0.1M HCl and 0.1 `MH_(2)SO_(4)`, each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 N NaCl solution.
The pH of the resulting mixture is

Options

A. 1.17
B. 1
C. 0.3
D. log 2 - log 3

Answer

To find the pH of the resulting mixture after mixing 0.1M HCl and 0.1M H₂SO₄, we can follow these steps:

### Step 1: Calculate the moles of HCl and H₂SO₄
- **HCl**:
- Molarity = 0.1 M
- Volume = 2 ml = 0.002 L
- Moles of HCl = Molarity × Volume = 0.1 mol/L × 0.002 L = 0.0002 moles

- **H₂SO₄**:
- Molarity = 0.1 M
- Volume = 2 ml = 0.002 L
- Moles of H₂SO₄ = Molarity × Volume = 0.1 mol/L × 0.002 L = 0.0002 moles

### Step 2: Determine the contribution of H⁺ ions from each acid
- **HCl**:
- Each mole of HCl produces 1 mole of H⁺.
- Therefore, moles of H⁺ from HCl = 0.0002 moles.

- **H₂SO₄**:
- Each mole of H₂SO₄ produces 2 moles of H⁺.
- Therefore, moles of H⁺ from H₂SO₄ = 2 × 0.0002 moles = 0.0004 moles.

### Step 3: Calculate the total moles of H⁺ ions
- Total moles of H⁺ = Moles from HCl + Moles from H₂SO₄
- Total moles of H⁺ = 0.0002 moles + 0.0004 moles = 0.0006 moles.

### Step 4: Calculate the total volume of the solution
- Initial volume of HCl = 2 ml
- Initial volume of H₂SO₄ = 2 ml
- Volume of NaCl solution added = 2 ml
- Total volume = 2 ml + 2 ml + 2 ml = 6 ml = 0.006 L.

### Step 5: Calculate the concentration of H⁺ ions
- Concentration of H⁺ = Total moles of H⁺ / Total volume
- Concentration of H⁺ = 0.0006 moles / 0.006 L = 0.1 M.

### Step 6: Calculate the pH of the solution
- pH = -log[H⁺]
- pH = -log(0.1) = 1.

### Final Answer
The pH of the resulting mixture is **1**.

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