100 mL of 0.2 N HCI is added to 100 mL of 0.18 N NaOH and whole volume is made one litre. What will be the pH of resulting solution.

Answer

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

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

1. **Calculate moles of HCl:**
- Normality (N) of HCl = 0.2 N
- Volume (V) of HCl = 100 mL = 0.1 L
- Moles of HCl = Normality × Volume = 0.2 N × 0.1 L = 0.02 moles

2. **Calculate moles of NaOH:**
- Normality (N) of NaOH = 0.18 N
- Volume (V) of NaOH = 100 mL = 0.1 L
- Moles of NaOH = Normality × Volume = 0.18 N × 0.1 L = 0.018 moles

### Step 2: Determine the limiting reagent and the remaining moles

- Since HCl and NaOH react in a 1:1 ratio:
- Moles of HCl = 0.02 moles
- Moles of NaOH = 0.018 moles

- NaOH is the limiting reagent because it is present in a smaller amount.

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

- Moles of HCl remaining after reaction:
- Moles of HCl consumed = Moles of NaOH = 0.018 moles
- Remaining moles of HCl = 0.02 moles - 0.018 moles = 0.002 moles

### Step 4: Calculate the concentration of H⁺ ions

- The total volume of the solution after mixing is 1 L.
- Concentration of H⁺ ions = Remaining moles of HCl / Total volume
- Concentration of H⁺ = 0.002 moles / 1 L = 0.002 M

### Step 5: Calculate the pH of the solution

- pH = -log[H⁺]
- pH = -log(0.002)
- pH ≈ 2.69

### Final Answer:
The pH of the resulting solution is approximately **2.69**.

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