What is the pH of a solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL 0.2 M `H_(2)SO_(4)` ?(A) 0.74 (B)7.4 (C)4.68 (D)0.468

Options

A. 0.74
B. 7.4
C. 4.68
D. 0.468

Explanation

To find the pH of the solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL of 0.2 M H₂SO₄, we can follow these steps:

### Step 1: Calculate the number of moles of H⁺ from HCl
- The number of moles of HCl can be calculated using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
- Given that the molarity of HCl is 0.1 M and the volume is 10 mL (which is 0.01 L):
\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 0.01 \, \text{L} = 0.001 \, \text{moles} \]

### Step 2: Calculate the number of moles of H⁺ from H₂SO₄
- H₂SO₄ is a strong acid that dissociates into 2 H⁺ ions, so its n-factor is 2.
- The number of moles of H₂SO₄ can be calculated similarly:
\[ \text{Number of moles of H₂SO₄} = \text{Molarity} \times \text{Volume (in liters)} \times \text{n-factor} \]
- Given that the molarity of H₂SO₄ is 0.2 M and the volume is 40 mL (which is 0.04 L):
\[ \text{Number of moles of H₂SO₄} = 0.2 \, \text{mol/L} \times 0.04 \, \text{L} \times 2 = 0.016 \, \text{moles} \]

### Step 3: Calculate the total number of moles of H⁺
- Now, we can find the total number of moles of H⁺ ions from both acids:
\[ \text{Total moles of H⁺} = \text{moles from HCl} + \text{moles from H₂SO₄} = 0.001 + 0.016 = 0.017 \, \text{moles} \]

### Step 4: Calculate the total volume of the solution
- The total volume of the mixed solution is:
\[ \text{Total volume} = 10 \, \text{mL} + 40 \, \text{mL} = 50 \, \text{mL} = 0.050 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions
- The concentration of H⁺ ions in the solution can be calculated as:
\[ \text{Concentration of H⁺} = \frac{\text{Total moles of H⁺}}{\text{Total volume (in liters)}} = \frac{0.017 \, \text{moles}}{0.050 \, \text{L}} = 0.34 \, \text{M} \]

### Step 6: Calculate the pH of the solution
- The pH of the solution can be calculated using the formula:
\[ \text{pH} = -\log[\text{H⁺}] \]
- Substituting the concentration of H⁺:
\[ \text{pH} = -\log(0.34) \approx 0.468 \]

### Final Answer
The pH of the solution is approximately **0.468**. Thus, the correct option is (D) 0.468.

---