What will be the solubility product of `Ca(OH)_2` if its solubility is `sqrt3`.

Options

A. 3
B. `3sqrt3`
C. `12sqrt3`
D. 27

Answer

To determine the solubility product (Ksp) of calcium hydroxide, \( Ca(OH)_2 \), given its solubility as \( \sqrt{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
Calcium hydroxide dissociates in water as follows:
\[ Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2OH^{-} (aq) \]

2. **Define Solubility**:
Let the solubility of \( Ca(OH)_2 \) be \( S \). Given that \( S = \sqrt{3} \), we can express the concentrations of the ions at equilibrium:
- The concentration of \( Ca^{2+} \) will be \( S \).
- The concentration of \( OH^{-} \) will be \( 2S \) because two hydroxide ions are produced for each formula unit of calcium hydroxide that dissolves.

3. **Substitute the Values**:
Substitute \( S \) and \( 2S \) into the expression for the solubility product (Ksp):
\[ Ksp = [Ca^{2+}][OH^{-}]^2 \]
Substituting the values we have:
\[ Ksp = S \cdot (2S)^2 \]

4. **Calculate Ksp**:
Now, calculate \( Ksp \):
\[ Ksp = S \cdot 4S^2 = 4S^3 \]
Substitute \( S = \sqrt{3} \):
\[ Ksp = 4(\sqrt{3})^3 \]

5. **Simplify the Expression**:
Calculate \( (\sqrt{3})^3 \):
\[ (\sqrt{3})^3 = 3^{3/2} = 3 \cdot \sqrt{3} \]
Therefore,
\[ Ksp = 4 \cdot 3 \cdot \sqrt{3} = 12\sqrt{3} \]

### Final Answer:
The solubility product \( Ksp \) of \( Ca(OH)_2 \) is \( 12\sqrt{3} \).

---