Question

\(100\ \text{mL}\) of \(0.1M\ HCl\) and \(100\ \text{mL}\) of \(0.05M\ H_2SO_4\) 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:

For acid mixture problems, first calculate total moles of \(H^+\), then divide by final volume.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the pH of an acidic mixture after dilution. We need to calculate the total amount of \(H^+\) ions contributed by each strong acid and then find the final concentration.
Step 2: Detailed Explanation:

Step A: Calculate millimoles of \(H^+\) from \(HCl\).
\(HCl\) is a monoprotic acid.
Millimoles = Molarity \(\times\) Volume = \(0.1 \times 100 = 10 \text{ mmol}\).

Step B: Calculate millimoles of \(H^+\) from \(H_2SO_4\).
\(H_2SO_4\) is a diprotic acid (releases \(2 H^+\) per molecule).
Millimoles of \(H_2SO_4 = 0.05 \times 100 = 5 \text{ mmol}\).
Millimoles of \(H^+ = 2 \times 5 = 10 \text{ mmol}\).

Step C: Find total millimoles of \(H^+\) in the mixture.
Total \(H^+ = 10 \text{ (from HCl)} + 10 \text{ (from } H_2SO_4) = 20 \text{ mmol}\).
This is equivalent to 0.02 moles.

Step D: Find the final molarity after dilution.
Final Volume (\(V\)) = 2.0 Liters.
Molarity of \(H^+ = \text{Moles} / \text{Volume} = 0.02 / 2.0 = 0.01 M\).
This can be written as \(10^{-2} M\).

Step E: Calculate the pH.
\(pH = -\log[H^+] = -\log(10^{-2}) = 2\).

Step 3: Final Answer:
The pH of the resulting solution is 2.