LSM 1101 Biochemistry of Biomolecules
2013/2014 Semester II
Practical 1
Name
:
Ang Chin Khye Gareth (A0122077M)
Ang Ju Hui (A0131255R)
Bench
:
1
TA
:
Goh Kah Yee
1. Introduction
The titration of a strong base (NaOH) against a weak acid (Histidine monohydrochoride ) enables us to know of properties of the weak acid (histidine). Such properties include the pKa of the acid and the pH range of the buffering capacity. One way to relate pH to pka would be through the Henderson – Hasselbalch equation as stated below. pH = pKa + log ([Conjugate Base]/[Acid])
Using this equation and the mid-point of the buffering capacity of the acid, we will be
…show more content…
(e) Method 2 ( finding the pH when pH = pKa, at 0.5mol and 1.5 mol of NaOH) will be a better method. As at 0.5 mole and 1.5 mole of NaOH, the mixture will contain 50% of acid and 50% of base hence pH = pKa. Even if there is an error in the NaOH prepared, it will not affect the estimated pKa of histidine as compared to using method 1 whereby the pKa estimate is done through observing the center of symmetry of the graph (point of inflexion). In the case of the latter, it is greatly affected if there is an error in the NaOH solution prepared. The error in the NaOH solution prepared will cause the point of inflexion to either shift left or right along the x-axis.
(f)(i) No. of mole of NaOH in 5ml= 5/1000 X 0.05 = 0.00025mol No. of moles of histidine left= 0.0004 – 0.00025 = 0.00015 Using pH = pKa + log ([Conjugate Base]/[Acid]) pH = 6.1 + log(0.00025/0.00015) = 6.32 (f)(ii) No. of mole of NaOH in 12ml = 12/1000 X 0.05 = 0.0006mol No. of moles of OH- left after reacting with histidine = 0.0006 – 0.0004 = 0.0002 No. of mole of (histidine)- after reacting with the remaining OH- = 0.0004-0.0002 = 0.0002 Using pH = pKa + log ([Conjugate Base]/[Acid]) pH = 9.3 + log(0.0002/0.0002) = 9.3
(f) Comparing with experimental values, When 5ml of NaOH is added for calculated data, pH = 6.32 When 5ml of NaOH is added for experimental data, pH = 6.3 When 12ml of NaOH is added for
(.1063 KIO31) (1 mol KIO214 g) x 6 mol S2O41 mol KIO= .00298.04150= .259 M
pH was recorded every time 1.00 mL of NaOH was added to beaker. When the amount of NaOH added to the beaker was about 5.00 mL away from the expected end point, NaOH was added very slowly. Approximately 0.20 mL of NaOH was added until the pH made a jump. The pH was recorded until it reached ~12. This was repeated two more times. The pKa of each trial are determined using the graphs made on excel.
In your laboratory notebook sum these two reactions to find the stoichiometric factor that relates moles of
Mole of chlorine : 1.0217g - .221g - .3946 g = .4061 g of chlorine
9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?
The protein molecules in many foods provide the amino acid building blocks required by our own cells to produce new proteins. To determine whether a sample contains protein, a reagent called Biuret solution is used. Biuret solution contains copper ions. However, the chemical state of the copper ions in Biuret solution causes them to form a chemical complex with the peptide bonds between amino acids (when present), changing the color of the solution. Biuret solution is normally blue, but changes to pink when short peptides are present and to violet when long polypeptides are present.
The purpose of this experiment was to determine the pKa, Ka, and molar mass of an unknown acid (#14). The pKa was found to be 3.88, the Ka was found to be 1.318 x 10 -4, and the molar mass was found to be 171.9 g/mol.
1. Suppose that you have an unlimited supply of copper (II) sulfate to react with iron. How many moles of copper would be
Ka is the acid dissociation constant and [HA] is the concentration of the weak acid . Strong acids usually completely dissociate and has a Ka value greater than 1. Weak acids don’t dissociate completely and have a Ka value much smaller than 1. pKa values are often used for weak acids due to being able to work with whole numbers
To begin, three sets ofabout 0.3000g of KHP are weighed out on an analytical balance. Put the three sets of KHP into three separate, labeled flasks. All three sets of the KHP is then dissolved with approximately 50mL of deionized water. Next, a buret is used to start the actual titration. Buret is initially filled to 0.00mL mark with the NaOH solution, this is recorded as initial volume. Next, add 2-3 drops of phenolphthalein indicator into each of the three flasks containing KHP. A magnetic stir bar is then added to the first flask, and placed above a stir plate. Everything is positioned under the buret. Stirrer is put on medium speed and the titration can start. Slowly release the NaOH into the KHP flask. As the end point is reached, a pink color will be seen in the flask. When the lightest pink possible remains in the solution for more than 30 seconds titration is complete. The final volume is recorded, and the same steps are taken for the other two sets of KHP solution. Finally, blank titration is completed to determine deviation.
weak bases). After ranking the pH of these solutions, you will then test your predictions in the laboratory.
5. In reaction three, the number of moles of NaOH can be calculated from the concentration of the solution (1.0M = 1.0mole/L) and the volume used. The calculation is below. Enter the result into Data Table 2.
In this lab, the purpose was to determine the stability of a substance after adding an acid or a base. The results claim that liver and buffer are the most resistance to change in pH. Looking at figure 3, buffer and liver both maintain a stable pH even with the addition of an acid or base. However, potato and water have less buffer in them since their pHs did change. In figure 3, the potato acid’s pH level decreased by two, and the potato base’s pH level increased by two. The level of pH of a water acid decreased by 4, while the water base’s pH increased by 5. These results all tie to the fact that buffer is a substance that maintains a stable pH; the presence of buffer in organisms help maintain homeostasis by binding or releasing hydrogen
4. To utilize the titration results to calculate the molarity of the hydrochloric acid and the
3. A few drops of 6 M acetic acid were added until it became basic.