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Chapter 15: Problem 94

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 \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 \mathrm{M} ?\)

Short Answer

Expert verified
In a solution with \(\left[\mathrm{H}^{+}\right] = 1.0 M\), the alanine ion with a net positive charge (+1) will predominate. In a solution with \(\left[\mathrm{OH}^{-}\right] = 1.0 M\), the alanine ion with a net negative charge (-1) will predominate.

Step by step solution

01

Calculate the pH and pOH of the given solutions

First, we will calculate the pH and pOH of the solutions with [H+]=1.0M and [OH-]=1.0M. For the solution with [H+]=1.0M: pH = -log([H+]) = -log(1.0) = 0 Now find the pOH of this solution using the relationship: pH + pOH = 14 0 + pOH = 14 pOH = 14 For the solution with [OH-]=1.0M: pOH = -log([OH-]) = -log(1.0) = 0 Now find the pH of this solution using the relationship: pH + pOH = 14 pH + 0 = 14 pH = 14
02

Determine the pKa and pKb of alanine

Given the \(K_{a} = 4.5 \times 10^{-3}\) and \(K_{b} = 7.4 \times 10^{-5}\), we can find the pKa and pKb values. pKa = -log(\(K_{a}\)) = -log(\(4.5 \times 10^{-3}\)) ≈ 2.35 pKb = -log(\(K_{b}\)) = -log(\(7.4 \times 10^{-5}\)) ≈ 4.13
03

Compare pH, pKa, and pKb for the solution with [H+]

In a solution with pH=0, we can make the following comparisons: - Since pH < pKa, the carboxylic acid group will be protonated and present in the cationic form. - Since pH < pKb, the amino group will be protonated and also present in the cationic form. Therefore, in the solution with [H+]=1.0 M, the predominant ion of alanine will have both the carboxylic acid and amino groups protonated, resulting in a net positive charge (+1).
04

Compare pH, pKa, and pKb for the solution with [OH-]

In a solution with pH=14, we can make the following comparisons: - Since pH > pKa, the carboxylic acid group will be deprotonated and present in the anionic form. - Since pH > pKb, the amino group will be deprotonated and present in the anionic form. Therefore, in the solution with [OH-]=1.0 M, the predominant ion of alanine will have both the carboxylic acid and amino groups deprotonated, resulting in a net negative charge (-1). Answer: In a solution with [H+]=1.0 M, the alanine ion with a net positive charge (+1) will predominate. In a solution with [OH-]=1.0 M, the alanine ion with a net negative charge (-1) will predominate.

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Most popular questions from this chapter

A friend asks the following: “Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH2 to form A2. Thus the amount of acid (HA) is decreased, and the amount of base (A2) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base (A2). Thus how can we claim that a buffered solution resists changes in the pH of the solution?” How would you explain buffering to this friend?

What volumes of 0.50 \(\mathrm{M} \mathrm{HNO}_{2}\) and 0.50 \(\mathrm{M}\) NaNO, must be mixed to prepare 1.00 \(\mathrm{L}\) of a solution buffered at \(\mathrm{pH}=3.55 ?\)

Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HClO}_{4}$ by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at $\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}} \text { for } \mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)$
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