Equilibrium constants at 25 ∘C are

Substance Kc Value of Kc
CaCO3 Ksp 4.5×10−9
H2CO3 Ka1 4.3×10−7
Ka2 5.6×10−11

What is the molar solubility of marble (i.e., [Ca2+] in a saturated solution) in acid rainwater, for which pH=4.20?

3 answers

CaCO3 ==> Ca^2+ + CO3^2- Ksp = (Ca^2+)(CO3^2-)
H2CO3 ==> H^+ + HCO3^- k1 = ....
HCO3^- ==> H^+ + CO3^3- k2 = ....
pH = 4.20; (H^+) = 6.31E-5

Solubility = (Ca^2+) = S
S = (CO3^2-)+(HCO3^-)+(H2CO3)

Find (CO3^2-) in terms of S.

k2 = (H^+)(CO3^2-)/(HCO3^-)
Plug in k2 and H^+ and solve for HCO3^- so that (HCO3^-) = some number x CO3^2-

Then k1 = (H^+)(HCO3^-)/(H2CO3)
Plug in H^+ and HCO3^- and solve for H2CO3 so that
(H2CO3) = some number x (CO3^2-).

Then go back to S = CO3^2- + HCO3^- + H2CO3 and plug all of that in to obtain
S = some number x (CO3^2-)
Then CO3^2-) = S/some number
Into Ksp, plug S in for Ca; plug S/some number in for CO3^2-; substitute for Ksp, solve for S and you have it.
Post your work if you get stuck. By the way, there is a quicker way to do this and it works on this problem but it doesn't work generall on all problems like it.

Thanks! I got it now

So what is the answer?