The solubility of `CaCO_(3)` is `7mg//L`. Calculate the `K_(sp)` of `BaCO_(3)` whne `Na_(2)CO_(3)` is added slowely a solution containing equimolar concentration of `Ca^(2+)` and `Ba^(2+)` and no precipitate is formed until `90%` of `Ba^(2+)` has beem precipitated as `BaCO_(3)`.

Explanation

Mw of `CaCO_(3) = 100 g mol^(-1)`
`[CaCO_(3) = (7 xx 10^(-3))/(100)`
`Q_(sp)` for `CaCO_(3) = ((7xx10^(-3))/(100))^(2) = 49xx10^(-10)`
When only `Ba^(2+)` is `90%` precipitated then only `CaCO_(3)` starts precipitation. Then, let solution contains `x M` of `Ca^(2+)` and `Ba^(2+)` each.
`[Ca^(2+)] [CO_(3)^(2-)] = 49 xx 10^(-10)`
`[CP_(3)^(2-)] =(49xx10^(-10))/(x)`
`[Ba^(2+)] = (x xx100)/(100)`
`(90% Ba^(2+)` precipitated `10% Ba^(2+)` left)
`K_(sp)` of `BaCO_(3) = [Ba^(2+)] [CO_(3)^(2-)]`
` = (x xx10)/(100) xx (49xx10^(-10))/(x) = 4.9 xx 10^(-10)`