Following solutions were prepared by mixing different volumes of NaOH and HCI of different concentrations:
(i) `60 mL M/10 HCl + 40 mL M/10 NaOH`
(ii) `55 mL M/10 HCl + 45 mL M/10 NaOH`,
(iii) `75 mL M/5 HCl + 25 mL M/5 NaOH`
(iv) `100 mL M/10 HCl + 100 mL M/10 NaOH`
pH of which one of them will be equal to 1?

Options

A. ii
B. i
C. iv
D. iii

Answer

To determine which of the given solutions has a pH equal to 1, we will analyze each option step by step.

### Step 1: Analyze the first solution
**Solution (i): 60 mL M/10 HCl + 40 mL M/10 NaOH**

1. **Calculate moles of HCl:**
- Volume of HCl = 60 mL = 0.060 L
- Concentration of HCl = M/10 = 0.1 M
- Moles of HCl = Volume × Concentration = 0.060 L × 0.1 M = 0.006 moles = 6 mmol

2. **Calculate moles of NaOH:**
- Volume of NaOH = 40 mL = 0.040 L
- Concentration of NaOH = M/10 = 0.1 M
- Moles of NaOH = Volume × Concentration = 0.040 L × 0.1 M = 0.004 moles = 4 mmol

3. **Determine the remaining moles of HCl after neutralization:**
- Moles of HCl remaining = 6 mmol - 4 mmol = 2 mmol

4. **Calculate the total volume of the solution:**
- Total volume = 60 mL + 40 mL = 100 mL = 0.1 L

5. **Calculate the concentration of H⁺ ions:**
- Concentration of H⁺ = Remaining moles of HCl / Total volume = 2 mmol / 0.1 L = 0.02 M

6. **Calculate pH:**
- pH = -log[H⁺] = -log(0.02) = 1.7 (approximately)

### Step 2: Analyze the second solution
**Solution (ii): 55 mL M/10 HCl + 45 mL M/10 NaOH**

1. **Calculate moles of HCl:**
- Volume of HCl = 55 mL = 0.055 L
- Moles of HCl = 0.055 L × 0.1 M = 0.0055 moles = 5.5 mmol

2. **Calculate moles of NaOH:**
- Volume of NaOH = 45 mL = 0.045 L
- Moles of NaOH = 0.045 L × 0.1 M = 0.0045 moles = 4.5 mmol

3. **Determine the remaining moles of HCl after neutralization:**
- Moles of HCl remaining = 5.5 mmol - 4.5 mmol = 1 mmol

4. **Calculate the total volume of the solution:**
- Total volume = 55 mL + 45 mL = 100 mL = 0.1 L

5. **Calculate the concentration of H⁺ ions:**
- Concentration of H⁺ = 1 mmol / 0.1 L = 0.01 M

6. **Calculate pH:**
- pH = -log(0.01) = 2

### Step 3: Analyze the third solution
**Solution (iii): 75 mL M/5 HCl + 25 mL M/5 NaOH**

1. **Calculate moles of HCl:**
- Volume of HCl = 75 mL = 0.075 L
- Moles of HCl = 0.075 L × 0.2 M = 0.015 moles = 15 mmol

2. **Calculate moles of NaOH:**
- Volume of NaOH = 25 mL = 0.025 L
- Moles of NaOH = 0.025 L × 0.2 M = 0.005 moles = 5 mmol

3. **Determine the remaining moles of HCl after neutralization:**
- Moles of HCl remaining = 15 mmol - 5 mmol = 10 mmol

4. **Calculate the total volume of the solution:**
- Total volume = 75 mL + 25 mL = 100 mL = 0.1 L

5. **Calculate the concentration of H⁺ ions:**
- Concentration of H⁺ = 10 mmol / 0.1 L = 0.1 M

6. **Calculate pH:**
- pH = -log(0.1) = 1

### Step 4: Analyze the fourth solution
**Solution (iv): 100 mL M/10 HCl + 100 mL M/10 NaOH**

1. **Calculate moles of HCl:**
- Volume of HCl = 100 mL = 0.1 L
- Moles of HCl = 0.1 L × 0.1 M = 0.01 moles = 10 mmol

2. **Calculate moles of NaOH:**
- Volume of NaOH = 100 mL = 0.1 L
- Moles of NaOH = 0.1 L × 0.1 M = 0.01 moles = 10 mmol

3. **Determine the remaining moles of HCl after neutralization:**
- Moles of HCl remaining = 10 mmol - 10 mmol = 0 mmol

4. **Calculate the total volume of the solution:**
- Total volume = 100 mL + 100 mL = 200 mL = 0.2 L

5. **Calculate the concentration of H⁺ ions:**
- Concentration of H⁺ = 0 mmol / 0.2 L = 0 M

6. **Calculate pH:**
- pH = 7 (neutral solution)

### Conclusion
From the above calculations, we find that the only solution with a pH equal to 1 is the third solution (iii): `75 mL M/5 HCl + 25 mL M/5 NaOH`.