The `K_(a)` value 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 miaxture 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_(sp)` 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 `yM`.
`{:(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_(sp) "of" CaCO_(3) = [Ca^(2+)] [CO_(3)^(2-)] = (x +y)x`
`K_(sp) "of" CaC_(2)O_(4) = [Ca^(2+)] [C_(2)O_(4)^(2-)] = (x+y)y`
`:. 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))`
`:. x = 3.615y`
Substituting the value of `x` in equation (i) or (ii)
`3.615 y(3.15y+y) = 4.7 xx 10^(-9)`
`3.615 xx 4.615y^(2) = 4.7 xx 10^(-9)`
`:. 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`