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

A sample of hard water contains 100 ppm of CaSO_(4).What minimum fract

1.	A sample of hard water contains 100 ppm of CaSO_(4).What minimum fraction of water shouldbe evaporated off so that solid CaSO_(4) begins to separate out ? K_(sp)for CaSO_(4)is 9.0xx10^(-6).
Answer» Solution :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 .