A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.

To determine the concentration of the acetic acid (CH₃COOH) solution, we'll use its pH value: \(pH = 2.68\). We know that: \[pH = -\log_{10}[H^+]\] Now, we'll solve for the concentration of hydrogen ions, [H⁺]: \[2.68 = -\log_{10}[H^+]\] \[10^{-2.68} = [H^+]\] Now, we find the molar concentration of hydrogen ions, [H⁺]: \[[H^+] \approx 2.09 \times 10^{-3} \mathrm{M}\] Since acetic acid is a monoprotic weak acid, we will assume that the concentration of acetic acid is equal to the concentration of hydrogen ions: \[[\mathrm{CH₃COOH}] \approx 2.09 \times 10^{-3} \mathrm{M}\]

We are to calculate the volume of 0.0975 M KOH solution that would neutralize 25.0 mL of the acetic acid solution. We'll use stoichiometry to determine this volume based on the neutralization reaction: \[\mathrm{CH₃COOH} + \mathrm{KOH} \rightarrow \mathrm{CH₃COOK} + \mathrm{H₂O}\] From the balanced equation, we see that 1 mole of CH₃COOH reacts with 1 mole of KOH. Let's now find the moles of CH₃COOH present in 25.0 mL of the acetic acid solution: moles of CH₃COOH = (Concentration of CH₃COOH) × (Volume of CH₃COOH) moles of CH₃COOH = \( (2.09 \times 10^{-3} \mathrm{M})(0.025 \mathrm{L})\) moles of CH₃COOH = \(5.23 \times 10^{-5} \mathrm{mol}\) Now, we'll find the volume of KOH solution required to neutralize the given amount of CH₃COOH: Volume of KOH (in L) = \(\frac{\text{moles of CH₃COOH}}{\text{Concentration of KOH}}\) Volume of KOH (in L) = \(\frac{5.23 \times 10^{-5}\mathrm{mol}}{0.0975 \mathrm{M}} \approx 5.36 \times 10^{-4} \mathrm{L}\) To convert this volume to milliliters, we'll multiply by 1,000: Volume of KOH (in mL) \(= 5.36 \times 10^{-4} \mathrm{L} \times 1000\mathrm{~\frac{mL}{L}} \approx 0.536 \mathrm{mL}\)

The volume of 0.0975 M KOH solution required to reach the equivalence point in the titration of 25.0 mL of the acetic acid solution with a pH of 2.68 is approximately 0.536 mL.