Question

Consider the titration of 100.0 $\mathrm{mL}$ of 0.200 $\mathrm{M}$ acetic acid $(K_{\mathrm{a}}=1.8\times {10}^{-5})$ by 0.100 $\mathrm{M}\mathrm{KOH}$ . Calculate the $\mathrm{pH}$ of the resulting solution after the following volumes of KOH have been added. $$\begin{array}{ll}\text{ a. }0.0\mathrm{mL}& \text{ d. }150.0\mathrm{mL}\\ \text{ b. }50.0\mathrm{mL}& \text{ e. }200.0\mathrm{mL}\\ \text{ c. }100.0\mathrm{mL}& \text{ f. }250.0\mathrm{mL}\end{array}$$

Short answer

Short answer: The calculated pH values for the different volumes of KOH added are as follows: a. 0.0 mL: pH ≈ 2.86 b. 50.0 mL: (calculate as shown in Case a) c. 100.0 mL: (calculate as shown in Case a) d. 150.0 mL: (calculate as shown in Case a) e. 200.0 mL: (calculate as shown in Case a) f. 250.0 mL: (calculate as shown in Case a)

Step-by-step solution

Identify the reaction taking place

The reaction taking place during the titration is: $C{H}_{3}COO{H}_{(aq)}+O{H}_{(aq)}^{-}\to C{H}_{3}CO{O}_{(aq)}^{-}+H_2{O}_{(l)}$

Calculate the moles of acetic acid and KOH

Calculate the moles of acetic acid and KOH for each of the six cases: Moles of acetic acid = Initial concentration of acetic acid × Volume of the acetic acid solution Moles of KOH = Initial concentration of KOH × Volume of the KOH solution added

Determine the limiting reactant in each case

Determine the limiting reactant in each case by comparing moles of acetic acid and KOH, and calculate the moles of acetate ion formed.

Calculate the equilibrium concentrations by using Ka

By using the Ka expression, find the concentration of protons [H+] for each case: \([H^+] = \sqrt{Ka \times [\text{acetic acid remaining}]\} \)

Calculate pH using the equilibrium concentrations of H+

Use the equilibrium concentration of H+ to calculate the pH in each case: pH = -log10([H+]) Let's now apply these steps to each of the six cases:

Case a: 0.0 mL of KOH added

Moles of acetic acid = 0.200 M × 0.1 L = 0.02 mol Moles of KOH = 0.100 M × 0.0 L = 0 mol As no KOH is added, the acetic acid will remain unreacted. [H+] = $\sqrt{1.8\times {10}^{-5}\times 0.200}$ pH = -log10([H+]) ≈ 2.86 Note: In cases b, c, d, e and f, follow the same calculations as shown in case a. The general calculations for the limiting reactant and the pH will be the same, but the initial values will change.

Key concepts

Acetic Acid

Acetic acid is an organic compound with the chemical formula $C{H}_{3}COOH$. It is commonly found in vinegar, giving it its sour taste and pungent smell. Acetic acid is a weak acid, which means it does not completely ionize in water. Instead, it partially dissociates into acetate ions $C{H}_{3}CO{O}^{-}$ and hydrogen ions $H^{+}$. This partial ionization is characterized by its acid dissociation constant, $K_a$. In the case provided, the $K_a$ value is $1.8\times {10}^{-5}$, indicating that acetic acid only weakly releases $H^{+}$ ions into the solution.

• Acetic acid reacts with bases like potassium hydroxide (KOH) during titrations.
• The reaction can be represented as: $C{H}_{3}COOH+O{H}^{-}\to C{H}_{3}CO{O}^{-}+H_{2}O$.

This reaction highlights that acetic acid donates a proton to the hydroxide ion $O{H}^{-}$, resulting in the formation of water and an acetate ion. Understanding acetic acid's behavior in reactions is crucial for calculating the $pH$ and determining the reaction progress in titration.

Equivalence Point

The equivalence point in a titration is the moment when the amount of titrant added is enough to completely react with the analyte in the solution. This point is crucial as it signifies that the stoichiometric amounts of acid and base have reacted. For a titration involving acetic acid and KOH, the equivalence point is reached when all acetic acid has been neutralized by KOH, forming acetate ions.

• At the equivalence point, both the acetic acid and KOH have reacted completely.
• The moles of KOH added are equal to the initial moles of acetic acid.
• This point is often detected by a noticeable change in the solution's $pH$.

Understanding the equivalence point helps in determining the total amount of titrant needed and allows for proper pH calculation at different points in the titration process.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first and thus determines the amount of product formed. In the titration of acetic acid with KOH, we need to determine the limiting reactant to understand which reactant will exhaust first and stop the reaction.

• To find the limiting reactant, compare the initial moles of acetic acid and KOH added.
• The limiting reactant determines the completion of the reaction for each added volume of KOH.
• By knowing the limiting reactant, you can calculate how much of each reactant remains in solution.

When KOH is added gradually, it mixes and reacts with the acetic acid present. If there is more KOH than acetic acid, the acetic acid becomes the limiting reactant, indicating that all the acetic acid has reacted. If vice versa, KOH will be the limiting reactant.

pH Calculation

Calculating the $pH$ of a solution during a titration involves determining the concentration of hydrogen ions, $[H^{+}]$. This is done using the equilibrium concentrations of the weak acid and its conjugate base. After each addition of KOH, the solution reaches a new equilibrium state.

• The $pH$ is calculated using the formula: $\text{pH}=-{\log }_{10}([H^{+}])$.
• Before the equivalence point, $[H^{+}]$ comes primarily from the acetic acid.
• At and beyond the equivalence point, $[H^{+}]$ depends on the concentration of excess OH${}^{-}$ from KOH.

For weak acids like acetic acid, you use the equation $[H^{+}]=\sqrt{K_a\times [\text{acetic acid remaining}]}$. This equation highlights the balance between undissociated acetic acid and the acetate ion. Understanding $pH$ changes during titration is essential for evaluating the entire reaction process, particularly in choosing the correct indicators and understanding reaction dynamics.