MHT CET · Chemistry · Ionic Equilibrium
The solubility product of the sparingly soluble salt \(A B_2\) is \(2.56 \times 10^{-10}\) at 298 K. Calculate its solubility in \(\mathrm{mol~} \mathrm{dm}^{-3}\) at the same temperature ?
- A \(1 \times 10^{-4}\)
- B \(2 \times 10^{-2}\)
- C \(4 \times 10^{-4}\)
- D \(3 \times 10^{-2}\)
Answer & Solution
Correct Answer
(C) \(4 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\(K_{sp} = [A^{2+}][B^-]^2 = s(2s)^2 = 4s^3\) \(2.56 \times 10^{-10} = 4s^3\)
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