Question

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

Answer 1

(I). Calculation of molar solubility of `CaSO_(4)`
`CaSO_(4)(s)hArr Ca^(2+) (aq) +SO_(4)^(2-) (aq)`
Let the solubility of salt in water be S
`[Ca^(2+) (aq)] =S " and " [SO_(4)^(2-) (aq)] =S`
`K_(sp) =[Ca^(2+) (aq)][SO_(4)^(2-)(aq)]`
`9.1xx 10^(-6) =S xx S`
`S=(9.1 xx 10^(-6))^(1//2) =3.02 xx 10^(-3) M=3.02 xx 10^(-3) mol L^(-1)`
(II) Calculation of amount of water reaquired
Molar mass of `CaSO_(4) =(40 + 32 +64) =136 g mol^(-1)`
mass of `CaSO_(4) =( 3.02 xx 10^(-3) mol L^(-1)) xx (136 g mol^(-1)) =0.411 g L^(-1)`
Volume of water required to dissolve 0.41 g of `CaSO_(4) =1L`
Volume of water requried to dissolve 1.0 g of `CaSO_(4) =((1L) xx (1.0 g))/((0.411 g)) = 2.43 L`