If the solubility of `Mg(OH)_2` in water is `S mol L^-1` then its `K_sp` will be

Options

A. `S^3`
B. `4 S^3`
C. `27 S^3`
D. `8 S^3`

Answer

To find the solubility product constant \( K_{sp} \) of \( Mg(OH)_2 \) given its solubility \( S \) in water, we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \( Mg(OH)_2 \) be \( S \) mol/L. This means that when \( Mg(OH)_2 \) dissolves, it produces:
- \( S \) mol/L of \( Mg^{2+} \)
- \( 2S \) mol/L of \( OH^{-} \) (since two hydroxide ions are produced for every 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 concentrations from Step 2 into the \( K_{sp} \) expression:
\[ K_{sp} = [S][2S]^2 \]

### Step 5: Simplify the expression
Calculating \( [2S]^2 \):
\[ [2S]^2 = 4S^2 \]
Now substituting this back into the \( K_{sp} \) expression:
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Final Result
Thus, the solubility product constant \( K_{sp} \) for \( Mg(OH)_2 \) is:
\[ K_{sp} = 4S^3 \]