Question

The emf of the cell $Ag\left|AgI\left|KI\right(0.05M\left)\right|\right|AgN{O}_3\left(0.05M\right)|Ag$ is 0.788 V. Calculate the solubility product of AgI.

• A. $K_{SP}=1.16\times {10}^{-16}$
• B. $K_{SP}=1.16\times {10}^{-14}$
• C. $K_{SP}=1.16\times {10}^{-18}$
• D. None of the above

Answer 1

The correct option is A $K_{SP}=1.16\times {10}^{-16}$

$KI$ is a strong electrolyte, hence,

$[I^{-}{]}_{LHS}=0.05M$

$Agl\left(s\right)$ is sparingly soluble . If we manage to calculate $A{g}^{+}\left(Ag\right)$ in LHS half-cell, $K_{sp}$ can be calculated at LHS half cell

$Ag\left(s\right)\to \underset{LHS}{A{g}^{+}\left(xM\right)}+e^{-}{E}_{OX}^0=-8.80$

at RHS half cell

$\frac{A{g}^{+}\underset{RHS}{\left(0.05M\right)}+e^{-}\to Ag\left(s\right)}{A{g}^{+}\left(0.05M\right)\to A{g}^{+}\left(xM\right)}\begin{array}{c}{E}_{red}^0=0.80V\\ E_{cell=0.00V}^0\end{array}$

$Q=\frac{[A{g}^{+}{]}_{LHS}}{[A{g}^{+}{]}_{RHS}}=\frac{x}{0.05}$

$LHS\left[A{g}^{+}\right]\left[I^{-}\right]=K_{sp}\left[Agl\right]$

$\therefore [A{g}^{+}{]}_{LHS}=x=\frac{K_{sp}\left(Agl\right)}{\left[I^{-}\right]}$

$=\frac{K_{sp}\left(Agl\right)}{0.05}$

$\therefore Q=\frac{K_{sp}\left(Agl\right)}{(0.05{)}^2}$

$\therefore F_{cell}=E_{cell}^0-\frac{0.0591}{n}\log Q$

$0.788=0-\frac{0.0591}{1}\log Q$

$\log Q=-13.3333-14.6667$

$\therefore Q=4.46416\times {10}^{-14}$

$\therefore K_{sp}=4.6416\times {10}^{-4}\times (0.05{)}^2$

$=1.16\times {10}^{-16}$