Question

Compare the g/L solubility of CaCO3 in pure water, a H3O+ (aq) solution with a PH of 3.00 and a H30+(aq) solution with a pH of 1.00.
Ksp for CaCO3 is 3.36 x 10^-9
CaCO3 (s) + H3O (aq) = Ca (aq) + HCO3 (aq) + H2O (l) (equilibrium equation)
Keq= 71

I knew how to do this for pure water and got
5.8x10-3 g/L
But i don't know how to do this for
Solution with a pH of 3.00 and solution with a pH of 1.00

Answer 1

I would do that this way.
K = 71 = (Ca^2+)(HCO3^-)/(H^+)
71 = (x)(x)/0.001
and solve for x = about 0.27M and convert that to g/L = about 27 g/L.

Answer 2

I just tried that and it told me the answer was incorrect. It gave a hint saying
In acidic solution, use the reaction given for the solvation of CaCO3 in acidic solution. Construct a concentration table that allows for the calculation of final concentration as if the reaction first goes to completion and then returns to equilibrium. The initial hydronium ion concentration is calculated from the pH.

I don't really get what it is telling me to do.

Answer 3

I did that first and didn't like the answer I got. Then I looked in my quant book and found a slightly different method for which I obtain 0.267M solubility. With that as an answer I looked to see how to get it and lo and behold came up with the exact same answer and it looked legit. So I posted.