Data collected during the titration of a 20.0 mL sample of a 0.10 M solution of a monoprotic acid wi — Acids and Bases Chemistry Question

STEPS:

1. Identify the equivalence point of the titration: The equivalence point on a titration curve of a weak acid with a strong base is represented by the point of maximum slope (the center of the near-vertical region of the pH curve). On the provided graph, this sharp vertical inflection occurs at exactly 20.0 mL of added NaOH.
2. Calculate the concentration of the NaOH titrant: At the equivalence point of a titration involving a monoprotic acid, the moles of added base ($\text{NaOH}$) must be stoichiometrically equal to the initial moles of the acid in the sample:
$$\text{Moles of acid} = \text{Moles of NaOH}$$
Using the molarity and volume relationship ($M_A V_A = M_B V_B$):
$$(0.10\text{ M}) \times (20.0\text{ mL}) = M_{\text{NaOH}} \times (20.0\text{ mL})$$
Solving for the molarity of NaOH:
$$M_{\text{NaOH}} = \frac{0.10\text{ M} \times 20.0\text{ mL}}{20.0\text{ mL}} = \mathbf{0.10\text{ M}}$$
This narrows down the correct choice to either Option B or Option D.
3. Identify the half-equivalence point: The half-equivalence point occurs when exactly half of the volume of strong base required to reach the equivalence point has been added:
$$V_{\text{half-equivalence}} = \frac{V_{\text{equivalence}}}{2} = \frac{20.0\text{ mL}}{2} = \mathbf{10.0\text{ mL}}$$
4. Relate the half-equivalence point pH to the $p\text{K}_a$: At this halfway point, exactly half of the weak acid ($\text{HA}$) molecules have reacted with the added $\text{OH}^-$ to form the conjugate base ($\text{A}^-$), resulting in equal concentrations of the weak acid and its conjugate base remaining in solution ($[\text{HA}] = [\text{A}^-]$). Substituting this equality into the Henderson^-Hasselbalch equation:
$$\text{pH} = p\text{K}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$
$$\text{pH} = p\text{K}_a + \log(1) \implies \text{pH} = p\text{K}_a$$
5. Determine the approximate $p\text{K}_a$ value from the titration curve: Find $10.0\text{ mL}$ on the horizontal axis (Volume of NaOH) and read the corresponding pH value on the vertical axis. The curve passes slightly below the $5.0$ grid line, at a pH of approximately 4.7. Therefore:
$$p\text{K}_a \approx \mathbf{4.7}$$
Combining both pieces of information ($p\text{K}_a \approx 4.7$ and $[\text{NaOH}] = 0.10\text{ M}$), we confirm that Option B is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While the $p\text{K}_a$ of $4.7$ is read correctly, the concentration of $\text{NaOH}$ is calculated incorrectly as $0.050\text{ M}$. This error would occur if a student incorrectly assumed that only $10.0\text{ mL}$ (the half-equivalence volume) was required to reach the equivalence point, or if they made a dilution math error.
• Option C is incorrect: A $p\text{K}_a$ value of $9.3$ represents the basic pH at (or slightly past) the equivalence point, rather than the pH at the half-equivalence point where $\text{pH} = p\text{K}_a$. Additionally, the concentration of $\text{NaOH}$ is incorrectly calculated as $0.050\text{ M}$.
• Option D is incorrect: While the concentration of $\text{NaOH}$ of $0.10\text{ M}$ is calculated correctly, the $p\text{K}_a$ is incorrect. A value of $9.3$ represents the basic pH at the equivalence point, not the pH at the half-equivalence point where $\text{pH} = p\text{K}_a$.