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

Options

A. `4.1xx10^(-5)`
B. `5.1xx10^(-5)`
C. `8.1xx10^(-5)`
D. `8.1xx10^(-7)`

Answer

B

Explanation

`Na_(2)CO_(3) to 2Na^(+) + CO_(3)^(2-)`
`[Na_(2)CO_(3)]=[CO_(3)^(2-)]=1xx10^(-4)M`
`BaCO_(3) to Ba^(2+) + CO_(3)^(2-)`
`K_(sp(BaCO_(3)))=[Ba^(2+)][CO_(3)^(2-)]`
`[Ba^(2+)]=(K_(sp)(BaCO_(3)))/([CO_(3)^(2-)]`
`[Ba^(2+)]=(5.1xx10^(-9))/(1XX10^(-4))`
`[Ba^(2+)]=5.1xx10^(-5)M`
Hence, at `5.1xx10^(-5)` concentration of `Ba^(2+)`, a precipitate will begin to form.