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 find the pH of the solution obtained by mixing 50 mL of 2N HCl and 50 mL of 1N NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Normality} = 0.050 \, \text{L} \times 2 \, \text{N} = 0.1 \, \text{moles} \]

- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Normality} = 0.050 \, \text{L} \times 1 \, \text{N} = 0.05 \, \text{moles} \]

### Step 2: Determine the limiting reactant
- HCl and NaOH react in a 1:1 ratio:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Since we have 0.1 moles of HCl and 0.05 moles of NaOH, NaOH is the limiting reactant.

### Step 3: Calculate the remaining moles of HCl after the reaction
- After the reaction, the moles of HCl left will be:
\[ \text{Remaining HCl} = 0.1 \, \text{moles} - 0.05 \, \text{moles} = 0.05 \, \text{moles} \]

### Step 4: Calculate the total volume of the solution
- The total volume after mixing:
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

### Step 5: Calculate the concentration of HCl in the final solution
- The concentration of HCl in the final solution:
\[ \text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Total Volume (L)}} = \frac{0.05 \, \text{moles}}{0.1 \, \text{L}} = 0.5 \, \text{N} \]

### Step 6: Calculate the pH of the solution
- Since HCl is a strong acid, it completely dissociates in solution. The pH can be calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] = -\log[0.5] \]
- Using the logarithm provided in the question, we can calculate:
\[ \text{pH} = -\log[5 \times 10^{-1}] = -(\log 5 + \log 10^{-1}) = -0.7 + 1 = 0.3 \]

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

---