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).

A hard water sample has 131 ppm CaSO_(4). What fraction of the water m

1.	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).
Answer» Solution :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.