Solve the following
Question:

The solubility of $\mathrm{CdSO}_{4}$ in water is $8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility in $0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution is______________. (Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \mathrm{M}$ )

Solution:

(64)

In pure water, $K_{s p}=S^{2}=\left(8 \times 10^{-4}\right)^{2}$

$=64 \times 10^{-8}$ In $0.01 \mathrm{MH}_{2} \mathrm{SO}_{4}$

$\mathrm{K}_{\mathrm{sp}}=\mathrm{x}(\mathrm{x}+0.01)$

$=64 \times 10^{-8}$

$\mathrm{x}+0.01 \cong 0.01 \mathrm{M}$

So, $x(0.01)=64 \times 10^{-8}$

$\mathrm{x}=64 \times 10^{-6} \mathrm{M}$

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