Question

Solubility of \(\mathrm{Mg}(\mathrm{OH})_2\) is ' S ' \(\mathrm{mol} / \mathrm{dm}^3\) its solubility product is equal to

• A. \(\mathrm{S}^{2}\)
• B. \(27\text{ }\mathrm{S}^{4}\)
• C. \(4\text{ }\mathrm{S}^{3}\)
• D. \(4\text{ }\mathrm{S}^{2}\)

Hint:

Write the dissociation equilibrium and express both ion concentrations in terms of S.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^{-}\)
If the solubility is S, then \([\text{Mg}^{2+}] = S\) and \([\text{OH}^{-}] = 2S\).

Step 2: Calculation
\(K_{sp} = [\text{Mg}^{2+}][\text{OH}^{-}]^2 = S\times(2S)^2\)
\(= 4S^3\).

Step 3: Compare the options
The expression \(4S^3\) is in the options.

Final Answer:
The solubility product is four times S cubed. \[ \boxed{\text{(C) }4\text{ S}^{3}} \]