Question

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. For calcium sulphate, `K_(sp)=9.1xx10^(-6)`.

Answer 1

`CaSO_(4)(s)hArrCa^(2+) (aq) + SO_(4)^(2-) (aq)`
If s is the solubility of `CaSO_(4)` in moles `L^(-1)`, then `K_(sp)=[Ca^(2+)]xx[SO_(4)^(2-)]=s^(2)`
or `s=sqrt(K_(sp))=sqrt(9.1xx10^(-6))=3.02xx10^(-3) "mol" L^(-1) = 3.02xx10^(-3) xx 136 g L^(-1) = 0.411 g L^(-1)` [Molar mass of `CaSO_(4) = 136 g "mol"^(-1)`)
Thus, for dissolving 0.411 g, water required = 1 L
`:. ` For dissolving 1 g, water required `=(1)/(0.411) L = 2.43 L`.