The graph above shows the titration curve that resulted when a sample of 0.1 M monoprotic acid was t — Acids and Bases Chemistry Question

STEPS:

1. Identify the equivalence point from the graph: Locate the steep, near-vertical rise in the titration curve, which represents the rapid pH transition that occurs when the moles of added titrant ($\text{NaOH}$) stoichiometrically equal the moles of analyte (monoprotic acid). This sharp inflection point occurs at exactly $50\text{ mL}$ of $\text{NaOH}$ added.
2. Determine the volume at the half-equivalence point: For a weak acid titration, the half-equivalence point is reached when exactly half of the volume of strong base required to reach the equivalence point has been added:
$$V_{\text{half}} = \frac{V_{\text{equivalence}}}{2} = \frac{50\text{ mL}}{2} = 25\text{ mL}$$
3. Relate the half-equivalence point to $\text{p}K_a$: According to the Henderson^-Hasselbalch equation:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$
At the half-equivalence point, exactly half of the weak acid molecules ($\text{HA}$) have been neutralized into conjugate base ions ($\text{A}^-$), resulting in equal concentrations of the weak acid and its conjugate base ($[\text{HA}] = [\text{A}^-]$). Because the ratio is 1, the $\log(1)$ term becomes 0, and the equation simplifies to:
$$\text{pH} = \text{p}K_a$$
4. Read the corresponding pH value from the curve: Find $25\text{ mL}$ of $\text{NaOH}$ added on the horizontal (x) axis, trace vertically up to the curve, and then trace horizontally to the vertical (y) axis to find the pH. The curve sits directly at a pH of $4.0$ at this point, indicating that the $\text{p}K_a$ of the acid is closest to $4.0$, which matches Option B.

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WHY_OTHERS_WRONG:

• A is incorrect: A pH of 3.0 represents the pH early in the titration (at around $7\text{ mL}$ of added $\text{NaOH}$) before the buffer capacity is fully established and before reaching the half-equivalence point.
• C is incorrect: A pH of 6.0 represents the pH late in the buffer region (at around $47\text{ mL}$ of added $\text{NaOH}$), where the weak acid is nearly depleted and the solution is rapidly losing its buffering capacity.
• D is incorrect: A pH of 8.0 is the pH at the equivalence point ($50\text{ mL}$ of added $\text{NaOH}$). Because a weak acid is titrated with a strong base, the resulting conjugate base hydrolyzes with water to produce a basic equivalence point (pH > 7).
• E is incorrect: A pH of 12.0 represents the pH in the post-equivalence region (at around $55\text{ mL}$ of added $\text{NaOH}$), where the pH of the system is dominated by the excess unreacted strong base.