[THC]

20 ml 0,50 M Na2SO4 is added to a mixture of 40 ml 0,10 M BaCl2 and 40 ml 0,10 M CaCl2.

Determine [Ba^2+] og [Ca^2+] at equlibrium.

Solubilities:
L(BaSO4) = 1,5*10^-9 M^2
L(CaSO4) = 6,1*10^-5 M^2.

So I found the concentrations
c(Ba^2+) = 0.040 M
c(Ca^2+) = 0.040 M
c(SO4^2.) = 0.100 M

But what now? I know that both salts precipitates since the concentrations of the ions exceeds the required value. But I'm not sure how to calculate the concentrations at equilibrium since I don't know how much of the salts are in solid state.

[AWK]

Assume completeness of precipitation. From stoichiometry you can find an excess of sulfate anions. Finaly calculate concentration of Ca²⁺ and Ba²⁺ at the presnce of the excess of sulfates.

[THC]

Assume completeness of precipitation. From stoichiometry you can find an excess of sulfate anions. Finaly calculate concentration of Ca²⁺ and Ba²⁺ at the presnce of the excess of sulfates.

So for BaSO4:
1,5*10^-9 M^2 =(0.040 M - x)*(0.100 M -x) => x ~ 0,040 M.

And Ba²⁺ is therefore 0,040 M - x = 2.5*10^-8 M.

The solution guide says the answer is Ba²⁺ = 6,6*10^-8 M.

[AWK]

No.
You have 0.004 mole of solid BaSO4. Its solubility depends on the excess concentration of sulfate anion.

[THC]

No.
You have 0.004 mole of solid BaSO4. Its solubility depends on the excess concentration of sulfate anion.

Ah! I get it now. Thank you, my results matches those in the solution guide now.