Question

Solubility of Marble? What is the solubility of Ca^2+ in normal rainwater, for which pH = 5.6?

Answer 1

What you have here is a common-ion effect. Ignoring the involvement of the pH for the moment, you'd have a Ksp/solubility question. That is, you'd contemplate the dissolution reaction of calcium carbonate in water:\[{\rm CaCO}_3(s) \rightarrow {\rm Ca}^{2+}(aq) + {\rm CO}_{3}^{2-}(aq)\]For which:\[K = [{\rm Ca}^{2+}][{\rm CO}_{3}^{2-}] = K_{sp}({\rm CaCO}_3) = 4.8 \times 10^{-9}.\]You could probably solve just that problem, right? Let x = [Ca+2] = [CO3-2] at equilibrium, and away you go.

The problem here is that you have an additional reaction that takes CO3-2 out of solution as soon as it's formed, and that is the acid-base reaction between CO3-2 (a good base) and H+ in the water:\[{\rm CO}_{3}^{2-}(aq) + {\rm H}^+(aq) \rightarrow {\rm HCO}_{3}^{-}(aq)\]For which\[K = \frac{[{\rm HCO}_{3}^{-}]}{[{\rm H}^+][{\rm CO}_{3}^{2-}]} = \frac{1}{K_a({\rm HCO}_{3}^{-})} = \frac{1}{5.0 \times 10^{-11}} = 2.0 \times 10^{10}\]To solve the combined equilibrium problems, you need to solve both K equations simultaneously.

If you let x = [Ca+2] at equilibrium, and y = [HCO3-] at equilibrium, then [CO3-2] will be x - y. Think carefully about that, so you are sure you know why. Now you have two equations in two unknowns:\[x(x-y) = K_{sp}\]and\[\frac{y}{(10^{- {\rm pH}})(x-y)}= \frac{1}{K_a}\] Also be sure you understand why I was able to write [H+] as 10^-pH.

Now it's just algebra. to simplify, let me define\[r = \frac{10^{- {\rm pH}}}{K_a}\]and\[s = K_{sp}\] Then our two equations are\[x(x-y) = s\] and \[\frac{y}{x-y} = r\]Solving the second equation for y I get\[y = \frac{r}{1 + r} x\]Substituting that into the first equation I get \[x^2 = s(1+r)\]Taking the square root and remembering what everything stands for we have:\[[{\rm Ca}^{2+}] = \sqrt{K_{sp}\left(1 + \frac{10^{- {\rm pH}}}{K_a}\right)}\]When I plug in the numbers I get:\[[{\rm Ca}^{2+}] = 0.016 M\]You'd want to go back and calculate [HCO3-] and [CO3-2], and make sure they satisfy the Ksp and Ka equations, to be sure I haven't made some algebra or math mistakes.

Answer 2

OK

Answer 3

le me check

Answer 4

no the answer is not correct

Answer 5

Well, I may have made an algebra or math booboo somewhere. It will be a good exercise to go through my solution very carefully and try to find out where -- then you'll know how I did it!

Answer 6

Hmm, also another source I just consulted has Ksp for CaCO3 at 3.36 x 10^-9. So you'll want to be sure the constants (Ksp,Ka) are those used in your course.