Question

Maximum heat Is released In mixing of

• A. 400 mL 0.3 M HCI + 300 mL 0.1 M NaOH
• B. 200 mL 0.3 M H2SO4 + 200 mL 0.6 M NaOH
• C. 100 mL 0.8 M HCI + 500 mL 0.1 M Ca(OH)2
• D. 500 mL 0.4 M HCI + 100 mL 0.2 M NaOH

Answer 1

Answer: (B)

B. 200 mL 0.3 M H2SO4 + 200 mL 0.6 M NaOH

Answer 2

The heat released during the neutralization of a strong acid and a strong base is approximately constant per mole of water formed (around -57.3 kJ/mol). Therefore, maximum heat release corresponds to the maximum moles of water formed. We calculate the moles of $H^+$ ions from the acid and $OH^-$ ions from the base for each option. The number of moles of water formed is determined by the limiting reactant, which is the minimum of the moles of $H^+$ and $OH^-$.

• Option A: Moles of $H^+$ from HCl = $0.4 \text{ L} \times 0.3 \text{ M} = 0.12 \text{ mol}$. Moles of $OH^-$ from NaOH = $0.3 \text{ L} \times 0.1 \text{ M} = 0.03 \text{ mol}$. Moles of $H_2O$ formed = $\min(0.12, 0.03) = 0.03 \text{ mol}$.
• Option B: Moles of $H^+$ from $H_2SO_4$ = $0.2 \text{ L} \times 0.3 \text{ M} \times 2 = 0.12 \text{ mol}$. Moles of $OH^-$ from NaOH = $0.2 \text{ L} \times 0.6 \text{ M} = 0.12 \text{ mol}$. Moles of $H_2O$ formed = $\min(0.12, 0.12) = 0.12 \text{ mol}$.
• Option C: Moles of $H^+$ from HCl = $0.1 \text{ L} \times 0.8 \text{ M} = 0.08 \text{ mol}$. Moles of $OH^-$ from $Ca(OH)_2$ = $0.5 \text{ L} \times 0.1 \text{ M} \times 2 = 0.10 \text{ mol}$. Moles of $H_2O$ formed = $\min(0.08, 0.10) = 0.08 \text{ mol}$.
• Option D: Moles of $H^+$ from HCl = $0.5 \text{ L} \times 0.4 \text{ M} = 0.20 \text{ mol}$. Moles of $OH^-$ from NaOH = $0.1 \text{ L} \times 0.2 \text{ M} = 0.02 \text{ mol}$. Moles of $H_2O$ formed = $\min(0.20, 0.02) = 0.02 \text{ mol}$.

Comparing the moles of water formed (0.03 mol, 0.12 mol, 0.08 mol, 0.02 mol), Option B yields the maximum moles of water (0.12 mol), and hence the maximum heat release.