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By assuming sulphide ion ( $S^{2 -}$ ) hydrolyses completely into $H S^{- 1}$ but further hydrolysis of $H S^{- 1}$ is insignificant. If molar solubility of PbS in water found to be $6.0 \times 10^{- 10} m o l . l i t^{- 1},$ the solubility product (K_sp) of PbS at same temperature is: $[K_{w} (H_{2} O) = 10^{- 14}, K a_{1} (H_{2} S) = 10^{- 7}, K a_{2} (H_{2} S) = 10^{- 14}]$

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$3.6 \times 10^{- 29} m o l^{2} . l i t^{- 2}$

$2.16 \times 10^{- 28} m o l^{2} . l i t^{- 2}$

$3.6 \times 10^{- 28} m o l^{2} . l i t^{- 2}$

$2.16 \times 10^{- 29} m o l^{2} . l i t^{- 2}$

answer is B.

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Detailed Solution

$S^{2 -} + H_{2} O ⇌ H S^{- 1} + O H^{-}$

Hydrolysis constant = $\frac{K w}{K a_{2}} = \frac{[H S^{- 1}] [O H^{-}]}{[S^{2 -}]}$

$∴ [S^{2 -}] = {(6 \times 10^{- 10})}^{2}$

$K_{S P} (P b S) = {[P b^{2 +}]}^{1} {[S^{2 -}]}^{1} = (6 \times 10^{- 10}) {(6 \times 10^{- 10})}^{2} = 2.16 \times 10^{- 28} M^{2}$

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