Calculate the pH of following solutions.
`N//50 HCl`

Answer

To calculate the pH of a \( \frac{N}{50} \) HCl solution, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Normality and Molarity**:
- Given that the solution is \( \frac{N}{50} \) HCl, we can convert this to molarity. Since HCl is a strong acid and completely dissociates, the normality (N) is equal to the molarity (M) for HCl.
- Therefore, \( \frac{N}{50} = \frac{1}{50} \) N = \( \frac{1}{50} \) M.

2. **Convert to Molar Concentration**:
- \( \frac{1}{50} \) M can be expressed as:
\[ \frac{1}{50} = 0.02 \, \text{M} = 2 \times 10^{-2} \, \text{M} \]

3. **Dissociation of HCl**:
- HCl dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
- This means that the concentration of \( \text{H}^+ \) ions produced will be equal to the concentration of HCl, which is \( 2 \times 10^{-2} \, \text{M} \).

4. **Calculate the pH**:
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- Substituting the concentration of \( \text{H}^+ \):
\[ \text{pH} = -\log(2 \times 10^{-2}) \]

5. **Break Down the Logarithm**:
- We can separate the logarithm:
\[ \text{pH} = -\log(2) - \log(10^{-2}) \]
- Since \( \log(10^{-2}) = -2 \):
\[ \text{pH} = -\log(2) + 2 \]

6. **Use the Value of \( \log(2) \)**:
- The value of \( \log(2) \) is approximately \( 0.3010 \):
\[ \text{pH} = -0.3010 + 2 \]
- Therefore:
\[ \text{pH} = 1.699 \]

7. **Final Result**:
- Rounding to two decimal places, the pH of the \( \frac{N}{50} \) HCl solution is approximately:
\[ \text{pH} \approx 1.70 \]