A sample of hard water contains 100 ppm of `CaSO_(4)`. What minimum fraction of water should be evaporated off so that solid `CaSO_(4)` begins to separate out ? `K_(sp) ` for `CaSO_(4) ` is `9.0xx10^(-6)`.

Explanation

Maximum solubility of `CaSO_(4)` in water can be calculated from its `K_(sp)` value as follows :
`S=sqrt(K_(sp))=sqrt(9.0xx10^(-6))=3.0=10^s(-3) ` mol `L^(-1)`
Suppose the volume of water taken = V litre
As `CaSO_(4)` present is 100 ppm, i.e., 100 g per `10^(6) g ` of water, therefore, `CaSO_(4)` present in V litres `(V xx 10^(3)g)` of water
`=(100)/(10^(6))xxV xx 10^(3) g = 0.1 V g = (0.1 V)/(136)` moles (Molar mass of `CaSO_(4) = 136` g `"mol"^(-1)`)
After evaporation, suppose volume of water left = V' litre
Thus, V' litre of water will now contain `= (0.1 V) /(136) ` moles of `CaSO_(4)`.
This should be equal to the maximum solubility in moles `L^(-1)`.
`:. (0. V)/(136) xx (1)/(V')= 3.0 xx 10^(-3) or V' = (0.1 V)/(136 xx 3 xx 10^(-3))=0.245 V `
`:. ` Volume of water evaporated `= V - V' = V - 0.245 V = 0.755 V`
i.e., 75.5 % of water should be evaporated off .