Lab report Essay
The Acid Dissociation Constant, KaName: Candice JosephDate: May 1, 20071. PurposeThe purpose of this experiment was to determine equilibrium constant Ka for dissociation of acetic acid in aqueous solution.
This will be done by measuring pH of aqueous solution of different concentrations of acetic acid. Besides, the class average value of Ka will be determined. Subsequently, pH of vinegar solution will be measured and used to determine concentration of acetic acid in vinegar.2.
IntroductionAcetic acid being a weak acid, dissociates only partially in aqueous solution as per the equationHC2H3O2(aq) H+(aq) + C2H3O2-(aq) ……(1)The equilibrium expression for this reaction will be Ka = …….
.(2)Here, the subscript eq stands for equilibrium.When acetic acid is added into distilled water equation (1) gives[H+(aq)]eq = [C2H3O2-(aq)]eq …….
. (3)Because, acetic acid is a weak acid, it dissociates into H+ and C2H3O2- to very small extent and therefore, [HC2H3O2(aq)]eq ~ [HC2H3O2(aq)]i ; here I stands for initial ………(4)Also, % ionization = x 100% ……… (5)As long as percentage ionization is less than 5%, the assumption in equation (4) is valid.
pH is a log scale representation of [H+(aq)] in an aqueous solution. It is defined aspH = -log[H+(aq)] ……….
. (6)From equation (6) follows, [H+(aq)] = 10-pH ……… (7)In this experiment, aqueous solution of different concentrations of acetic acid was prepared and pH of these solutions was measured. Then using equation (7), [H+(aq)] was calculated and using equations (3) and (2), Ka of acetic acid was calculated for different concentrations of acetic acid in aqueous solution. The average value of the entire class was taken to determine the average Ka value of acetic acid.
Finally, pH of a vinegar solution was measured and using the average Ka value obtained in this experiment was used to determine concentration of acetic acid in the vinegar solution.3. Materials:IBM compatible computer, Serial box interface, Logger pro, Vernier pH amplifier and pH electrode, 100 mL beaker, 2.0M HC2H3O2, distilled water, 100 mL volumetric flask, pipets, pipet bulb, thermometer.
4. Procedure:1. Lab coat, gloves and safety glasses were used as safety measure during the experiment as the experiment involved handling of hazards chemical.2.
Approximately 50 mL of distilled water was poured into 100 mL volumetric flask.3. Volume of 2.0M acetic acid needed for preparing 100 mL solution of acetic acid of the assigned concentrations 0.
2M, 0.3M, 0.4M and 0.5M was calculated.
4. 100 mL acetic acid solution of assigned concentrations was prepared using pipet, pipet bulb and 100 mL volumetric flask.5. The computer was prepared for recording pH of the acetic acid solution.
6. pH of the acetic acid solution was prepared by first rinsing the pH electrode in 30 mL of the assigned solution, throwing away the solution and measuring the pH in a fresh solution of the assigned concentration. Temperature of the solution was also measured.7.
pH of the vinegar was also measured in similar manner.5. Results and Analysis:Measured value of pH of 2.0M acetic acid solution and calculated value of Ka of 2.
0M acetic acid solution is presented in table below. The calculation is presented in Appendix.Table 1: Measured value of pH of 2.0M acetic acid solution and calculated Ka valueAssigned Concentration2.
0MVolume of 2.0m acetic acidNot ApplicableMeasured pH2.35.01×10-3M5.
01×10-3M2.0MKa1.26×10-5Temperature25oC% ionization0.25%Is assumption Valid?YesThe value of measured pH and calculated and Ka values are presented in table 2, below.
The sample calculation is presented in Appendix. 1. Table 2: Measured pH and calculated and Ka values for different concentrations of acetic acid solutionAcetic Acid Concentration (M)Measured pH(M)Calculated Ka0.22.
From table 2, it can be seen that as the concentration of acetic acid solution increasesa) pH Decreasesb) Increasesc) Ka Remains same within experimental error3. Measured pH of vinegar 2.34. Molarity of acetic acid in vinegar solution = 2.
81M (calculation is shown below).From experiments and calculations we have pH = 2.3 and Ka = 8.92×10-6From equation (7) [H+(aq)] = 10-pH = 10-2.
3 = 5.01×10-3From equation (3) [C2H3O2-(aq)]= [H+(aq)] = 5.01×10-3From equation (2) Ka = Appendix:Calculation of Ka from measured pH for a given concentration1. Acetic acid concentration 2.0 M Measured pH = 2.3From equation (7) [H+(aq)] = 10-pH = 10-2.3 = 5.01×10-3From equation (3) [C2H3O2-(aq)]= [H+(aq)] = 5.01×10-3From equation (2) Ka =Similar calculations were made for other concentrations.
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Lab report Essay. (2017, Mar 16). Retrieved from https://graduateway.com/lab-report-10/