If the solubility of an aqueous solution of `Mg(OH)_(2)` be X ` moll t^(-1)` than `k_(sp)` of `Mg(OH)_(2)`is:

Options

A. `4X^(3)`
B. `108X^(5)`
C. `27X^(4)`
D. 9X

Explanation

To find the solubility product constant (\(K_{sp}\)) of \(Mg(OH)_2\) given that its solubility is \(X\) moles per liter, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \(Mg(OH)_2\) in water can be represented as:
\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \(Mg(OH)_2\) be \(X\) moles per liter. This means that at equilibrium:
- The concentration of \(Mg^{2+}\) ions will be \(X\) moles per liter.
- The concentration of \(OH^{-}\) ions will be \(2X\) moles per liter (since two hydroxide ions are produced for each formula unit of \(Mg(OH)_2\)).

### Step 3: Write the expression for \(K_{sp}\)
The solubility product constant (\(K_{sp}\)) is given by the expression:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]

### Step 4: Substitute the concentrations into the \(K_{sp}\) expression
Substituting the values from Step 2 into the \(K_{sp}\) expression:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 = (X)(2X)^2 \]

### Step 5: Simplify the expression
Now, simplify the expression:
\[ K_{sp} = X \cdot (2X)^2 = X \cdot 4X^2 = 4X^3 \]

### Conclusion
Thus, the solubility product constant \(K_{sp}\) of \(Mg(OH)_2\) is:
\[ K_{sp} = 4X^3 \]

### Final Answer
The correct answer is \(4X^3\).

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