pH value of which of the following is not equal to one

Options

A. 0.1 M `HNO_(3)`
B. 0.05 M `H_(2) SO_(4)`
C. 0.1 M `CH_(3)COOH`
D. `50 cm^(3)` of `0.4 M HCl + 50 cm^(3)` of 0.2 M NaOH

Explanation

To determine which of the following options has a pH value not equal to one, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding pH Calculation**:
The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
where \([\text{H}^+]\) is the concentration of hydrogen ions in moles per liter.

2. **Option 1: 0.1 M HNO3**:
- HNO3 is a strong acid and completely dissociates in solution:
\[ \text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^- \]
- Therefore, the concentration of \([\text{H}^+]\) is 0.1 M.
- Calculating pH:
\[ \text{pH} = -\log(0.1) = 1 \]
- **Conclusion**: pH = 1 (not the answer).

3. **Option 2: 0.05 M H2SO4**:
- H2SO4 is also a strong acid and completely dissociates:
\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \]
- Therefore, the concentration of \([\text{H}^+]\) becomes:
\[ 2 \times 0.05 = 0.1 \text{ M} \]
- Calculating pH:
\[ \text{pH} = -\log(0.1) = 1 \]
- **Conclusion**: pH = 1 (not the answer).

4. **Option 3: 0.1 M CH3COOH**:
- CH3COOH (acetic acid) is a weak acid and does not completely dissociate:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
- The dissociation constant (Ka) for acetic acid is relatively small, indicating that only a fraction of the acid dissociates.
- Therefore, the concentration of \([\text{H}^+]\) will be less than 0.1 M, resulting in a pH greater than 1.
- **Conclusion**: pH > 1 (this is the answer).

5. **Option 4: 50 cm³ of 0.4 M HCl + 50 cm³ of 0.2 M NaOH**:
- Calculate the milli-equivalents of HCl:
\[ \text{milli-equivalents of HCl} = 0.4 \times 50 = 20 \]
- Calculate the milli-equivalents of NaOH:
\[ \text{milli-equivalents of NaOH} = 0.2 \times 50 = 10 \]
- Remaining acid after neutralization:
\[ 20 - 10 = 10 \text{ milli-equivalents} \]
- The concentration of \([\text{H}^+]\) after neutralization:
\[ \text{Total volume} = 50 + 50 = 100 \text{ cm}^3 = 0.1 \text{ L} \]
\[ [\text{H}^+] = \frac{10 \text{ milli-equivalents}}{100 \text{ cm}^3} = 0.1 \text{ M} \]
- Calculating pH:
\[ \text{pH} = -\log(0.1) = 1 \]
- **Conclusion**: pH = 1 (not the answer).

### Final Answer:
The pH value of **Option 3: 0.1 M CH3COOH** is not equal to one.