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 (K_sp) of magnesium hydroxide (Mg(OH)₂) given its solubility (S mol L⁻¹), we can follow these steps:

### Step 1: Write the dissociation equation
When magnesium hydroxide dissolves in water, it dissociates as follows:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Express the solubility in terms of S
Let the solubility of Mg(OH)₂ be S mol L⁻¹. At equilibrium, the concentrations of the ions will be:
- \([\text{Mg}^{2+}] = S\)
- \([\text{OH}^-] = 2S\) (since 2 moles of OH⁻ are produced for every mole of Mg(OH)₂ that dissolves)

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

### Step 4: Substitute the equilibrium concentrations into the K_sp expression
Substituting the concentrations we found in Step 2 into the K_sp expression:
\[ K_{sp} = (S)(2S)^2 \]

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

### Conclusion
Thus, the solubility product \( K_{sp} \) of magnesium hydroxide is:
\[ K_{sp} = 4S^3 \]

### Final Answer
The correct answer is \( K_{sp} = 4S^3 \).

---