Question

pka = 9.76 (chemical structure with \\(\stackrel{\oplus}{\text{nh}_3}\\) and \\(\text{cooh}\\), pka = 2.33) at ph 6.1 \\(\
ightarrow\\) ?
options:
a. (structure with \\(\text{nh}_2\\) and \\(\text{cooh}\\))
b. (structure with \\(\text{nh}_2\\) and \\(\stackrel{\ominus}{\text{coo}}\\))
c. (structure with \\(\stackrel{\oplus}{\text{nh}_3}\\) and \\(\text{cooh}\\))
d. (structure with \\(\stackrel{\oplus}{\text{nh}_3}\\) and \\(\stackrel{\ominus}{\text{coo}}\\))
choices: \\(\circ\\) a \\(\circ\\) b \\(\circ\\) c \\(\circ\\) d

Explanation:

Step1: Analyze the pKa values and pH

The molecule has two ionizable groups: the carboxylic acid (pKa = 2.33) and the ammonium group (pKa = 9.76). The solution pH is 6.1.

For the carboxylic acid group (pKa = 2.33), since pH (6.1) > pKa (2.33), the acid will deprotonate, forming the carboxylate ion ($\ce{-COO^-}$).

For the ammonium group (pKa = 9.76), since pH (6.1) < pKa (9.76), the ammonium group will remain protonated ($\ce{-NH_3^+}$).

Step2: Match with the options

• Option A: The ammonium group is deprotonated ( $\ce{-NH_2}$) and the carboxylic acid is protonated ($\ce{-COOH}$) – incorrect, as pH < pKa of ammonium (should be protonated) and pH > pKa of acid (should be deprotonated).
• Option B: The ammonium group is deprotonated ( $\ce{-NH_2}$) – incorrect, as pH < pKa of ammonium.
• Option C: The carboxylic acid is protonated ($\ce{-COOH}$) – incorrect, as pH > pKa of acid (should be deprotonated).
• Option D: The carboxylic acid is deprotonated ($\ce{-COO^-}$) and the ammonium group is protonated ($\ce{-NH_3^+}$) – matches the analysis.

Answer:

D. $\ce{NH_3^+}$ -containing and $\ce{COO^-}$ -containing structure (the structure in option D)