The dissociation constant of a weak acid is `10^(-6)` . Then the `P^(H)` of 0.01 Nof that acid is

Options

A. 2
B. 7
C. 8
D. 4

Answer

To find the pH of a 0.01 N solution of a weak acid with a dissociation constant (K_a) of \(10^{-6}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Dissociation constant, \(K_a = 10^{-6}\)
- Concentration of the acid, \(C = 0.01 \, N\)

2. **Calculate \(pK_a\):**
- The relationship between \(K_a\) and \(pK_a\) is given by:
\[ pK_a = -\log(K_a) \]
- Substituting the value of \(K_a\):
\[ pK_a = -\log(10^{-6}) = 6 \]

3. **Use the formula for pH of a weak acid:**
- The formula to calculate the pH of a weak acid is:
\[ pH = \frac{1}{2} pK_a - \log \left( \frac{C}{2} \right) \]
- Here, \(C\) is the concentration of the acid.

4. **Substitute the values into the pH formula:**
- First, calculate \(\frac{C}{2}\):
\[ \frac{C}{2} = \frac{0.01}{2} = 0.005 \]
- Now substitute \(pK_a\) and \(\frac{C}{2}\) into the pH formula:
\[ pH = \frac{1}{2} \times 6 - \log(0.005) \]

5. **Calculate \(-\log(0.005)\):**
- We can express \(0.005\) as \(5 \times 10^{-3}\):
\[ -\log(0.005) = -\log(5 \times 10^{-3}) = -\log(5) - \log(10^{-3}) = -\log(5) + 3 \]
- Approximating \(-\log(5) \approx -0.7\):
\[ -\log(0.005) \approx 3 - 0.7 = 2.3 \]

6. **Final calculation of pH:**
- Substitute back into the pH equation:
\[ pH = 3 - 2.3 = 4.7 \]
- However, since we are using the approximation, we can round it to:
\[ pH \approx 4 \]

### Conclusion:
The pH of the 0.01 N solution of the weak acid is approximately **4**.