Principle used

The temperature rise on mixing acid and base depends on the heat released: $q = n\,|\Delta H_{\text{neut}}|$ and $\Delta T = \frac{q}{mc}$ where:

• $n$ = moles of water formed (or moles neutralized),
• $|\Delta H_{\text{neut}}|$ is larger for strong acid + strong base (about $57\,\text{kJ mol}^{-1}$),
• for weak acid + strong base, the heat released is smaller because some energy is used to ionize the weak acid.

Since all solutions are dilute aqueous solutions, we take density $\approx 1\,\text{g mL}^{-1}$ and specific heat nearly same for all. So compare: $\Delta T \propto \frac{\text{heat released}}{\text{total volume}}$

Evaluate each option

Option A: $50\,\text{mL HCl}$ and $20\,\text{mL NaOH}$

Both are $1\,\text{M}$.

Moles of HCl: $n_{\text{HCl}} = 1 \times 0.050 = 0.050\,\text{mol}$ Moles of NaOH: $n_{\text{NaOH}} = 1 \times 0.020 = 0.020\,\text{mol}$

Limiting reagent = NaOH, so moles neutralized: $n = 0.020\,\text{mol}$

Heat released: $q_A = 0.020 \times 57 = 1.14\,\text{kJ}$

Total volume: $V_A = 50 + 20 = 70\,\text{mL}$

So, $\Delta T_A \propto \frac{1.14}{70}$

Option B: $30\,\text{mL HCl}$ and $30\,\text{mL NaOH}$

Moles of HCl: $n_{\text{HCl}} = 1 \times 0.030 = 0.030\,\text{mol}$ Moles of NaOH: $n_{\text{NaOH}} = 1 \times 0.030 = 0.030\,\text{mol}$

Complete neutralization: $n = 0.030\,\text{mol}$

Heat released: $q_B = 0.030 \times 57 = 1.71\,\text{kJ}$

Total volume: $V_B = 60\,\text{mL}$

So, $\Delta T_B \propto \frac{1.71}{60}$

Option C: $45\,\text{mL CH}_3\text{COOH}$ and $25\,\text{mL NaOH}$

Weak acid + strong base, so heat of neutralization is less than $57\,\text{kJ mol}^{-1}$.

Moles of acetic acid: $n_{\text{acid}} = 1 \times 0.045 = 0.045\,\text{mol}$ Moles of NaOH: $n_{\text{base}} = 1 \times 0.025 = 0.025\,\text{mol}$

Limiting reagent = NaOH, hence $n = 0.025\,\text{mol}$

If it were strong acid, heat would be: $0.025 \times 57 = 1.425\,\text{kJ}$ but actual heat is less than this.

Total volume: $V_C = 70\,\text{mL}$

Hence, $\Delta T_C < \frac{1.425}{70}$

Option D: $30\,\text{mL CH}_3\text{COOH}$ and $30\,\text{mL NaOH}$

Moles of acid: $n_{\text{acid}} = 0.030\,\text{mol}$ Moles of base: $n_{\text{base}} = 0.030\,\text{mol}$

So, $n = 0.030\,\text{mol}$

Again weak acid + strong base, so $q_D < 0.030 \times 57 = 1.71\,\text{kJ}$

Total volume: $V_D = 60\,\text{mL}$

Hence, $\Delta T_D < \frac{1.71}{60}$

Comparison

• Option B has strong acid + strong base, so maximum heat per mole.
• It also has complete neutralization of $0.030$ mol in only $60\,\text{mL}$ total solution.
• Option D has same volumes and same neutralized moles, but weaker acid gives less heat.
• Options A and C either produce less heat or have larger total volume.

Therefore, the largest increase in temperature occurs for: $\boxed{\text{B}}$

Check with stored answer

Stored correct answer: B

My derived answer: B

So they agree.