What volume of 0.2M `Ba(OH)_(2)` must be added to 300 mL of 0.08 M HCl solution to get a solution in which the molarity of hydroxyl `(OH^(-))` ions is 0.8 M?

Options

A. 375 mL
B. 300 mL
C. 225 mL
D. 150 mL

Answer

To solve the problem, we need to determine the volume of 0.2 M \( \text{Ba(OH)}_2 \) that must be added to 300 mL of 0.08 M HCl solution to achieve a hydroxyl ion concentration of 0.8 M.

### Step-by-Step Solution:

1. **Identify the reaction**:
The reaction between \( \text{Ba(OH)}_2 \) and \( \text{HCl} \) can be represented as:
\[ \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]
From this reaction, we see that 1 mole of \( \text{Ba(OH)}_2 \) produces 2 moles of \( \text{OH}^- \) ions.

2. **Calculate the moles of \( \text{HCl} \)**:
The concentration of \( \text{HCl} \) is 0.08 M, and the volume is 300 mL (or 0.3 L).
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.08 \, \text{mol/L} \times 0.3 \, \text{L} = 0.024 \, \text{mol} \]

3. **Determine the moles of \( \text{OH}^- \) needed**:
We want the final concentration of \( \text{OH}^- \) ions to be 0.8 M in the total volume after adding \( \text{Ba(OH)}_2 \). The total volume after adding \( V \) mL of \( \text{Ba(OH)}_2 \) will be \( 300 + V \) mL or \( 0.3 + \frac{V}{1000} \) L.
\[ \text{Moles of } \text{OH}^- = \text{Concentration} \times \text{Total Volume} = 0.8 \, \text{mol/L} \times \left(0.3 + \frac{V}{1000}\right) \, \text{L} \]

4. **Calculate the moles of \( \text{OH}^- \) produced by \( \text{Ba(OH)}_2 \)**:
If \( V \) mL of \( \text{Ba(OH)}_2 \) is added, the moles of \( \text{Ba(OH)}_2 \) added will be:
\[ \text{Moles of } \text{Ba(OH)}_2 = 0.2 \, \text{mol/L} \times \frac{V}{1000} \, \text{L} \]
Since each mole of \( \text{Ba(OH)}_2 \) produces 2 moles of \( \text{OH}^- \):
\[ \text{Moles of } \text{OH}^- = 2 \times 0.2 \times \frac{V}{1000} = 0.0004V \]

5. **Set up the equation**:
The total moles of \( \text{OH}^- \) from both sources (from \( \text{HCl} \) and \( \text{Ba(OH)}_2 \)) must equal the moles needed:
\[ 0.0004V - 0.024 = 0.8 \left(0.3 + \frac{V}{1000}\right) \]

6. **Solve the equation**:
Expanding the right side:
\[ 0.0004V - 0.024 = 0.24 + 0.0008V \]
Rearranging gives:
\[ 0.0004V - 0.0008V = 0.24 + 0.024 \]
\[ -0.0004V = 0.264 \]
\[ V = \frac{0.264}{0.0004} = 660 \, \text{mL} \]

7. **Final calculation**:
However, we need to consider the moles of \( \text{HCl} \) that will react with \( \text{Ba(OH)}_2 \). The moles of \( \text{HCl} \) will be:
\[ 0.024 - 0.0004V = 0 \]
Solving gives:
\[ V = 150 \, \text{mL} \]

### Final Answer:
The volume of 0.2 M \( \text{Ba(OH)}_2 \) that must be added is **150 mL**.