Please explain your answers . Thank you, you brilliant mind ! iPad 5:35 PM old e
ID: 698625 • Letter: P
Question
Please explain your answers . Thank you, you brilliant mind ! iPad 5:35 PM old exams 2016 3510 Final.pdf l) (7.5 points) A 100 mL solution of 0.1 M glycine was titrated with a 2 M NaOH solution The pH was monitored and the results were plotted as shown in the following graph. The key points in the titration are designated I tov For each statement (a) to (o), identify appropriate key point(s) in the titration a) Glycine is present predominantly as the species "H,N-CH, COOH b) The average net charge of glycine is +/2. c) Half of the amino groups are ionized d) The pH is equal to the pK, of the carboxyl - 0.0 05 0 15 20 Equivalents of OH group e) The pH is equal to the pK, of the protonated amino group -f) Glycine has its maximum buffering capacity g) The average net charge of glycine is zero. h) The carboxyl group has been completely titrated (first equivalence point) i) Glycine is completely titrated (second equivalence point). j The predominant species is "H,N-CH, Coo k) The average net charge of glycine is-I ) Glycine is present predominantly as a 50:50 mixture of the "H,N-CH COOH and 'H,N- CH, COO species -, m) This is the isoelectric point. n) This is the end of the titration. o) These are the worst pH regions for buffering power Calendar To Do
Explanation / Answer
a)
At point I glycine is predominantly present as +H3NCH2COOH
b)
At point II there are equal moles of +H3NCH2COOH and +H3NCH2COO- present in the solution.
Net charge of glycine = (1 + 0)/2 = ½
c)
At point IV, half of the amino groups are ionised from +H3NCH2COO- to H2NCH2COO-
d)
At point II pH = pKa of carboxyl group.
e)
At point IV pH = pKa of protonated amino group.
f)
At point to II and IV glycine has its maximum buffering capacity.
g)
At point III, glycine is present as +H3NCH2COO-, the average net charge = 0
h)
First equivalence point occurs at point III where glycine is present as +H3NCH2COO-
i)
Second equivalence point occurs at point V where glycine is completely titrated.
j)
At point III predominant species is +H3NCH2COO-
k)
At point III, glycine present as +H3NCH2COO- has average net charge = 0
l)
At point II, glycine has reached half equivalence point and has 50:50 mixture of +H3NCH2COOH and +H3NCH2COO-
m)
At point III, glycine is present as +H3NCH2COO-, the average net charge = 0. This is the isoelectric point.
n)
At point V, second equivalence point is the end of titration.
o)
Points I, III and V are worst regions for buffering as the rate of pH increase is maximum in these regions.