Question

100 mL of 0.1M $HCl$ and 100 mL of 0.05 M $H_{2}SO_{4}$ are mixed and the solution is diluted to 2.0 L by adding water. The pH of the resulting solution is

• A. 1
• B. 3
• C. 2
• D. 4

Hint:

Don't forget that $H_{2}SO_{4}$ is diprotic—it gives two $H^{+}$ ions per molecule!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The pH of a solution depends on the total concentration of hydrogen ions ($H^+$). When mixing strong acids, we calculate the total moles of $H^+$ and divide by the final total volume.
Step 2: Key Formula or Approach:
1. Moles of $H^+$ from $HCl = M \times V$ (in L).
2. Moles of $H^+$ from $H_2SO_4 = 2 \times M \times V$ (since it is dibasic).
3. $[H^+] = \frac{\text{Total moles of } H^+}{\text{Total Volume in L}}$.
4. $pH = -\log[H^+]$.
Step 3: Detailed Explanation:

Moles from $HCl = 0.1 \times 0.1 = 0.01 \text{ moles}$.
Moles from $H_2SO_4 = 2 \times (0.05 \times 0.1) = 2 \times 0.005 = 0.01 \text{ moles}$.
Total moles of $H^+ = 0.01 + 0.01 = 0.02 \text{ moles}$.
Total final volume = $2.0 \text{ L}$.
$[H^+] = \frac{0.02}{2.0} = 0.01 = 10^{-2} \text{ M}$.
$pH = -\log(10^{-2}) = 2$.
Step 4: Final Answer:
The pH of the resulting solution is 2.