Animated Solution for Chemistry - Ionic Equilibrium: Comprehension Passage

Imagine you are a chemical detective, and you are handed two distinct experiments. In the first experiment, you mix a strong acid ($\text{HCl}$) with a strong base ($\text{NaOH}$). In the second, you swap the strong acid for a weak one (${\text{CH}}_3\text{COOH}$). Both experiments take place in identical insulated beakers with the exact same total volume of $200\text{ mL}$.

Why is this volume detail so crucial? Because it tells us that the physical environment—the mass of the solution and the calorimeter itself—remains perfectly constant. This means the total heat capacity ($C_{\text{total}}$) of the system is identical for both reactions.

Let's dive into the first experiment. We mix $100\text{ mL}$ of $1.0\text{ M HCl}$ with $100\text{ mL}$ of $1.0\text{ M NaOH}$. This gives us exactly $0.1\text{ moles}$ of water formed. We are given a golden piece of information: the enthalpy of neutralization for a strong acid and strong base is always $-57.0\text{ kJ/mol}$.

This means the heat released ($Q_1$) is simply the moles of water formed multiplied by the magnitude of this enthalpy:

This $5700\text{ J}$ of heat caused a temperature rise of $5.7{}^{\circ }\text{C}$. By using the fundamental calorimetry equation $Q=C_{\text{total}}\Delta T$, we can unlock the total heat capacity of our system:

Armed with our total heat capacity, we move to the second experiment. Here, we mix $100\text{ mL}$ of $2.0\text{ M}$ acetic acid ($200\text{ mmol}$) with $100\text{ mL}$ of $1.0\text{ M NaOH}$ ($100\text{ mmol}$). Since $\text{NaOH}$ is the limiting reagent, exactly $100\text{ mmol}$ (or $0.1\text{ moles}$) of water will form again.

The temperature rise this time is $5.6{}^{\circ }\text{C}$. Let's calculate the heat released ($Q_2$):

Now, what is the enthalpy of neutralization for this weak acid? We divide the heat released by the moles of water formed:

Notice something? This value ($-56.0\text{ kJ/mol}$) is less negative than the strong acid's $-57.0\text{ kJ/mol}$. Why? Because weak acids don't give up their protons easily. A portion of the neutralization energy must be spent to break the $\text{H-A}$ bond. This is known as the enthalpy of dissociation.

Our journey isn't over. We need to find the pH of the solution after the second experiment. Let's take an inventory of what's left in the beaker. We started with $200\text{ mmol}$ of acetic acid and $100\text{ mmol}$ of $\text{NaOH}$. After the reaction, the $\text{NaOH}$ is completely consumed, leaving us with $100\text{ mmol}$ of unreacted acetic acid and $100\text{ mmol}$ of newly formed sodium acetate.

A weak acid coexisting with its conjugate base? That is the textbook definition of an acidic buffer! To find its pH, we summon the Henderson-Hasselbalch equation:

First, we calculate the $\text{p}{K}_a$ from the given $K_a$ of $2.0\times 10^{-5}$:

Since the millimoles of salt and acid are perfectly equal ($100\text{ mmol}$ each), their ratio is $1$, and the $\log (1)$ is zero.

The elegance of this problem lies in how it seamlessly weaves thermodynamics into ionic equilibrium, proving that in chemistry, everything is connected.