if equal volumes of 0.1 M `H_2SO_4` and 0.1 M HCl are mixed then the pH of resulting solution will be (log 15 = 1.176)

Options

A. 0.82
B. 1.64
C. 2.82
D. 1.82

Answer

To find the pH of the resulting solution when equal volumes of 0.1 M H₂SO₄ and 0.1 M HCl are mixed, we can follow these steps:

### Step 1: Determine the moles of H⁺ ions from H₂SO₄
- **Given:** Molarity of H₂SO₄ = 0.1 M, Volume = V (in liters)
- **Calculation:**
- Moles of H₂SO₄ = Molarity × Volume = 0.1 × V = 0.1V moles
- Each molecule of H₂SO₄ produces 2 H⁺ ions.
- Therefore, moles of H⁺ from H₂SO₄ = 2 × 0.1V = 0.2V moles.

### Step 2: Determine the moles of H⁺ ions from HCl
- **Given:** Molarity of HCl = 0.1 M, Volume = V (in liters)
- **Calculation:**
- Moles of HCl = Molarity × Volume = 0.1 × V = 0.1V moles.
- Each molecule of HCl produces 1 H⁺ ion.
- Therefore, moles of H⁺ from HCl = 0.1V moles.

### Step 3: Calculate the total moles of H⁺ ions in the solution
- **Total moles of H⁺ ions:**
- Total H⁺ = Moles from H₂SO₄ + Moles from HCl
- Total H⁺ = 0.2V + 0.1V = 0.3V moles.

### Step 4: Calculate the total volume of the resulting solution
- **Total volume:**
- Since equal volumes of both acids are mixed, the total volume = V + V = 2V.

### Step 5: Calculate the concentration of H⁺ ions
- **Concentration of H⁺ ions:**
- Concentration = Total moles of H⁺ / Total volume
- Concentration of H⁺ = (0.3V) / (2V) = 0.15 M.

### Step 6: Calculate the pH of the solution
- **Using the formula for pH:**
- pH = -log[H⁺]
- pH = -log(0.15) = -log(15 × 10⁻²) = -[log(15) + log(10⁻²)]
- pH = -[log(15) - 2] = 2 - log(15).
- Given log(15) = 1.176, we have:
- pH = 2 - 1.176 = 0.824.

### Step 7: Finalize the answer
- The closest option to 0.824 is 0.82.

### Answer:
The pH of the resulting solution is approximately **0.82**.