Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100\(M\) lactic actid $\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right.\( \)\mathrm{p} K_{\mathrm{a}}=3.86 )$ is titrated with 0.100 \(\mathrm{M}\) NaOH solution. Calculate the pH after the addition of $0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL},\( \)20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\( \)26.0 \mathrm{mL}, 28.0 \mathrm{mL}$ , and 30.0 \(\mathrm{mL}\) of the NaOH. Plot the results of your calculations as \(\mathrm{pH}\) versus milliliters of NaOH added.

Calculate the initial moles of lactic acid in the \(25.0 \,\mathrm{mL}\) solution by using the given information that it has \(0.100 \,\mathrm{M}\) concentration: Moles of lactic acid (acid) = Initial Volume of the solution (in liters) × Initial Concentration of lactic acid (in \(\mathrm{M}\)) Moles of lactic acid (acid) = \(0.025 \times 0.100\) Moles of lactic acid (acid) = \(0.00250 \,\mathrm{mol}\)

Recall that the NaOH concentration is \(0.100\,\mathrm{M}\). Moles of NaOH will be calculated as follows: Moles \(\mathrm{NaOH}\) = Volume of NaOH added (in liters) × Concentration of NaOH (in \(\mathrm{M}\)) Calculate the moles of NaOH for each given point: \(0.0\,\mathrm{mL}, 4.0\,\mathrm{mL}, 8.0\,\mathrm{mL}, 12.5\,\mathrm{mL}, 20.0\,\mathrm{mL}, 24.0\,\mathrm{mL}, 24.5\,\mathrm{mL}, 24.9\,\mathrm{mL}, 25.0\,\mathrm{mL}, 25.1\,\mathrm{mL}, 26.0\,\mathrm{mL}, 28.0\,\mathrm{mL}\), and \(30.0\,\mathrm{mL}\).

For each point in the titration, determine whether it occurs before the equivalence point, at the equivalence point, or after the equivalence point. - If before the equivalence point (i.e., moles of \(\mathrm{NaOH}\) < moles of lactic acid), we have to calculate the concentration of lactic acid (acid) and use its \(\mathrm{p} K_{\mathrm{a}}\) value to find the \(\mathrm{pH}\). - At the equivalence point (i.e., moles of \(\mathrm{NaOH}\) = moles of lactic acid), the \(\mathrm{pH}\) is determined by the concentration of the produced salt. - After the equivalence point (i.e., moles of \(\mathrm{NaOH}\) > moles of lactic acid), we have to calculate the concentration of the excess \(\mathrm{OH^-}\) ions and use it to find the \(\mathrm{pH}\). Calculate the pH accordingly for each given point.

After calculating the pH at each given point, create a plot with milliliters of NaOH added (x-axis) against pH (y-axis). This will result in a titration curve that shows how the pH changes as more \(\mathrm{0.100\,\mathrm{M}}\) NaOH is added to the lactic acid solution.