Thanks DrBob for answering the question! I was assuming it was something similar to that, but my confusion is that a follow up question is asking what happens to the dissolved Ca2+ if the pH is reduced to 2, so the pH has some effect, but I don't know what.
I assume this is the CaC2O4 problem. If the pH is reduced to 2, that makes the H^+ MUCH larger (by a factor of 10,000,000 which is really MUCH larger) and the solubility of the CaC2O4 is increased significantly. The reason is that you add a competing reaction (actually two reactions). You have this originally,
CaC2O4(s) ==> Ca^2+(aq) + C2O4^2-(aq)
and at a pH 9 the H^+ is too small to make much difference. BUT, when pH is increased to 2 you have this.
C2O4^2- + H^+ ==> HC2O4^- which has a k2 when written the other way PLUS this
HCO4^- + H^+ ==> H2C2O4 which is k1 when written the other way. Since both k1 and k2 are relatively small, the addition of H^+ forces the C2O4^2- and HC2O4^- to form more H2C2O4 and that means the C2O4^- from the original CaC2O4 is diminished. To compensate for that (remember Ksp is constant), more CaC2O4 dissolves. The bottom line is that CaC2O4 is more soluble in acid solution than in neutral or basic solution. The same thing happens with other slightly soluble salts such as CaSO4 (because of k2 for H2SO4) and similar slightly soluble salts.