I talked to my professor, and he said the following: Grams of glycine is determi
ID: 496290 • Letter: I
Question
I talked to my professor, and he said the following:
Grams of glycine is determined using the known number of grams of glycine starting material in solution. This is then divided by the number of moles of 1 eq NaOH (the amt of NaOH necessary to fully deprotonate 1 of the functional groups which also equals the amt of NaOH necessary for conversion between pka1 and pka2). This equals the experimentally determined MW of glycine, and this is then compared to the known MW of glycine.You will follow the same pathway using your experimental data to determine the MW of the unknown amino acid. Using this data as well as the estimated pKa values from your graph you can then estimate the identity of your unknown.
The amount of glycine used in the the first graph was 0.2 and for the 2nd graph I used 0.2504. The concentration of NaOH we used was 0.9655. Would it be normal for me to find an experimental MW of 142.8 for glycine, which is almost double the known MW? I used the 0.2 g of glycine used and found 1 eq of NaOH to be 0.0014 moles (0.9655 M NaOH * 0.0015 L titrated at equivalence point = 0.0014 moles NaOH = 0.0014 moles glycine). Any help on this equation would be helpful!
l am currently working on a lab report and am confused by the questions involved: (1) Plot two pH titration graphs (computer generated; one for glycine, one for the unknown); mark the pKas on each graph. (2) Calculate the molecular weight for glycine from the titration curve. Show your work. (3) Calculate the molecular weight for the unknown from the titration curve. Show your work. I have already done both of the graphs in the titration, shown by the following 2 graphs, labeled which is which at the top:Explanation / Answer
The pKa or dissociation constant is a measure of the strength of an acid or a base. It is the pH at which one-half of the acid present has reacted with the base.
When the curve reaches the first of two flat regions, the point where one-half of the carboxyl group is deprotonated is reached.
Note: the midpoint of the flat region is considered as pKa. Please check your pKa values.
At the point of inflexion, the carboxyl group is completely deprotonated and the isoelectric point is reached. At this point, the amino acid exists in zwitterionic form. So, the number equivalents of NaOH is equal to the number of equivalents of glycine.
The isoelectric point of an amino acid can be calculated according to the following equation:
pI = {pKa (-COOH) + pKa (-NH3)}/2
= (2.38 + 8.74)/2 = 5.56
at equivalance point; volume of NaOH = (5 mL + 6.5 mL) /2 = 5.75 mL
no. of moles of NaOH = Molarity x volume in L
= 0.9655 M x 0.00575 L
= 0.00555 moles
No. moles of NaOH = No. moles of Glycine = 0.00555 moles = wt/ mol.wt
we have wt = 0.2 g
therefore;
0.00555 moles = 0.2g/ mol.wt
mol.wt = 0.2/0.00555
mol. wt of glycine = 36 grams
Similarly; for unknown sample
pI = {pKa (-COOH) + pKa (-NH3)}/2
= (2.2 + 9.34)/2 = 5.77
at equivalance point; volume of NaOH = (8.4 mL + 11.5 mL) /2 = 9.95 mL
no. of moles of NaOH = Molarity x volume in L
= 0.9655 M x 0.00995 L
= 0.00960 moles
No. moles of NaOH = No. moles of unknown = 0.00960 moles = wt/ mol.wt
we have wt = 0.2504 g
therefore;
0.00960 moles = 0.2504g/ mol.wt
mol.wt = 0.2504/0.00960
mol. wt of glycine = 26 grams