Question

The solubility-product constant (Ksp) for CaCO3 at 25°C is 3.4 ✕ 10⁻⁹. What is the molar solubility of this substance in 0.32 M CaCl₂ at 25°C?

a) 1.7 ✕ 10⁻⁹ M
b) 5.3 ✕ 10⁻⁹ M
c) 9.6 ✕ 10⁻⁹ M
d) 12.8 ✕ 10⁻⁹ M

Answer 1

The molar solubility of CaCO3 in 0.32 M CaCl2 is 1.06 × 10–10 M at 25°C, which is calculated considering the common ion effect and the Ksp of CaCO3. The available answer options do not match this value, suggesting a discrepancy in the provided data.

Step-by-step explanation:

To calculate the molar solubility of CaCO3 in a 0.32 M CaCl2 solution, we begin by acknowledging that CaCO3 dissociates into Ca2+ and CO32- in water with the solubility-product constant (Ksp) of 3.4 × 10–9. The presence of CaCl2 means there is a common ion effect, which will affect the solubility. Based on the Ksp expression, Ksp = [Ca2+][CO32-], we can infer that the initial concentration of Ca2+ due to CaCl2 will suppress the dissolution of CaCO3. Therefore, the molar solubility (s) of CaCO3 will be the concentration of added CO32- at equilibrium:

Ksp = [Ca2+ + s][s]

As Ca2+ is in excess (≈0.32 M), we assume s is negligible in comparison. Thus, we simplify the expression to:

Ksp = [0.32][s]

Solving for s gives us the molar solubility of CaCO3 in a solution with CaCl2:

3.4 × 10–9 = (0.32) s

s = ±1.06 × 10–10 M

This value does not match any of the answer choices given, so it's possible that either the question has a typo in the Ksp value, or the answer choices provided are incorrect.