Answer:
The correct answer is 1 x 10⁻⁵
Explanation:
In order to solve the problem, we can use the Henderson-Hasselbach equation to find the ratio alanine-COOH/alanine-COO⁻ (amino acid species with carboxylic acid protonated) and the ratio alanine-NH₂/alanine-NH₃⁺ (amino acid species with amino group neutral):
Henderson-Hasselbach equation: [tex]pH= pKa + log \frac{[A^{-} ]}{[HA]}[/tex]
For carboxylic acid group (pKa= 3) at pH 7:
[tex]7 = 3 + log \frac{[Ala-COO^{-} ]}{[Ala-COOH]}[/tex]
[tex]4= log \frac{[ Ala-COO^{-} ]}{[Ala-COOH]}[/tex]
[tex]10^{4} = \frac{[Ala-COO^{-} ]}{[Ala-COOH ]}[/tex]
We need the ratio of species with protonated carboxylic acid group, so we need the inverse:
[tex]10^{-4}= \frac{[Ala-COOH ]}{[Ala-COO^{-} ] }[/tex]
For amino group (pKa= 8) at pH 7:
[tex]7= 8 + log \frac{[Ala-NH_{2} ]}{[Ala-NH_{3} ^{+} ]}[/tex]
[tex]-1= log \frac{[ Ala-NH_{2} ]}{[Ala-NH_{3} ^{+} ]}[/tex]
[tex]10^{-1} = \frac{[Ala-NH_{2} ]}{[Ala-NH_{3} ^{+} ]}[/tex]
Finally, to find the ratio of neutral species we multiply the ratios:
[tex]\frac{[Ala-COOH ]}{[Ala-COO^{-} ]}[/tex] X [tex]\frac{[Ala-NH_{2} ] }{[ Ala-NH_{3} ^{+} ]}[/tex]= 10⁻⁴ x 10⁻¹= 10⁻⁵
