MHT CET · Chemistry · Ionic Equilibrium
Solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(s \mathrm{~mol} \mathrm{~L}^{-1}\). The solubility product \(\left(K_{s p}\right)\) under the same condition is
- A \(4 s^{3}\)
- B \(3 s^{4}\)
- C \(4 s^{2}\)
- D \(s^{3}\)
Answer & Solution
Correct Answer
(A) \(4 s^{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{Ca}(\mathrm{OH}) & \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ K_{s p} &=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \\ &=(s)(2 s)^{2}=4 s^{3} \end{aligned}\)
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