50 ml water is added to a 50 ml solution of `Ba(OH)_(2)` of strength 0.01 M. The pH value of the resulting solution will be

Options

A. 8
B. 10
C. 12
D. 6

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the moles of Ba(OH)₂ in the original solution.
Given:
- Volume of Ba(OH)₂ solution = 50 mL = 0.050 L
- Concentration of Ba(OH)₂ = 0.01 M

Moles of Ba(OH)₂ = Concentration × Volume
\[ \text{Moles of Ba(OH)}_2 = 0.01 \, \text{mol/L} \times 0.050 \, \text{L} = 0.0005 \, \text{mol} \]

### Step 2: Calculate the total volume after adding water.
- Volume of water added = 50 mL = 0.050 L
- Total volume = Volume of Ba(OH)₂ solution + Volume of water
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.100 \, \text{L} \]

### Step 3: Calculate the new concentration of Ba(OH)₂ after dilution.
New concentration after dilution:
\[ \text{New Concentration} = \frac{\text{Moles of Ba(OH)}_2}{\text{Total Volume}} = \frac{0.0005 \, \text{mol}}{0.100 \, \text{L}} = 0.005 \, \text{M} \]

### Step 4: Determine the concentration of OH⁻ ions.
Ba(OH)₂ dissociates as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \]
From the dissociation, 1 mole of Ba(OH)₂ produces 2 moles of OH⁻ ions. Therefore:
\[ \text{Concentration of OH}^- = 2 \times \text{Concentration of Ba(OH)}_2 = 2 \times 0.005 \, \text{M} = 0.010 \, \text{M} \]

### Step 5: Calculate the pOH of the solution.
Using the concentration of OH⁻ ions:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.010) = 2 \]

### Step 6: Calculate the pH of the solution.
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 2 = 12 \]

### Final Answer:
The pH value of the resulting solution is **12**.

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