`CaCO _(3) and BaCO _(3)` have solubility product values `1xx10^(-8) and 5 xx10^(-9)` respectively. If water is shaken up with both solids till equilibrium is reached, the concentration of `CO_(3)^( 2-)`ion is

Options

A. `1.5 xx 10^(-8)`
B. `1.225 xx 10^(-4)`
C. `2.25 xx 10^(-9)`
D. `2.5 xx 10^(-8)`

Answer

B

Explanation

Since `K_(sp)` of `CaCO_(3)` and `BaCO_(3)` are very close. So concentration of any species cannot be neglected. Let the solubility of `CaCO_(3)` and `BaCO_(3)` are x and y M.
`therefore CaCO_(3) hArr underset(x)(Ca^(2+)) + underset(x)(CO_(3)^(2-)) , BaCO_(3) hArr underset(y)(Ba^(2+)) + underset(y)(CO_(3)^(2-))`
Total `[CO_(3)^(2-)] = (x+y)`
`therefore K_(sp)` of `CaCO_(3) = [Ca^(2+)][CO_(3)^(2-)] = (x+y), K_(sp)` of `BaCO_(3) = [Ba^(2+)][CO_(3)^(2-)] = y(x+y)`
`therefore x (x+y) = 10^(-8), y(x+y) = 5 xx 10^(-9)`
`(5 xx 10^(-9))/(10^(-8)) = (y(x+y))/(x(x+y)), therefore (y)/(x) = 5 xx 10^(-1) rArr y = 0.5 x`
Substitute the value of `y, x (x+0.5x) = 10^(-8), x = 0.8 xx 10^(-4), y = 0.5 xx 0.8 xx 10^(-4)`
`therefore [ Ca^(2+)] = x = 0.8 xx 10^(-4) therefore [Ba^(2+)] = y = 0.4 xx 10^(-4), [CO_(3)^(2-)] = (x+y) = 1.2 xx 10^(-4)`