Question

The solubility product constant of CaSO4 is 2.4 x 10^-5 mol2dm-6, K sp of Ca(OH)2 is 5.5 x 10^-6 mol3dm-9 and K sp of CaCO3 is 8.7 x 10^-9 mol2dm-6. How can I calculate the concentration of Ca2+ in the solution?

Answer 1

I got `["Ca"^(2+)] = "0.0042 M"`, less than what you would get when placing just `"CaSO"_4` in solution (`"0.0049 M"`), or just `"Ca"("OH")_2` in solution (`"0.0111 M"`).

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Well, if they're all in solution at the same time, why not construct a composite equilibrium?

`"CaSO"_4(s) rightleftharpoons "Ca"^(2+)(aq) + "SO"_4^(2-)(aq)`
`"Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH"^(-)(aq)`
`ul("CaCO"_3(s) rightleftharpoons "Ca"^(2+)(aq) + "CO"_3^(2-)(aq)`
`"CaSO"_4(s) + "Ca"("OH")_2(s) + "CaCO"_3(s) rightleftharpoons 3"Ca"^(2+)(aq) + "CO"_3^(2-)(aq) + "SO"_4^(2-)(aq) + 2"OH"^(-)(aq)`

`beta = K_(sp1)K_(sp2)K_(sp3)`

`= 2.4 xx 10^(-5) cdot 5.5 xx 10^(-6) cdot 8.7 xx 10^(-9) = 1.15 xx 10^(-18)`

`= ["Ca"^(2+)]^3["CO"_3^(2-)]["SO"_4^(2-)]["OH"^(-)]^2`

We define `s` for the concentration of whatever in solution has a coefficient of `1`. Thus,

`1.15 xx 10^(-18) = (3s)^3(s)(s)(2s)^2`

`= 108s^7`

Therefore,

`color(blue)(["Ca"^(2+)]) = 3s = 3(beta/108)^(1//7)`

`= 3 cdot ((1.15 xx 10^(-18))/108)^(1//7) "M"`

`=` `color(blue)("0.0042 M")`