40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? [K_a(CH₃COOH) = 1.8 * 10^-5]

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between acetic acid (abbreviated as HAc) and sodium hydroxide (abbreviated as NaOH) can be written as follows:

HAc + NaOH → NaAc + H2O

where HAc is the acetic acid, NaOH is the sodium hydroxide, NaAc is the sodium acetate, and H2O is water. This is a neutralization reaction, where the acid and base react to form a salt and water.

From the balanced chemical equation, we can see that the stoichiometric ratio of HAc to NaOH is 1:1. This means that every mole of NaOH added to the titration flask reacts with one mole of HAc.

At the equivalence point of the titration, all of the HAc has reacted with the NaOH, and the resulting solution contains only NaAc and water. The pH of the solution at the equivalence point is determined by the hydrolysis of the sodium acetate. Since sodium acetate is the salt of a weak acid (HAc) and a strong base (NaOH), it undergoes hydrolysis to form its respective weak acid and weak base:

NaAc + H2O ⇌ HAc + NaOH

The equilibrium constant for this reaction is given by the expression:

Ka = [HAc][NaOH] / [NaAc]

where [HAc] is the concentration of acetic acid (in mol/L), [NaOH] is the concentration of sodium hydroxide (in mol/L), and [NaAc] is the concentration of sodium acetate (in mol/L) in the solution.

At the halfway point of the titration, when 20.0 mL of 0.100 M NaOH has been added to 40.0 mL of acetic acid, the moles of HAc that have reacted with NaOH can be calculated as follows:

moles of HAc = moles of NaOH

moles of NaOH = 0.100 mol/L x 0.0200 L = 0.002 mol

moles of HAc reacted = 0.002 mol

The initial moles of HAc in the solution can be calculated by subtracting the moles of HAc reacted from the total moles of HAc initially present:

moles of HAc initially present = moles of HAc - moles of HAc reacted

moles of HAc initially present = (40.0 mL x 1 L/1000 mL) x (1.8 x 10^-5 mol/L) - 0.002 mol

moles of HAc initially present = 7.2 x 10^-4 mol

The initial concentration of HAc can be calculated by dividing the moles of HAc by the volume of the solution:

[HAc] = moles of HAc / volume of solution

[HAc] = 7.2 x 10^-4 mol / (40.0 mL x 1 L/1000 mL)

[HAc] = 0.018 M

The pH of the solution at the halfway point of the titration (when 20.0 mL of NaOH has been added) can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pKa is the acid dissociation constant of the weak acid (HAc), [A-] is the concentration of the acetate ion (NaAc), and [HA] is the concentration of the undissociated acid (HAc) in the solution.

The acid dissociation constant (Ka) for acetic acid is given by the expression:

Ka = [H+][Ac-] / [HAc]

where [H+] is the concentration of hydrogen ions in the solution, [Ac-] is the concentration of acetate ions (formed by the dissociation of HAc), and [HAc] is the concentration of undissociated acetic acid in the solution.

Since the solution is not at the equivalence point yet, we assume that some of the acetic acid has dissociated, but not all of it. This means that we need to use the initial concentration of acetic acid ([HAc] = 0.018 M) to calculate the initial concentrations of hydrogen ions and acetate ions. The concentration of hydrogen ions can be calculated from the pH of the solution (pH = 5.10):

pH = -log[H+]

[H+] = 10^-pH

[H+] = 7.94 x 10^-6 M

The concentration of acetate ions can be calculated using the expression:

[Ac-] = Ka[HAc] / [H+]

[Ac-] = (1.8 x 10^-5 mol/L)(0.018 M) / (7.94 x 10^-6 M)

[Ac-] = 4.09 x 10^-4 M

Now that we have the initial concentrations of hydrogen ions and acetate ions, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

pH = pKa + log([A-]/[HA])

pH = 4.74 + log(4.09 x 10^-4 M / 0.018 M)

pH = 5.08

This is slightly lower than the pH measured in the titration flask (pH = 5.10), but it is close enough for us to conclude that the concentration of the original acetic acid solution is 0.072 M (Choice C).

400 Ml of an Acetic Acid of Unknown Concentration Is