The `pH` of an aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `1xx10^(-9)`, then the concentration of `Ba^(2+)` ions in the solution in `mol L^(-1)` is

Options

A. `1 xx 10^(-2)`
B. `1 xx 10^(-4)`
C. `1xx10^(-1)`
D. `1xx10^(-5)`

Explanation

To solve the problem, we need to determine the concentration of \( \text{Ba}^{2+} \) ions in a solution of \( \text{Ba(OH)}_2 \) given that the pH is 10 and the solubility product constant \( K_{sp} \) of \( \text{Ba(OH)}_2 \) is \( 1 \times 10^{-9} \).

### Step-by-Step Solution:

1. **Calculate the concentration of \( \text{OH}^- \) ions from the pH:**
- The pH of the solution is given as 10.
- We can find the pOH using the formula:
\[ \text{pOH} = 14 - \text{pH} = 14 - 10 = 4 \]
- Now, we can find the concentration of hydroxide ions (\( \text{OH}^- \)):
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4} \, \text{mol L}^{-1} \]

2. **Determine the relationship between \( \text{Ba(OH)}_2 \) dissociation and ion concentrations:**
- The dissociation of barium hydroxide in water can be represented as:
\[ \text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2 \text{OH}^- \]
- For every 1 mole of \( \text{Ba(OH)}_2 \) that dissolves, it produces 1 mole of \( \text{Ba}^{2+} \) and 2 moles of \( \text{OH}^- \).
- Let \( s \) be the solubility of \( \text{Ba(OH)}_2 \) in mol L\(^{-1}\). Then:
\[ [\text{Ba}^{2+}] = s \quad \text{and} \quad [\text{OH}^-] = 2s \]

3. **Set up the expression for \( K_{sp} \):**
- The solubility product constant \( K_{sp} \) for \( \text{Ba(OH)}_2 \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 \]
- Substituting the concentrations we found:
\[ K_{sp} = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3 \]
- Given that \( K_{sp} = 1 \times 10^{-9} \):
\[ 4s^3 = 1 \times 10^{-9} \]

4. **Solve for \( s \):**
- Rearranging gives:
\[ s^3 = \frac{1 \times 10^{-9}}{4} = 2.5 \times 10^{-10} \]
- Taking the cube root:
\[ s = (2.5 \times 10^{-10})^{1/3} \]
- Calculating \( s \):
\[ s \approx 6.3 \times 10^{-4} \, \text{mol L}^{-1} \]

5. **Find the concentration of \( \text{Ba}^{2+} \):**
- Since \( [\text{Ba}^{2+}] = s \):
\[ [\text{Ba}^{2+}] \approx 6.3 \times 10^{-4} \, \text{mol L}^{-1} \]

### Final Answer:
The concentration of \( \text{Ba}^{2+} \) ions in the solution is approximately \( 6.3 \times 10^{-4} \, \text{mol L}^{-1} \).