1. Identify the ionizable groups in aspartic acid

   Aspartic acid has three ionizable groups:

   • $\alpha$-carboxyl group: $pK_{a1} = 1.88$
   • side-chain carboxyl group: $pK_{aR} = 3.65$
   • $\alpha$-amino group: $pK_{a2} = 9.60$ (not needed for pI here)

2. Write the ionic forms as pH increases

   • At very low pH: fully protonated, net charge $+1$
   • First deprotonation at $pK_{a1} = 1.88$: net charge becomes $0$
   • Second deprotonation at $pK_{aR} = 3.65$: net charge becomes $-1$

   So, the zwitterion (net zero charge) exists between the two acidic dissociations $1.88$ and $3.65$.

3. Formula for isoelectric point of acidic amino acids

   For acidic amino acids like aspartic acid, $pI = \frac{pK_{a1} + pK_{aR}}{2}$

4. Calculate

   $pI = \frac{1.88 + 3.65}{2} = \frac{5.53}{2} = 2.765$

   Therefore, $pI \approx 2.77$

5. Check options

   • A: $1.88$ ❌
   • B: $3.65$ ❌
   • C: $5.74$ ❌
   • D: $2.77$ ✅

Hence, the correct answer is Option D.