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Determination of the Dissociation Constant of a Weak Acid; K . Written by: Gianni Johnson , Chiara Palacios , Maya Golic , Jazmin Perez Experimental Team includes: Gianni Johnson , Chiara Palacios, Maya Golic , Jazmin Perez Experiment performed on: 10/19/2023

Purpose The purpose behind doing this experiment was to perform a titration experiment using the standardized NaOH solution we made in a previous lab. With the NaOH solution, we titrated it into an unknown acid to assist us in finding its dissociation constant and then identify it with that information. Another objective was to find the molar concentration of the weak acid. The learning goals that were associated with this experiment were to use the pH meter, and be able to calibrate it, to record the pH of the unknown acid during titration, and another goal was to quantitatively determine the equilibrium constants. To achieve these goals we titrated the NaOH solution into the unknown acid over a course of 4 different trials and, using a labquest, we recorded the pH of the acid as we titrated the NaOH. Then, using the data we got from the four trials we found the half-equivalence point from the equivalence point. The equivalence point represents the pKa and from that value we can find the Ka. Every acid has a Ka that is unique to that acid, so that value will help us identify which acid we are dealing with as well as its concentration. pPK_=— log[K ] Methods Throughout this whole experiment, we are finding the dissociation constant of an unknown weak acid and finding the molar concentration of the weak acid. Four trials were needed in order to perform the experiment. Our solution of unknown acid was prepared by adding 20.00 mL of an unknown acid to a 250 mL beaker. We were instructed to set up a 50 mL burette and fill it up to 0 mL with the standardized NaOH solution that we made in a previous lab. We then had to set up the Labquest and pH meter. We calibrated the pH meter for stability. We used two different solutions set up to

two different pH’s, 4.0 and 7.0, and used them to calibrate the pH indicators. After we calibrated, we then started to titrate the standardized NaOH into the unknown acid. In the first titration we kept adding 1.0 mL of our NaOH solution. Each time we keep adding NaOH to the unknown acid, the pH went up and was recorded in the Labquest. As we kept adding we swirled it and pressed “keep” in the LabQuest to record the pH. We stopped titrating until the pH reached 10 and the graph curved and leveled out. For the second titration we started out by adding 1.0 mL of the NaOH solution. We started adding 0.5 mL around the ranges of the half equivalence point and then again around the ranges of the equivalence point. Using our data from the four trials we identified our half-equivalence and equivalence point to find the values of the pKa and the Ka. Data Figure 1: Titration curve for trial #1 Unknown Acid Titration Trial #1

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Figure 2: Plot of the first derivative of the titration curve for trial #1. Derivative of Unknown Acid Titration Trial #1 25 Figure 3: Titration curve for trial #2 Unknown Acid Titration Trial #2

Figure 4: Plot of the first derivative of the titration curve for trial #2 Derivative of Unknown Acid Titration Trial #2 2i5 Figure 5: Titration curve for trial #3 Unknown Acid Titration Trial #3

Figure 6: Plot of the first derivative of the titration curve for trial #3 Derivative of Unknown Acid Titration Trial #3 25 Figure 7: Titration curve for trial #4 Unknown Acid Titration Trial #4

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Figure 8: Plot of the first derivative of the titration curve for trial #4 Derivative of Unknown Acid Titration Trial #4 Table 1: Determination of the acid dissociation constant Calculations 25 Trial #1 Trial #2 Trial #3 Trial #4 - Volume of NaOH at 16.66 17.47 18.5 18.75 equivalence point (mL) Volume of NaOH at 2 7.78 8.735 9.25 9.375 equivalence point (mL) pH at 2 equivalence point 3.14 3.09 3.06 3.06 pK, 3.14 3.09 3.06 3.06

K, 7.24 x 10 *|8.13 x 10*|8.71 x 10™* 8.71 x 10 Table 2: Determination of the concentration of the unknown acid [ Trial #1 Trial #2 Trial #3 Trial #4 = Volume of unknown acid 20 20 20 20 Molarity of Standardized 0.0935 0.0935 0.0935 0.0935 NaOH Volume of NaOH at 16.66 17.47 18.5 18.75 equivalence point Molarity of unknown acid 0.0778 0.0817 0.0865 0.0877 B Average | 0.0834 Molarity (M) L | Sample calculations: 1. 10075 — kq (-3.09) _ 10 Ka 8.13 x 10 M1V1 - MZVZ 0.0935 x 17.47 =M, X 20

M2 = 0.0817 M Questions: (show all work for computational questions for full credit!) 1. From your list of common weak acids, which weak acid do you think the unknown was? We believe that the weak acid used was hydrofluoric acid. Hydrofluoric Acid has a Ka value of 6.8 X 10 . The average Ka that we got was 8.19 x 10 . HF is closest in Ka value to our average. 2. Identify 2-3 sources of possible error in this experiment. Some of the possible sources of errors in this experiment could have been over-pouring the correct amount of NaOH stored from the very first experiment, titrating it into the unknown solution. The correct amount varied from 1.0mL to 0.5mL during the titration process. This could have increased our pH value causing an incorrect graph with the points showing a sudden peak too soon before we reach the certain amount of volume needed or pH. Another source of error could have been mixing up the concentrations of the calibrating liquid to calibrate our data to the corresponding solution.

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. What would be the pH of a solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00mL of 0.125 M NaOH? Assume additive volumes. Number of moles in HCI = Molarity X Volume in liter = 0.250M x 0.030L = 0.0075 moles Numbers of moles in NaOH 0.125M x 0.030L = 0.00375 moles Molesof H' = Moles of HCL — Molesof NaOH = 0.0075 moles — 0.030 moles = (0.00375 moles Total volume 30.00mL + 30.00mL = 60mlL = 0.060L +, __ Moles [H ] ~ Volume + 0.00375Moles [H ] - 0.060L [H'] =0.0625 10

pH = — log[H'] pH = — log[0.0625] pH = 1.2 . What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid? Use a K, for acetic acid of 1.85x107. [Acid] = 0.50M [Conjugate base] = 0.75M Ka =1.85 x10 " pKa =— logKa pKa =— log(1.85 x 10) pKa = 4.73 _ conjugate base pH = pKa + log( g ) 0.75 pH = 4.73 + log(4=, pH = 4.91 11

5. Calculate the pH of a solution prepared by mixing 15.0 mL of.10 M NaOH and 30.0 mL of 0.1 M benzoic acid (K,=6.5x10%). Initial Benzoic = 30.0mL X 0.1 = 3m Mol of Benzoic Acid Initial NaNOH = 15 x 0.1 = 1.5mMolof NaOH Benzoic reaction left: 3 —1.5=1.5m Mol Final Volume of the solution: 30mL + 15mL = 45mL Final Benzoic left in solution: 1.5/45 = 0.0333M pH = pKa + log([salt]/[acid]) =— 10g(6.5 x 10"°) + log(0.0333/0.033) pH = 4.19 12

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