The pH of a solution obtained by mixing 50 mL of 2N HCI and 50 mL of 1 N NaOH is [log 5 = 0.7]

Options

A. 1.7
B. 1.3
C. 0.7
D. 0.3

Answer

To solve the problem of finding the pH of a solution obtained by mixing 50 mL of 2N HCl and 50 mL of 1N NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate Millimoles of HCl:**
- Given: Volume of HCl (V1) = 50 mL, Normality (N1) = 2N
- Millimoles of HCl = Volume (mL) × Normality (N) = 50 mL × 2N = 100 millimoles

2. **Calculate Millimoles of NaOH:**
- Given: Volume of NaOH (V2) = 50 mL, Normality (N2) = 1N
- Millimoles of NaOH = Volume (mL) × Normality (N) = 50 mL × 1N = 50 millimoles

3. **Determine the Limiting Reactant:**
- HCl and NaOH will react in a 1:1 ratio.
- 100 millimoles of HCl will react with 50 millimoles of NaOH.
- After the reaction, the remaining HCl = 100 millimoles - 50 millimoles = 50 millimoles.

4. **Calculate Moles of H⁺ Remaining:**
- Moles of H⁺ remaining = 50 millimoles = 50 × 10⁻³ moles.

5. **Calculate Total Volume of the Solution:**
- Total volume = Volume of HCl + Volume of NaOH = 50 mL + 50 mL = 100 mL = 100 × 10⁻³ L.

6. **Calculate Concentration of H⁺:**
- Concentration of H⁺ = Moles of H⁺ / Total Volume (L)
- Concentration of H⁺ = (50 × 10⁻³ moles) / (100 × 10⁻³ L) = 0.5 M or 5 × 10⁻¹ M.

7. **Calculate pH:**
- pH = -log[H⁺]
- pH = -log(5 × 10⁻¹)
- pH = -1 + log(5) (using the property of logarithms)
- Given log(5) = 0.7, thus:
- pH = -1 + 0.7 = -0.3.

8. **Final Result:**
- Therefore, the pH of the solution is 0.3.

### Final Answer:
The pH of the solution is **0.3**.

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