Properties of Solutions

Solutions n Saturated n Unsaturated n Supersaturated n (Demo sodium acetate)

Solution Concentration n A 10.7 m soln of NaOH has d = 1.33 g/cm 3. Calc. n a) mole fraction of NaOH n b) the weight % of NaOH n c) the M of the soln

answers n a) n b) 30.0% NaOH n c) 9.98 M

Temperature and Solubility n If delta H is positive (endothermic) n If delta H is negative (exothermic)

Energy Changes in the Solution Process (slide ) n Three steps in the solution process n 1)solute particles are separated n delta H 1 = positive n 2) solvent particles are separated n delta H 2 = positive n 3) solvent and solute mix n delta H 3 = negative

n Heat of solution = H 1 + H 2 + H 3 n If H 3 > H 1 +H 2 the soln is exothermic - solute dissolves and solution warms n If H 3 < H 1 +H 2 the soln is endo - solute dissolves and soln cools n If H 3 <<< H 1 +H 2 the solute may not dissolve

Ionic Solutes - heat of hydration n Process same as last slide n H soln = H solute + H hydration n Ion size and charge determines hydration energy - charge density

n Which has the higher charge density? n a) Na + or Cs + b) Sr 2+ or Rb + n Which has the larger H hydration ? n a) Mg 2+ or Ba 2+ b) Mg 2+ or Na + n What generalization can we make about size, charge and solubility?

n NH 4 Cl diss. in water - endothermic n a) is H lattice for NH 4 Cl larger than H hydration ?? n ans: lattice must be > hydration n b) Given the answer to a) why does the solution form?? n ans: entropy

Factors Affecting Solubility n Molecular Structure - list generalizations n Ex. polar sub. dissolve polar

n which of the following will result in a more concentrated solution Explain n KNO 3 in water or KNO 3 in CCl 4 ?

Factors Affecting Solubility n Gas Pressure - Henry’s Law - n S gas = K H x P gas n Ex. degasing chlorine

n The PP of CO 2 inside a soda btl. is adjusted to 4 atm. What is the solubility of CO 2 ? K H CO 2 = 3 x mol/L*Atm n ans: 0.1 mol/liter

Colligative Properties Vapor Pressure of Solns n Raoult’s Law n VP soln = X solvent x VP solvent n X = mole fraction

n Calculate the Vapor Pressure when mol of glycerol (C 3 H 8 O 3, 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 Solution n mole fraction of solvent = n 27.4 mol H 2 O = n mol glycerol n VP = 92.5 torr x = 92.0 torr

Colligative Properties n boiling point elevation (phase diagram) n freezing point depression n delta T = K b mi k b = C n delta T = K f mi K f = 1.86 C

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

Problem Solution n mole of antifreeze = n 1.00 x 10 3 g/62.07 = 16.1 mol n molality = 16.1 mol/ kg solvent n = 3.62

molality problem solution n T b = K b x molality x i n = C/m x 3.62 m = 1.85 C n Therefore BP = C

Collig. Prop. MM calc n 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 K f for benzene and appl. to unknown)

Answer n K f(benz) --> T f = m x K f xi n 5.1 C = 1 m x K f K f = 5.1 C/m n then for unknown n 0.51 C = X (5.1C/m)X = 0.1 mol/Kg n next n 0.1 mol/kg (0.5kg)=0.05 mol unknown

ans cont’d n finally n 20g unknown/0.05 mol = 400 g/mol

Colligative Properties n antifreeze n making ice cream n salt on the streets

Lab: Let’s make Ice Cream

Osmotic Pressure n Pressure = MRTi n Isotonic solutions (IV drip) n 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).