Pka Lab Report PDF

Preview

Summary

Lab Report...

Description

pKa & Ka Experiment #: 8 Chem 221 Lab: Section #: 1 Group #: 2.7 Date: 11/06/19

I certify that I have read and have approved the submission of this group lab report:

Print Team Member 1

Taylor Borowicz

Team Member 2

Caylie Graeber

Team Member 3

Nathan Manni

Team Member 4

Yuliet Monatukwa

Signature

Abstract: The purpose of this experiment was to determine the pK a and K a of a weak acid. This was accomplished by graphing a titration curve mL by mL as a mystery solute was titrated by a .1 M (experimentally 0.098 M) solution of NaOH. Through the derivatives of this curve, it was found that the pK a and K a of this weak acid were 4.62 and 2.399𝞊-5, respectively. By use of the molarity of NaOH and the chemical equation H X(aq ) + N aOH (aq ) → H 2 O + N aX (aq ) , along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Given the actual molar mass, 69.49 g/mol, it can be concluded that this weak acid is Hydroxylamine Hydrochloride, with molecular formula ClH4NO. Introduction: The objective of this experiment is to determine the molar mass, pK a , and K a of a weak acid based on a titration curve. An acid is a compound that can donate a proton, and the strength of an acid can be determined by the compound’s ability to donate a proton. 1 The strength of an acid is determined by the pH or negative log of the hydronium ion concentration. A  titration curve is a graphical representation of the pH of a solution during a titration. At the equivalence point of the titration, the moles of the weak acid and the base added are equal and it corresponds with a color change of the indicator. Furthermore, the equivalence point is the inflection point of the titration curve. At an inflection point, the second derivative changes from positive to negative and the first derivative has a maximum. At half of the equivalence point, the pH is equivalent to the pK a because the molar concentration of the resulting conjugate base is equal to the molar concentration of the acid are equal. K a is the acidity constant and equals the molar concentration of hydronium times the molar concentration of the conjugate base divided by the molar concentration of the weak acid. 2 Taking the negative antilog of the p K a to the tenth power, results in theK a of the unknown weak acid. Multiplying the volume of NaOH used to titrate to equivalence by the molarity of NaOH and the mole ratio of NaOH and the weak acid produces the moles of acid in solution. Dividing the mass of the acid by this number produces the molar mass. Experimental Method: The first step of this experiment was to conduct a test to calibrate the pH probe. This was done by placing the probe in buffer solutions of known pH. The probe was removed from the storage solution and rinsed with distilled water. The probe was placed in two known solutions, one with a pH of 10 and the other with a pH of 4 to calibrate the sensor. The probe was rinsed between each test. The second step of this experiment was to make a standardized NaOH

solution. This is done because solid NaOH absorbs moisture and the mass recorded isn't pure NaOH for this reason. 1 g of NaOH was dissolved in 500 mL of distilled water. This solution was made using two volumetric flasks. Than a cleaned buret was filled to the 0 mL mark with the prepared NaOH solution. Than three samples of KHP which is a pure solid acid are measured. The masses of the samples collected were .26 g, .21 g, and .24 g. These samples were each dissolved in 50 mL of distilled water. Two drops of phenolphthalein were added to each solution as an indicator of when the solution was at a certain pH. The initial volume of the NaOH solution in the buret was recorded before each titration. NaOH was added to each KHP solution at different rates until the solution turned a light pink color. The volume change was recorded for each solution. Calculations were made to determine the average NaOH concentrations in each solution. This was done by using the KHP mass used and the volume of NaOH used to titrate. This concluded the preliminary steps of the lab. The third part of this experiment was to make a 50 mL solution using distilled water and 0.2 g of an unknown acid. This solution was then titrated with the prepared NaOH solution by adding 1 mL of NaOH to the unknown acid solution at a time, and the pH was recorded after every mL. This was done to make a graph of pH vs mL of NaOH added. This data and its first and second derivatives were used to determine the pH and volume of NaOH where the pKa and the Ka were at equilibrium. The pH was used to find the pKa and the Ka, and the mL of NaOH added were used to determine the molar mass of the weak acid. Table 1: Data Points Results: To determine the pH of the solution at equilibrium, this graph was created by taking the pH of the solution at each mL titrated using .1 M NaOH solution. By graphing the derivatives on the same graph, it was determined that the pH of the solution at equilibrium was 9.24. By dividing this number by two, it is determined that the pKa is 4.62,  equal to the pH of the weak acid. 10-4.62  is therefore the Ka value. Using  the molarity of NaOH and the chemical equation H X(aq ) + N aOH (aq ) → H 2 O + N aX (aq ) , along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Description: These are the points used to make Graph 1, with the point closest to the equivalence point highlighted.

Graph 1: The Titration Curve and its First and Second Derivatives

Description: This graph was used to calculate the equivalence point of the solution.

Table 2: Data Sample pH at pKa # Equilibrium 4

9.24

Ka (10-4.62)

4.62 2.399𝞊-5

Solute Mass (g)

Volume NaOH (mL)

0.20

32

Molarity Molar Mass NaOH (M) Solute (g) 0.098

63.78

Description: This data was measured before or during the reaction or calculated as described above.

Discussion: The experiment was completed twice, as on the first try there was an error with the pH electrode. The most probable issue is that the probe was not calibrated to the right extent; therefore, on the first try, the results were inclusive with the data. The pH electrode was recalibrated and a more accurate answer was yielded from this try on the experiment. The answer for the pH of the unknown acid was close to the actual pH of the acid, this is due to the pH electrode not being calibrating perfectly and technical and human error. Although, the data derived from the experiment was close to the correct answer, it was not exactly right. To determine the concentration of sodium hydroxide, three samples with 0.2 to 0.3 g of KHP were each dissolved into approximately 50 mL of water. Then sodium hydroxide was titrated into the solution of KHP and water. Since there were three different solutions, the concentration, due to

our experiment, was more accurate. The calculations were done by two people with calculators; therefore, the calculations were precise. The titration was done very meticulously, which means it was less likely to create an error for the calculations; however, there is human error which can be taken as a factor in titration, as well as probe inaccuracies despite repeat calibration. Through the experiment, it was determined that the acid was a weak acid with an equivalence point pH of 9.24. This acid is partially dissociated in water, meaning that both the undissociated acid and its dissociation products being present, in solution, in equilibrium with each other. It also means that the acid is going to have a lower pKa constant. By using the molarity of sodium hydroxide, mass of the solute HX, and the chemical equation H X(aq ) + N aOH (aq ) → H 2 O + N aX (aq ) it was then determined that the molar mass was 63.78 g/mol. Because the molarity and mass are derived from the experiment, there is a chance of error in the result of the molar mass, as the instruments used in general chemistry laboratories are often imprecise. Conclusion: The purpose of this experiment was to use titrations to determine the pKa and Ka of a weak acid. A solution of NaOH was made to be titrated with an unknown weak acid. Preliminary steps were done to standardize the NaOH solution and to calibrate the pH electrode. The titration found that the equilibrium pH was 9.24. This was used to find the pKa of 4.62 and the Ka of 10-4.62. This value is important because the Ka is the dissociation strength of the acid. The molar mass of the acid was calculated using experimental measurements to be 63.78 g/mol. Given the accurate molar mass, 69.49 g/mol, it can be concluded that this weak acid is Hydroxylamine Hydrochloride, with molecular formula ClH4NO. References: Atkins, P.; Jones, L.; Laverman, L. Chemistry Principles: The Quest for Insight , 7th ed.; W.H. Freeman and Company: New York, NY, 2016; pp 264. Whitney, M. Chemistry 221L: Laboratory Experiments for Fundamentals of Chemistry , 1st ed.; Michael P. Whitney: Grand Forks, ND, 2018; pp 31....