Question

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.

Short answer

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.

Step-by-step solution

Calculate the concentration of the acetic acid solution from its pH

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: \[[\mathrm{CH₃COOH}] \approx 2.09 \times 10^{-3} \mathrm{M}\]

Use stoichiometry to determine the volume of KOH needed to neutralize the acetic acid

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) \(= 5.36 \times 10^{-4} \mathrm{L} \times 1000\mathrm{~\frac{mL}{L}} \approx 0.536 \mathrm{mL}\)

Report the required volume of KOH

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.

Key concepts

Acetic Acid

Acetic acid is a weak, monoprotic acid that is commonly found in vinegar. Its chemical formula is \( \text{CH}_3\text{COOH} \). Being a weak acid means it does not completely dissociate in water. Instead, it partially ionizes into acetate ions (\( \text{CH}_3\text{COO}^- \)) and hydrogen ions (\( \text{H}^+ \)). This partial dissociation is important when calculating concepts like pH or working with reactions such as titrations.
Acetic acid is a key player in various chemical calculations, particularly due to its weak nature. It is typically characterized by a pungent smell and sour taste and has various applications, from household vinegar to industrial processes.
Understanding acetic acid's properties and behavior is crucial in chemistry, as it helps us predict how it will react in different scenarios, especially in titration, where precise reaction control is necessary.

pH Calculation

The pH of a solution indicates its acidity or basicity, calculated using the concentration of hydrogen ions \( \left( [H^+] \right) \). For any solution, pH is determined by the formula:
\[ pH = -\log_{10}[H^+] \]
This logarithmic relationship means that even a small change in hydrogen ion concentration results in a significant pH change. Acetic acid, being a weak acid, partially dissociates in solution. So, to calculate its concentration, we first determine [H\(^+\)] from its pH value.
In the given exercise with acetic acid having a pH of 2.68, the concentration of \([H^+]\) was calculated using:
\[ [H^+] = 10^{-2.68} \approx 2.09 \times 10^{-3} \, \text{M} \]
This calculation is pivotal in deriving other metrics like the volume of titrant needed, as it directly translates into the amount of acetic acid present in the solution.
Through understanding pH calculations, we gain insights not just into the strength of an acid, but also into its concentration and potential chemical behavior in reactions.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It is essentially a concept of balancing equations to understand how much of each element is involved. For chemical titrations, stoichiometry is integral as it helps us figure out the precise amount of a titrant needed to reach an equivalence point.
In the context of our exercise, we have acetic acid reacting with potassium hydroxide:\[ \text{CH}_3\text{COOH} \,+\, \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O} \].
Here, one mole of acetic acid reacts with one mole of KOH. This one-to-one ratio simplifies our calculations:

• Calculate moles of acetic acid using its concentration and volume.
• Use the stoichiometric ratio to find moles of KOH needed, equivalent to moles of acetic acid.

This understanding allows chemists to control reaction extents and outcomes efficiently, pivotal in chemical preparation and industrial applications.

Equivalence Point

The equivalence point in titration marks where the amount of titrant added is stoichiometrically equal to the quantity of the substance being titrated. It implies complete neutralization of the analyte by the titrant.
In our case, reaching the equivalence point means the moles of KOH added are equal to the moles of acetic acid present initially. With acetic acid being a weak acid and KOH a strong base, the equivalence point does not imply a neutral pH of 7, but rather the completion of the reaction, after which both components are completely reacted.
A correctly calculated equivalence point ensures accurate titration results, providing insights into the concentration of unknown solutions. This ensures precision in preparing chemical solutions and studying reaction mechanisms.
Understanding the equivalence point is fundamental in titration exercises, helping achieve accurate experimental results while naturally extending into real-world chemical applications.