Which of the following mixing of 1 M base and 1 M acid leads to the largest increase in temperature?

For the same molarity, the rise in temperature $$\Delta T$$ on mixing an acid with a base depends on two points:
  • heat evolved, $$q = n_{\text{limiting}}\;|\Delta H_{\text{neut}}|$$
  • total mass of the final solution, $$m = (\text{total volume}) \times \rho$$ (take $$\rho \approx 1\,\text{g ml}^{-1}$$).
The temperature rise is obtained from the calorimetry relation $$q = m\,C_p\,\Delta T$$ with $$C_p \approx 4.18\,\text{J g}^{-1}\text{K}^{-1}$$ for dilute aqueous solutions.

Standard enthalpy of neutralisation
• strong acid + strong base: $$|\Delta H_{\text{neut}}| \approx 57\,\text{kJ mol}^{-1}$$.
• weak acid (e.g. $$CH_3COOH$$) + strong base: the acid must first ionise, so the net heat evolved is smaller, about $$51{-}55\,\text{kJ mol}^{-1}$$.
Thus a strong acid-strong base pair always releases the greater amount of heat per mole neutralised.

Let us evaluate each option. All solutions are $$1\,\text{M}$$, so $$1\,\text{ml} \equiv 1\,\text{mmol}$$.

Case A: $$30\,\text{ml}\;CH_3COOH + 30\,\text{ml}\;NaOH$$
Moles acid $$= 0.030$$, moles base $$= 0.030$$ ⇒ complete neutralisation of $$0.030\,\text{mol}$$.
Heat evolved $$q_A = 0.030 \times 55\,\text{kJ} = 1.65\,\text{kJ} = 1650\,\text{J}$$.
Total volume $$= 60\,\text{ml}$$ ⇒ mass $$m = 60\,\text{g}$$.
$$\Delta T_A = \dfrac{1650}{60 \times 4.18} \approx 6.6^\circ\text{C}$$.

Case B: $$45\,\text{ml}\;CH_3COOH + 25\,\text{ml}\;NaOH$$
Moles acid $$= 0.045$$, moles base $$= 0.025$$ ⇒ base is limiting, $$0.025\,\text{mol}$$ neutralised.
Heat evolved $$q_B = 0.025 \times 55\,\text{kJ} = 1.38\,\text{kJ} = 1380\,\text{J}$$.
Total volume $$= 70\,\text{ml}$$ ⇒ mass $$m = 70\,\text{g}$$.
$$\Delta T_B = \dfrac{1380}{70 \times 4.18} \approx 4.7^\circ\text{C}$$.

Case C: $$30\,\text{ml}\;HCl + 30\,\text{ml}\;NaOH$$
Both strong, moles acid $$= 0.030$$, moles base $$= 0.030$$ ⇒ $$0.030\,\text{mol}$$ neutralised.
Heat evolved $$q_C = 0.030 \times 57\,\text{kJ} = 1.71\,\text{kJ} = 1710\,\text{J}$$.
Total volume $$= 60\,\text{ml}$$ ⇒ mass $$m = 60\,\text{g}$$.
$$\Delta T_C = \dfrac{1710}{60 \times 4.18} \approx 6.8^\circ\text{C}$$.

Case D: $$50\,\text{ml}\;HCl + 20\,\text{ml}\;NaOH$$
Moles acid $$= 0.050$$, moles base $$= 0.020$$ ⇒ base is limiting, $$0.020\,\text{mol}$$ neutralised.
Heat evolved $$q_D = 0.020 \times 57\,\text{kJ} = 1.14\,\text{kJ} = 1140\,\text{J}$$.
Total volume $$= 70\,\text{ml}$$ ⇒ mass $$m = 70\,\text{g}$$.
$$\Delta T_D = \dfrac{1140}{70 \times 4.18} \approx 3.9^\circ\text{C}$$.

Comparing: $$\Delta T_C \gt \Delta T_A \gt \Delta T_B \gt \Delta T_D$$. The greatest temperature rise occurs for Case C, mixing equal volumes of the strong acid $$HCl$$ and strong base $$NaOH$$.

Hence, the correct choice is Option C.