Question

The solubility product $\left(K_{sp}mo{l}^{3}d{m}^{-9}\right)$ of $M{X}_2$ at 298 K based on the information available for 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 $4\times {10}^{-15}$

$M/M^{2+}$ (saturated solution of salt $M{X}_2\left)||M^{2+}\right(0.001M)$ emf of concentration cell,

$E_{cell}=\frac{0.059}{n}log\frac{[M^{2+}{]}_{RHS}}{[M^{2+}{]}_{LHS}}$

$0.059=\frac{0.059}{2}log\frac{\left[0.001\right]}{\left[M^{2+}\right]}$

$\therefore [M^{2+}{]}_{LHS}={10}^{-5}M$

Let solubility of salt be S mol/litre

thus $\underset{\left(S\right)}{M{X}_2}\to \underset{S}{M^{2+}}+\underset{2S}{2X^{-}}$

$\therefore K_{sp}=4s^3=4\times ({10}^{-5}{)}^3=4\times {10}^{-15}$