HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2-(aq) 1. The dissociation of ethanoic acid, HC2H3O2(aq), is — Equilibrium Chemistry Question
Problem Context
HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2-(aq)
- The dissociation of ethanoic acid, HC2H3O2(aq), is represented above. A student is given the task of determining the value of Ka for HC2H3O2(aq) using two different experimental procedures.
In a separate experimental procedure, the student titrates 10.0 mL of the 2.000 M HC2H3O2(aq) with an NaOH(aq) solution of unknown concentration. The student monitors the pH during the titration. The following titration curve was created using the experimental data presented in the table.
[VISUAL]
Model Answer
M1V1 = M2V2
Vi = (0.115 M)(100.0 mL) / 2.000 M = 5.75 mL
1 point is earned for the correct volume. || Use the buret to deliver 5.75 mL of 2.000 M HC2H3O2 to the 100 mL volumetric flask. Then add distilled water from the wash bottle to the flask (adding the last few drops with an eyedropper) until the volume of liquid in the flask is at the calibration mark.
1 point is earned for dispensing from the buret.
1 point is earned for diluting the solution to the calibration mark of the volumetric flask.
Model Answer
pH = 2.92 ⇒ [H3O+] = 10^-2.92 = 0.0012 M
Ka = [H3O+][C2H3O2-] / [HC2H3O2]
Since [H3O+] = [C2H3O2-], then
Ka = (0.0012)(0.0012) / (0.115 - 0.0012) = (0.0012)^2 / 0.114 = 1.3 x 10^-5
1 point is earned for correct conversion of pH to [H3O+].
1 point is earned for a value of Ka consistent with the student's value of [H3O+]. || Percent dissociation = [C2H3O2-]/[HC2H3O2]0 * 100 = 0.0012 / 0.115 * 100 = 1.0%
1 point is earned for the correct percent dissociation.
Model Answer
HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l)
1 point is earned for the correct equation.
Model Answer
From the pH curve, the equivalence point occurs at 14.0 mL.
10.0 mL * (2.000 mol HC2H3O2 / 1000 mL) = 0.0200 mol HC2H3O2(aq)
0.0200 mol HC2H3O2(aq) * (1 mol NaOH / 1 mol HC2H3O2) = 0.0200 mol NaOH
0.0200 mol NaOH / 0.0140 L solution = 1.43 M NaOH(aq)
1 point is earned for determining the moles of acid.
1 point is earned for determining the molar concentration of the base.
Model Answer
At the half-equivalence point (~7.0 mL) the pH of the solution is equal to the pKa of the acid. The antilog of the negative pH is equal to the value of Ka.
1 point is earned for a correct explanation (numerical explanation not required).