The solubility of ${P b S O}_{4}$ in water is $0.03 g / L$ . Calculate the solubility product of ${P b S O}_{4}$ .
How many mol CuI $(K = {\times 10}^{- 12}$ ) will dissolve in $1.0 L$ of $0.010 M N a l$ solution?

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Solution

$P b S O_{4} ⇌ {P b}^{2 +} + {S O_{4}}^{2 -}$
$K_{s p}$ for the above reaction will be-
$K_{s p} = [{P b}^{2 +}] [{S O_{4}}^{2 -}]$
Solubility of $P b S O_{4} = 0.03 g / L$
Molecular weight of $P b S O_{4} = 303.26 g / m o l$
Concentration of dissolved $P b S O_{4} = \frac{solubility}{mol. wt.} = \frac{0.03}{303.26} = 9.89 \times 10^{- 5} m o l / L$
$∴ K_{s p} = {(9.89)}^{2} \times 10^{- 10} = 9.78 \times 10^{- 11}$

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