Amino Acid Charge Analysis at pH 7.0

To determine the net charge of the amino acid at pH 7.0, we compare the pH value to the given $pK_a$ values for each ionizable group: $pK_{COOH} = 2.19$, $pK_{NH2} = 9.67$, and $pK_R = 4.25$.

The ionization state follows these rules:

• When pH > $pK_a$, the group is deprotonated.
• When pH < $pK_a$, the group is protonated.

Applying these rules at pH 7.0:

• Carboxyl group (-COOH): Since pH (7.0) > $pK_{COOH}$ (2.19), the group is deprotonated, carrying a charge of -1 (i.e., -COO⁻).
• Amino group (-NH₂): Since pH (7.0) < $pK_{NH2}$ (9.67), the group is protonated, carrying a charge of +1 (i.e., -NH₃⁺).
• Side chain R group: Since pH (7.0) > $pK_R$ (4.25), the proton-donating group in the R group is deprotonated, carrying a charge of -1 (R⁻).

The net charge is the sum of the individual charges:

Net Charge = (Charge of -NH₃⁺) + (Charge of -COO⁻) + (Charge of R⁻)

Net Charge = (+1) + (-1) + (-1) = -1

Therefore, the majority of the molecules will have a net charge of -1 at pH 7.0.

Analysis of Other Options

Evaluating the remaining options:

• Option 2: At pH 4.25, which is the $pK_R$, the R group is equally likely to be protonated or deprotonated. The -COOH group (pH > $pK_{COOH}$) is deprotonated (-1 charge), and the -NH₂ group (pH < $pK_{NH2}$) is protonated (+1 charge). The net charge is not 0 at pH 4.25. The isoelectric point ($pI$), where net charge is 0, is approximately $(pK_{COOH} + pK_R) / 2 = (2.19 + 4.25) / 2 = 3.22$.
• Option 3: At pH 3.22, the pH is less than $pK_R$ (4.25). This means the R group will be predominantly in its protonated form (R-H), not deprotonated.
• Option 4: During titration with a base, deprotonation occurs in order of increasing acidity (lowest $pK_a$ first). The $pK_a$ values are 2.19 ($pK_{COOH}$), 4.25 ($pK_R$), and 9.67 ($pK_{NH2}$). Thus, deprotonation begins with the -COOH group, not the R group.