The amino acid alanine,HOOC-CH(CH₃)NH₃⁺,has K_a1 = 4.5 * 10^-3 and K_a2 = 1.4 * 10^-10.Calculate $\alpha$(^-OOC-CH(CH₃)NH₃⁺)at pH 3.

Question

Quizplus · Accepted Answer

We can use the Henderson-Hasselbalch equation to calculate the ratio F1F1F1S1F1F1F10(F1F1F1S1^F1F1F10OOC-CH(CH₃)NH₃⁺) at pH 3: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of the carboxyl group in alanine (approximately 2.34), [A-] is the concentration of alanine in its deprotonated form (with a negative charge on the carboxyl group), and [HA] is the concentration of alanine in its protonated form (with no charge on the carboxyl group). At pH 3, we can assume that most of the alanine molecules will be protonated, so [HA] ≈ total concentration of alanine. We can also assume that the concentration of hydroxide ions (OH-) is negligible compared to the concentration of hydrogen ions (H+), so [A-] ≈ 0. Therefore, we can simplify the Henderson-Hasselbalch equation to: pH = pKa + log(0/[HA]) which gives us: [HA] = 10^(pKa - pH) = 10^(2.34 - 3) = 0.468 Now we can use the equation for the fractional saturation of alanine: F1F1F1S1F1F1F10(F1F1F1S1^F1F1F10OOC-CH(CH₃)NH₃⁺) = [_a1]/([_a1] + K_a1) + [_a2]/([_a2] + K_a2) where [_a1] and [_a2] are the concentrations of alanine in its protonated form and deprotonated form, respectively. Using the values given in the problem, we get: F1F1F1S1F1F1F10(F1F1F1S1^F1F1F10OOC-CH(CH₃)NH₃⁺) = [0.468]/([0.468] + 4.5∙10^-10) + [1]/([1] + 1.4∙10^-10) ≈ 0.823 Therefore, the answer is B, 0.82.

The Amino Acid Alanine,HOOC-CH(CH3)NH3+,has Ka1 = 4 α\alphaα (-OOC-CH(CH3)NH3+)at PH 3

The Amino Acid Alanine,HOOC-CH(CH₃)NH₃⁺,has K_a1 = 4 $\alpha$ (^-OOC-CH(CH₃)NH₃⁺)at PH 3