40.0 ml of an acetic acid solution 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 equation for the reaction between acetic acid and sodium hydroxide is:

CH3COOH + NaOH → CH3COONa + H2O

Since the reaction is 1:1, we can use the following equation to calculate the concentration of the acetic acid solution:

MaVa = MbVb

where Ma is the concentration of the acetic acid solution, Va is the volume of the acetic acid solution used, Mb is the concentration of the NaOH solution, and Vb is the volume of the NaOH solution used.

At the halfway point of the titration, we know that half of the acetic acid has been neutralized:

MaVa = MbVb/2

We also know that at this point, the pH of the solution is equal to the pKa of acetic acid, which is 4.76. The Henderson-Hasselbalch equation can be used to calculate the ratio of the concentrations of the conjugate base (acetate) to the acid (acetic acid):

pH = pKa + log([acetate]/[acetic acid])

5.10 = 4.76 + log([acetate]/[acetic acid])

0.34 = log([acetate]/[acetic acid])

Antilog(0.34) = [acetate]/[acetic acid]

2.15 = [acetate]/[acetic acid]

Since the concentration of the acetate ion is equal to the concentration of NaOH that has been added, we can calculate the concentration of NaOH at the halfway point:

0.100 M NaOH x 20.0 mL = 0.00200 moles NaOH

Since half of the acetic acid has been neutralized, the moles of acetic acid remaining in the solution is equal to the moles of NaOH added:

0.00200 moles NaOH = Ma x 20.0 mL/2

Ma = 0.072 M

400 Ml of an Acetic Acid Solution of Unknown Concentration