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

To find the pH and pOH of the solutions, we can use the formulas: For H+ concentration: pH = -log10[H+] For OH- concentration: pOH = -log10[OH-] Since pH + pOH = 14, we can determine the pOH and pH of the solutions respectively. Given: \([H+] = 1.0 M\) pH = -log10(1.0) = 0 pOH = 14 - pH = 14 - 0 = 14 Given: \([OH^-] = 1.0 M\) pOH = -log10(1.0) = 0 pH = 14 - pOH = 14 - 0 = 14 The pH of the first solution is 0 and the second solution's pH is 14.

In order to compare the pH to the acidity and basicity of alanine, we will need to find the pKa and pKb values using the given Ka and Kb. pKa = -log10(Ka) pKb = -log10(Kb) Given: Ka = \(4.5 \times 10^{-3}\) and Kb = \(7.4 \times 10^{-5}\) For alanine carboxylic acid group: pKa = -log10(\(4.5 \times 10^{-3}\)) = 2.35 For alanine amino group: pKb = -log10(\(7.4 \times 10^{-5}\)) = 4.13

Now that we have found the pH, pOH, pKa, and pKb values, we can determine the predominant ions of alanine in the given conditions. In the first solution with pH = 0: Since the pH is significantly lower than the pKa value (0 < 2.35), the carboxylic acid group will lose a proton, creating a positively charged ion. The ion will be in the form NH3+CH2COO-. In the second solution with pH = 14: Since the pH is significantly higher than the pKb value (14 > 4.13), the amino group will gain a proton. Also, since the pH is much higher than the pKa value, the carboxylic acid group will lose a proton. The ion will be in the form NH2CH2COO-. Conclusion: In the solution with \(\left[\mathrm{H}^{+}\right]=1.0 M\), the predominant ion of alanine will be NH3+CH2COO-. In the solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M\), the predominant ion of alanine will be NH2CH2COO-.