Solutions for Chemical Activities Exercises
1.
a) since this is a +1, -1 electrolyte, μ = [KNO3] = 0.20 mM
b)
2.
3.
(a)
activity coefficients decrease with increasing ionic strength
(b)
activity coefficients decrease with increasing ionic charge
(c)
activity coefficients decrease with decreasing hydrated radius
All values gotten from table of activity coefficients.
For
μ = 0.050M, γ(SO42-) = 0.445
μ = 0.10 M, γ(HS-) = 0.76
μ = 0.025M,
4.
Determine the solubility of AgSCN in the presence of the common ion SCN-.
AgSCN(s)
X Ag+(aq) + SCN-(aq)
init
0
0.060
equil
x
0.060 + x
where x equals the increase in the [Ag+] as a result of the dissolving process.
at equil, Ksp = 1.1 x 10-12 = [Ag+]γAg+[SCN-]γSCNSince Ksp << 1 the contribution of AgSCN to the ionic strength is negligible.
Since KSCN is a +1, -1 electrolyte, μ = [KSCN] = 0.060 M
for μ = 0.060 M, Solutions for Chemical Activities Exercises
for μ = 0.060 M,
Substitute into the reaction quotient and solve for x.
Ksp = 1.1 x 10-12 = (x)(0.790)(0.060 + x)(0.800) • (x)(0.790)(0.060)(0.800)
assumes x << 0.060
Y
5.
x = 2.90 x 10-11 M = [Ag+]
Kw = 1.0 x 10-14 = [H3O+]γH3O+[OH-]γOH-
at equilibrium [H3O+] = [OH-] = x
Since LiBr is a +1, -1 electrolyte, μ = [LiBr] = 0.050 M
for μ = 0.050 M
γH+ = 0.86
γOH- = 0.81
Therefore, Kw = 1.0 x 10-14 = (x)(0.86)(x)(0.81)
Y x = 1.20 x 10-7 = [H3O+]
pH = -log (1.20 x 10-7 )(0.86) = 6.99
6.
pH = -log aH+ = -log Kw/aOHaOH- = [OH-]γOHfor 0.010 M NaOH + 0.012 M LiNO3, μ = 0.022 M (since all electrolytes involve
+1, -1 ions)
for μ = 0.022 M,
pH = -log (1.0 x 10-14)/(0.010)(0.873) = 11.94
Without using activities, pH = -log [H+] = -log Kw/[OH-] = -log (1.0 x 10-14)/(0.010) =
12.00 Solutions for Chemical Activities Exercises
7.
Phenol (C6H5OH) is a weak acid and will not contribute significantly to the ionic
strength. The ionic strength of the solution based on the LiBr electrolyte is
0.050 M.
C6H5OH(aq) + H2O(l)
init
º H3O+(aq) + C6H5O-(aq)
~10-7 M
0.0500 M
equil 0.0500 - x
~x
for μ = 0.0500 M
0
γH+ = 0.86
x
γC6H5O- = 0.835
γC6H5OH = 1
Y x = 2.65 x 10-6 M = [C6H5O-] = [H3O+] and pH = -log (2.65 x 10-6 )(0.86) = 5.64
8.
The mixture is a HC2H3O2 /C2H3O2- buffer.
HC2H3O2 + H2O
init:
0.195 M
equil: ~0.195 M
X H3O+ + C2H3O2~10-7 M
0.040 M
x
~0.040 M Solutions for Chemical Activities Exercises
Since HC2H3O2 is a weak acid, the only important source of ions in the buffer
solution is NaC2H3O2.
[NaC2H3O2] = 0.040 M
Y μ = 0.040 M Y γC2H3O2- = 0.902 - 0.062 = 0.840
γHC2H3O2 • 1 (since HC2H3O2 is uncharged)
Y pH = 3.99
Solutions for Chemical Activities Exercises
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