The `pK_(a)` of HCOOH is 3.8 and `pK_(b)` of `NH_(3)` is 4.8, find the pH of aqueous solution of 1M `HCOONH_(4)`.

Explanation

To find the pH of a 1M solution of HCOONH₄ (ammonium formate), we can use the formula for the pH of a solution made from a weak acid and its conjugate base. The relevant formula is:

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

Where:
- \( pK_w = 14 \) (at 25°C)
- \( pK_a \) is the dissociation constant of the weak acid (HCOOH in this case).
- \( pK_b \) is the dissociation constant of the weak base (NH₃ in this case).

### Step-by-step Solution:

1. **Identify the given values**:
- \( pK_a \) of HCOOH = 3.8
- \( pK_b \) of NH₃ = 4.8
- \( pK_w \) = 14

2. **Substitute the values into the formula**:
\[ \text{pH} = \frac{1}{2} \left( 14 + 3.8 - 4.8 \right) \]

3. **Calculate the expression inside the parentheses**:
\[ 14 + 3.8 - 4.8 = 14 + 3.8 - 4.8 = 14 - 1 = 13 \]

4. **Divide by 2**:
\[ \text{pH} = \frac{13}{2} = 6.5 \]

5. **Final answer**:
The pH of the 1M solution of HCOONH₄ is **6.5**.