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: Understand the dissociation of calcium carbonate.
The dissociation of calcium carbonate (\( \text{CaCO}_3 \)) is given by:
\[ \text{CaCO}_3 \rightleftharpoons \text{Ca}^{2+} + \text{CO}_3^{2-} \]
The solubility product \( K_{\text{sp}} \) is the product of the concentrations of the ions in equilibrium. For calcium carbonate, we have:
\[ K_{\text{sp}} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \]

Step 2: Set up the relation.
Let the solubility of calcium carbonate be \( s = 6.4 \times 10^{-5} \, \text{mol dm}^{-3} \). Then:
\[ [\text{Ca}^{2+}] = s \quad \text{and} \quad [\text{CO}_3^{2-}] = s \]
Thus, the solubility product is:
\[ K_{\text{sp}} = (6.4 \times 10^{-5})^2 = 5.06 \times 10^{-10} \]

Step 3: Final conclusion.
Thus, the solubility product of calcium carbonate at 298 K is:
\[ \boxed{5.06 \times 10^{-10}} \]