Use the diagram below to answer the following questions. glycine.PNG 1.) Which p

Question

Use the diagram below to answer the following questions. glycine.PNG 1.) Which points on the diagram represent the equivalence points? Select all that apply 2.) Which points on the diagram represent the half-equivalence points? Select all that apply 3.) At what point is alanine present predominantly as a mixture of +H3N-CH2-COOH and +H3N-CH2-COO-? 4.) The isoelectric point, or pl, is where the net charge on the amino acid is zero. At what point is this? 5.) The titration curve was generated by titrating a 5.00-mL sample of 0.200 M alanine with a 0.100 M solution of NaOH. How much NaOH solution is required to reach the first equivalence point? 6.) What is the pKa of the COOH group of alanine? 7.) What is the pKa of the NH3+ group of alanine? Which structure(s) of alanine below is/are predominant at point A? Select all that apply amino acids structures.PNG Which structure(s) of alanine below is/are predominant at point D? Select all that apply

Explanation / Answer

Alanine is an amino acid. The pKa of COOH and NH3+, are 2.3 and 9.7, respectively. As per given titration curve it begin titration at a low pH (~1). There are two distinct waves present as it is a diprotic acid. Carboxylic acid being stronger acid (having low pKa!), will deprotonate first. When addition of NaOH is equivalent to all COOH of alanine then there will be a first equivalence and molecule will carry no charge. Even a small addition of NaOH will increase the pH of system. Then aftter the first equivalence, further addtion of NaOH will neutralize NH3+. After addition of 2 equivalents of NaOH, complete NH3+ will be neutralized. and further addition of even small amount of NaOH will give drastic increase in pH. This will be second equivalence point.
1) Hence, first equivalence point is represented by "C".
2) B and D are half equivalence points.
3) At first half equivalence points represented by "B" will have predominant mixture of +H3NCH2-COOH and +H3NCH2-COOH.
4) At the first equivalence point in this titration, when all of the lower pKa acid has been neutralized. The pH at which this occurs can be calculated as approximately the average of the pKa's of the COOH and NH3+, (2.3 + 9.7)/2 = ~6. At this pH, the alanine will carry no net charge. The carboxyl groups will carry a negative charge but the amine group will still be fully protonated and carry a positive charge. This is the isoelectric point of alanine. So PI is indicated by "C".
5) M1 = 0.2M, V1 = 5mL, and M2 = 0.1M, V2=?, we know, M1V1 = M2V2.
V2 = (0.2*5)/0.1 = 10mL. Thus to reach at first equivalnce we will require 10mL of 0.1M NaOH.
6) pKa of COOH = 2.3
7) pKa of NH3+ = 9.7
8) At point A, pH is very low (below 2), hence both COOH and NH3+ will be protonated. This gives predominant presence of structure B.
9) Point D is a second half equivalnce point. At this point NH3+ and NH2 will be present on alanine molecule along with COO-.
Thus at point D there will be mixture of Strcture C and D will be present preidominantly.

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