InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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?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» Correct Answer - A 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` |
|