Question

The electrochemical cell shown below is a concentration cell, $M|M^{2+}$ (Saturated solution of a sparingly soluble salt, $M{X}_2\left)||M^{2+}\right(0.001mold{m}^{-3})M.$ The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at 298K is 0.059

The solubility product $(K_{sp}:mo{l}^{3}d{m}^{-9})$ of $M{X}_2$ at 298 K based on the information available the given concentration cell is (take $2.303\times R\times 298/F=0.059V)$

• A. $1\times {10}^{-15}$
• B. $4\times {10}^{-15}$
• C. $1\times {10}^{-12}$
• D. $4\times {10}^{-12}$

Answer 1

The correct option is B $4\times {10}^{-15}$

The solubility equilibrium for $M{X}_2$ is
$M{X}_2\left(s\right)\rightleftharpoons M^{2+}\left(aq\right)+2X^{-}\left(aq\right)$
Solubility product, $K_{sp}=\left]\right[M^{2+}\left]\right[X^{-}{]}^2$
$={10}^{-5}\times 2\times {10}^{-5}{)}^2=4\times {10}^{-15}$
[$\because$ In saurated solution of $M{X}_2,\left[X^{-}\right]=2\left[M^{2+}\right]]$