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Determination of the pKas for Glycine

Paper Type: Free Essay Subject: Chemistry
Wordcount: 2984 words Published: 24th Jan 2018

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Name: Aleksandra Redlinska

Lab Partner: Adrianna Czerlonko

Lab Instructor: Jenq-Kuen Huang

Lab TA: Chandra K.Ailneni

Objectives:

The purpose of this experiment is to titrate glycine and determine its pKa values. This is done to show that glycine is a dipolar ion. This experiment will help explain how pKa values and pH are closely related and will help show the importance of pH in amino acids.

Background:

Amino acids consist of an amino group (-NH3+) and a carboxyl group (-COOH). When they are in an acidic solution, the amino group becomes protonated and the carboxyl group gets dissociated. When in a solution with a neutral pH, amino acids will usually be dipolar and the amino group will be protonated while the carboxyl group becomes deprotonated. When amino acids are in basic solution, the protonated amino group is the only one that changes, by losing a proton (Biochemistry). The equation used to find the equilibrium constant, Ka, of a reaction is the products divided by the reactants. The Ka can then be used to find the pKa by taking the negative log of the Ka. The Ka value can be used to figure out whether the acid will dissociate or not and how strong an acid is. Glycine is an amino acid which contains both acidic and basic pKas. This shows that it can react to changes in the pH. By adding NaOH to glycine, the proton ion will dissociate. This dissociation will occur due to the titration done using the glycine solution.

Materials:

  • 0.1 N Sodium hydroxide (60 mL)
  • Glycine (0.6 g or 8.0 meq)
  • 0.1 N Hydrochloric acid (60 mL)
  • Deionized water
  • 100 mL volumetric flask
  • Two burettes
  • Four 250 mL beakers
  • Funnel
  • Graduated cylinder
  • pH meter

Procedure:

Glycine (0.6 g) was weighed out and put in a 100 ml flask. Water was added to the flask to make 100 mL of solution. The sample (25 mL) was transferred to a 250 mL beaker and distilled water was added (100 mL). The solution was titrated with 0.1 N NaOH (30 mL in 1.0 mL increments). The pH and volume of base were measured after each 1.0 mL of NaOH added and the data was recorded and plotted. Distilled water (125 mL) was placed in a 250 mL beaker. The solution was once again titrated with 0.1 N NaOH (30 mL in 1.0 mL increments). The pH and volume of base were once again collected and plotted. The original glycine solution (25 mL) was transferred to a 250 mL beaker and distilled water (100 mL) was added. The solution was titrated with 0.1 N HCl (30 mL in 1.0 mL increments). The total volume of acid and the pH were recorded. Distilled water (125 mL) was placed in a 250 mL beaker and the solution was titrated with 0.1 N HCl (30 mL in 1.0 mL increments). The volume of the acid and the pH was recorded and plotted.

Results:

Table 1. Titration with NaOH.

mL 0.1 N NaOH added

pH of 25 mL glycine

pH of 25 mL water control

0

5.90

5.65

1

7.40

9.39

2

7.85

9.80

3

8.05

10.00

4

8.23

10.14

5

8.30

10.25

6

8.46

10.35

7

8.54

10.45

8

8.65

10.50

9

8.70

10.55

10

8.75

10.60

11

8.85

10.65

12

8.90

10.68

13

8.99

10.72

14

9.05

10.75

15

9.13

10.79

16

9.25

10.83

17

9.33

10.86

18

9.45

10.89

19

9.60

10.93

20

9.75

10.96

21

9.86

11.00

22

9.99

11.03

23

10.11

11.05

24

10.20

11.07

25

10.26

11.10

26

10.33

11.13

27

10.42

11.15

28

10.46

11.18

29

10.52

11.20

30

10.57

11.25

Table 2. Titration with HCl.

mL 0.1 N HCl added

pH of 25 mL glycine

pH of 25 mL water control

0

5.25

4.45

1

3.98

3.37

2

3.72

3.19

3

3.57

3.05

4

3.45

2.96

5

3.37

2.87

6

3.30

2.82

7

3.23

2.77

8

3.18

2.73

9

3.12

2.69

10

3.07

2.66

11

3.02

2.63

12

2.98

2.60

13

2.95

2.58

14

2.91

2.55

15

2.88

2.53

16

2.85

2.51

17

2.82

2.49

18

2.80

2.47

19

2.77

2.45

20

2.75

2.44

21

2.72

2.42

22

2.70

2.41

23

2.68

2.40

24

2.66

2.38

25

2.65

2.37

26

2.63

2.36

27

2.61

2.35

28

2.60

2.34

29

2.59

2.33

30

2.57

2.31

Calculations:

Determining pKa1 (HCl titration)

At pH = 2.5, 30 mL of 0.1 N HCl was used for the glycine titration and 16 mL was used for water.

30 mL – 16 mL = 14 mL

14 mL = 0.014 L

0.014 L was used to titrate glycine at pH 2.5

The normality of HCl used was 0.1 N

N = #eq/L

0.1 N = X/0.014 L

X = 0.0014 eq

0.0014 eq = 1.4 meq

It takes 1.4 meq of HCl to convert glycine to an acidic form

The amount of glycine at the start of the titration was:

(8 meq)(25 mL/100 mL) = 2 meq

The amount of glycine left after the titration was:

2 meq – 1.4 meq = 0.6 meq

Formula:

pH = pKa1 + log [conjugate base]/ [acid]

Conjugate base = 0.6 meq

Acid = 1.4 meq

2.5 = pKa1 + log [0.6]/[1.4]

2.5 = pKa1 – 0.368

pKa1 = 2.868

meq of glycine used at the start of the titration was 2 meq

Normality of HCl used was: 0.1 N

pH

Vol. of HCl added in mL

meq of HCl added = meq of +NH3CH2CO2H formed

meq of +NH3CH2CO2 remaining

pKa1

2.5

14 mL

1.4 meq

2 – 1.4 = 0.6 meq

2.868

2.6

16 mL

1.6 meq

2 – 1.6 = 0.4 meq

3.202

2.7

13 mL

1.3 meq

2 – 1.3 = 0.7 meq

2.969

Avg. pH = 2.6

Avg. pKa1 = 3.013

           

Determining pKa2 (NaOH titration)

At pH = 9.4, 18 mL of 0.1 N NaOH was used for the glycine titration and 1 mL was used for the water.

18 mL – 1 mL = 17 mL

17 mL = 0.017 L

0.017 L was used to titrate glycine at pH 9.4

The normality of HCl used was 0.1 N

N = #eq/L

0.1 N = X/0.017 L

X = 0.0017 eq

0.0017 eq = 1.7 meq

It takes 1.7 meq of NaOH to convert glycine to an acidic form

The amount of glycine at the start of the titration was:

(8 meq)(25 mL/100 mL) = 2 meq

The amount of glycine left after the titration was:

2 meq – 1.7 meq = 0.3 meq

Formula:

pH = pKa2 + log [conjugate base]/ [acid]

Conjugate base = 1.7 meq

Acid = 0.3 meq

9.4 = pKa2 + log [1.7]/[0.3]

9.4 = pKa2 + 0.753

pKa2 = 8.647

meq of glycine used at the start of the titration was 2 meq

Normality of NaOH used was: 0.1 N

pH

Vol. of NaOH added in mL

meq of NaOH added = meq of NH2CH2CO2 formed

meq of +NH3CH2CO2 remaining

pKa2

9.4

17 mL

1.7 meq

2 – 1.7 = 0.3 meq

8.647

9.8

19 mL

1.9 meq

2 – 1.9 = 0.1 meq

8.521

10.1

19 mL

1.9 meq

2 – 1.9 = 0.1 meq

8.521

Avg. pH = 9.8

Avg. pKa2 = 8.563

           

Discussion and Conclusion:

The average pKa1 calculated for the titration using HCl was 3.013. The true pKa 1 for an acidic amino acid is 2.3. The average pKa2 calculated for the titration using NaOH was 8.863. The true pKa2 for a basic amino acid is 9.6. This difference could have been caused by dirty glassware or improper measuring of the 0.1 increments of HCl and NaOH using the burette. The error could have also been due to the absence of an analytical scale to make precise measurements of glycine and the pH meter not being cleaned thoroughly enough. Since there was not enough time for each group to complete both parts of the experiment using HCl and NaOH, each group only did one of the titrations. Since my group did not have time to do the HCl titration, we got the results from another group. This could have also caused an error because we did not monitor how precise the other group was.

This experiment shows that the pKas of amino acids are important. They could be used to calculate the pH and strength of the amino acid. This observation is displayed in the data above and the graph. This experiment helped me understand how pKa and pH are related.

References:

Berg, Jeremy M., John L. Tymoczko, and Lubert Stryer.Biochemistry. 6th Edition, W.H. Freeman and Company, New York, (2002).

Cronk, J. (2012, January 1). BIOCHEMISTRY Dictionary. Retrieved September 23, 2014, from http://guweb2.gonzaga.edu/faculty/cronk/biochem/dictionary.cfm?letter=front

Cronk, J. (2012, January 1). CHEM 440 lectures. Retrieved September 23, 2014, from http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/L05-index.cfm

Huang, Jack. Experiment 2: “Determination of the pKas for Glycine” Chem 421 Laboratory. Sep 15, 2014.

Answers to Questions:

1. The pKa associated with a typical aliphatic carboxylic acid is 2.3 and the pKa associated with a typical aliphatic amine is 9.6. The pKas associated with an alpha-carboxyl group in an amino acid are between 2 and 3. The pKas associated with a gamma-carboxyl group in an amino acid are between 4.3 and 4.5. The pKas associated with an alpha-amino group in an amino acid are between 9 and 10. The pKa associated with an epsilon-amino group in an amino acid is around 10. The pKas associated with an imidazole group in an amino acid are between 6 and 7. The pKas associated with a guanidine group in an amino acid is around 12.5. The pKas for these groups are taken away once the amino acids are incorporated into proteins. The groups form bonds and the only pKa values that can still be seen are the ones that are at the C and N terminus of the chain.

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2. The top equation is correct because the top equation contains a Zwitterion. This Zwitterion would cause it to be able to react with HCl (a strong base) and NaOH (a strong acid). The bottom equation shows no charges on the nitrogen or oxygen, making the reaction not able to happen. My data supports this because when adding HCl, the pH went down and when adding NaOH, the pH went up. This shows that ions were taken away and added causing deprotonation and protonation of the glycine.

3. The structure of Lys-Ala-Asp is:

The N-terminal is the NH3+ on the very left side and the C-terminal is the OH on the right side. The net charge of this tripeptide at pH 2 is +2. The net charge at pH 6 is 0. The net charge at pH 13 is -2.

 

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