Question

\(200\ \mathrm{mL}\) of \(0.01\ \mathrm{M}\ \mathrm{HCl}\) is mixed with \(400\ \mathrm{mL}\) of \(0.01\ \mathrm{M}\ \mathrm{H_2SO_4}\). The \(\mathrm{pH}\) of the mixture is _________.
Given: \(\log2=0.30,\ \log3=0.48,\ \log5=0.70,\ \log7=0.84,\ \log11=1.04\)

• A. \(1.14\)
• B. \(1.78\)
• C. \(2.34\)
• D. \(3.02\)

Hint:

Count total moles of H⁺ (HCl gives 1 per molecule, H₂SO₄ gives 2) and divide by the total volume.

Answer 1

The Correct Option is B

Moles of \(\mathrm{H^+}\) from HCl: \(0.200\ \mathrm{L}\times0.01=2\times10^{-3}\ \mathrm{mol}\).
Moles of \(\mathrm{H^+}\) from \(\mathrm{H_2SO_4}\) (assuming complete dissociation, 2 \(\mathrm{H^+}\) per molecule): \(2\times0.400\ \mathrm{L}\times0.01=8\times10^{-3}\ \mathrm{mol}\).
Total \(\mathrm{H^+}\) \(=10\times10^{-3}=10^{-2}\ \mathrm{mol}\); total volume \(=600\ \mathrm{mL}=0.6\ \mathrm{L}\).
\([\mathrm{H^+}]=\dfrac{10^{-2}}{0.6}=\dfrac{1}{60}\ \mathrm{M}\).
\(\mathrm{pH}=-\log\dfrac1{60}=\log60=\log2+\log3+\log10=0.30+0.48+1=1.78\).
Answer: (2) 1.78