It is my understanding that calcium carbonate dissolves faster in lower pH environments. I am interested to know how higher PH levels affect the rate of dissolution. So I have broken this question down into 2 parts:
Calcium carbonate dissolves by falling apart into ions (calcium cations and carbonate anions):
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)}$$
The rate of dissolution does not depend on pH. However, there is a back reaction, which depends on the concentration of the calcium cation and the carbonate anion. The carbonate anion concentration, in turn, depends on the pH because you have carbonate, bicarbonate and dissolved carbon dioxide in fast equilibrium:
$$\ce{CO3^2-(aq) + 2 H+(aq) <=> HCO3^-(aq) + H+(aq) <=> H2CO3(aq) <=> CO2(aq) + H2O(l) }$$
You can take a look at what happens to carbonate at different pH values here (it is called speciation).
The bottom line is that in the pH range above 10 or 11, pH does not matter for carbonate (although the calcium ion might precipitate as hydroxide). When you go more acidic than pH 10, for every pH step, about 10 times less carbonate exists because it turns into bicarbonate of carbonic acid/carbon dioxide. So the back reaction is slower, potentially speeding up the net forward reaction.
Whether the back reaction has an effect on the net rate of dissolution depends on what is in the water before you add the solid calcium carbonate. It also depends on how much you add.
Given two specific scenarios (the entire composition of the system), a quantitative answer would be possible.
