Example #1: Calculate the solubility product of AgI at 25.0 °C, given the following data:
Reduction half-reaction E° (V) AgI(s) + e¯ ---> Ag(s) + I¯ -0.15 I_{2}(s) + 2e¯ ---> 2I¯ -0.54 Ag^{+} + e¯ ---> Ag(s) 0.80
Solution:
1) The chemical equation for AgI dissolving is:
AgI(s) ⇌ Ag^{+}(aq) + I¯(aq)
and the K_{sp} expression is:
K_{sp} = [Ag^{+}] [I¯]
2) The equations to use are:
E° (V) AgI(s) + e¯ ---> Ag(s) + I¯ -0.15 Ag(s) ---> Ag^{+} + e¯ -0.80 Yielding E° = -0.95 V
3) Use the Nernst Equation:
E_{cell} = E° - (0.0591 / n) log K0 = -0.95 - (0.0591 / 1) log K
0.95 / -0.0591 = log K
log K = -16.07
K = 8.51 x 10¯^{17}
Note: some instructors might insist that you round the answer off to two significant figures. On this site, the K_{sp} is listed as 8.52 x 10¯^{17}.
Note also that you never have to use the K_{sp} expression to calculate anything. The K_{sp} is determined directly from the electrochemical data.
Example #2: Using the following reduction potentials, calculate the solubility product for AgCN at 298 K:
Ag^{+} + e¯ ---> Ag E = 0.80 V AgCN + e¯ ---> Ag + CN¯ E = -0.01 V
Solution:
1) Reverse first half-reaction:
Ag ---> Ag^{+} + e¯ E = -0.80 V AgCN + e¯ --> Ag + CN¯ E = -0.01 V
2) Add:
AgCN ---> Ag^{+} + CN¯ E = -0.81 V
3) Use -RT ln K = -nFE
(8.314) (298) ln K = (1) (96485) (-0.81)ln K = -31.544
K_{sp} = 2.0 x 10^{-14}