Question 28.Int.1: L-Threonine is one of the essential amino acids. It is found...
L-Threonine is one of the essential amino acids. It is found in animal protein (for example, eggs and milk) but is missing in some grains, such as rice. It is now common practice to fortify grains with the essential amino acids that they lack. The structure of L-threonine is shown below.
\begin{matrix} \underset{|}{OH} \quad\quad\quad\quad \ \ \underset{||}{O} \\ CH_3-CH-CH-C-OH \\ \ \ \overset{|}{NH_2} \\ \end{matrix}
pK_{a_1} = 2.15 \quad pK_{a_2} = 9.12
Use the above data to determine the principal form of L-threonine and the direction that it will migrate under the influence of an electric field in a gel that is 0.25 M NaH_2PO_4 and 0.50 M Na_2HPO_4.
Analyze
The key to this problem is to recognize that the NaH_2PO_4/Na_2HPO_4 mixture is a buffer that controls the pH of the gel. Once the pH of the gel is established we can identify the ionization state for the bulk of the L-threonine and hence the direction it will migrate.
Use pK_{a_2} from Table 16.4 and the Henderson-Hasselbalch equation to calculate the pH of the gel.
pH = p K_{ a _2}+\log \left(\left[ HPO _4{ }^{2-}\right] /\left[ H _2 PO _4{ }^{-}\right]\right)
and
pH = 7.20 + log(0.50/0.25) = 7.50
The structures of the three ionic forms in which L-threonine can appear are
\begin{matrix} Acidic solution: & & Isoelectric point: \\ CH_3CH(OH)CHCOOH & & CH_3CH(OH)CHCOO^- \\ \quad\quad \overset{|}{NH_3{ }^+} & \quad\quad\quad & \quad\quad \ \overset{|}{NH_3{ }^+} \end{matrix}
\begin{matrix} Basic solution: \\ CH_3CH(OH)CHCOO^- \\ \quad\quad \ \overset{|}{NH_2} \end{matrix}
The pH at the first equivalence point in the titration of a weak diprotic acid is given by equation (17.5). Applied to L-threonine, the result is
\text { for } H _2 PO _4{ }^{-}: \quad pH =\frac{1}{2}\left( p K_{ a _1}+ p K_{ a _2}\right)=\frac{1}{2}(2.15+7.20)=4.68 (17.5)
pH =\frac{ p K_{ a _1} + p K_{ a _2}}{2}=\frac{2.15 + 9.12}{2} = 5.64The predominant species at this first equivalence point of pH = 5.64 is the dipolar zwitterion, which would not migrate at all in an electric field. However, in the gel medium of pH = 7.50 the hydroxide ion concentration is 100 times as great. The predominant species is the anion with charge -1. This form of the L-threonine migrates toward the positive electrode, the anode.
Assess
This problem suggests two ways to establish the isoelectric point of an amino acid. The first, used here, is to calculate pI by combining pK_a values of a polyprotic amino acid. Another way is to find the pH of a gel medium in which the amino acid does not migrate in the presence of an applied electric field, hence pH = pI. Additionally, migration through an electric field in a gel (gel electrophoresis) can be used to separate amino acid mixtures.
| TABLE 16.4 Ionization Constants of Some Polyprotic Acids | ||||
| Acid | Ionization Equilibria | Ionization Constants, K | pK |  |
| \text{Hydrosulfuric}^a | H _2 S + H _2 O \rightleftharpoons H _3 O ^{+}+ HS ^{-} | K_{a_1} = 1.0 \times 10^{-7} | pK_{a_1} = 7.00 | |
| HS ^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ S ^{2-} | K_{a_2} = 1 \times 10^{-19} | pK_{a_2} = 19.00 | ||
| \text{Carbonic}^b | H _2 CO _3+ H _2 O \rightleftharpoons H _3 O ^{+}+ HCO _3{ }^{-} | K_{a_1} = 4.4 \times 10^{-7} | pK_{a_1} = 6.36 | |
| HCO _3^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ CO _3{ }^{2-} | K_{a_2} = 4.7 \times 10^{-11} | pK_{a_2} = 10.33 | ||
| Citric | H _3 C _6 H _5 O _7+ H _2 O \rightleftharpoons H _3 O ^{+}+ H _2 C _6 H _5 O _7{ }^{-} | K_{a_1} = 7.5 \times 10^{-4} | pK_{a_1} = 3.12 | |
| H _2 C _6 H _5 O _7{ }^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ HC _6 H _5 O _7{ }^{2-} | K_{a_2} = 1.7 \times 10^{-5} | pK_{a_2} = 4.77 | ||
| HC _6 H _5 O _7{ }^{2-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ C _6 H _5 O _7{ }^{3-} | K_{a_3} = 4.0 \times 10^{-7} | pK_{a_3} = 6.40 | ||
| Phosphoric | H _3 PO _4+ H _2 O \rightleftharpoons H _3 O ^{+}+ H _2 PO _4{ }^{-} | K_{a_1} = 7.1 \times 10^{-3} | pK_{a_1} = 2.15 | |
| H _2 PO _4{ }^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ HPO _4{ }^{2-} | K_{a_2} = 6.3 \times 10^{-8} | pK_{a_2} = 7.20 | ||
| HPO _4{ }^{2-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ PO _4{ }^{3-} | K_{a_3} = 4.2 \times 10^{-13} | pK_{a_3} = 12.38 | ||
| Oxalic | H _2 C _2 O _4+ H _2 O \rightleftharpoons H _3 O ^{+}+ HC _2 O _4{ }^{-} | K_{a_1} = 5.6 \times 10^{-2} | pK_{a_1} = 1.25 | |
| HC _2 O _4{ }^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ C _2 O _4{ }^{2-} | K_{a_2} = 5.4 \times 10^{-5} | pK_{a_2} = 4.27 | ||
| \text{Sulfurous}^c | H _2 SO _3+ H _2 O \rightleftharpoons H _3 O ^{+}+ HSO _3{ }^{-} | K_{a_1} = 1.3 \times 10^{-2} | pK_{a_1} = 1.89 | |
| HSO _3{ }^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ SO _3{ }^{2-} | K_{a_2} = 6.2 \times 10^{-8} | pK_{a_2} = 7.21 | ||
| \text{Sulfuric}^d | H _2 SO _4+ H _2 O \rightleftharpoons H _3 O ^{+}+ HSO _4{ }^{-} | K_{a_1} = \text{ very large} | pK_{a_1} < 0 | |
| HSO _4{ }^{-}+ H _2 O \rightleftharpoons H _3 O ^{+}+ SO _4{ }^{2-} | K_{a_2} = 1.1 \times 10^{-2} | pK_{a_2} = 1.96 | ||
_{}^{a}\text{The} value for K_{a_2} of H_2S most commonly found in older literature is about 1 \times 10^{-14}, but current evidence suggests that the value is considerably smaller.
_{}^{b}H_2CO_3 cannot be isolated. It is in equilibrium with H_2O and dissolved CO_2. The value given for K_{a_1} is actually for the reaction
CO _2(aq)+2 H _2 O \rightleftharpoons H _3 O ^{+}+ HCO _3{ }^{-}
Generally, aqueous solutions of CO_2 are treated as if the CO_2(aq) were first converted to H_2CO_3, followed by ionization of the H_2CO_3.
_{}^{c}H_2SO_3 is a hypothetical, nonisolatable species. The value listed for K_{a_1} is actually for the reaction
SO _2 (aq)+2 H _2 O \rightleftharpoons H _3 O ^{+}+ HSO _3{ }^{-}
_{}^{d}H_2SO_4 is completely ionized in the first step.