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

Concept: For strong acids, total \([H^+]\) is calculated from total moles of \(H^+\) divided by total volume.

Step 1: Moles of \(H^+\) from \(HCl\): \[ M\times V=0.1\times 0.1=0.01 \]

Step 2: Moles of \(H_2SO_4\): \[ 0.05\times 0.1=0.005 \] Since one mole of \(H_2SO_4\) gives \(2\) moles of \(H^+\): \[ \text{Moles of }H^+=2\times 0.005=0.01 \]

Step 3: Total moles of \(H^+\): \[ 0.01+0.01=0.02 \]

Step 4: Final volume: \[ 2.0\ L \] \[ [H^+]=\frac{0.02}{2.0}=0.01=10^{-2} \]

Step 5: Calculate pH. \[ pH=-\log[H^+] \] \[ pH=-\log(10^{-2}) \] \[ pH=2 \] Therefore, \[ \boxed{2} \]