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, follow these steps:

### Step 1: Determine the dissociation of Ba(OH)₂
Ba(OH)₂ is a strong base and dissociates completely in water:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

### Step 2: Calculate the concentration of OH⁻ ions
For every 1 mole of Ba(OH)₂ that dissociates, it produces 2 moles of OH⁻ ions. Therefore, if the concentration of Ba(OH)₂ is 0.001 M, the concentration of OH⁻ ions will be:
\[ [\text{OH}^-] = 2 \times [\text{Ba(OH)}_2] = 2 \times 0.001 \, \text{M} = 0.002 \, \text{M} \]

### Step 3: Calculate the pOH
To find the pOH, use the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.002) \]

Calculating this gives:
\[ \text{pOH} \approx 2.7 \]

### Step 4: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the value of pOH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 2.7 = 11.3 \]

### Final Answer
The pH of the 0.001 M Ba(OH)₂ solution is approximately **11.3**.

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