What will be the pH of a solution prepared by mixing 100ml of 0.02M `H_(2)SO_(4)` with 100ml of 0.05 M HCl solution ?

Options

A. More than 3
B. Less than 1
C. More than 1 and less than 2
D. More than 2 and less than 2

Answer

To find the pH of a solution prepared by mixing 100 ml of 0.02 M sulfuric acid (H₂SO₄) with 100 ml of 0.05 M hydrochloric acid (HCl), we can follow these steps:

### Step 1: Calculate the number of moles of H₂SO₄
- Volume of H₂SO₄ solution = 100 ml = 0.1 L
- Concentration of H₂SO₄ = 0.02 M

Number of moles of H₂SO₄ = Concentration × Volume
= 0.02 mol/L × 0.1 L = 0.002 moles = 2 millimoles (mmol)

### Step 2: Calculate the number of moles of HCl
- Volume of HCl solution = 100 ml = 0.1 L
- Concentration of HCl = 0.05 M

Number of moles of HCl = Concentration × Volume
= 0.05 mol/L × 0.1 L = 0.005 moles = 5 millimoles (mmol)

### Step 3: Determine the total moles of H⁺ ions produced
- H₂SO₄ dissociates to produce 2 moles of H⁺ ions for every mole of H₂SO₄:
- From 2 mmol of H₂SO₄, we get 2 × 2 = 4 mmol of H⁺ ions.

- HCl dissociates to produce 1 mole of H⁺ ions for every mole of HCl:
- From 5 mmol of HCl, we get 5 mmol of H⁺ ions.

Total moles of H⁺ = 4 mmol (from H₂SO₄) + 5 mmol (from HCl) = 9 mmol

### Step 4: Calculate the total volume of the mixed solution
- Total volume = Volume of H₂SO₄ + Volume of HCl
= 100 ml + 100 ml = 200 ml = 0.2 L

### Step 5: Calculate the concentration of H⁺ ions in the mixed solution
Concentration of H⁺ = Total moles of H⁺ / Total volume
= 9 mmol / 200 ml
= 0.045 M

### Step 6: Calculate the pH of the solution
pH = -log[H⁺]
= -log(0.045)
≈ 1.346

### Conclusion
The pH of the solution prepared by mixing 100 ml of 0.02 M H₂SO₄ with 100 ml of 0.05 M HCl is approximately **1.346**.

---