Question

The solubility of calcium carbonate at 298 K is \( 6.4 \times 10^{-5} \, \text{mol dm}^{-3} \). Calculate the value of solubility product at the same temperature.

• A. \( 5.06 \times 10^{-10} \)
• B. \( 4.096 \times 10^{-9} \)
• C. \( 3.05 \times 10^{-10} \)
• D. \( 2.8 \times 10^{-9} \)

Hint:

For sparingly soluble salts, the solubility product is calculated by squaring the solubility of the salt in water.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the molar solubility (\( S \)) and need to calculate the solubility product (\( \text{K}_{\text{sp}} \)) for \( \text{CaCO}_3 \).
Step 2: Key Formula or Approach:
Dissociation: \( \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \)
For an AB type salt: \( \text{K}_{\text{sp}} = S^2 \).
Step 3: Detailed Explanation:
Given molar solubility \( S = 6.4 \times 10^{-5} \text{ mol/L} \).
Calculate \( \text{K}_{\text{sp}} \):
\[ \text{K}_{\text{sp}} = (6.4 \times 10^{-5})^2 \] \[ \text{K}_{\text{sp}} = (6.4)^2 \times 10^{-10} \] \[ \text{K}_{\text{sp}} = 40.96 \times 10^{-10} \] \[ \text{K}_{\text{sp}} = 4.096 \times 10^{-9} \] Step 4: Final Answer:
The solubility product is \( 4.096 \times 10^{-9} \).