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In an aqueous solution $10^{- 2} M {Na}_{2} {SO}_{4}$ and $10^{- 2} M NaI$ are present. Now pure $Pb {({NO}_{3})}_{2}$ is added gradually then calculate concentration of ${SO}_{4}^{2 -}$ when ${PbI}_{2}$ start precipitating in solution $[K_{s p} ({PbI}_{2}) = 10^{- 9}$ and $K_{s p} ({PbSO}_{4}) = 10^{- 8}$ ]

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$10^{- 3} M$

$10^{- 2} M$

$10^{- 6} M$

$10^{- 5} M$

answer is B.

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Detailed Solution

$[{Pb}^{2 +}]$ when ${PbI}_{2}$ starts to ppt. $= \frac{10^{- 9}}{{(10^{- 2})}^{2}} = 10^{- 5} M$

$∴ [{SO}_{4}^{2 -}]$ when ${PbSO}_{4}$ starts to ppt. $= \frac{10^{- 8}}{10^{- 5}} = 10^{- 3} M$

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