Help me with this. My answer is not coming

Question

The values of Ksp of CaCO3 and CaC2O4 are 4.7 × 10–9 and 1.3 × 10–9 respectively at 25°C.

If the mixture of these two is washed with water, what is the concentration of Ca2+ ions in water?

Answer 1

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

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! 

 

Answer 2

After writing the dissolution equations and solving for x, you will find out that the concentration of calcium ion is:

 

[Ca2+] = (Ksp)^1/2

 

Since the total concentration of Ca2+ will be a combination of the two salts, you need to determine [Ca2+] for each salt and the total concentration will be approximately the sum of both values.

 

NOTE: the above is simply an approximation and the true value will depend on the common-ion effect. However, since both Ksp values are very similar, our approximation will not be too inaccurate.