A 0.35 g sample of Li(s) is placed in an Erlenmeyer flask containing 100 mL of water at 25°C. A ball — Acids and Bases Chemistry Question

STEPS:

1. Identify the type of titration: The curve represents the titration of a weak monoprotic acid with a strong base (lithium hydroxide, $\text{LiOH}$, produced from the reaction of $\text{Li}$ with water). This is indicated by the moderately acidic starting pH ($\approx 3.4$), the flat buffering region, and a basic pH at the equivalence point.
2. Locate the equivalence point on the pH curve: The equivalence point is represented by the midpoint of the sharp, nearly vertical region of the titration curve. Looking at the horizontal axis, this steep rise is centered at exactly $25.0\text{ mL}$ of base added.
3. Determine the volume at the half-equivalence point: The half-equivalence point (or midpoint of the buffer region) is the point at which exactly half of the volume of strong base required to reach the equivalence point has been added:
$$V_{\text{half}} = \frac{25.0\text{ mL}}{2} = \mathbf{12.5\text{ mL}}$$
4. Relate the half-equivalence point to the $\text{p}K_a$: At this exact midpoint, exactly half of the initial weak acid ($\text{HA}$) has been neutralized to form its conjugate base ($\text{A}^-$), meaning their concentrations are equal:
$$[\text{HA}] = [\text{A}^-]$$
Applying the Henderson^-Hasselbalch equation:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = \text{p}K_a + \log(1) \implies \mathbf{\text{pH} = \text{p}K_a}$$
Thus, at the half-equivalence point, the measured pH of the solution is equal to the $\text{p}K_a$ of the weak acid.
5. Estimate the pH value from the graph: Find $12.5\text{ mL}$ on the horizontal axis (just to the right of the $10.0\text{ mL}$ line) and trace it vertically to the curve. The corresponding pH on the vertical axis is approximately $4.7$. Among the given choices, this value is closest to $5$, confirming Option B as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect (4): A pH of 4 corresponds to a volume of only about $5\text{ mL}$ of base added. At this early stage of the titration, the weak acid has barely been neutralized, so $[\text{HA}] > [\text{A}^-]$ and the $\text{pH}$ is still significantly below the $\text{p}K_a$ value.
• Option C is incorrect (8): A pH of 8 represents the pH at the equivalence point of this titration (where the curve rises sharply at $25.0\text{ mL}$). At the equivalence point, all of the weak acid has been converted into its conjugate base, making the solution basic due to the hydrolysis of the conjugate base, which does not equal the $\text{p}K_a$.
• Option D is incorrect (12): A pH of 12 is only reached far past the equivalence point (after adding about $28\text{ mL}$ or more of base). At this point, the pH of the mixture is dominated entirely by the excess unreacted strong hydroxide base in the solution.