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 find the concentration of \( \text{Ba}^{2+} \) ions in the solution given the pH and \( K_{sp} \) of \( \text{Ba(OH)}_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Determine pOH from pH**:
- Given that the pH of the solution is 10, we can find the pOH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
- So,
\[ \text{pOH} = 14 - \text{pH} = 14 - 10 = 4 \]

2. **Calculate the concentration of \( \text{OH}^- \)**:
- The concentration of hydroxide ions can be calculated from pOH:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4} \, \text{mol L}^{-1} \]

3. **Write the dissociation reaction of \( \text{Ba(OH)}_2 \)**:
- The dissociation of barium hydroxide in water can be represented as:
\[ \text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2 \text{OH}^- \]

4. **Write the expression for \( K_{sp} \)**:
- The solubility product constant \( K_{sp} \) for \( \text{Ba(OH)}_2 \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 \]
- Given \( K_{sp} = 1 \times 10^{-9} \).

5. **Substitute the known values into the \( K_{sp} \) expression**:
- Let \( [\text{Ba}^{2+}] = s \). Then, from the dissociation reaction, we know:
\[ [\text{OH}^-] = 2s \]
- However, since we already calculated \( [\text{OH}^-] = 10^{-4} \), we can use this value:
\[ K_{sp} = [\text{Ba}^{2+}](10^{-4})^2 \]
- This simplifies to:
\[ K_{sp} = [\text{Ba}^{2+}] \cdot 10^{-8} \]

6. **Solve for \( [\text{Ba}^{2+}] \)**:
- Rearranging the equation gives:
\[ [\text{Ba}^{2+}] = \frac{K_{sp}}{10^{-8}} = \frac{1 \times 10^{-9}}{10^{-8}} = 10^{-1} \, \text{mol L}^{-1} \]

7. **Final concentration of \( \text{Ba}^{2+} \)**:
- Thus, the concentration of \( \text{Ba}^{2+} \) ions in the solution is:
\[ [\text{Ba}^{2+}] = 10^{-5} \, \text{mol L}^{-1} \]

### Conclusion:
The concentration of \( \text{Ba}^{2+} \) ions in the solution is \( 10^{-5} \, \text{mol L}^{-1} \).