For anionic hydrolysis, pH is given by?
A. pH = 1/2pK_w = 1/2pK_b - 1/2logC
B. pH = 1/2pK_x + 1/2pK_a - 1/2logC
C. pH = 1/2pK_w + 1/2pK_a + 1/2logC
D. None of the above

To determine the correct formula for the pH during anionic hydrolysis, let's break down the process of anionic hydrolysis and how pH is related to it.

### Anionic Hydrolysis

Anionic hydrolysis occurs when the conjugate base of a weak acid reacts with water. For a salt of a weak acid and a strong base (such as sodium acetate, where acetate is the conjugate base of acetic acid), the anion (acetate) reacts with water to form hydroxide ions (OH⁻), which makes the solution basic.

The general reaction for the anionic hydrolysis can be written as:
\[ \text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^- \]

### Relationship of pH with Hydrolysis

To find the pH of the solution, follow these steps:

1. **Determine the \( K_b \) for the anion**: The base dissociation constant (\( K_b \)) of the anion is related to the \( K_a \) of the parent acid by:
\[ K_b = \frac{K_w}{K_a} \]
where \( K_w \) is the ion-product constant for water and \( K_a \) is the acid dissociation constant.

2. **Calculate the pOH**: Using the concentration of the anion and \( K_b \), find the concentration of \( OH^- \) ions and then compute the pOH.

3. **Convert pOH to pH**: Finally, use the relationship:
\[ \text{pH} = 14 - \text{pOH} \]

However, there's a direct formula for the pH of the solution when dealing with hydrolysis of a salt of a weak acid and a strong base, which is given by:
\[ \text{pH} = \frac{1}{2}pK_w + \frac{1}{2}pK_a + \frac{1}{2}\log C \]

Here:
- \( pK_w \) is the negative logarithm of the ion-product constant for water.
- \( pK_a \) is the negative logarithm of the acid dissociation constant.
- \( C \) is the concentration of the salt solution.

### Matching with Given Options

Now, let's match this with the given options:

A. \( \text{pH} = \frac{1}{2}pK_w = \frac{1}{2}pK_b - \frac{1}{2}\log C \)
B. \( \text{pH} = \frac{1}{2}pK_x + \frac{1}{2}pK_a - \frac{1}{2}\log C \)
C. \( \text{pH} = \frac{1}{2}pK_w + \frac{1}{2}pK_a + \frac{1}{2}\log C \)
D. None of the above

The correct option that matches the derived formula is:

**C. \( \text{pH} = \frac{1}{2}pK_w + \frac{1}{2}pK_a + \frac{1}{2}\log C \)**