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?

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"#).

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")#