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Q.

The solubility of $P b S O_{4}$ in water is $0 .303 g / l$ at $25^{0} C$ . It’s solubility product at that temperature is $10^{- x} M^{2}$ then $x = ?$
(Given GMW of ${PbSO}_{4} = 303 g$ ).

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Detailed Solution

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solubility $= 0 .303 g / l = \frac{0 .303}{303} mol / l = 10^{- 3} moles / l$
$∴ K_{sp} = {(S)}^{2} = {(10^{- 3})}^{2} = 10^{- 6} M^{2}$

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