A weak acid of dissociation constant `10^(-5)` is being titrated with aqueous NaOH solution . The pH at the point of one third of neutralization of the acid will be

Options

A. `5 + log - log 3`
B. `5- log 2`
C. `5- log 3`
D. `5- log 6`

Explanation

To find the pH at the point of one third of neutralization of a weak acid with a dissociation constant \( K_a = 10^{-5} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding One Third Neutralization**:
- Let the initial concentration of the weak acid be \( C \).
- At one third neutralization, \( \frac{1}{3}C \) of the acid has been neutralized by NaOH, leaving \( \frac{2}{3}C \) of the weak acid remaining.
- The amount of salt formed (the conjugate base) is \( \frac{1}{3}C \).

2. **Concentration of Acid and Salt**:
- Concentration of the weak acid (HA) remaining: \( [HA] = \frac{2}{3}C \)
- Concentration of the salt (A⁻) formed: \( [A^-] = \frac{1}{3}C \)

3. **Using the Henderson-Hasselbalch Equation**:
- The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[A^-]}{[HA]}\right) \]

4. **Calculating \( pK_a \)**:
- Given \( K_a = 10^{-5} \):
\[ pK_a = -\log(K_a) = -\log(10^{-5}) = 5 \]

5. **Substituting Values into the Equation**:
- Substitute \( [A^-] = \frac{1}{3}C \) and \( [HA] = \frac{2}{3}C \) into the equation:
\[ \text{pH} = 5 + \log\left(\frac{\frac{1}{3}C}{\frac{2}{3}C}\right) \]
- Simplifying the logarithm:
\[ \text{pH} = 5 + \log\left(\frac{1}{2}\right) \]

6. **Calculating the Logarithm**:
- We know that \( \log\left(\frac{1}{2}\right) = -\log(2) \):
\[ \text{pH} = 5 - \log(2) \]

7. **Final Answer**:
- The pH at the point of one third neutralization of the weak acid is:
\[ \text{pH} = 5 - \log(2) \]