Question

what is solubility, in mol/L, of MgCO3 in a 0.65 mol/L solution of MgCL2 if the ksp of MgCO3 is 2.5*10^-5

Answer 1

MgCO3 ==> Mg^2+ + CO3^2-
....x......x........x
where x = solubility in mols/L.

........MgCl2 --> Mg^2+ + 2Cl^-
I........0.65.....0..........0
C.......-0.65....0.65.....2*0.65
E .......0.........0.65....1.30

Substitute from the above ICE charts into Ksp for MgCO3 and solve for x.
(Mg^2+) = x+0.65 (that's x from MgCO3 and 0.65 from MgCl2.)
(CO3^2-) = x