`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is

Options

A. `10^(-6)M^(3)`
B. `4xx10^(-6)M^(3)`
C. `5xx10^(-7)M^(3)`
D. `5xx10^(-6)M^(3)`

Answer

To find the solubility product (Ksp) of barium hydroxide (Ba(OH)₂) given that the pH of its saturated solution is 12, we can follow these steps:

### Step-by-Step Solution

1. **Determine pOH from pH**:
- We know that pH + pOH = 14.
- Given pH = 12, we can find pOH:
\[ pOH = 14 - pH = 14 - 12 = 2 \]

**Hint**: Remember that pH and pOH are related through the equation pH + pOH = 14.

2. **Calculate the hydroxide ion concentration [OH⁻]**:
- The relationship between pOH and hydroxide ion concentration is given by:
\[ pOH = -\log[OH⁻] \]
- Rearranging this gives:
\[ [OH⁻] = 10^{-pOH} = 10^{-2} = 0.01 \, \text{M} \]

**Hint**: Use the formula for converting pOH to concentration: [OH⁻] = 10^(-pOH).

3. **Write the dissociation equation for Ba(OH)₂**:
- Barium hydroxide dissociates in water as follows:
\[ Ba(OH)₂ \rightleftharpoons Ba^{2+} + 2OH^{-} \]
- From the stoichiometry of the reaction, for every 1 mole of Ba(OH)₂ that dissolves, it produces 1 mole of Ba²⁺ and 2 moles of OH⁻.

4. **Relate the solubility (S) to the concentrations**:
- Let the solubility of Ba(OH)₂ be S. Then:
- [Ba²⁺] = S
- [OH⁻] = 2S (because there are 2 moles of OH⁻ for every mole of Ba(OH)₂)

5. **Set up the expression for Ksp**:
- The solubility product Ksp is given by:
\[ Ksp = [Ba^{2+}][OH^{-}]^2 \]
- Substituting the concentrations in terms of S:
\[ Ksp = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

6. **Find the value of S**:
- From our earlier calculation, we know that [OH⁻] = 0.01 M, which means:
\[ 2S = 0.01 \implies S = \frac{0.01}{2} = 0.005 \, \text{M} \]

**Hint**: Remember that the concentration of OH⁻ is double the solubility of Ba(OH)₂.

7. **Calculate Ksp**:
- Now substitute S into the Ksp expression:
\[ Ksp = 4S^3 = 4(0.005)^3 = 4(1.25 \times 10^{-7}) = 5 \times 10^{-7} \]

**Hint**: When calculating Ksp, ensure you cube the solubility correctly and multiply by the appropriate coefficient.

### Final Answer
The value of the solubility product (Ksp) of Ba(OH)₂ is:
\[ Ksp = 5 \times 10^{-7} \, \text{mol}^3/\text{L}^3 \]