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Valency and Oxidation State

The solubilit...

Question

The solubility product of $M A$ is $2 \times 10^{- 12}$ at $25^{\circ} C$ . A solution is 0.1 M in $M (N O_{3})_{2}$ and it is saturated with 0.01 M $H_{2} A$ , then the minimum pH that must be maintained to start precipitation of $M A$ is (Given: $K_{a 1} (H_{2} A) = 4 \times 10^{- 7}; K_{a 2} (H_{2} A) = 5 \times 10^{- 11}$ )

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Solution

$M A ⇌ M^{2 +} + A^{2 -}; K_{s p} = 2 \times 10^{- 12}$
$2 \times 10^{- 12} = [0.1] [A^{2 -}]$ -----(1)
From the reaction,
$H_{2} A ⇌ 2 H^{+} + A^{2 -}; K_{a} = K_{a 1} \times K_{a 2}$
$[A^{2 -}] = K_{a 1} . K_{a 2} \frac{[H_{2} A]}{[H^{+}]^{2}}$ ----(2)
Putting (2) in (1)
$[H^{+}]^{2} = \frac{0.1 \times 2 \times 10^{- 17} \times 10^{- 2}}{2 \times 10^{- 12}}$
$[H^{+}] = 10^{- 4}$
pH= 4

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