The `K_(a)`  values of `CaCO_(3)` and `CaC_(2)O_(4)` in water are `4.7 xx 10^(-9)` and `1.3 xx 10^(-9)`,  respectively, at `25^(@)C`. If a mixture of two is washed with `H_(2)O`,  what is `Ca^(2+)` ion concentration in water?

Options

A. `7.746 xx 10^(-5)`
B. `5.831 xx 10^(-5)`
C. `6.856 xx 10^(-5)`
D. `3.606 xx 10^(-5)`

Answer

A

Explanation

Since `K_(s p)` of `CaCO_(3)` and `CaC_(2)O_(4)` are very close, so concentration of any species cannot be neglected.
Let the solubilities of `CaCO_(3)` and `CaC_(2)CO_(4)` are x and y M.
`{:(CaCO_(3),hArr,Ca^(2+)+,CO_(3)^(2-),),(,,x,x,),(CaC_(2)O_(4),hArr,Ca^(2+) +,CO_(3)^(2-),),(,,y,y,):}`
Total `[Ca^(2+)] = x + y`
`K_(s p)` of `CaCO_(3) = [Ca^(2+)][CO_(3)^(2-)] = (x+y)x`
`K_(s p)` of `CaC_(2)O_(4) = [Ca^(2+)][C_(2)O_(4)^(2-)] = (x+y)y`
`therefore x(x+y) = 4.7 xx 10^(-9)` .......(i)
`y(x+y) = 1.3 xx 10^(-9)` ....(ii)
`(x)/(y) = (4.7 xx 10^(-9))/(1.3 xx 10^(-9))`
`therefore x = 3.615 y`
Substituting the value of x in equation (i) or (ii)
`3.615y(3.615y + y) = 4.7 xx 10^(-9)`
`3.615 xx 4.615 y^(2) = 4.7 xx 10^(-9)`
`therefore y = ((4.7 xx 10^(-9))/(16.68))^(1//2) = 1.67 xx 10^(-5)M`
`x = 6.03 xx 10^(-5)M`
`[Ca^(2+)] = x + y = (6.03 xx 10^(-5) + 1.67 xx 10^(-5))`
`= 7.707 xx 10^(-5)M`