvent and the gas becomes less soluble.

Everyday example:

Pvap of solutionsA non-volatile solute LOWERS the vapor pressure of the

solventMolecules of the solute block the surface of the solvent,

making it harder for the solvent molecules to escape into the gas phase.

The number of particles is directly proportional to the decrease in the amount of vapor pressure, so strong electrolytes (which completely dissociate into ions) have a greater effect on the Pvap.

A volatile solute INCREASES the vapor pressure of the solvent

Raoult’s law

RAOULT’S LAWPsoln = Xsolvent (P°solvent)

Psoln = observed vapor pressure

Xsolvent = mole fraction of the solvent

P°solvent = the pressure of the solvent alone

Raoult’s law practiceGlycerine, C3H8O3 – a non volatile solute

What is the Psoln made by adding 164 g of glycerine to 338

ml of water at 39.8°C? Vapor pressure of water is 54.74 torr at this temperature. Density of water is 0.992 g/ml

Raoult’s law with electrolytes52.9 g CuCl2 (a strong electrolyte) is added to 800.0 ml of

water at 52.0°C. Vapor pressure of water at this temperature is 102.1 torr and the density is .987 g/ml.

Hint: Write the dissociation reaction to determine the total number of moles of ions in solution.

Finding MM using Raoult’s law29.6 °C P°H20 = 31.1 torr 86.7 grams of an unknown non-volatile, non-electrolyte

is added to 350.0 g of water and the P soln = 28.6 torr. What is the molar mass of the substance?

VOLATILE SOLUTESContribute to vapor pressure

Ptotal = Psolute + Psolvent

Ptotal = (Xsolute )(P°solute) + (Xsolvent)(P°solvent)

Practice problemWhat is the vapor pressure when 58.9 g of hexame is

mixed with 44.0 grams of benzene at 60.0°C?P° Hexane C6H14 60.0°C is 573 torrP° benzene C6H6 60.0°C is 391 torr

COLLIGATIVE PROPERTIESBoiling point elevationFreezing point depressionOsmotic pressure

A colligative property only depends on the number, not the identity, of the solute particles.

BP ELEVATIONA non-volatile solute increases the bp of the liquid

(since the overall Psoln is lowered)Recall that boiling occurs when Patm = Psoln

bp elevation depends on the molal concentration of the solute

If the bp elevation is known, the molar mass of a solute may be determined.

BP ELEVATIONΔT = i Kb msolute

ΔT = bp elevationi = van t Hoff factor = number of ions that result

from complete dissociationKb is the bp elevation constant for a solvent

Kb H20 = 0.51 °C kg/mol

msolute = moles solute/ kg solvent

Practice problemWhat is the boiling point elevation if 31.65 g of NaCl is

added to 220.0 ml of water at 34.0 °C if the density of water is 0.994 g/ml. Assume complete dissociation of NaCl.

FREEZING POINT DEPRESSIONThe solvent must be cooled to a lower temperature

to form crystals since solute particles are blocking the solid formation

ΔT = i Kf msolute

Kf H20 = -1.86 °C kg/mol

Practice problemHow many grams of glycerin (C3H8O3) must be added to

350.0 grams of water to lower the freezing point to -3.84 °C?

Vant Hoff factor ii expected = the number of moles of ions that results

from complete dissociation of a solute.Ex. NaCl Na+ + Cl-

i = 2But, i actually = 1.9Why is it lower?

Some ions in solution pair up momentarily – “ion pairing”i = moles of particles in solution/ moles of solute

dissolved

iIon pairing is greater in solutions with higher charged

particles since the ions will have a greater attraction for each other.

Ex. FeCl3

i expected = 4i observed = 3.4

OSMOSISOsmosis – flow of a solvent into a solution through a

semi-permeable membrane due to differences in solute concentration. Flows into a region with higher solute concentration.

hypertonic – higher pressure than the surrounding solution due to higher solute concentration.

hypotonic – lower solute concentration and pressure than the surrounding solution

Isotonic – solutions with identical osmotic pressures.

Osmotic pressure

Lid with pressure sensors

Lowpressure

Highpressure

OSMOTIC PRESSUREOsmotic pressure – the pressure that just stops osmosis.Measured in atmCan be used to find molar mass

More accurate than bp or fp detemination

= i M R T = osmotic pressure (in atm)

i = vant Hoff factorM = molarity of the solute

R = 0.0821 Latm/mol KT = Kelvin temperature

Practice problemThe osmotic pressure of a solution of 26.5 mg of

aspartame per L is 1.70 torr @ 30.0 °C. What is the molar mass of aspartame?

Everyday exampleDesalination of water by reverse osmosis.Apply a pressure greater than the osmotic pressure to a

solution of salt water and the pure water will pass through the semi-permeable membrane towards the pure solvent side.