MHT CET · Chemistry · Ionic Equilibrium
The solubility product of \(\mathrm{Mg}(\mathrm{OH})_2\) is \(1.8 \times 10^{-11}\) at \(298 \mathrm{~K}\). What is its solubility in \(\mathrm{mol} \mathrm{dm}^{-3}\) ?
- A \(1.650 \times 10^{-4}\)
- B \(2.120 \times 10^{-4}\)
- C \(3.184 \times 10^{-4}\)
- D \(4.550 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(1.650 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{Mg}(\mathrm{OH})_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Mg}_{\text {(aq) }}^{2+}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} \\ & \text {Here, } x=1, \mathrm{y}=2 \\ \therefore \quad & \mathrm{K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+\mathrm{y}}=(1)^1(2)^2 \mathrm{~S}^{1+2}=4 \mathrm{~S}^3 \\ \therefore \quad & \mathrm{S}=\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{1.8 \times 10^{-11}}{4}} \\ & =\sqrt[3]{4.5 \times 10^{-12}}=1.650 \times 10^{-4}\end{array}\)
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