$$ \begin{array}{|c|c|} \hline
\text{Acid} & \text{Volume of} \\
\text{Solution} & \ce{NaOH} \text{ Added (mL)} \\ \hline \hline
\text{A} & 40 \\ \hline
\text{B} & 75 \\ \hline
\text{C} & 115 \\ \hline
\text{D} & 200 \\ \hline
\end{array}
$$
To maximize the yield in a certain manufacturing process, a solution of a weak monoprotic acid that has a concentration
between 0.20 \(M\) and 0.30 \(M\) is required. Four 100. mL samples of the acid at different concentrations are each
titrated with a 0.20 \(M \ce{NaOH}\) solution. The volume of \(\ce{NaOH}\) needed to reach the end point for each sample is given
in the table above. Which solution is the most suitable to maximize the yield?
To reach end point, the number of moles of hydronium ions and hydroxide ions are equivalent.
We first calculate the range of hydronium ions available.
Since the acid is monoprotic, the moles of acid is equal to the moles of hydronium:
Lower value
$$ 0.20 M \text{ Acid} = \frac{\pu{0.20 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.20 mol}}{\pu{1 L}}\times \pu{0.100 L} = \pu{0.02 mol} $$
Upper value
$$ 0.30 M \text{ Acid} = \frac{\pu{0.30 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.30 mol}}{\pu{1 L}}\times \pu{0.100 L} = \pu{0.03 mol} $$
We can do a similar calculation for the base to see how many hydoxide ions are present.
Solution A
$$ 0.20 M \text{ Hydroxide} = \frac{\pu{0.20 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.20 mol}}{\pu{1 L}}\times \pu{0.040 L} = \pu{0.008 mol} $$
Solution C
$$ 0.20 M \text{ Hydroxide} = \frac{\pu{0.20 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.20 mol}}{\pu{1 L}}\times \pu{0.115 L} = \pu{0.023 mol} $$
Solution B
$$ 0.20 M \text{ Hydroxide} = \frac{\pu{0.20 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.20 mol}}{\pu{1 L}}\times \pu{0.075 L} = \pu{0.015 mol} $$
Solution D
$$ 0.20 M \text{ Hydroxide} = \frac{\pu{0.20 mol}}{\pu{1 L}} $$
$$ \frac{\pu{0.20 mol}}{\pu{1 L}}\times \pu{0.200 L} = \pu{0.040 mol} $$
Solution C provides an end point in which the number of moles of hydroxide ions is equal to an acceptable number of moles of hydronium ions.