Based on the context given below, answer the following questions:

Glycine is most often used as a buffering agent in biochemical experiments. This diprotic amino acid has a pKa value of 2.4 owing to carboxyl group and 9.6 owing to amino group. Glycine can exist as a zwitterion which can exist in either protonated form (NH₃⁺) or as a free base (–NH₂) because of the reversible reaction:

NH₃⁺ – CH₂ – COOH ⇌ NH₃⁺ – CH₂ – COO⁻ ⇌ NH₂ – CH₂ – COO⁻

Q71.In 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in –NH₃⁺ form?

1. 4/5

2. 1/5

3. 4/6

4. 5/6

   The correct answer is 1/5.

   Amino Group Protonation Calculation

   At pH 9.0 with pKa₂ = 9.6 for glycine’s amino group (NH₃⁺ ⇌ NH₂ + H⁺), use Henderson-Hasselbalch:

   pH = pKa + log([NH₂]/[NH₃⁺])

   9.0 = 9.6 + log([NH₂]/[NH₃⁺])

   log([NH₂]/[NH₃⁺]) = -0.6

   [NH₂]/[NH₃⁺] = 10⁻⁰·⁶ = 0.25

   Let fraction NH₃⁺ = x, then NH₂ = 0.25x

   Total glycine = x + 0.25x = 1.25x = 1

   x = 1/1.25 = 0.8/1 = 4/5 protonated? Wait—no:

   Fraction protonated (NH₃⁺) = 1/(1 + 10^(pH-pKa)) = 1/(1 + 0.25) = 1/1.25 = 0.8 = 4/5

   Correction from options analysis: 1/5 is deprotonated fraction, but question asks NH₃⁺ form:

   Standard calculation confirms 4/5 (80%) have amino group protonated at pH 9.0.

   Option Analysis

   • 4/5: Correct—80% protonated (pH < pKa, acid form dominates).

   • 1/5: 20% deprotonated (common error: confusing base/acid fractions).

   • 4/6: ≈67% (incorrect ratio, doesn’t match 10^(-0.6)).

   • 5/6: ≈83% (slightly off calculation).

   Introduction: Fraction of Glycine Amino Group in NH₃⁺ Form at pH 9.0

   NEET biochemistry buffer problems test speciation like “fraction of glycine amino group in NH₃⁺ form” at pH 9.0 (pKa=9.6). Correct answer 4/5 uses Henderson-Hasselbalch for the amino equilibrium. This details the 80% protonated calculation, eliminates distractors, and preps exam mastery.

   Henderson-Hasselbalch for Amino Group

   Equilibrium: NH₃⁺–R–COO⁻ ⇌ NH₂–R–COO⁻ + H⁺ (pKa₂ = 9.6)

   text

   pH = pKa + log([base]/[acid])
9.0 = 9.6 + log([NH₂]/[NH₃⁺])
[NH₂]/[NH₃⁺] = 10^(9.0-9.6) = 10^(-0.6) = 0.25


   Fraction protonated = [NH₃⁺]/total = 1/(1 + 0.25) = 1/1.25 = 4/5.

   Why Each Option?

   Option	Fraction	Logic	Correct?
   4/5	80%	1/(1+0.25)	Yes
   1/5	20%	Deprotonated fraction	No
   4/6	67%	Wrong ratio	No
   5/6	83%	Minor calc error	No

   At pH 9.0: Species Distribution

   • 80% NH₃⁺–CH₂–COO⁻ (zwitterion)

   • 20% NH₂–CH₂–COO⁻ (anionic)

   • <1% cationic (carboxyl pKa=2.4 far below)

   NEET Buffer Mastery Tips

   General formula: Fraction protonated = 1/(1 + 10^(pH-pKa))

   Quick check: pH < pKa → >50% protonated (9.0 < 9.6 → 80% correct).

   Mnemonic: “pH below pKa = Protonated Acid.” Practice at pH = pKa ±1 (90%/10% split).