Question

What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`?
(For calcim sulphae , `K_(sp) is 9.1xx10^(-6))`.

Answer 1

`CaSO_(4)(s) hArr Ca^(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)xx136 g L^(-1)`
`=0.411 g L^(-1)`
`(Mw CaSO_(4)=136 g mol^(-1))`
Thus, for dissolveing `0.411 g`, water required `=1L`
`:.` For dissolving 1 g, water required `=1/0.411L=2.43 L`