When NaOH solution is gradually added to the solution of weak acid (HA), the pH of the solution is found to be 5.0 at the addition of 10 and 6.0 at further addition of 10 ml of same NaOH. (Total volume of NaOH =20ml). Calculate `pK_(a)` for HA. `[log2 = 0.3]`

Explanation

To solve the problem, we need to calculate the \( pK_a \) of the weak acid \( HA \) given the pH values at two different stages of NaOH addition.

### Step-by-Step Solution:

1. **Identify the given information:**
- pH after adding 10 ml of NaOH: \( 5.0 \)
- pH after adding another 10 ml of NaOH (total 20 ml): \( 6.0 \)

2. **Use the Henderson-Hasselbalch equation:**
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
where \( [\text{A}^-] \) is the concentration of the salt (conjugate base) and \( [\text{HA}] \) is the concentration of the weak acid.

3. **Set up the equations for both pH values:**
- For the first pH (5.0):
\[ 5.0 = pK_a + \log \left( \frac{10C_2}{C_1V_1 - 10C_2} \right) \quad \text{(1)} \]
- For the second pH (6.0):
\[ 6.0 = pK_a + \log \left( \frac{20C_2}{C_1V_1 - 20C_2} \right) \quad \text{(2)} \]
Here, \( C_2 \) is the concentration of NaOH, and \( C_1 \) is the concentration of the weak acid \( HA \). \( V_1 \) is the initial volume of the weak acid solution.

4. **Subtract equation (1) from equation (2):**
\[ 6.0 - 5.0 = \log \left( \frac{20C_2}{C_1V_1 - 20C_2} \right) - \log \left( \frac{10C_2}{C_1V_1 - 10C_2} \right) \]
This simplifies to:
\[ 1.0 = \log \left( \frac{20C_2 (C_1V_1 - 10C_2)}{10C_2 (C_1V_1 - 20C_2)} \right) \]
\[ 10 = \frac{20(C_1V_1 - 10C_2)}{10(C_1V_1 - 20C_2)} \]
\[ 10(C_1V_1 - 20C_2) = 2(C_1V_1 - 10C_2) \]
Expanding and simplifying gives:
\[ 10C_1V_1 - 200C_2 = 2C_1V_1 - 20C_2 \]
\[ 8C_1V_1 = 180C_2 \]
\[ C_1V_1 = 22.5C_2 \quad \text{(3)} \]

5. **Substituting equation (3) back into either equation (1) or (2):**
Using equation (1):
\[ 5.0 = pK_a + \log \left( \frac{10C_2}{22.5C_2 - 10C_2} \right) \]
Simplifying gives:
\[ 5.0 = pK_a + \log \left( \frac{10C_2}{12.5C_2} \right) \]
\[ 5.0 = pK_a + \log \left( \frac{10}{12.5} \right) \]
\[ 5.0 = pK_a + \log \left( 0.8 \right) \]
Using the provided value \( \log 2 = 0.3 \):
\[ \log 0.8 = \log \left( \frac{8}{10} \right) = \log 8 - \log 10 = 3 \log 2 - 1 = 3(0.3) - 1 = 0.9 - 1 = -0.1 \]
Thus:
\[ 5.0 = pK_a - 0.1 \]
\[ pK_a = 5.0 + 0.1 = 5.1 \]

### Final Answer:
\[ pK_a = 5.1 \]