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MHT CET · Chemistry · Ionic Equilibrium

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

  1. A \(5.06 \times 10^{-10}\)
  2. B \(4.096 \times 10^{-9}\)
  3. C \(3.05 \times 10^{-10}\)
  4. D \(2.8 \times 10^{-9}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4.096 \times 10^{-9}\)

Step-by-step Solution

Detailed explanation

\(K_{sp} = s^2\) \(K_{sp} = (6.4 \times 10^{-5})^2\)