LSM 1101 Lab Report 1 V Essay

2015 Words9 Pages
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
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(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

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