How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

The given concentration of the lactic acid solution is \(0.1000 \mathrm{M}\). So, Initial concentration of lactic acid, \([\mathrm{HA}]_{\mathrm{initial}} =0.1000 \mathrm{M}\)

We are given that the \(\mathrm{pH}=3.75\). Now, we use the Henderson-Hasselbalch equation: We have, \(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-_{\mathrm{eq}}]}{[\mathrm{HA}_{\mathrm{eq}}]}\) Rearranging for \([\mathrm{A}^-_{\mathrm{eq}}]\), $ [\mathrm{A}^-_{\mathrm{eq}}]= [\mathrm{HA}_{\mathrm{eq}}] \cdot 10^{\left(\mathrm{pH}-\mathrm{p}K_\mathrm{a}\right)} $ Now, we can calculate the moles of the conjugate base \(\mathrm{A}^-\) formed: moles of \(\mathrm{A}^-\) formed \(= \mathrm{moles} \;[\mathrm{HA}]_{\mathrm{initial}} - \mathrm{moles}\; [\mathrm{HA}]_{\mathrm{eq}}\) The moles of \(\mathrm{OH}^-\) ions required \(= \mathrm{moles} \; \mathrm{A}^-\).

Using the concentration of \(\mathrm{NaOH}\), we can calculate the volume of the \(\mathrm{NaOH}\) solution required to form the desired buffer solution: Volume of \(\mathrm{NaOH}\) solution \(= \dfrac{\mathrm{moles} \; \mathrm{OH}^-}{[\mathrm{OH}^-]}\) Now, we plug in the values to find the volume of the \(\mathrm{NaOH}\) solution. First, let's find the moles of \(\mathrm{HA}_{\mathrm{eq}}}\): \([\mathrm{HA}_{\mathrm{eq}}] = \left( 0.1000 - \rm{moles \; {A}^-}\right)\phantom{.}\mathrm{M}\) Now, we can use the Henderson-Hasselbalch equation to find \(\mathrm{moles} \; \mathrm{A}^-\): $ [\mathrm{A}^-_{\mathrm{eq}}]= [\mathrm{HA}_{\mathrm{eq}}] \cdot 10^{\left(\mathrm{pH}-\mathrm{p}K_\mathrm{a}\right)} $ \(\Longrightarrow \mathrm{moles} \; \mathrm{A}^- = \frac{0.1000 - \mathrm{moles} \; \mathrm{A}^-}{\mathrm{L}} \cdot 10^{\left(3.75-\log 1.38 \cdot 10^{-4}\right)}\) Solving the equation, we get: \(\mathrm{moles} \; \mathrm{A}^- = 6.989 \cdot 10^{-4}\;\mathrm{mol}\) Finally, we calculate the volume of \(\mathrm{NaOH}\) solution: Volume of \(\mathrm{NaOH}\) solution \(= \dfrac{6.989 \cdot 10^{-4}\;\mathrm{mol}}{1.000 \phantom{.}\mathrm{M}}\) \(\Longrightarrow V_\mathrm{NaOH} = 6.99\;\mathrm{µL}\) Therefore, \(6.99\;\mathrm{µL}\) of \(1.000\;\mathrm{M}\;\mathrm{NaOH}\) solution must be added to the \(25.00\;\mathrm{mL}\) of the \(0.1000\;\mathrm{M}\) lactic acid solution to produce a buffer with \(\mathrm{pH}=3.75\).