pH of a solution of salt of weak acid and weak base is :
`pH=1/2pK_w+1/2pK_a-1/2pK_b`
and that of weak acid and strong base is
`pH=1/2pK_w+1/2pK_a+1/2logc`
For a salt of weak acid and weak base having `K_a=K_b`, the pH at `25^@C` will be

Options

A. 7
B. 14
C. 0
D. 1

Explanation

To find the pH of a solution of a salt formed from a weak acid and a weak base where \( K_a = K_b \), we can use the provided formula for the pH of such a solution:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b \]

### Step 1: Identify the relationship between \( K_a \) and \( K_b \)
Since it is given that \( K_a = K_b \), we can denote both as \( K \). Therefore, we have:

\[ pK_a = -\log K \quad \text{and} \quad pK_b = -\log K \]

### Step 2: Substitute \( pK_a \) and \( pK_b \) into the pH formula
Now substituting \( pK_a \) and \( pK_b \) into the pH formula:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} (-\log K) - \frac{1}{2} (-\log K) \]

### Step 3: Simplify the equation
Since \( -\log K \) appears in both terms, they will cancel each other out:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} (-\log K) + \frac{1}{2} \log K \]
\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} \cdot 0 \]
\[ \text{pH} = \frac{1}{2} pK_w \]

### Step 4: Calculate \( pK_w \) at \( 25^\circ C \)
At \( 25^\circ C \), \( pK_w \) is typically \( 14 \):

\[ \text{pH} = \frac{1}{2} \times 14 = 7 \]

### Final Result
Thus, the pH of the solution of the salt of a weak acid and weak base where \( K_a = K_b \) at \( 25^\circ C \) is:

\[ \text{pH} = 7 \]