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In an aqueous solution 10^(-2) M Na(2)SO...

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_(sp) (PbI_(2))=10^(-9)` and `K_(sp)(PbSO_(4))=10^(-8)`].

A

`10^(-2)` M

B

`10^(-3)` M

C

`10^(-6)` M

D

`10^(-5)` M

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The correct Answer is:
b
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