When a salt reacts with water resulting into formation of acidic or basic solution, the process is referred to as salt hydrolysis. The pH of salt solution can be calculated using the following equations.
`pH = ( 1)/(2) ( p K_(w) + pK_(a) + log C)` for salt of weak acid and strong base.
`pH = (1)/(2) ( pK_(w)- pK_(b) - log C ) ` for salt of weak base and strong acid.
`pH = (1)/(2) ( pK_(w) + pK_(a) - pK_(b))` for salt of weak acid and weak base
equal volume of 0.1M solution of weak acid HA is titrated with 0.1M NaOH solution till the end point `pK_(a)` for acid is 6 and degree of hydroglysis is less compared to 1.The pH of the resultant solution at the end point is

Options

A. 8.4335
B. 9.3495
C. 7.2565
D. 8.9755

Answer

To solve the problem, we need to determine the pH of the solution formed when a weak acid (HA) is titrated with a strong base (NaOH) at the endpoint. The given values are:

- Concentration of weak acid (HA) = 0.1 M
- Concentration of NaOH = 0.1 M
- \( pK_a \) of the weak acid = 6
- \( pK_w \) (for water at 25°C) = 14
- Degree of hydrolysis is less than 1.

### Step-by-Step Solution:

1. **Understand the Reaction**:
When the weak acid (HA) reacts with the strong base (NaOH), they form a salt (A⁻) and water:
\[ HA + NaOH \rightarrow A^- + H_2O \]

2. **Determine the Concentration of the Salt**:
Since equal volumes of 0.1 M HA and 0.1 M NaOH are mixed, at the endpoint, all the acid will be converted to its salt.
- If we assume the volume of HA is \( V \) mL, then the moles of HA = \( 0.1 \times \frac{V}{1000} \) and the same for NaOH.
- After the reaction, the total volume will be \( 2V \) mL.
- The concentration of the salt (A⁻) formed will be:
\[ [A^-] = \frac{0.1 \times V/1000}{2V/1000} = \frac{0.1}{2} = 0.05 \text{ M} \]

3. **Use the pH Formula for Weak Acid and Strong Base**:
The formula for calculating pH for a salt formed from a weak acid and a strong base is:
\[ pH = \frac{1}{2} (pK_w + pK_a + \log C) \]
where:
- \( pK_w = 14 \)
- \( pK_a = 6 \)
- \( C = 0.05 \)

4. **Calculate the Logarithm**:
Calculate \( \log C \):
\[ \log(0.05) = \log(5 \times 10^{-2}) = \log(5) + \log(10^{-2}) \approx 0.699 - 2 = -1.301 \]

5. **Substitute Values into the pH Formula**:
Substitute the values into the pH formula:
\[ pH = \frac{1}{2} (14 + 6 + (-1.301)) \]
\[ pH = \frac{1}{2} (20 - 1.301) = \frac{1}{2} (18.699) \approx 9.3495 \]

6. **Final Result**:
Thus, the pH of the resultant solution at the endpoint is approximately:
\[ pH \approx 9.35 \]