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

Options

A. 1.17
B. `1.0`
C. 0.3
D. log 2 - log 3

Explanation

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

### Step 1: Calculate the moles of HCl and H₂SO₄
1. **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.002 \, \text{L} = 0.0002 \, \text{mol} \]

2. **Moles of H₂SO₄**:
H₂SO₄ is a diprotic acid, meaning it can donate two protons (H⁺ ions). Therefore, the effective moles of H⁺ from H₂SO₄ will be:
\[ \text{Moles of H₂SO₄} = 0.1 \, \text{mol/L} \times 0.002 \, \text{L} = 0.0002 \, \text{mol} \]
Since each H₂SO₄ can donate 2 H⁺ ions:
\[ \text{Moles of H⁺ from H₂SO₄} = 2 \times 0.0002 \, \text{mol} = 0.0004 \, \text{mol} \]

### Step 2: Calculate the total moles of H⁺
Now, we can find the total moles of H⁺ in the mixture:
\[ \text{Total moles of H⁺} = \text{Moles of HCl} + \text{Moles of H⁺ from H₂SO₄} = 0.0002 \, \text{mol} + 0.0004 \, \text{mol} = 0.0006 \, \text{mol} \]

### Step 3: Calculate the total volume of the solution
The total volume after mixing is:
\[ \text{Total Volume} = 2 \, \text{ml (HCl)} + 2 \, \text{ml (H₂SO₄)} + 2 \, \text{ml (NaCl)} = 6 \, \text{ml} = 0.006 \, \text{L} \]

### Step 4: Calculate the concentration of H⁺ in the mixture
Now we can calculate the concentration of H⁺ ions in the resulting mixture:
\[ \text{Concentration of H⁺} = \frac{\text{Total moles of H⁺}}{\text{Total Volume}} = \frac{0.0006 \, \text{mol}}{0.006 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the pH of the solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H⁺}] \]
Substituting the concentration of H⁺:
\[ \text{pH} = -\log(0.1) = 1 \]

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