`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_(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 `yM`.
`:. {:(CaCO_(3) hArr,Ca^(2+)+,CO_(3)^(2-)),(,x,x):}`
`{:(BaCO_(3) hArr,Ba^(2+)+,CO_(3)^(2-)),(,y,y):}`
Total `[CO_(3)^(2-)] = (x +y)`
`:. K_(sp)` of `CaCO_(3) = [Ca^(2+)] [CO_(3)^(2+)] = x(x+y)`
`K_(sp)` of `BaCO_(3) = [Ba^(2+)] [CO_(3)^(2-)] = y (x+y)`
`:. x(x+y) = 10^(-8) ...(i)`
`y(x+y) = 5 xx 10^(-9) ...(ii)`
`(5xx10^(-9))/(10^(-8)) = (y(x+y))/(x(x+y))`
`:. (y)/(x) = 5 xx 10^(-1)`
`y = 0.5 x`
Substitute the value of `y` in (i) or (ii).
`x(x+ 0.5x) = 10^(-8)`.
`x = 0.8 xx 10^(-4)`
`y = 0.5 xx 0.8 xx 10^(-4)`
`:. [Ca^(2+)] = x = 0.8 xx 10^(-4)`
`:. [Ba^(2+)] = y = 0.4 xx 10^(-4)`
`[CO_(3)^(2-)] = (x+y) = 1.2 xx 10^(-4)`.