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Chapter 21: Problem 133

All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{a}=4.5 \times 10^{-3}\) and the amino group has \(K_{b}=7.4 \times 10^{-5}.\) Three 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 \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\), the predominant ion of alanine is \(NH_3^+CH_2COOH\), and in a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 \mathrm{M}\), the predominant ion of alanine is \(NH_2CH_2COO^-\).

Step by step solution

01

1. Converting Ka and Kb to pKa and pKb

First, convert the given \(K_a\) and \(K_b\) values to pKa and pKb by using the equations: \(pK_{a} = -\log_{10}{K_{a}}\) and \(pK_{b} = -\log_{10}{K_{b}}\) For alanine, \(pK_{a} = -\log_{10}{(4.5 \times 10^{-3})} = 2.35\) \(pK_{b} = -\log_{10}{(7.4 \times 10^{-5})} = 4.13\)
02

2. Calculate pH values for both solutions

This can be done using the equations: \(pH = -\log_{10}{[\mathrm{H}^{+}]}\) for the first solution, and \(pOH = -\log_{10}{[\mathrm{OH}^{-}]}\) and \(pH + pOH = 14\) for the second solution. For the first solution: \(pH = -\log_{10}{(1.0 \mathrm{M})} = 0\) For the second solution: \(pOH = -\log_{10}{(1.0 \mathrm{M})} = 0\) Since \(pH + pOH = 14\), \(pH = 14\)
03

3. Compare pH values with pKa and pKb to determine predominant ions

Now, we'll compare the pH values of the solutions to the pKa and pKb values. For the first solution with \(pH = 0\): Since \(pH < pK_a\), the carboxylic acid group will donate a proton, and alanine will be in its cationic form: \(NH_3^+CH_2COOH\). For the second solution with \(pH = 14\): Since \(pH > pK_b\), the amino group will gain a proton, and alanine will be in its anionic form: \(NH_2CH_2COO^-\). So, in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\), the predominant ion of alanine is \(NH_3^+CH_2COOH\), and in a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 \mathrm{M}\), the predominant ion of alanine is \(NH_2CH_2COO^-\).

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