50 ml of `H_(2),O` is added to 50 ml of `1xx 10^(-3)` M barium hydroxide solution. What is the `P^(H)` of the resulting solution?

Options

A. `3.0`
B. 3.3
C. `11.7`
D. `11.0`

Answer

To find the pH of the resulting solution when 50 ml of water is added to 50 ml of 1 x 10^(-3) M barium hydroxide solution, we can follow these steps:

### Step 1: Calculate the moles of barium hydroxide (Ba(OH)₂)
Given:
- Volume of Ba(OH)₂ solution = 50 ml = 0.050 L
- Molarity of Ba(OH)₂ = 1 x 10^(-3) M

Using the formula:
\[ \text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of Ba(OH)}_2 = 1 \times 10^{-3} \, \text{mol/L} \times 0.050 \, \text{L} = 5 \times 10^{-5} \, \text{mol} \]

### Step 2: Calculate the moles of OH⁻ produced
Barium hydroxide dissociates as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]
This means that 1 mole of Ba(OH)₂ produces 2 moles of OH⁻. Therefore, the moles of OH⁻ produced will be:
\[ \text{Moles of OH}^- = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 5 \times 10^{-5} = 1 \times 10^{-4} \, \text{mol} \]

### Step 3: Calculate the final volume of the solution
The total volume after adding 50 ml of water to 50 ml of Ba(OH)₂ solution is:
\[ \text{Total Volume} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml} = 0.1 \, \text{L} \]

### Step 4: Calculate the concentration of OH⁻ in the final solution
Using the formula:
\[ \text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Total Volume}} \]
\[ \text{Concentration of OH}^- = \frac{1 \times 10^{-4} \, \text{mol}}{0.1 \, \text{L}} = 1 \times 10^{-3} \, \text{M} \]

### Step 5: Calculate the pOH of the solution
Using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
\[ \text{pOH} = -\log(1 \times 10^{-3}) = 3 \]

### Step 6: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

### Final Answer
The pH of the resulting solution is **11**.

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