Question

The pH of a mixture containing 400 mL of 0.1 $MH_2SO_4$ and 400 mL of 0.1 M NaOH will be approximately

• A. 1.30
• B. 7.00
• C. 1.00
• D. 0.70

Answer 1

Answer: (A)

1.30

Answer 2

1. Calculate initial millimoles: $H_2SO_4 = 0.1 \text{ M} \times 400 \text{ mL} = 40$ mmol, $NaOH = 0.1 \text{ M} \times 400 \text{ mL} = 40$ mmol.
2. Reaction: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. Thus, 40 mmol $NaOH$ reacts with $40/2 = 20$ mmol $H_2SO_4$.
3. Excess $H_2SO_4 = 40 \text{ mmol} - 20 \text{ mmol} = 20$ mmol. Total volume = 800 mL.
4. Molarity of excess $H_2SO_4 = \frac{20 \text{ mmol}}{800 \text{ mL}} = 0.025$ M.
5. Assuming complete dissociation of $H_2SO_4$, $[H^+] = 2 \times [H_2SO_4] = 2 \times 0.025 \text{ M} = 0.05$ M.
6. pH = -log$[H^+]$ = -log(0.05) $\approx$ 1.30.