The pH of 0.001 M `Ba(OH)_(2)` solution will be

Options

A. 2
B. 8.4
C. 11.3
D. 2.7

Explanation

To find the pH of a 0.001 M Ba(OH)₂ solution, we can follow these steps:

### Step 1: Determine the concentration of hydroxide ions (OH⁻)
Barium hydroxide, Ba(OH)₂, dissociates in water as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

From the dissociation equation, we can see that 1 mole of Ba(OH)₂ produces 2 moles of OH⁻ ions. Therefore, if we have a 0.001 M solution of Ba(OH)₂, the concentration of OH⁻ ions will be:
\[ \text{[OH}^-] = 2 \times 0.001 \, \text{M} = 0.002 \, \text{M} \]

### Step 2: Calculate the pOH
The pOH is calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the concentration of OH⁻ ions:
\[ \text{pOH} = -\log(0.002) \]

Using a calculator, we find:
\[ \text{pOH} \approx 2.69 \]

### Step 3: Calculate the pH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]

Now, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2.69 \]
\[ \text{pH} \approx 11.31 \]

### Conclusion
The pH of a 0.001 M Ba(OH)₂ solution is approximately **11.31**.

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