Question

30 cc of $\frac{M}{3}$ HCl, 20 cc of $\frac{M}{2}$ HNO$_{3}$ and 40 cc of $\frac{M}{4}$ NaOH solutions are mixed and the volume was made up to $1~\text{dm}^{3}$. The pH of the resulting solution is

• A. 8
• B. 2
• C. 1
• D. 3

Hint:

$pH = -\log[H^+]$. Always calculate the excess component after neutralization.

Answer 1

The Correct Option is B

Step 1: Calculate Milliequivalents
Total \( \mathrm{H}^{+} \) milliequivalents \( = (30 \times \frac{1}{3}) + (20 \times \frac{1}{2}) = 10 + 10 = 20 \). Total \( \mathrm{OH}^{-} \) milliequivalents \( = 40 \times \frac{1}{4} = 10 \).
Step 2: Find \( \mathrm{H}^{+} \) Concentration
Net \( \mathrm{H}^{+} \) left \( = 20 - 10 = 10 \) milliequivalents. \( [\mathrm{H}^{+}] = \frac{10}{1000} = 10^{-2} \text{ g ions/dm}^{3} \).
Step 3: Calculate pH
\( \mathrm{pH} = -\log[\mathrm{H}^{+}] = -\log(10^{-2}) = 2 \).
Final Answer: (b)