If solubility of calcium hydroxide is `sqrt(3)`, then its solubility product will be

Options

A. 27
B. 3
C. 9
D. `12sqrt(3)`

Explanation

To find the solubility product (Ksp) of calcium hydroxide (Ca(OH)₂) given its solubility, we can follow these steps:

### Step 1: Write the dissociation equation
When calcium hydroxide dissolves in water, it dissociates into calcium ions (Ca²⁺) and hydroxide ions (OH⁻):
\[ \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define solubility
Let the solubility of calcium hydroxide be \( S \) moles per liter. Given that the solubility is \( \sqrt{3} \), we have:
\[ S = \sqrt{3} \]

### Step 3: Determine the concentrations of ions
From the dissociation equation, we can see that:
- The concentration of calcium ions, \([Ca^{2+}]\), will be \( S \).
- The concentration of hydroxide ions, \([OH^-]\), will be \( 2S \) (since 2 moles of OH⁻ are produced for every mole of Ca(OH)₂ that dissolves).

Substituting \( S \):
\[ [Ca^{2+}] = S = \sqrt{3} \]
\[ [OH^-] = 2S = 2\sqrt{3} \]

### Step 4: Write the expression for Ksp
The solubility product constant (Ksp) is given by the formula:
\[ K_{sp} = [Ca^{2+}][OH^-]^2 \]

### Step 5: Substitute the concentrations into the Ksp expression
Now, substituting the values we found:
\[ K_{sp} = [Ca^{2+}][OH^-]^2 = (\sqrt{3}) \times (2\sqrt{3})^2 \]

Calculating \( (2\sqrt{3})^2 \):
\[ (2\sqrt{3})^2 = 4 \times 3 = 12 \]

Now substituting this back into the Ksp expression:
\[ K_{sp} = \sqrt{3} \times 12 \]

### Step 6: Simplify the expression
Thus, we have:
\[ K_{sp} = 12\sqrt{3} \]

### Final Answer
The solubility product \( K_{sp} \) of calcium hydroxide is:
\[ K_{sp} = 12\sqrt{3} \]

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