An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid, HC2H3O2(aq), and 5 — Acids and Bases Chemistry Question

STEPS:

1. Determine the initial state of the buffer: The buffer is prepared by mixing equal volumes ($50.\text{ mL}$) of equal concentrations ($0.20\text{ M}$) of acetic acid ($\text{HC}_2\text{H}_3\text{O}_2$, a weak acid) and sodium acetate ($\text{NaC}_2\text{H}_3\text{O}_2$, which dissociates into the conjugate base $\text{C}_2\text{H}_3\text{O}_2^-$).
* Initial moles of weak acid:
$$n_{\text{acid}} = 0.050\text{ L} \times 0.20\text{ M} = \mathbf{0.010\text{ mol}}$$
* Initial moles of conjugate base:
$$n_{\text{base}} = 0.050\text{ L} \times 0.20\text{ M} = \mathbf{0.010\text{ mol}}$$
* Since the concentrations of the weak acid and conjugate base are initially equal, the starting pH of this buffer is exactly equal to its $\text{p}K_a$, which is $4.7$.
2. Analyze the addition of the strong base: A $5.0\text{ mL}$ sample of $0.10\text{ M}$ $\text{NaOH}$ is added. The strong base dissociates completely to yield hydroxide ions ($\text{OH}^-$):
* Moles of added $\text{OH}^-$:
$$n_{\text{OH}^-} = 0.0050\text{ L} \times 0.10\text{ M} = \mathbf{0.00050\text{ mol}}$$
3. Determine the neutralization reaction: Hydroxide ions are a strong base and will react completely with the weak acid in the buffer to form its conjugate base and water:
$$\text{HC}_2\text{H}_3\text{O}_2(aq) + \text{OH}^-(aq) \rightarrow \text{C}_2\text{H}_3\text{O}_2^-(aq) + \text{H}_2\text{O}(l)$$
4. Calculate final moles of each species after neutralization:
* Final moles of weak acid ($\text{HC}_2\text{H}_3\text{O}_2$):
$$0.010\text{ mol} - 0.00050\text{ mol} = \mathbf{0.0095\text{ mol}}$$
* Final moles of conjugate base ($\text{C}_2\text{H}_3\text{O}_2^-$):
$$0.010\text{ mol} + 0.00050\text{ mol} = \mathbf{0.0105\text{ mol}}$$
* Because both species are in the same total volume, the species with more moles has the higher concentration. Therefore, $\text{C}_2\text{H}_3\text{O}_2^-$ is the acetate species present in greater concentration (this narrows our choices down to C or D).
5. Determine the final pH using the Henderson^-Hasselbalch equation:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]}\right)$$
Since the concentration of the conjugate base is greater than the concentration of the weak acid, the ratio $\frac{[\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} > 1$. The logarithm of any number greater than 1 is positive, meaning:
$$\text{pH} = 4.7 + (\text{positive number}) \implies \mathbf{\text{pH} > 4.7}$$
*(Alternatively, adding any amount of base to a buffer system must cause its pH to increase from its starting value of 4.7).* This identifies Option D as the correct pairing.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This option claims that $\text{HC}_2\text{H}_3\text{O}_2$ remains in greater concentration and that the $\text{pH} < 4.7$. Adding $\text{NaOH}$ consumes $\text{HC}_2\text{H}_3\text{O}_2$ and produces $\text{C}_2\text{H}_3\text{O}_2^-$, meaning the acid's concentration must decrease below that of its conjugate base. Additionally, adding base to a buffer causes the pH to rise, making it greater than 4.7, not less.
• Option B is incorrect: While this option correctly identifies that the pH must rise above 4.7 due to the addition of a base, it incorrectly identifies $\text{HC}_2\text{H}_3\text{O}_2$ as the species in greater concentration.
• Option C is incorrect: Although this option correctly identifies $\text{C}_2\text{H}_3\text{O}_2^-$ as the species in greater concentration, it incorrectly states that the $\text{pH} < 4.7$. Adding base shifts the buffer ratio towards the conjugate base, which mathematically forces the pH to rise above the $\text{p}K_a$ of 4.7.