100 ml of 0.04 N HCl aqueous solution is mixed with 100 ml of 0.02 N NaOH solution. The pH of the resulting solution is

Options

A. `1.0`
B. 1.7
C. `2.0`
D. 2.3

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the moles of HCl and NaOH
- **HCl**:
- Normality (N) = 0.04 N
- Volume (V) = 100 ml = 0.1 L
- Moles of HCl = Normality × Volume = 0.04 N × 0.1 L = 0.004 moles = 4 millimoles

- **NaOH**:
- Normality (N) = 0.02 N
- Volume (V) = 100 ml = 0.1 L
- Moles of NaOH = Normality × Volume = 0.02 N × 0.1 L = 0.002 moles = 2 millimoles

### Step 2: Determine the limiting reactant
- The reaction between HCl and NaOH is a 1:1 reaction:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- We have 4 millimoles of HCl and 2 millimoles of NaOH.
- Since NaOH is less than HCl, NaOH is the limiting reactant.

### Step 3: Calculate the remaining moles of HCl after the reaction
- Moles of HCl remaining = Initial moles of HCl - Moles of NaOH used
- Moles of HCl remaining = 4 millimoles - 2 millimoles = 2 millimoles

### Step 4: Calculate the total volume of the solution
- Total volume = Volume of HCl + Volume of NaOH = 100 ml + 100 ml = 200 ml = 0.2 L

### Step 5: Calculate the concentration of H⁺ ions in the resulting solution
- Concentration of H⁺ ions = Moles of HCl remaining / Total volume
- Concentration of H⁺ ions = 2 millimoles / 0.2 L = 0.01 moles/L = 10^-2 M

### Step 6: Calculate the pH of the resulting solution
- pH = -log[H⁺]
- pH = -log(10^-2) = 2

### Final Answer:
The pH of the resulting solution is **2**.

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