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A solution is 0.10 M Ba(NO(3))(2) and 0....

A solution is 0.10 M `Ba(NO_(3))_(2)` and 0.10 M `Sr(NO_(3))_(2.)` If solid `Na_(2)CrO_(4)` is added to the solution, what is `[Ba^(2+)]` when `SrCrO_(4)` begins to precipitate?
`[K_(sp)(BaCrO_(4))=1.2xx10^(-10),K_(sp)(SrCrO_(4))=3.5xx10^(-5)]`

A

`7.4xx10^(-7)`

B

`2.0xx10^(-7)`

C

`6.1xx10^(-7)`

D

`3.4xx10^(-7)`

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