4613.65)

Colligative properties depend on the concentration of particles in solution.

PA = XA PA, )P = Xsolute Psolvent,)Tf = i Kf m, )Tb = i Kb m, A = i M R TIn the last 3 equations, the i is the number of particles in the formula for thesubstance.

i = # solute particles in solution

- integer for ideal behavior

i = 1 for a non-dissociating or non-ionizing compound (nonelectrolyte) = # particles (ions) resulting from the formula unit

a) A 0.10 m aqueous NaCl solution has a higher boiling point than a 0.10 maqueous C6H12O6 solution. This is because the NaCl is ionic and when put inH2O it dissociates (comes apart) to form ions. It forms 2 ions, Na+ and Cl&,for every NaCl. Thus, 0.10 moles of NaCl will give 0.20 moles of ions(particles) in solution. The C6H12O6 is a molecular non-ionizing solute andstays together as a single particle when it dissolves (as most, but not all,molecular substances do). So 0.10 moles of C6H12O6 will give 0.10 moles ofsolute particles in solution. The boiling point elevation of a solution isdirectly related to the total moles of dissolved particles (b.p. increases). TheNaCl solution will have a higher b.p. since its concentration of particles isgreater than that of the C6H12O6 Ideally, the 0.10 m NaCl solution shouldhave twice the effect as the 0.10 m C6H12O6 solution (the b.p. elevationshould be twice as great).

4713.65) (cont.)

b) Calculate the b.p. of the two solutions. Must first calculate the b.p. elevation,

)Tb = i Kb mKb = 0.51C/m for H2O

0.10 m NaCl: )Tb = (2)*(0.51C/m)*(0.10 m) = 0.20 m * 0.51C/m = 0.102C

Tb = 100.102C = 100.10C

0.10 m C6H12O6: )Tb = (1)*(0.51C/m)*(0.10 m) = 0.10 m * 0.51C/m = 0.051C

Tb = 100.051C = 100.10C

In this case, the change is so small there really isnt any difference in the boilingpoints of the two solutions (to the correct number of sig. fig.). There would be asignificantly larger difference in the freezing points and osmotic pressures of thetwo solutions.

c) For solutions of electrolytes the experimental change in b.p. is smaller andthe b.p. is lower than that calculated (ideal value based on the b.p. elevationequation). (There are also differences in the theoretical and experimentalvalues of the other colligative properties.) Why is this?In solutions of strong electrolytes, such as NaCl, the ions completelydissociate in solution. However, the electrostatic attractions between the ionsin solution can still occur and they form ion pairs. This ion association (ionpairing) reduces the effective number of particles in solution. Since theeffective concentration of particles is lower than that calculated, theexperimental colligative property will be different than the calculated value(based on the ideal concentration of particles). You can determine whatscalled the vant Hoff factor, i, the effective number of particles (which I referto as iobs). This observed i ( iobs) is less than that of the ideal i, iideal (basedon the # ions from the formula unit) and approaches iideal as the solutionbecomes more dilute (ions are less likely to find each other and form ionpairs). For NaCl, iideal = 2, iobs = 1.87 (0.100 m) and 1.97 (0.001 m) solutions.

dec. conc.iobs -----------------> iideal

4813.67)

More particles ===> greater )Tf(Higher conc part.) ===> (lower f.p.)

)Tf = i Kf mstated = Kf (i mstated) = Kf mparticlesReally only need to calculate the mparticles

Greater mpart ===> greater )Tf ===> lower f.p.Glucose: i = 1 (nonelectrolyte - does NOT dissociate into ions)

i mstated = (1)(0.120 m) = 0.120 mpart

LiBr: i = 2 LiBr (aq) > Li+ (aq) + Br! (aq)

i mstated = (2)(0.050 m) = 0.10 mions (the ions are the particles)

Zn(NO3)2: i = 3 Zn(NO3)2 (aq) > Zn2+ (aq) + 2 NO3! (aq)

i mstated = (3)(0.050 m) = 0.15 mions (the ions are the particles)

0.05 m LiBr < 0.120 m glucose < 0.05 m Zn(NO3)2---------------------------------------------------->

dec. f.p.

Could calc f.p. (Kf (H2O) = 1.86 C/m) - longer & no need to for this question

Compound mstated i mpart )Tf(C) Tf(C)Glucose 0.120 1 0.120 0.22 -0.22LiBr 0.050 2 0.10 0.19 -0.19Zn(NO3)2 0.050 3 0.15 0.28 -0.28

This all assumes ion-pairing for the ionic solutes is negligible- i.e. ideal ionic solutions formed in which i = # particles in the formula)

4913.70) Want f.p. and b.p. of the following solutions

For f.p. depression and b.p. elevation

)Tf = i Kf mstated )Tb = i Kb mstatedi = # particles = 1 for nonionizing (nonelectrolytes) = # ions from formula of cmpd for ideal ionic soln

a) 0.30 m glucose in C2H5OH (ethanol)

For ethanol: Kf = 1.99C/m Kb = 1.22C/mn.f.p. = -114.6C n.b.p. = 78.4C

glucose is nonionizing i = 1

1) Calc. f.p. of soln

)Tf = (1)(1.99C/m)(0.30 m) = 0.597CNote: Tf.p. soln = Tf.p. solvent ! )Tf (f.p. depression - lowering)

Tf.p. soln = -114.6C ! 0.597C = -115.197C = -115.2C

2) Calc. b.p. of soln

)Tb = (1)(1.22C/m)(0.30 m) = 0.366CNote: Tb.p. soln = Tb.p. solvent ! )Tb (b.p. elevation - raising)

Tb.p. soln = 78.4C + 0.366C = 78.766C = 78.8C

5013.70) (cont.)

5113.70) (cont.)

5213.71)

Determine the number of grams of ethylene glycol (C2H6O2) that must be added to1.00 kg of H2O to produce a solution that freezes at -5.00C.

Use f.p. depression to determine the molality of the C2H6O2 solution and from thatthe mass of C2H6O2.

)Tf = 5.00C (The normal f.p. of H2O is 0.00C))Tf = i Kf mstated i = 1 (nonionizing nonelectrolyte)

)Tf 5.00Cm = ------ = ------------- = 2.6881 m

Kf 1.86C/m

2.6881 mol C2H6O2 62.06 g C2H6O2? g C2H6O2 = 1.00 kg H2O x ------------------------- x --------------------- 1 kg H2O 1 mol C2H6O2

= 166.84 g C2H6O2

= 167 g C2H6O2

5313.73)

5413.76)

5513.79)

5613.80)

5713.80) (cont.)