12 10 8 PH 6 4 2 0 11.30 9.60 5.97 2.34 (11) 0.5 (IV) 1,0 1.5 OH (equivalents) (m) This is the isoelectric point. (n) Th

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12 10 8 Ph 6 4 2 0 11 30 9 60 5 97 2 34 11 0 5 Iv 1 0 1 5 Oh Equivalents M This Is The Isoelectric Point N Th 1 (44.85 KiB) Viewed 109 times

12 10 8 PH 6 4 2 0 11.30 9.60 5.97 2.34 (11) 0.5 (IV) 1,0 1.5 OH (equivalents) (m) This is the isoelectric point. (n) This is the end of the titration. 2.0 2. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mL solution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and the results were plotted as shown in the graph. The key points in the titration are designated I to V. For each of the statements (a) to (o). identify the appropriate key point in the titration and justify your choice. (0) These are the worst pH regions for buffering power. (a) Glycine is present predominantly as the species 'H,N-CH₂COOH. (b) The average net charge of glycine is +½. (e) The pH is equal to the pKa 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). (c) Half of the amino groups are ionized. (d) The pH is equal to the pKa of the carboxyl group (i) Glycine is completely titrated (second equivalence point). (j) The predominant species is +H3N CH2 COO (k) The average net charge of glycine is -1. (1) Glycine is present predominantly as a 50:50 mixture of +H3N-CH2-COOH and +H3N-CH2-COO-.