Consider the following cell: $M (s) | M X_{2} (s), X^{-} (0.2 M) | | M^{2 +} (0.01 M) | M (s)$
The EMF of the cell at 298 K is 0.236V
[ $M X_{2} (s)$ being the sparingly soluble salt]
Pick out the correct options of the following:

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The solubility of $M X_{2} i n 0.2 M X^{-} i s 1.0 \times 10^{- 10} M$

The solubility product of $M X_{2} i s 2.0 \times 10^{- 11} M^{3}$

The solubility product of $M X_{2} i s 4.0 \times 10^{- 12} M^{3}$

The solubility of $M X_{2}$ in pure water is $1.0 \times 10^{- 4} M$

answer is B, C, D.

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Detailed Solution

$E_{c e l l} = \frac{0.059}{2} \log \frac{{[M^{2 +}]}_{r i g h t}}{{[M^{2 +}]}_{l e f t}} 0.236 = \frac{0.059}{2} \log \frac{[0.01]}{{[M^{2 +}]}_{l e f t}} {[M^{2 +}]}_{l e f t} = 1 \times 10^{- 10} M ∴ {Solubility of MX}_{2} i n 0.2 M X = 1 \times 10^{- 10} M K_{S P} o f M X_{2} = [M^{2 +}] {[X^{-}]}^{2} = 1 \times 10^{- 10} \times {(0.2)}^{2} = 4 \times 10^{- 12} M^{3} {Solubility of MX}_{2} i n w a t e r = {(\frac{K_{s p}}{4})}^{1 / 3} = {(\frac{4.0 \times 10^{- 12}}{4})}^{1 / 3} = 1.0 \times 10^{- 4} M$