Molar Solubility of CaCO3 in Pure Water and Na2CO3 Solution

What is the molar solubility of CaCO3 in (a) pure water and (b) 0.10 M Na2CO3?

A) s=9.3×10^-5, B) s=8.7×10^-8, C) s=9.3×10^-5, D) s=8.7×10^-8

Answer:

The molar solubility of CaCO3 in pure water is 9.3×10^-5, and in a 0.10 M Na2CO3 solution is 8.7×10^-8.

Final answer: The molar solubility of CaCO3 in pure water can be found through solubility and equilibrium principles, whereas in a 0.10 M Na2CO3 solution, the molar solubility will be lower due to the common ion effect.

Explanation:

The molar solubility of CaCO3 both in pure water and in a solution of 0.10 M Na2CO3 can be calculated using solubility data and ICE (Initial, Change, Equilibrium) tables. For the equation CaCO3 <-> Ca2+ + CO32-, molar solubility is equivalent to the equilibrium concentrations of Ca2+ and CO32-. Given the chemical equilibrium and the solubility product, we can calculate the molar solubility. For (a) pure water, the molar solubility is similar to the example's result of 1.45 × 10^-4 M for a certain compound. It is applicable in the same manner for CaCO3 in pure water. For (b) a solution of 0.10 M Na2CO3, due to common ion effect, the molar solubility of CaCO3 will decrease. Therefore, the molar solubility in a solution of 0.10M Na2CO3 will be less than in pure water.