`CaCO_(3)` and  `BaCO_(3)` have solubility product values `1 xx 10^(-8)` and `5 xx 10^(-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. None of these.

Answer

B

Explanation

Since `K_(s p)` 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_(s p)` of `CaCO_(3) = [Ca^(2+)][CO_(3)^(2+)] = x(x+y)`
`K_(s p)` of `BaCO_(3) = [Ba^(2+)][CO_(3)^(2-)] = y(x+y)`
`therefore x(x+y) = 10^(-8)` ........(i)
`y(x+y) = 5 xx 10^(-9)` .....(ii)
`(5 xx 10^(-9))/(10^(-8)) = (y(x+y))/(x(x+y))`
`therefore (y)/(x) = 5 xx 10^(-1)`
`y = 0.5 x`
Substitute the value of y in (i) or (ii).
`x(x+0.5 x) = 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)`