Question

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)`)
A. `4.1xx10^(-5) M`
B. `5.1xx10^(-5) M`
C. `8.1xx10^(-8) M`
D. `8.1xx10^(-7) M`

Answer 1

Correct Answer - B
`Na_(2)CO_(3)hArr2Na^(+)+CO_(3)^(2-)`
`[Na_(2)CO_(3)]=[CO_(3)^(2-)=1xx10^(-4) M`
`BaCO_(3)hArrBa^(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+)`, precipitate will begin to form.