Question

The purpose of the lab was to find the equilibrium constant of the rxn of acetic acid and ethanol to yield ethyl acetate and water. Hydrochloric acid is used as a catalyst. We have to determine the equilibrium concentrations of all the species (water included) present at equilibrium. In test tubes, different concentrations of the equilibrium system were prepared then titrated with NaOH to determine the total acid content of the system at equilibrium. From this you will be able to determine the concentration of all species at equilibrium and the equilibrium constant.

To determine the total concentration of acid at equilibrium, which is equal to the concentration of HCl and acetic acid. First of all you need to determine the concentration of the catalyst HCl, that is present at equilibrium. The first table is data from the titration of 5.00mL and 6 M HCl by NaOH. I'm lost on how to find the number of moles of HCl. I used the C1*V1=C2*V2, to find the molarity of NaOH, and dimensional analysis to find the moles of HCl but I don't think I did it right.

The second table is the titration of a soultion by NaOH. The contents of the solution were 5.00mL 6M HCl, 2.00mL ethanol, and 3.00mL of glacical acetic acid. I'm lost on how to find the molarity of NaOH, the moles of NaOH, the moles of acid at equivalence point, and average moles of acetic acid at equilibrium.

Please help me with this. :)

Initial burette reading final huvette reading vol of NaOH used Imolarity of Naot moles of NaOH added moles of Hel at Tequivalence point average moles of HCI Run / Run 2 20.20 ml 1 33.10ml 133.10mL 46.04 mL 12.90mL L 12.94 mL 2.33M 2,32M 0.489 mol 0.490 mol The avg. # of moles of Hel obtained above should be subtracted from the total of moles of acid that you will calculate in bottles A,B,C 1 Run 0.86 mb 31.14 m2 30.28 ML Run 2 0.88 mL 34.64 mL 33.76 mL linitial burrete reading_ final burrete reading te volume of Naot used Molarity of NaOH soc moles of Nach added. Moles of acid at I equivalence point. ed average maies of acid lang mol acetic acid Lat equilibrium

Answer 1

Solution: modes fnce added Hut Naon. Nach + H2O Miles of fill = miles of Naon = (yo's ix Volkyn- 2.33 mx 12.9 m2 .. = 30.057 mmolo (Run-1) = . .= 2.33 X 12.54. ml (Run-2) 300150 mmal. average molds of 14 = 30.057 +30.150 2 = 300104 ommal 0.0 3010y mo. Determination of Equilibrium constant . Mg-8-06995 +420 chjioon. + Gingon - Mess K = [chécouns] [iro] (cascoon] [C, Mson] initeil Ethand = mil acetic acid = GM na = 3. mL 5ml fcs Scanned with CamScanner:

LLLLLLLL AURI W AWILI - for RUN-1 modurity of Naon = Volum of Naon used = 31.14 - 0186 = 30.28 mil! males of Naon added = 30.29X 2.33m = 70.5524 in modo moles of acetic acid = 0.07055 mol = molesch Noon- meluanel = 70.5524-30.104) in mal = ug.4484 mmodo 0.04044 moi. for RUN- males of Naon added = 2.33m x 33.76 me = 78.6608 mmol males of auchcauid = moles of Naon - males of ni = (70.6608 - 30.104) = 48.5568 m malo = 0.04855 mol average hole facelicaed - 40.4494 + 48,5568 = 44.5026 nmad = 0.0445 Meli Madering of Nach = 2.33m malley of Naon added =0.070 And 0.07866 euro • moles of and at Equivalence point = moles of Naon = 0.0705 and 0.07866 areyage males o acid = 0.0705 +0.07866 = 0.07458 muf Average mod acetic acid = 0.0445 molo Run-2 2