1Chapter 14 — Solutions

Overview of Chapter 14• Solutions• Concentrations:

MolarityMolalityMole fraction

• Colligative PropertiesFreezing point depressionBoiling point elevation

• Osmosis

Questions to consider:

• How much salt do we need to add to ice to melt it?

• Why is it bad to drink sea water?

• How can you purify sea water?

Today’s Topics

• Boiling point elevation

• Freezing point depression

• Osmosis

Boiling Point Elevation

Elevation in BP = ∆Tbp = Kbp•m(where Kbp is characteristic of solvent)

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

Change in Boiling PointIf we dissolve 62.1 g of ethylene glycol (1.00 mol) in250. g of water, what is the boiling point of thesolution?

Kbp = +0.512 oC/molal for water

Solution1. Calculate solution molality = 4.00 m2. ∆Tbp = Kbp • m ∆Tbp = +0.512 oC/molal (4.00 molal) ∆Tbp = +2.05 oC Boiling Point = 102.05 oC

Freezing Point Depression

The freezing point of a solution is lowerthan that of the pure solvent.

FP depression = ∆Tfp = Kfp•m

Pure waterPure water Solution ofSolution ofEthylene glycolEthylene glycol

and waterand water

2Chapter 14 — Solutions

Calculate the freezing point of a 4.00 molalglycol/water solution.

Kfp = -1.86 oC/molal

Solution∆Tfp = KFP • m = (-1.86 oC/molal)(4.00 m)∆Tfp = -7.44 oCRecall that ∆Tbp = +2.05 ˚C for this solution.

Freezing Point Depression Boiling Point Elevation and Freezing Point Depression

∆T = K•m•ii = van’t Hoff factor = number of particles

produced per formula unit.Compound Theoretical Value of iEthylene glycol 1NaCl 2CaCl2 3

Q: How much NaCl must be dissolved in 4.00 kg ofwater to lower the freezing point to -10.00 oC?(Kfp = -1.86 oC/molal)

SolutionFirst, get the concentration from ∆Tfp = Kfp • m • i

Here, i = 2 for NaCl. For every mole of NaCl salt, we getNa+ and Cl- ions, so we get 2 moles of particles.

-10.00 oC = (-1.86 oC/molal) • Concentration • 2

Concentration = 2.69 molal NaCl

Freezing Point DepressionQ: How much NaCl must be dissolved in 4.00 kg ofwater to lower the freezing point to -10.00 oC?(Kfp = -1.86 oC/molal)

SolutionConcentration = 2.69 molal NaCl

2.69 moles NaCl / kg water x (58.44 g/mole NaCl) =157 g NaCl / kg water

157 g NaCl / kg water x 4 kg water =629 g NaCl

Freezing Point Depression

Boiling and freezing point constants Osmosis

Dissolving theshell in vinegar

Egg in cornsyrup

Egg in purewater

3Chapter 14 — Solutions

Osmosis Osmosis

The semipermeable membrane allows only themovement of solvent molecules.

Solvent molecules move from pure solvent tosolution in an attempt to make both have the

same concentration of solute.

Driving force is entropy

Osmotic Pressure, ∏Equilibrium is reached whenpressure produced by extrasolution counterbalancespressure of solvent moleculesmoving through themembrane.OSMOTIC PRESSURE, ∏ ∏ = cRT

(c is conc. in mol/L)

Osmoticpressure

Osmosis

Osmosis of solventfrom one solution toanother can continueuntil the solutions areISOTONIC — theyhave the sameconcentration.

Osmosis and Living Cells

No water in or out Water goesout of cell Water goes into cell

Reverse Osmosis: Water Desalination

Water desalination plant in Tampa