[VISUAL] To maximize the yield in a certain manufacturing process, a solution of a weak monoprotic a — Acids and Bases Chemistry Question

STEPS:

1. Understand the chemistry of an acid-base titration: In a titration of a weak monoprotic acid ($\text{HA}$) with a strong base ($\text{NaOH}$), the equivalence point (or stoichiometric end point) is reached when the moles of base added are exactly equal to the moles of acid initially present in the sample.
2. Set up the stoichiometric relationship: Because the acid is monoprotic, it donates one proton ($\text{H}^+$) per molecule, reacting with hydroxide ions ($\text{OH}^-$) in a $1:1$ molar ratio:
$$M_{\text{acid}} \times V_{\text{acid}} = M_{\text{base}} \times V_{\text{base}}$$
3. Identify the values and constraints given in the problem:
* Volume of each acid sample titrated ($V_{\text{acid}}$) = $100.\text{ mL}$ (or $0.100\text{ L}$)
* Molarity of the sodium hydroxide titrant ($M_{\text{base}}$) = $0.20\text{ M}$
* Target acid concentration ($M_{\text{acid}}$) = between $0.20\text{ M}$ and $0.30\text{ M}$
4. Rearrange the equation to solve for the initial molarity of each acid sample:
$$M_{\text{acid}} = \frac{M_{\text{base}} \times V_{\text{base}}}{V_{\text{acid}}}$$
Substituting our known constants:
$$M_{\text{acid}} = \frac{0.20\text{ M} \times V_{\text{base}}}{100.\text{ mL}}$$
This simplifies to a direct relationship:
$$M_{\text{acid}} = 0.0020 \times V_{\text{base}}\text{ (where } V_{\text{base}} \text{ is in mL)}$$
5. Calculate the concentration of each acid solution using the experimental data table:
* Solution A ($V_{\text{base}} = 40\text{ mL}$):
$$M_{\text{acid}} = 0.0020 \times 40 = \mathbf{0.080\text{ M}}$$
* Solution B ($V_{\text{base}} = 75\text{ mL}$):
$$M_{\text{acid}} = 0.0020 \times 75 = \mathbf{0.15\text{ M}}$$
* Solution C ($V_{\text{base}} = 115\text{ mL}$):
$$M_{\text{acid}} = 0.0020 \times 115 = \mathbf{0.23\text{ M}}$$
* Solution D ($V_{\text{base}} = 200\text{ mL}$):
$$M_{\text{acid}} = 0.0020 \times 200 = \mathbf{0.40\text{ M}}$$
6. Compare the calculated concentrations to the required range: Only Solution C has an initial concentration ($0.23\text{ M}$) that falls securely within the required manufacturing range of $0.20\text{ M}$ to $0.30\text{ M}$. This identifies Option C as the most suitable choice to maximize the yield.

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WHY_OTHERS_WRONG:

• Option A is incorrect (Solution A): A titration volume of $40\text{ mL}$ indicates that the acid concentration is only $0.080\text{ M}$. This is too dilute and fails to meet the minimum required threshold of $0.20\text{ M}$.
• Option B is incorrect (Solution B): A titration volume of $75\text{ mL}$ indicates that the acid concentration is $0.15\text{ M}$. This concentration is also below the necessary range.
• Option D is incorrect (Solution D): A titration volume of $200\text{ mL}$ indicates that the acid concentration is $0.40\text{ M}$. This exceeds the maximum allowable concentration of $0.30\text{ M}$.