Match Column -I with Column -II `:`
Column -I Column-II
(A) Ph of water ( p ) `( 1)/( 2) p K_(w)`
( B ) Ph of a salt of strong acid and strong base ( q) `pH = (1)/(2) [ pK_(w) + pK_(a) + log c ]`
(C ) Ph of a salt of weak acid and strong base ( r ) `pH = ( 1)/( 2) [ pK _(w) + pK_(a) - pK_(b)]`
(D) Ph of a salt of weak acid and weak base (s) 7
where `K_(a) , K_(b)` are dissociation constants of weak acid and weak base and `K_(w) = ` Ionic product of water.

Answer

To solve the matching question between Column-I and Column-II, we will analyze each item in Column-I and find the corresponding expression in Column-II.

### Step-by-Step Solution:

1. **(A) pH of water**:
- The pH of pure water at 25°C is 7, which is a neutral pH.
- This corresponds to the expression \( p = \frac{1}{2} pK_w \), where \( pK_w \) is 14 at 25°C.
- Therefore, **(A) matches with (p)**.

2. **(B) pH of a salt of strong acid and strong base**:
- A salt formed from a strong acid and a strong base does not undergo hydrolysis, and thus the pH remains neutral at 7.
- This also corresponds to \( s \) (which is 7).
- Therefore, **(B) matches with (s)**.

3. **(C) pH of a salt of weak acid and strong base**:
- A salt of a weak acid and a strong base will have a basic solution. The pH can be calculated using the formula:
\[ pH = \frac{1}{2} [pK_w + pK_a + \log c] \]
- This corresponds to expression \( q \).
- Therefore, **(C) matches with (q)**.

4. **(D) pH of a salt of weak acid and weak base**:
- The pH of a solution formed by a salt of a weak acid and a weak base is given by:
\[ pH = \frac{1}{2} [pK_w + pK_a - pK_b] \]
- This corresponds to expression \( r \).
- Therefore, **(D) matches with (r)**.

### Final Matches:
- (A) → (p)
- (B) → (s)
- (C) → (q)
- (D) → (r)