Equilibrium Lab Report

Justin Jung 3/9/10 Determination of an Equilibrium Constant Dr. Butler, Section 913 Partners: Alison Atkins Objectives This lab was conducted to determine the equilibrium constant of a reaction. We used the following chemical equation: Ethyl acetate (EtAc) + Water 

        Procedures The following materials were used: 50mL buret, 5mL Mohr pipets, 125 or 250mL Erlenmeyers flasks with stoppers, 50mL graduated cylinder, 3M HCl, ethyl acetate, distilled water, and standard 1.000M NaOH. This was a two-week lab. In the first week, three solutions with different amounts of ethyl acetate and distilled water were prepared. In the second week, the 3M HCl solution was titrated with the standardized NaOH solution to determine the moles of HCl and water in the solution. The three solutions that were prepared in the first week were titrated with the NaOH solution as well. Data A. Titration of HCl Trial 1

Trial 2

Concentration of NaOH (M) 1.043 Initial volume of NaOH (mL) 0.1 Final volume of NaOH (mL) 15.2 Volume of NaOH used (mL) 15.1 Moles of NaOH (mol) 0.0157 Moles of HCl (mol) 0.0157 Average of HCl (mol) 0.0158 Sample Calculations: Trial 1: (1.043M NaOH)(15.1mL NaOH) = 0.0157mol NaOH = 0.0157mol HCl B. Titration of Equilibrium Solutions Solution A Concentration of NaOH (M) Initial volume of NaOH 0

15.2 30.45 15.25 0.01591 0.01591

Solution B 1.043

Solution C

2.5

22.2

(mL) Final Volume of NaOH 20.5 22.2 (mL) Volume of NaOH used 20.5 19.7 (mL) Moles of NaOH (mol) 0.0214 0.0205 Moles of acid (mol) 0.0214 0.0205 Sample Calculations: Solution A: (1.043M NaOH)(20.5mL NaOH) = 0.0214mol NaOH = 0.0214mol acid C. Equilibrium Constant Calculations Solution A Initial moles EtAc (mol) 0.0508 Initial moles H2O (mol) 0.259 Moles of acid (mol) 0.0214 Moles of HCl (mol) 0.0158 Moles of HAc (equil.) 0.0056 (mol) Moles of EtOH (equil.) 0.0056 (mol) Moles of EtAc (equil.) 0.0452 (mol) Moles of H2O (equil.) 0.253 (mol) KC 0.0027 Average KC Standard deviation Sample Calculations: Solution B: Initial moles EtAc: 4.00mL x 0.893g/mL = 3.57g 3.57g x 1mol/88.0g = 0.0406mol EtAc Initial moles H2O: 1.00mL x 1.00g/mL = 1.00g distilled water 1.00g x 1mol/18.0g = 0.0556mol distilled water 5.00mL HCl soln x 1.05g/mL HCl soln = 5.25g HCl solution 0.0158mol x 36.5g/mol = 0.577g HCl in HCl solution 5.25g  0.577g = 4.67g water 4.67g x 1mol/18.0g = 0.259mol water in HCl solution 0.0556mol + 0.259mol = 0.315mol H2O Moles HAc at equilibrium: 0.0205mol (Table B)  0.0158 (Table A) = 0.0047mol HAc

38.8 16.6 0.0173 0.0173

Solution B 0.0406 0.315 0.0205 0.0158 0.0047

Solution C 0.0305 0.370 0.0173 0.0158 0.0015

0.0047

0.0015

0.0359

0.0290

0.310

0.369

0.0020 0.0016 0.0013

0.00021

Moles EtOH at equilibrium: 0.00 + 0.0047mol = 0.0047mol EtOH Moles EtAc at equilibrium: 0.0406mol  0.0047mol = 0.0359mol EtAc Moles H2O at equilibrium: 0.315mol  0.0047mol = 0.310mol H2O KC = (0.0047mol HAc)(0.0047mol EtOH) / (0.0359mol EtAc)(0.310mol H 2O) = 0.0020 Results According to the equilibrium calculations for the three solutions, as the volume of e thyl acetate decreased and the volume of water increased, the equilibrium constants decreased. The average and standard deviation of the three equilibrium constants was affected greatly by the e quilibrium constant for Solution C, which was significantly small compared to the other two constants. The volume of NaOH that was used to titrate Solution C was 16.6; ethyl alcohol and acetic acid also had significantly le ss moles in Solution C. Although it is certain that the Solu tion C turned pink at that volume, the e quilibrium constant could have come out low because either the NaOH solution or Solution C was contaminated.