Question

What is the molar solubility of calcium sulfate in pure water?

Ksp=2.4x10-5 for calcium sulfate.
A. What is the molar solubility of calcium sulfate in pure water?

B. What is the mass solubility of calcium sulfate in pure water, expressed in g/L?

Answer 1

We assess the following equilibrium:

`CaSO_4(s) rightleftharpoons Ca^(2+) + SO_4^(2-)`

Explanation:

And `K_(sp)=([Ca^(2+)][SO_4^(2-)])/[[CaSO_4(s)]`

But since `[CaSO_4(s)]`, the concentration of a solid, is a meaningless concept, the equilibrium is governed solely by the ion product, i.e.

`K_(sp)=[Ca^(2+)][SO_4^(2-)]=2.4xx10^-5`.

If we say the solubility of `CaSO_4=S`, then,

`K_(sp)=[Ca^(2+)][SO_4^(2-)]=SxxS=S^2=2.4xx10^-5`

And thus `S=sqrt{2.4xx10^-5}` `=` `4.90xx10^-3*mol*L^-1`

Given that `CaSO_4` has a molar mass of `136.14*g*mol^-1`,

`S=136.14*g*cancel(mol^-1)xx4.90xx10^-3*cancel(mol)*L^-1~=0.7*g*L^-1`

Note that this problem assumes standard conditions of `298*K` and `1*atm` pressure. Under non-standard conditions, say at `398*K`, how do you think the solubility would evolve?