Q. For the given concentration cell $M|M^{2 +}$ $\left(\right.$ saturated solution of $\left(MX\right)_{2}\left.\right)\parallelM^{2 +}\left(\right.0.001molar\left.\right)$ $MX_{2}$ is a sparingly soluble salt) the EMF is $0.059volt$ at the temperature $298K.$ What is the solubility product of $MX_{2}$ at $298K$ ? $\left(\right.$ Consider $\frac{2 . 303 R \times 298}{ F}=0.059V\left.$

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A

$10^{- 15}$

B

$4\times 10^{- 15}$

C

$10^{- 12}$

D

$4\times 10^{- 12}$

Solution:

$E_{\text{cell }}=-\frac{0 . 059}{2}log\frac{\left[M^{2 +}\right]}{\left[\right. 0 . 001 \left]\right.}$
$\left[M^{2 +}\right]=10^{- 5}$
$\left(MX\right)_{2}\left(\right.s\left.\right)\rightleftharpoonsM^{2 +}+2X^{-}$
$\left[X^{-}\right]=2\times 10^{- 5}$
$K_{sp}=\left[M^{2 +}\right]\left[X^{-}\right]^{2}$
$=\left(10\right)^{- 5}\times \left(2 \times \left(10\right)^{- 5}\right)^{2}$
$=4\times 10^{- 15}$

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