
Julie S. answered 02/25/17
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I believe you need to address the equilibria separately, but combine the values for the Ca2+. In other words, let x be the amount of Ca2+ from the CaSO4, and let y be the amount of Ca2+ from the CaC2O4.
For simplicity and avoiding writing all of the superscripts and subscripts, I'm just going to say "Ca" for the calcium ion and CO3 for carbonate and C2O4 for oxalate.
So your first equilibrium
CaCO3 ↔ Ca + CO3 K = 4.7 × 10–9
If "x" molar of this dissolves, then [Ca] from this process will be x and [CO3] will be x.
CaC2O4 ↔ Ca + C2O4 1.3 × 10–9
So your first equilibrium
CaCO3 ↔ Ca + CO3 K = 4.7 × 10–9
If "x" molar of this dissolves, then [Ca] from this process will be x and [CO3] will be x.
CaC2O4 ↔ Ca + C2O4 1.3 × 10–9
If "y" molar of this dissolves, then [Ca] from this process will be y and [C2O4] will be y.
Since we are getting Ca dissolved from both solids and both equilibria, its concentration will be (x+y).
That would mean that [Ca][CO3] = (x + y)(x) = 4.7 × 10–9 and [Ca][C2O4] = (x + y) (y) = 1.3 × 10–9
From here I'm a bit stuck, unless you can assume that x will be larger than y, because the K value is larger for that equilibrium. Specifically 4.7 / 1.3 = 3.6 times as large. If we can assume that, then x = 3.6y and then you can get equations with one variable through substitution.
But I'm not 100% sure that is a valid assumption to make. It seems logical, but I haven't dealt with simultaneous equilibria of this nature in a long time. :(
Hope those thoughts help you, good luck!