Question

From the standard potentials

$$\begin{gathered} {\mathrm{Ag}}_2{\mathrm{SeO}}_4\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\rightleftharpoons 2\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{SeO}}_4^{2-},E^0=0.355\mathrm{V} \\ {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right),E^0=0.799\mathrm{V} \end{gathered}$$

calculate the solubility product constant for ${\mathrm{Ag}}_2{\mathrm{SeO}}_4$.

Short answer

The solubility product constant for ${\mathrm{Ag}}_2{\mathrm{SeO}}_4$is obtained as approximately $8.70\times {10}^{-15}$.

Step-by-step solution

Concept Introduction

There is an equilibrium constant required for the dissolution of a solid substance into an aqueous solution. This equilibrium constant is known as solubility product constant. The symbol K_sp is used to represent it.

Completing the reaction

The given reactions are -

$$\begin{gathered} {\mathrm{Ag}}_2{\mathrm{SeO}}_4\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\rightleftharpoons 2\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{SeO}}_4^{2-},E^0=0.355\mathrm{V} \\ {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right),E^0=0.799\mathrm{V} \end{gathered}$$

The overall reaction will be -

$$\begin{gathered} 2{\mathrm{Ag}}^{+}+{\mathrm{SeO}}_4^{2-}\rightleftharpoons {\mathrm{Ag}}_2{\mathrm{SeO}}_4 \\ {E}_{cell}^o=0.799\mathrm{V}-0.355\mathrm{V} \\ =0.444\mathrm{V} \end{gathered}$$

Calculate solubility product constant 

Let us apply the formula for$K$.

localid="1645716594081" $$\begin{gathered} \ln K=\frac{n{E}_{cell}^o}{25.693\times {10}^{-3}\mathrm{V}} \\ \ln K=\frac{2\times 0.444\mathrm{V}}{25.693\times {10}^{-3}\mathrm{V}} \\ \ln K=\frac{0.888\mathrm{V}}{25.693\times {10}^{-3}\mathrm{V}} \\ \ln K=34.68 \\ \Rightarrow K=e^{34.68} \\ \Rightarrow K=1.15\times {10}^{15} \end{gathered}$$

The formula for the solubility product constant is -

localid="1645716602158" $$\begin{gathered} Ksp=\frac{1}{K} \\ =\frac{1}{1.15\times {10}^{15}} \\ =8.70\times {10}^{-15} \end{gathered}$$