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Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=\) \(7.4 \times 10^{-5} .\) Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0\) \(\mathrm{M} ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0\) \(\mathrm {M} ?\)

Short Answer

Expert verified
In a solution with \([\mathrm{H}^{+}] = 1.0\, \mathrm{M}\), the anionic form (\(\mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COO}^{-}\)) of alanine would predominate. In a solution with \([\mathrm{OH}^{-}] = 1.0\, \mathrm{M}\), the cationic form (\(\mathrm{H}_{3} \mathrm{N}^{+}\mathrm{CH}_{2}\mathrm{COOH}\)) of alanine would predominate.

Step by step solution

01

Identify the three possible charged ions of alanine

Alanine has a carboxylic acid group (COOH) and an amino group (NH2). When it loses a proton from the carboxylic acid group, it becomes negatively charged (anionic). When it gains a proton at the amino group, it becomes positively charged (cationic). We also have the neutral form (zwitterion) where the amino group grabs a proton from the carboxylic acid group. Therefore, the three possible charged ions of alanine are: 1. Anionic form: \( \mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COO}^{-} \) 2. Cationic form: \( \mathrm{H}_{3} \mathrm{N}^{+}\mathrm{CH}_{2}\mathrm{COOH} \) 3. Zwitterionic form: \( \mathrm{H}_{3} \mathrm{N}^{+}\mathrm{CH}_{2}\mathrm{COO}^{-} \)
02

Determine the form of alanine in the presence of H+ ions

When alanine is in the presence of H+ ions, it can react as either an acid or a base. We must compare the Ka and Kb values to see which reaction predominates. Since the Ka value (\(4.5 \times 10^{-3}\)) is much larger than the Kb value (\(7.4 \times 10^{-5}\)), it is more likely that alanine will behave as an acid and lose a proton from the carboxylic acid group. This leads to the formation of the anionic form: \( \mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COOH} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} \) So, in a solution with \([\mathrm{H}^{+}] = 1.0\, \mathrm{M}\), the anionic form of alanine would predominate.
03

Determine the form of alanine in the presence of OH- ions

Now we need to determine the form of alanine in the presence of OH- ions. Since the Kb value (\(7.4 \times 10^{-5}\)) is much smaller than the Ka value (\(4.5 \times 10^{-3}\)), alanine will act as a base and gain a proton at the amino group. This leads to the formation of the cationic form: \( \mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COO}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{H}_{3} \mathrm{N}^{+}\mathrm{CH}_{2}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} \) So, in a solution with \([\mathrm{OH}^{-}] = 1.0\, \mathrm{M}\), the cationic form of alanine would predominate.
04

In conclusion,

In a solution with a high concentration of H+ ions (\([\mathrm{H}^{+}] = 1.0\, \mathrm{M}\)), the anionic form (\(\mathrm{H}_{2} \mathrm{N}\mathrm{CH}_{2}\mathrm{COO}^{-}\)) of alanine would predominate. In a solution with a high concentration of OH- ions (\([\mathrm{OH}^{-}] = 1.0\, \mathrm{M}\)), the cationic form (\(\mathrm{H}_{3} \mathrm{N}^{+}\mathrm{CH}_{2}\mathrm{COOH}\)) of alanine would predominate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alanine Structure
Alanine is one of twenty amino acids used to build proteins in our bodies and has a simple structure that plays a vital role in understanding its function. At its core, alanine consists of a central carbon (alpha carbon) to which four different groups are attached: a hydrogen atom, a carboxyl group (COOH), an amino group (NH2), and a methyl group (CH3).

The carboxyl group is what gives alanine its acidic properties, while the amino group provides its basic characteristics. The unique feature of amino acids, including alanine, is the presence of both acidic and basic functional groups, which enables them to participate in various chemical reactions. This dual functionality is essential for the formation of proteins and the biochemical activities they undertake.
Acid-Base Properties in Amino Acids
Amino acids, due to their amphoteric nature, can act as either acids or bases. This is because they contain both a basic amino group and an acidic carboxyl group. The behavior of amino acids in various pH environments is determined by the acid dissociation constant (Ka) of the carboxyl group and the base dissociation constant (Kb) of the amino group.

In the case of alanine, the acid-base properties become evident in different pH conditions. If the environment is acidic, the carboxyl group can lose a proton, resulting in a negatively charged anion. Conversely, in a basic environment, the amino group can gain a proton and form a positively charged cation. The exact nature of the charged species that will prevail depends on the pH of the solution relative to the pKa and pKb values of the amino acid, highlighting the importance of these constants in predicting the amino acid's behavior in different pH environments.
pKa and pKb in Amino Acids
Understanding the pKa and pKb values of amino acids is crucial when predicting their behavior in various pH environments. The pKa value is the negative base-10 logarithm of the acid dissociation constant, Ka, and indicates the pH level at which an amino acid's carboxyl group donates a proton. Similarly, pKb is related to the base dissociation constant, Kb, and signifies the pH level at which the amino group accepts a proton.

For alanine, with a Ka value of 4.5 x 10-3 and a consequent pKa that is relatively low, we can infer that the carboxyl group is a stronger acid compared to the basicity of the amino group with a Kb of 7.4 x 10-5. It implies that at a lower pH (higher concentration of H+ ions), alanine tends to lose a proton and exist predominantly in its anionic form. In contrast, at a higher pH (lower concentration of H+ ions; higher OH- ions), alanine is more likely to accept a proton at the amino group, prevailing in its cationic form.

These dissociation behaviors explain the existence of three forms of alanine: anionic, cationic, and the neutral zwitterion form, depending on the surrounding pH levels. Through titration experiments and pH measurements, we can determine the precise points at which alanine transitions between these forms, further illuminating the complex and dynamic nature of amino acids.

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Most popular questions from this chapter

Two drops of indicator HIn \(\left(K_{\mathrm{a}}=1.0 \times 10^{-9}\right),\) where HIn is yellow and \(\operatorname{In}^{-}\) is blue, are placed in \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HCl. a. What color is the solution initially? b. The solution is titrated with 0.10 \(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after \(200.0 \mathrm{mL}\) NaOH has been added?

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 M C_{2} H_{5} N H_{2} / 0.25 M C_{2} H_{5} N H_{3} C l\) b. \(0.25 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\) c. \(0.50 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\)

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.10 M\) acetic acid/0.25 \(M\) sodium acetate b. \(0.25 M\) acetic acid/0.10 \(M\) sodium acetate c. \(0.080 M\) acetic acid/0.20 \(M\) sodium acetate d. \(0.20 M\) acetic acid/0.080 \(M\) sodium acetate

One method for determining the purity of aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: $$\begin{aligned} &\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\\\&\text { Aspirin } \quad \frac{\text { Boil }}{10 \min } \underset{\text { Salicylate ion }}{\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{3}^{-}(a q)}+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ A sample of aspirin with a mass of 1.427 g was boiled in \(50.00 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH} .\) After the solution was cooled, it took \(31.92 \mathrm{mL}\) of \(0.289 \mathrm{M}\) HCl to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

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