The pH of a solution obtained by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` is

Options

A. `-log 2`
B. ` -log0.2`
C. 1
D. 2

Explanation

To find the pH of the solution obtained by mixing 50 mL of 0.4 N HCl and 50 mL of 0.2 N NaOH, follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH

1. **For HCl:**
- Normality (N) = 0.4 N
- Volume (V) = 50 mL = 50 × 10^(-3) L
- Number of moles of HCl = Normality × Volume in liters
- Number of moles of HCl = 0.4 N × 0.050 L = 0.02 moles

2. **For NaOH:**
- Normality (N) = 0.2 N
- Volume (V) = 50 mL = 50 × 10^(-3) L
- Number of moles of NaOH = Normality × Volume in liters
- Number of moles of NaOH = 0.2 N × 0.050 L = 0.01 moles

### Step 2: Determine the reaction between HCl and NaOH

- The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- From the calculations, we have:
- 0.02 moles of HCl
- 0.01 moles of NaOH

### Step 3: Calculate the remaining moles after neutralization

- Since NaOH is the limiting reagent (0.01 moles), it will neutralize an equal amount of HCl:
- Moles of HCl remaining = Initial moles of HCl - Moles of NaOH
- Moles of HCl remaining = 0.02 moles - 0.01 moles = 0.01 moles

### Step 4: Calculate the total volume of the solution

- Total volume = Volume of HCl + Volume of NaOH
- Total volume = 50 mL + 50 mL = 100 mL = 100 × 10^(-3) L

### Step 5: Calculate the molarity of the remaining HCl

- Molarity (M) = Number of moles / Volume in liters
- Molarity of remaining HCl = 0.01 moles / 0.100 L = 0.1 M

### Step 6: Determine the concentration of H⁺ ions

- Since HCl dissociates completely in solution:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
- Therefore, the concentration of H⁺ ions = 0.1 M

### Step 7: Calculate the pH of the solution

- pH = -log[H⁺]
- pH = -log(0.1) = 1.0

### Final Answer

The pH of the solution is **1.0**.

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