A hard water sample has `131 ppm CaSO_(4)`. What fraction of the water must be evporated in a container before solid `CaSO_(4)` begins to deposit. `K_(sp) of CaSO_(4) = 9.0 xx 10^(-6)`.

Explanation

Maximum solubility of `CaSO_(4)` in water
`S = sqrt(K_(sp)) = 3 xx 10^(-3)mol L^(-1)`
Let `V` litre of sample is taken, then `CaSO_(4)` present
`= (131 xx V xx 10^(3))/(10^(6)) g`
[`because` ppm = g of `CaSO_(4)` in `10^(6) g` of sample]
`= 131 xx 10^(-3) V g`
`= (131 xx 10^(-3)xxV)/(136)` mole in V L
If water is evaporated on heating so that just precipitation of `CaSO_(4)` occurs. Let `V_(1)L` of water is left, then
`(131 xx 10^(-3)xxV)/(136)` mol is present in `V_(1)L` solution is equal to `3 xx 10^(-3) xx V_(1)` mol
`:. (131 xx 10^(-3) xx V)/(136) = 3 xx 10^(-3) xx V_(1) :. V_(1) = 0.32V`
Thus, volume evaporated =` V - 0.32 V = 0.68V` or `68%` of water should be evaporated.