46
13.65)
Colligative properties depend on the concentration of particles in solution.
PA = XA P°A,
)Tf = i Kf m,
)P = Xsolute P°solvent,
)Tb = i Kb m,
A = iMRT
In the last 3 equations, the “i” is the number of particles in the formula for the
substance.
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 m
aqueous C6H12O6 solution. This is because the NaCl is ionic and when put in
H2O 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 and
stays 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 of
solute particles in solution. The boiling point elevation of a solution is
directly related to the total moles of dissolved particles (b.p. increases). The
NaCl solution will have a higher b.p. since its concentration of particles is
greater than that of the C6H12O6 Ideally, the 0.10 m NaCl solution should
have twice the effect as the 0.10 m C6H12O6 solution (the b.p. elevation
should be twice as great).
47
13.65) (cont.)
b) Calculate the b.p. of the two solutions. Must first calculate the b.p. elevation,
)Tb = i Kb m
Kb = 0.51°C/m for H2O
0.10 m NaCl:
)Tb = (2)*(0.51°C/m)*(0.10 m) = 0.20 m * 0.51°C/m
= 0.102°C
Tb = 100.102°C = 100.10°C
0.10 m C6H12O6: )Tb = (1)*(0.51°C/m)*(0.10 m) = 0.10 m * 0.51°C/m
= 0.051°C
Tb = 100.051°C = 100.10°C
In this case, the change is so small there really isn’t any difference in the boiling
points of the two solutions (to the correct number of sig. fig.). There would be a
significantly larger difference in the freezing points and osmotic pressures of the
two solutions.
c)
For solutions of electrolytes the experimental change in b.p. is smaller and
the b.p. is lower than that calculated (ideal value based on the b.p. elevation
equation). (There are also differences in the theoretical and experimental
values of the other colligative properties.) Why is this?
In solutions of strong electrolytes, such as NaCl, the ions completely
dissociate in solution. However, the electrostatic attractions between the ions
in solution can still occur and they form ion pairs. This ion association (ion
pairing) reduces the effective number of particles in solution. Since the
effective concentration of particles is lower than that calculated, the
experimental colligative property will be different than the calculated value
(based on the “ideal” concentration of particles). You can determine what’s
called the van’t Hoff factor, i, the effective number of particles (which I refer
to as iobs). This “observed” i ( iobs) is less than that of the ideal i, iideal (based
on the # ions from the formula unit) and approaches iideal as the solution
becomes more dilute (ions are less likely to “find” each other and form ion
pairs). For NaCl, iideal = 2, iobs = 1.87 (0.100 m) and 1.97 (0.001 m) solutions.
dec. conc.
iobs -----------------> iideal
48
13.67)
More particles ===> greater )Tf
(Higher conc part.) ===> (lower f.p.)
)Tf = i Kf mstated = Kf (i mstated) = Kf mparticles
Really 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:
LiBr (aq) —> Li+ (aq) + Br! (aq)
i=2
i mstated = (2)(0.050 m) = 0.10 mions (the ions are the particles)
Zn(NO3)2 (aq) —> Zn2+ (aq) + 2 NO3! (aq)
Zn(NO3)2: i = 3
i mstated = (3)(0.050 m) = 0.15 mions (the ions are the particles)
ˆ
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
LiBr
Zn(NO3)2
0.120
0.050
0.050
1
2
3
0.120
0.10
0.15
0.22
0.19
0.28
-0.22
-0.19
-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)
49
13.70) 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 mstated
i = # 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 = 1.99°C/m
n.f.p. = -114.6°C
glucose is nonionizing
Kb = 1.22°C/m
n.b.p. = 78.4°C
i=1
1) Calc. f.p. of soln
)Tf = (1)(1.99°C/m)(0.30 m) = 0.597°C
Note:
Tf.p. soln = Tf.p. solvent ! )Tf
(f.p. depression - lowering)
Tf.p. soln = -114.6°C ! 0.597°C = -115.197°C = -115.2°C
2) Calc. b.p. of soln
)Tb = (1)(1.22°C/m)(0.30 m) = 0.366°C
Note:
Tb.p. soln = Tb.p. solvent ! )Tb
(b.p. elevation - raising)
Tb.p. soln = 78.4°C + 0.366°C = 78.766°C = 78.8°C
50
13.70) (cont.)
51
13.70) (cont.)
52
13.71)
Determine the number of grams of ethylene glycol (C2H6O2) that must be added to
1.00 kg of H2O to produce a solution that freezes at -5.00°C.
Use f.p. depression to determine the molality of the C2H6O2 solution and from that
the mass of C2H6O2.
)Tf = 5.00°C
(The normal f.p. of H2O is 0.00°C)
)Tf = i Kf mstated
i = 1 (nonionizing nonelectrolyte)
)Tf
5.00°C
m = ------ = ------------- = 2.6881 m
Kf
1.86°C/m
2.6881 mol C2H6O2
62.06 g C2H6O2
? g C2H6O2 = 1.00 kg H2O x ------------------------- x --------------------1 mol C2H6O2
1 kg H2O
= 166.84 g C2H6O2
= 167 g C2H6O2
53
13.73)
54
13.76)
55
13.79)
56
13.80)
57
13.80) (cont.)