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4. Base titration of glycine. An aqueous solution of glycine (0.1 M, 100 ml, pH

ID: 980056 • Letter: 4

Question

4. Base titration of glycine. An aqueous solution of glycine (0.1 M, 100 ml, pH = 1.72) was titrated with 2M NaOH. The pH was monitored and the results are shown in the graph below. The key points in the titration are designated I to V. In each space provided below, write the key point (I to V) and briefy explain your choice. Glycine is present predominantly as the Key point Explain why species +H3N–CH2–COOH ____ The average net charge of glycine is +0.5 ____ Half of the amino groups are ionized The pH is equal to the pKa of the carboxyl ____ group The pH is equal to the pKa of the protonated ____ amino group ____ The isoelectric point of glycine The carboxyl group has been completely ____ titrated (first equivalence point) Glycine is completely titrated (second ____ equivalence point) ____ The predominant species is +H3N–CH2–COO- ____ Glycine has its maximum buffering capacity ____. Statement infront of blanks are to be designated as key point (1-5) as according to the figure and explain why?

2' 1.3 . 30 10 9.60 (tV 5.17 2.34 ) o.S 1. S 2.0

Explanation / Answer

Glycine is present predominantly as the Key point Explain why species +H3N–CH2–COOH ____ The average net charge of glycine is +0.5 ____ Half of the amino groups are ionized The pH is equal to the pKa of the carboxyl ____ group The pH is equal to the pKa of the protonated ____ amino group ____ The isoelectric point of glycine The carboxyl group has been completely ____ titrated (first equivalence point) Glycine is completely titrated (second ____ equivalence point) ____ The predominant species is +H3N–CH2–COO- ____ Glycine has its maximum buffering capacity ____. Statement infront of blanks are to be designated as key point (1-5) as according to the figure and explain why?

1) below the PI isolectric point since at low PH H + ions are more the OH- ions as a result COOH cant be deproteinated (II & III region)

2) between PI and pka 1 2.34 to due to partial deprotenation of COOH (II & III)

3)at II

4)at IV

5) 5.97 the net charge is zero here (III)

6)at iso electric point (III)

7) at PH above PI (III &IV)

8) at IV

9) at Pi ziwitter ion is predominated net charge is zero (III)

10) at the pi and between pka1 and pka2 it hass very less effect on PH (II &III & IV)

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