Question

The value of `K_(sp)` for `SrSO_4` is `2.8 times 10^-7`. What is the solubility of `SrSO_4` in moles per liter?

Answer 1

`"Solubility"` `~=` `100*mg*L^-1?`

Explanation:

Now we interrogate the solubility equilibrium.....

`SrSO_4(s) rightleftharpoons Sr^(2+) + SO_4^(2-)`

And `K_"sp"=[Sr^(2+)][SO_4^(2-)]`

We write `S="solubility of strontium sulfate."` And thus `S=[Sr^(2+)]`, and `S=[SO_4^(2-)]`...and we substitute these values into the solubility expression......

`K_"sp"=[Sr^(2+)][SO_4^(2-)]=SxxS=S^2`

And thus..............................

`S=sqrt(K_"sp")=sqrt(2.8xx10^-7)=5.29xx10^-4*mol*L^-1`.

And the gram solubility is thus.....

`5.29xx10^-4*mol*L^-1xx183.68*mol*L^-1=??`

If we attempted to dissolve the slat up in say `0.100*mol*L^-1` `"sodium sulfate"`, do you think the solubility would increase or decrease? Why?