The solubility of $C d S O_{4}$ in water is $8.0 \times 10^{- 4}$ mol L $^{- 1}$ . Its solubility in $0.01 M H_{2} S O_{4}$ solution is ______× $10^{- 6} m o l L^{- 1}$ . (Round off to the Nearest Integer).
Assume that solubility is much less than 0.01 M.

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Solution

Answer :64
$C d S O_{4} (s) ⇌ C d^{+ 2} S (a q) + S O_{4}^{2 -} S (a q)$

$S = 8 \times 10^{- 4} K_{s p} = S^{2} = 64 \times 10^{- 8}$

In $H_{2} S O_{4}$ ,
$C d S O_{4} (s) ⇌ C d^{2 +} S + S O_{4}^{2 -} S + 10^{- 2}$

$K_{s p} (C d S O_{4}) = S (S + 10^{- 2})$
$S + 10^{- 2} \approx 10^{- 2} ∵ solubility is much less than 0.01 M$

$K_{s p} (C d S O_{4}) = S \times 10^{- 2}$
$64 \times 10^{- 8} = S \times 10^{- 2}$
$S = 64 \times 10^{- 6}$

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