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

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At pH 3, alanine exists primarily in its fully protonated form, meaning the amino group (-NH3+) and the carboxyl group (-COOH) are both in their protonated states. The given pKa values are for the carboxyl group (pKa1 = 4.5) and the amino group (pKa2 = 1.4). To calculate the fraction of alanine molecules ($\alpha$) in the fully protonated form at pH 3, we use the Henderson-Hasselbalch equation for the carboxyl group, since its pKa is closest to the pH in question. The equation simplifies to $\alpha = \frac{1}{1 + 10^{pH-pKa1}}$. Substituting the given values, $\alpha = \frac{1}{1 + 10^{3-4.5}} = \frac{1}{1 + 10^{-1.5}} \approx 0.82$. This calculation assumes the dominant species at pH 3 is the one with both the amino and carboxyl groups protonated, which corresponds to the fully protonated form of alanine.

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