Relationship between the Titration Curve and the Acid-Base...

Relationship between the Titration Curve and the Acid-Base Properties of Glycine A $100 \mathrm{mL}$ solution of $0.1 \mathrm{M}$ glycine at $\mathrm{pH} 1.72$ was titrated with $2 \mathrm{M} \mathrm{NaOH}$ solution. The $\mathrm{pH}$ 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. (a) Glycine is present predominantly as the species $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}$ COOH. (b) The average net charge of glycine is $+\frac{1}{2}$ (c) Half of the amino groups are ionized. (d) The $\mathrm{pH}$ is equal to the $\mathrm{p} K_{\mathrm{a}}$ of the carboxyl group. (e) The $\mathrm{pH}$ is equal to the $\mathrm{p} K_{\mathrm{a}}$ 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 $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$ (k) The average net charge of glycine is -1 (l) Glycine is present predominantly as a 50: 50 mixture of $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}$ and $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$ $(\mathrm{m})$ This is the isoelectric point. (n) This is the end of the titration. (o) These are the worst pH regions for buffering power.

Relationship between the Titration Curve and the Acid-Base Properties of Glycine A $100 \mathrm{mL}$ solution of $0.1 \mathrm{M}$ glycine at $\mathrm{pH} 1.72$ was titrated with $2 \mathrm{M} \mathrm{NaOH}$ solution. The $\mathrm{pH}$ 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.
(a) Glycine is present predominantly as the species $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}$ COOH.
(b) The average net charge of glycine is $+\frac{1}{2}$
(c) Half of the amino groups are ionized.
(d) The $\mathrm{pH}$ is equal to the $\mathrm{p} K_{\mathrm{a}}$ of the carboxyl group.
(e) The $\mathrm{pH}$ is equal to the $\mathrm{p} K_{\mathrm{a}}$ 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 $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$
(k) The average net charge of glycine is -1
(l) Glycine is present predominantly as a 50: 50 mixture of $^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}$ and
$^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$
$(\mathrm{m})$ This is the isoelectric point.
(n) This is the end of the titration.
(o) These are the worst pH regions for buffering power.

Key Concepts

Acid?Base Titration Curve

An acid–base titration curve is a graphical representation of the pH of a solution as a titrant is added. It reveals how pH changes during neutralization and shows characteristic inflection points corresponding to the neutralization of specific ionizable groups in a molecule. In polyprotic systems, such as molecules with more than one titratable group, the curve contains regions that reflect individual deprotonation or protonation events.

Acid Dissociation Constant (pKa)

The pKa is the negative logarithm of the acid dissociation constant (Ka) and represents the pH at which a specific ionizable group is half protonated and half deprotonated. It is a critical parameter in understanding the acid–base behavior of a molecule since it marks the point where there is an equilibrium between the acidic and basic forms. In titration curves, half-equivalence points coincide with the pKa values of the respective ionizable groups.

Equivalence Points

Equivalence points are the stages in a titration where the amount of titrant added stoichiometrically equals the amount of acid or base present in the solution. In a polyprotic or amphoteric molecule, multiple equivalence points are observed, each corresponding to the complete neutralization of one ionizable group. They are key in determining the titration endpoints and the acid–base properties of the molecule.

Buffering Capacity

Buffering capacity refers to the ability of a solution to resist changes in pH upon the addition of small amounts of acid or base. It is maximized when significant amounts of the weak acid and its conjugate base are present, typically around the pKa value. In titration curves, similar regions around half-equivalence points indicate strong buffering because both the protonated and deprotonated forms coexist.

Isoelectric Point (pI)

The isoelectric point is defined as the pH at which a molecule, particularly an amphoteric compound such as an amino acid, carries no net electrical charge. At this point, the positive and negative charges balance each other out. It is a crucial concept for understanding the solubility and migration of biomolecules in electric fields, and it is often identified as a key feature on titration curves for such compounds.

Ionization States of Amphoteric Molecules

Amphoteric molecules have the ability to either donate or accept protons depending on the pH of the environment, resulting in multiple ionization states such as fully protonated, zwitterionic, and fully deprotonated forms. These states determine the molecule’s net charge and influence its chemical behavior and reactivity. Understanding the transitions between these forms is fundamental to interpreting titration curves and the acid–base properties of such molecules.

Transcript

00:01 In the atom, there are three subatomic particles.

00:05 These are going to be the protons in the nucleus with a positive charge, the neutrons, neutrons, in the nucleus with the neutral charge, and then the electrons, which have a negative charge.

00:28 The electrons are located in the shells surrounding the nucleus.

00:32 The outer shell is going to be the valent shell, which is an important factor to remember.

00:37 So in the nucleus, the number of protons, let me just separate this real quick, the number of protons is going to equal the atomic number.

00:48 Number of protons equals atomic.

00:56 So that's going to be a defining feature of which element it is.

01:01 Then the neutrons, you can count those in the nucleus as well to get the mass number, because mass number is equal to the protons plus the neutrons.

01:17 So the mass number can be variable, and this is going to be because the number of protons will stay the same, but the number of neutrons can vary.

01:27 So you can have the same atomic number, same atomic number, but you can have a different mass number.

01:43 Different mass number.

01:46 So if you have the same atomic number but different mass numbers, that's going to give you an isotope.

01:57 So then we have electrons.

02:00 I said electrons were in the space surrounding the nucleus.

02:03 The varying energy levels.

02:06 So we're going to look at the valence electrons, which are going to determine the types of chemical bonding that it can participate in.

02:18 So determines, let's say number of valence electrons determines types of chemical bonding.

02:32 Okay, so we're going to look at these two types of chemical bonds.

02:37 First, we have to ionic bonds and ionic bonds are going to it's going to be a transfer of electrons and i'm going to use this as a little shorthand for electrons okay so you can see the outer shell of sodium does not have it is not a full shell and neither is chloride in that outer shell.

03:09 So it would take quite a bit of energy to fill the shell of sodium.

03:15 So it would actually be easier to give that electron away.

03:22 And chloride, it would take more energy to remove all of those electrons on that outer shell rather than just taking that one electron from the sodium.

03:33 So this way, both of the ions, sodium and chloride are ions, and are able to have full outer shells by interacting with each other.

03:43 So the sodium giving gives and then the chloride takes.

03:51 Okay.

03:52 So then covalent bonds are a little bit different.

03:55 They're going to be a sharing of electrons instead of the transfer of electrons...

Please give Ace some feedback