[VISUAL] The acid-dissociation constants of HC3H5O3(aq) and CH3NH3+(aq) are given in the table above — Acids and Bases Chemistry Question

STEPS:

1. Understand the core concept of a buffer: An acid-base buffer consists of a weak acid and its conjugate base. It is highly resistant to pH changes when small amounts of strong acid or base are added. According to the Henderson^-Hasselbalch equation:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \quad \text{}$$
A buffer is most effective when the desired pH is close to the weak acid's $\text{p}K_a$ value (typically within $\pm 1$ pH unit).
2. Calculate the $\text{p}K_a$ for both conjugate acid systems: Using the table of acid-dissociation constants ($K_a$):
* Lactic acid ($\text{HC}_3\text{H}_5\text{O}_3$):
$$\text{p}K_a = -\log(K_a) = -\log(8.3 \times 10^{-4}) \approx 4 - \log(8.3) \approx \mathbf{3.1} \quad \text{}$$
* Methylammonium ion ($\text{CH}_3\text{NH}_3^+$):
$$\text{p}K_a = -\log(K_a) = -\log(2.3 \times 10^{-11}) \approx 11 - \log(2.3) \approx \mathbf{10.6} \quad \text{}$$
3. Select the appropriate buffer system: Since the target pH is approximately 3, we must choose the lactic acid/lactate system ($\text{p}K_a \approx 3.1$) because its $\text{p}K_a$ is extremely close to the target pH. This immediately eliminates Options A and D, which would form buffers with a pH near $10.6$.
4. Determine the stoichiometry required to form a buffer: To make a buffer from a weak acid, we can partially neutralize it with a strong base to generate its conjugate base in solution. A perfect buffer contains equal amounts of the weak acid and its conjugate base, meaning we must neutralize exactly half of the starting weak acid.
5. Analyze the chemical reaction in Option B:
* Initial moles of weak acid ($\text{HC}_3\text{H}_5\text{O}_3$):
$$n_{\text{acid}} = 100.\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} \times 0.1\text{ M} = \mathbf{0.010\text{ mol}} \quad \text{}$$
* Initial moles of strong base ($\text{NaOH}$):
$$n_{\text{base}} = 50.\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} \times 0.1\text{ M} = \mathbf{0.0050\text{ mol}} \quad \text{}$$
* Neutralization reaction:
$$\text{HC}_3\text{H}_5\text{O}_3(aq) + \text{OH}^-(aq) \rightarrow \text{C}_3\text{H}_5\text{O}_3^-(aq) + \text{H}_2\text{O}(l)$$
Because $\text{OH}^-$ is the limiting reactant, it reacts completely to convert $0.0050\text{ mol}$ of weak acid into its conjugate base:
* Remaining weak acid: $0.010\text{ mol} - 0.0050\text{ mol} = \mathbf{0.0050\text{ mol}}$
* Produced conjugate base ($\text{C}_3\text{H}_5\text{O}_3^-$): $\mathbf{0.0050\text{ mol}}$
6. Calculate the final pH of the mixture: Because the moles (and thus concentrations) of the weak acid and its conjugate base are exactly equal:
$$\text{pH} = \text{p}K_a + \log\left(\frac{0.0050\text{ mol}}{0.0050\text{ mol}}\right) = \text{p}K_a + \log(1) = \text{p}K_a \approx \mathbf{3.1}$$
This creates an ideal buffer with a pH of approximately 3, confirming Option B is correct.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This mixture contains $0.010\text{ mol}$ of the weak acid $\text{CH}_3\text{NH}_3^+$ and $0.0050\text{ mol}$ of the strong base $\text{NaOH}$. While it successfully neutralizes half of the weak acid to create an equimolar buffer of $\text{CH}_3\text{NH}_3^+$ and $\text{CH}_3\text{NH}_2$, the resulting pH is equal to the $\text{p}K_a$ of $\text{CH}_3\text{NH}_3^+$, which is $10.6$ rather than 3.
• Option C is incorrect: This mixture combines a weak base ($\text{NaC}_3\text{H}_5\text{O}_3$) and a strong base ($\text{NaOH}$). Mixing two bases does not create a conjugate weak acid-base system, so no buffer is formed. Furthermore, because of the excess unreacted strong base, the pH will be highly basic ($\text{pH} > 13$).
• Option D is incorrect: This mixture combines a weak acid ($\text{CH}_3\text{NH}_3^+$) and its conjugate weak base ($\text{CH}_3\text{NH}_2$) in equal amounts. While this is a buffer, the pH of this solution is equal to the $\text{p}K_a$ of the weak acid, which is $10.6$ rather than 3.