You are provided with `500mL` of hard water, containing `0.005 mol` of `CaC1_(2)` and two `H_(2)SO_(4)` samples of `0.001M` and `0.02M` concentration. Which one or both or none can be used for precipitating `Ca^(2+)` ions.
`K_(sp)` of `CaSO_(4) = 2.4 xx 10^(-4)`.

Explanation

`[Ca^(2+)] = 0.005mol (500 mL)^(-1)`
`= 0.01 mol (1000mL)^(-1)`
i. Using `0.001 M H_(2)So_(4)`: (When equal volumes of hard water and `H_(2)SO_(4)` are used).
`:. [Ca^(2+)]` after addition of `H_(2)SO_(4) = (0.01)/(2) = 0.005M`
`[SO_(4)^(2-)]` in `0.01M H_(2)SO_(4) = 0.001M`.
`[SO_(4)^(2-)]` in solution `= 0.001//2 = 0.0005M`
`:. Q_(sp)` or `(IP)` of `CaSO_(4) = [Ca^(2+)] [SO_(4)^(2-)]`
`= 0.005 xx 0.0005`
`= 2.5 xx 10^(-6)`
`Q_(sp) lt K_(sp) (2.5 xx 10^(-6) lt 2.4 xx 10^(-4))`.
Hence, no precipitation of `CaSO_(4)` will occur.
ii. Using `0.02M H_(2)SO_(4):-` (When equal volumes of hard water and `H_(2)SO_(4)`are used)
`:. [Ca^(2+)]` after addition of `H_(2)SO_(4) = (0.01)/(2) = 0.005M`
`[SO_(4)^(2-)]` in `0.01M H_(2)SO_(4) = 0.002M`.
`[SO_(4)^(2-)]` in solution `= (0.02)/(2) = 0.01M`
`:. Q_(sp)` or `(IP)` of `CaSO_(4) = [Ca^(2+)] [SO_(4)^(2-)]`
`= 0.005 xx 0.01`
` = 5 xx 10^(-5)`
`Q_(sp) lt K_(sp) (5 xx 10^(-5) lt 2.4 xx 10^(-4))`.
Hence no precipitation of `CaSO_(4)` will occur.