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 the solution prepared by mixing 100 ml of 0.02 M \( H_2SO_4 \) with 100 ml of 0.05 M HCl, we will follow these steps:

### Step 1: Calculate the moles of \( H_2SO_4 \)
- Molarity (M) = moles/volume (L)
- Moles of \( H_2SO_4 \) = Molarity × Volume
- Volume of \( H_2SO_4 \) = 100 ml = 0.1 L
- Molarity of \( H_2SO_4 \) = 0.02 M

\[ \text{Moles of } H_2SO_4 = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{moles} \]

### Step 2: Determine the contribution of \( H^+ \) ions from \( H_2SO_4 \)
- \( H_2SO_4 \) is a strong acid and dissociates completely in two steps:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
- Therefore, 1 mole of \( H_2SO_4 \) produces 2 moles of \( H^+ \).
- Moles of \( H^+ \) from \( H_2SO_4 \):
\[ \text{Moles of } H^+ \text{ from } H_2SO_4 = 2 \times 0.002 \, \text{moles} = 0.004 \, \text{moles} \]

### Step 3: Calculate the moles of \( HCl \)
- Molarity of \( HCl \) = 0.05 M
- Volume of \( HCl \) = 100 ml = 0.1 L

\[ \text{Moles of } HCl = 0.05 \, \text{mol/L} \times 0.1 \, \text{L} = 0.005 \, \text{moles} \]

### Step 4: Determine the total moles of \( H^+ \) ions
- Total moles of \( H^+ \):
\[ \text{Total } H^+ = \text{Moles from } H_2SO_4 + \text{Moles from } HCl = 0.004 + 0.005 = 0.009 \, \text{moles} \]

### Step 5: Calculate the total volume of the solution
- Total volume = Volume of \( H_2SO_4 \) + Volume of \( HCl \) = 100 ml + 100 ml = 200 ml = 0.2 L

### Step 6: Calculate the concentration of \( H^+ \) ions
\[ \text{Concentration of } H^+ = \frac{\text{Total moles of } H^+}{\text{Total volume in L}} = \frac{0.009 \, \text{moles}}{0.2 \, \text{L}} = 0.045 \, \text{M} \]

### Step 7: Calculate the pH of the solution
- pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(0.045) \]

Using a calculator:
\[ \text{pH} \approx 1.35 \]

### Conclusion
The pH of the solution is approximately **1.35**.

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