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Ionic equilibrium

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By Expert Faculty of Sri Chaitanya

Question

An aqueous solution of metal chloride MCl₂ (0.05 M) is saturated with H₂S (0.1 M). The minimum pH at which metal sulphide will be precipitated is ____.
$[K_{sp} MS = 5 \times 10^{- 21}, K_{1} (H_{2} S) = 10^{- 7}, K_{2} (H_{2} S) = 10^{- 14}$

Moderate

Solution

$K_{sp} MS = [M^{2 +}] [S^{2 -}]$
$5 \times 10^{- 21} = (0.05) [S^{2 -}]$
For precipitation $Q_{sp} = K_{sp}$
$∴ [S^{2 -}] = \frac{5 \times 10^{- 21}}{5 \times 10^{- 2}} = 10^{- 19}$
$For H_{2} S :$
$\begin{array}{l} H_{2} S ⇌ H^{\oplus} + {HS}^{⊖} (K_{1}) \\ {HS}^{⊖} ⇌ H^{\oplus} + S^{2 -} (K_{2}) \\ K_{1} \times K_{2} = \frac{{[H^{\oplus}]}^{2} [S^{2 -}]}{H_{2} S} \\ 10^{- 7} \times 10^{- 14} = \frac{{[H^{\oplus}]}^{2} [10^{- 19}]}{0 .1} \\ {[H^{\oplus}]}^{2} = 10^{- 3}, \end{array}$
$\Rightarrow [Taking negative logarithm both sides]$
$\begin{array}{l} - 2 \log [H^{\oplus}] = 3 \\ pH = \frac{3}{2} = 1 .5 \end{array}$

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