Calculate the heat evolved in the reaction between HCl and NaOH, given their initial concentrations and volumes, and the final temperature.

$$T_{initial} = \frac{T_{acid} + T_{alkali}}{2}$$ $$T_{initial} = \frac{25^\circ\text{C} + 25.2^\circ\text{C}}{2}$$ $$T_{initial} = \frac{50.2^\circ\text{C}}{2}$$ $$T_{initial} = 25.1^\circ\text{C}$$

$$\Delta T = T_{final} - T_{initial}$$ $$\Delta T = 38.8^\circ\text{C} - 25.1^\circ\text{C}$$ $$\Delta T = 13.7^\circ\text{C}$$

$$V_{total} = V_{HCl} + V_{NaOH}$$ $$V_{total} = 50 \text{ cm}^3 + 50 \text{ cm}^3$$ $$V_{total} = 100 \text{ cm}^3$$

Assumption: The density of the solution is $1 \text{ g/cm}^3$.

$$m = V_{total} \times \text{density}$$ $$m = 100 \text{ cm}^3 \times 1 \text{ g/cm}^3$$ $$m = 100 \text{ g}$$

Assumption: The specific heat capacity of the solution is 4.2 \text{ J g}^{-1} \text{ ^\circ C}^{-1}.

$$Q = mc\Delta T$$ Q = 100 \text{ g} \times 4.2 \text{ J g}^{-1} \text{ ^\circ C}^{-1} \times 13.7 \text{ ^\circ C} $$Q = 5754 \text{ J}$$ $$Q = 5.754 \text{ kJ}$$

Moles of HCl:

$$n_{HCl} = \text{Concentration} \times \text{Volume}$$ $$n_{HCl} = 2 \text{ mol/dm}^3 \times \frac{50}{1000} \text{ dm}^3$$ $$n_{HCl} = 2 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3$$ $$n_{HCl} = 0.1 \text{ mol}$$

Moles of NaOH:

$$n_{NaOH} = \text{Concentration} \times \text{Volume}$$ $$n_{NaOH} = 2 \text{ mol/dm}^3 \times \frac{50}{1000} \text{ dm}^3$$ $$n_{NaOH} = 2 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3$$ $$n_{NaOH} = 0.1 \text{ mol}$$

Since HCl and NaOH react in a 1:1 molar ratio, $0.1 \text{ mol}$ of water is formed.

$$\Delta H_{neut} = \frac{-Q}{\text{moles of water formed}}$$ $$\Delta H_{neut} = \frac{-5.754 \text{ kJ}}{0.1 \text{ mol}}$$ $$\Delta H_{neut} = -57.54 \text{ kJ/mol}$$

a) The heat of neutralization of hydrochloric acid with sodium hydroxide is $\boxed{\mathbf{-57.54 \text{ kJ/mol}}}$.

Assumptions made:

• The density of the solution is $1 \text{ g/cm}^3$.
• The specific heat capacity of the solution is 4.2 \text{ J g}^{-1} \text{ ^\circ C}^{-1}.
• No heat is lost to the surroundings or absorbed by the plastic cup.
• The reaction goes to completion.