The solubility of CdSO _ 4 in water is 8 . 0 × 10 ^ - 4 mol L ^ - 1...

In pure water,

${K}_{sp} = {S}^{2} = {\left(8 \times {10}^{- 4}\right)}^{2}$

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

In $0 . 01M{H}_{2}{SO}_{4}$

$\mathop{{H}_{2}{SO}_{4 ( aq )}}_{x} \to \mathop{2{H}_{( aq )}^{+}}_{x} + \mathop{{SO}_{4}^{2 -} ( aq . )}_{( x + 0 . 01 )}$

${K}_{sp} = x ( x + 0 . 01 )$

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

$x + 0 . 01 ≅ 0 . 01M$

So, $x ( 0 . 01 ) = 64 \times {10}^{- 8}$

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