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Solid Ba(NO(3))(2) is gradually dissolve...

Solid `Ba(NO_(3))_(2)` is gradually dissolven in `1.0xx10^(-4)` M `Na_(2)CO_(3)` solution. At what concentration of `Ba^(2+)` will a precipitate begin to form?
(`K_(sp)` for `BaCO_(3)=5.1xx10^(-9)`)

A

`4.1xx10^(-5)`M

B

`5.1xx10^(-5)`M

C

`8.1xx10^(-8)`M

D

) `8.1xx10^(-7)`M

Text Solution

Verified by Experts

The correct Answer is:
B
`[Ba^(2+)]=K_(sp)/([CO_(3)^(2-)])=(5.1xx10^(-9))/10^(-4) =5.1xx10^(-5)M`
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