What is the total volume of solution ( in ml), when 1M NaOH is required to be added in 100ml of 1M `CH_(3) COOH ( K_(a) = 10^(-5))` solution so that its pH becomes 6

Options

A. 190.91
B. 90.9
C. 100
D. 135.5

Answer

To solve the problem, we need to determine the total volume of the solution when 1M NaOH is added to 100ml of 1M acetic acid (CH₃COOH) to achieve a pH of 6.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Volume of acetic acid (CH₃COOH) = 100 ml
- Concentration of acetic acid = 1 M
- Concentration of NaOH = 1 M
- Desired pH = 6
- \( K_a \) of acetic acid = \( 10^{-5} \)

2. **Calculate the Moles of Acetic Acid:**
\[ \text{Moles of CH}_3\text{COOH} = \text{Concentration} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} = 100 \, \text{mmol} \]

3. **Use the Henderson-Hasselbalch Equation:**
The pH of a buffer solution is given by:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
First, calculate \( pK_a \):
\[ pK_a = -\log(K_a) = -\log(10^{-5}) = 5 \]
Now, substituting the values into the equation:
\[ 6 = 5 + \log\left(\frac{V}{100 - V}\right) \]

4. **Rearranging the Equation:**
\[ 1 = \log\left(\frac{V}{100 - V}\right) \]
Converting from logarithmic form:
\[ 10 = \frac{V}{100 - V} \]

5. **Cross-Multiplying to Solve for V:**
\[ 10(100 - V) = V \]
\[ 1000 - 10V = V \]
\[ 1000 = 11V \]
\[ V = \frac{1000}{11} \approx 90.91 \, \text{ml} \]

6. **Calculate the Total Volume of the Solution:**
\[ \text{Total Volume} = \text{Volume of CH}_3\text{COOH} + \text{Volume of NaOH} = 100 \, \text{ml} + 90.91 \, \text{ml} = 190.91 \, \text{ml} \]

### Final Answer:
The total volume of the solution when 1M NaOH is added to achieve a pH of 6 is approximately **190.91 ml**.