Solide `Ba(NO_(3))_(2)` is hradually dissolved in a `1.0xx10^(-4)M Na_(2)CO_(3)` solution. At what concentration of `Ba^(2+)` will a precipitate being to from ? `(K_(sp)` for `BaCO_(3)=5.1xx10^(-9))`

Options

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

Answer

C

Explanation

`Ba^(2+)(aq.)+CO_(3)^(2-)(aq.)hArrBaCO_(3)(s)`
`K_(sp)(BaCO_(3))=C_(Ba^(2+))C_(CO_(3)^(2-))`
or `C_(Ba^(2+))= (K_(sp))/(C_(CO_(3)^(2-))`
`= (5.1xx10^(-9))/(10^(-4))`
`= 5.1xx10^(-5)`
Precipitation occurs when ionic product just exceeds `K_(sp)`. Thus, precipitate begins to appear when the concentration of `Ba^(2+)` ions just exceeds `5.1xx10^(-5)M`.