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Homogeneous Mixture

The solubilit...

Question

The solubility product of a MA 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 $M A$ .
Given: $K_{a_{1}} (H_{2} A) = 4 \times 10 - 7$ at $25^{\circ} C$
$K_{a_{2}} (H_{2} A) = 5 \times 10 - 11$ at $25^{\circ} C$

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Solution

$M A; M^{2 +} + A^{2 -}; K_{s p} = 2 \times 10^{- 12}$
$2 \times 10^{- 12} = [0.1] [A^{2 -}] = [0.1] \frac{K_{A_{1}} K_{A_{2}} [H_{2} A]}{[H^{+}]^{2}}$
${[H^{+}]}^{2} = \frac{0.1 \times 2 \times 10^{- 17} \times 10^{- 2}}{2 \times 10^{- 12}}$
$[H^{+}]^{2} = 10^{- 20 + 12} = 10^{- 8}$
$[H^{+}] = 10^{- 4}$
$p H = 4$

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Similar questions

M A

2 \times 10^{- 12}

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