properties of solutions. solutions n saturated n unsaturated n supersaturated n (demo sodium...

Properties of SolutionsProperties of Solutions

SolutionsSolutionsSaturatedUnsaturatedSupersaturated (Demo sodium acetate)

Solution ConcentrationSolution Concentration

A 10.7 m soln of NaOH has d = 1.33 g/cm3. Calc.

a) mole fraction of NaOH b) the weight % of NaOH c) the M of the soln

answersanswers

a) 0.161 b) 30.0% NaOH c) 9.98 M

Temperature and SolubilityTemperature and Solubility

If delta H is positive (endothermic) If delta H is negative (exothermic)

Energy Changes in the Solution Energy Changes in the Solution ProcessProcess (slide 264 + 265) (slide 264 + 265)

Three steps in the solution process 1)solute particles are separated delta H1 = positive 2) solvent particles are separated delta H2 = positive 3) solvent and solute mix delta H3 = negative

Heat of solution = H1 + H2 + H3

If H3> H1 +H2 the soln is exothermic - solute dissolves and solution warms

If H3< H1 +H2 the soln is endo - solute dissolves and soln cools

If H3<<< H1 +H2 the solute may not dissolve

Ionic Solutes - heat of hydrationIonic Solutes - heat of hydration

Process same as last slide Hsoln = Hsolute + Hhydration

Ion size and charge determines hydration energy - charge density

Which has the higher charge density? a) Na+ or Cs+ b) Sr2+ or Rb+

Which has the larger Hhydration? a) Mg2+ or Ba2+ b) Mg2+ or Na+

What generalization can we make about size, charge and solubility?

NH4Cl diss. in water - endothermic

a) is Hlattice for NH4Cl larger than Hhydration??

ans: lattice must be > hydration

b) Given the answer to a) why does the solution form??

ans: entropy

Factors Affecting SolubilityFactors Affecting Solubility

Molecular Structure - list generalizations

Ex. polar sub. dissolve polar

which of the following will result in a more concentrated solution Explain

KNO3 in water or KNO3 in CCl4?

Factors Affecting SolubilityFactors Affecting Solubility

Gas Pressure - Henry’s Law - Sgas = KH x Pgas

Ex. degasing chlorine

The PP of CO2 inside a soda btl. is adjusted to 4 atm. What is the solubility of CO2? KH CO2 = 3 x 10-2 mol/L*Atm

ans: 0.1 mol/liter

Colligative PropertiesColligative PropertiesVapor Pressure of SolnsVapor Pressure of Solns

Raoult’s Law

VPsoln = Xsolvent x VPsolvent

X = mole fraction

Calculate the Vapor Pressure when 0.137 mol of glycerol (C3H8O3, MM = 92 amu) is added to 27.4 mol of water at 50 C. (VP of water at this T is 92.5 torr).

Problem SolutionProblem Solution

mole fraction of solvent = 27.4 mol H2O = 0.995 27.4 + 0.137 mol glycerol

VP = 92.5 torr x 0.995 = 92.0 torr

Colligative PropertiesColligative Properties

boiling point elevation(phase diagram)

freezing point depressiondelta T = Kbmi kb = 0.512 C

delta T = Kfmi Kf = 1.86 C

Boiling Point ElevationBoiling Point Elevation

If 1.00 kg of antifreeze (MM = 62.07 - non volatile, non dissociating) is added to 4450 g of water what is the boiling point of the solution?

Problem SolutionProblem Solution

mole of antifreeze = 1.00 x 10 3 g/62.07 = 16.1 mol

molality = 16.1 mol/ 4.450 kg solvent = 3.62

molality problem solutionmolality problem solution

Tb = Kb x molality x i

= 0.512 C/m x 3.62 m = 1.85 C

Therefore BP = 101.85 C

Collig. Prop. MM calcCollig. Prop. MM calc

1 mol of napthalene is diss. in 1000 g of benzene - the FP changes from 5.51C to 0.41 C. When 20. g of an unkwn is diss. in 500 g benzene the Fp of soln is 5.00C. What is the MM of the unkwn. (Hint: find Kf for benzene and appl. to unknown)

AnswerAnswer

Kf(benz) --> Tf = m x Kfxi

5.1 C = 1 m x Kf Kf = 5.1 C/m then for unknown 0.51 C = X (5.1C/m)X = 0.1 mol/Kg next 0.1 mol/kg (0.5kg)=0.05 mol unknown

ans cont’dans cont’d

finally 20g unknown/0.05 mol = 400 g/mol

Colligative PropertiesColligative Properties

antifreezemaking ice creamsalt on the streets

Lab: Let’s make Ice CreamLab: Let’s make Ice Cream

Osmotic PressureOsmotic Pressure

Pressure = MRTi Isotonic solutions (IV drip)Killing slugs(the idea here (which most north carolinians can

relate to) is that you can either put salt on them or put them in water. Either way their death is related to the osmotic flow of water in or out of the cell).