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?
- A \(5.06 \times 10^{-10}\)
- B \(4.096 \times 10^{-9}\)
- C \(3.05 \times 10^{-10}\)
- D \(2.8 \times 10^{-9}\)
Answer & Solution
Correct Answer
(B) \(4.096 \times 10^{-9}\)
Step-by-step Solution
Detailed explanation
\(K_{sp} = s^2\) \(K_{sp} = (6.4 \times 10^{-5})^2\)
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