Q. how to calculate isoelectric point

Q: How do I get the fractional protonation for an acid or base group?

For an acidic group, fraction deprotonated is \frac{1}{1 + 10^{\mathrm{p}K_- \mathrm{pH}}} . For basic group, fraction protonated is \frac{1}{1 + 10^{\mathrm{pH} - \mathrm{p}K_a}} .

Q: Which equation links pH and pKfor conjugate acid base pair?

Use Henderson-Hasselbalch. \mathrm{pH} = \mathrm{p}K_+ \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} . It relates pH to the ratio of deprotonated and protonated forms.

Answer

Identify all ionizable groups and their pKa values. Find the two pKa values that flank the neutral form of the molecule. The isoelectric point is the average of those two pKa values.

Using Henderson‑Hasselbalch to justify, at the point where the conjugate forms are equal the pH equals the pKa, so the pI is the mean of the two bordering pKa values.

\[ \mathrm{pI} \;=\; \frac{\mathrm{p}K_{a1} \;+\; \mathrm{p}K_{a2}}{2} \]

Example glycine with \mathrm{p}K_{a1}=2.34 and \mathrm{p}K_{a2}=9.60 gives

\[ \mathrm{pI} \;=\; \frac{2.34 \;+\; 9.60}{2} \;=\; 5.97 \]

Detailed Explanation

Definition. The isoelectric point, abbreviated as \( \mathrm{pI} \), is the pH at which a molecule (commonly an amino acid or peptide) has net zero electric charge. To calculate it you must know the acid dissociation constants (pK_a values) of all ionizable groups on the molecule, then determine which two pK_a values bracket the neutral form of the molecule, and finally take their average.

Step 1. Gather pK_a values. Identify every ionizable group and write its pK_a. For a simple amino acid these normally include the carboxyl group (pK_a1), the amino group (pK_a2), and, if present, an ionizable side chain (pK_a3, pK_a4, …).

Step 2. Determine the sequence of protonation states and the net charge of each microspecies. Start from very low pH (fully protonated form) and move to high pH (fully deprotonated form), removing one proton at each pK_a in order of increasing pK_a. After each deprotonation compute the net charge. The neutral species is the microspecies whose net charge equals zero.

Step 3. Identify the two pK_a values that flank the neutral species. Those are the pK_a that correspond to the protonation and the deprotonation steps immediately before and immediately after the neutral form.

Step 4. Compute the pI as the arithmetic mean of those two pK_a values. In formula form:

\[
\mathrm{pI} = \frac{\text{p}K_{a,\text{below}} + \text{p}K_{a,\text{above}}}{2}
\]

Notes. For amino acids with only two ionizable groups (no ionizable side chain, e.g. glycine) the neutral form lies between the two pK_a values, so the pI equals the average of those two pK_a values. For amino acids with an ionizable side chain (acidic or basic), determine which pair of successive pK_a values surrounds the neutral form; the pI is the average of that pair.

Worked example 1 — Glycine (no ionizable side chain). Typical pK_a values: pK_{a1} (COOH) = 2.34, pK_{a2} (NH3+) = 9.60. The neutral (zwitterion) form exists between these two pK_a values, so

\[
\mathrm{pI} = \frac{2.34 + 9.60}{2} = 5.97
\]

Worked example 2 — Lysine (basic side chain). Typical pK_a values: pK_{a1} (COOH) = 2.18, pK_{a2} (\alpha\text{-}NH_{3}^{+}) = 8.95, pK_{a3} (\varepsilon\text{-}NH_{3}^{+}) = 10.53. Starting from low pH the net charges are +2, +1, 0, then -1 as successive groups deprotonate. The neutral species lies between pK_{a2} and pK_{a3}, so

\[
\mathrm{pI} = \frac{8.95 + 10.53}{2} = 9.74
\]

Worked example 3 — Glutamic acid (acidic side chain). Typical pK_a values: pK_{a1} (COOH, \alpha) = 2.19, pK_{a2} (\text{side chain COOH}) = 4.25, pK_{a3} (NH3+) = 9.67. The neutral species is between pK_{a1} and pK_{a2}, therefore

\[
\mathrm{pI} = \frac{2.19 + 4.25}{2} = 3.22
\]

Summary checklist you can follow for any amino acid or peptide:

1. List all pK_a values in increasing order. 2. Starting at low pH, assign charges to each group and compute net charge after each deprotonation. 3. Find the microspecies with net charge zero. 4. Average the pK_a immediately below and immediately above that neutral microspecies to get the pI.

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Chemistry FAQs

What is the isoelectric point?

The isoelectric point, pI, is the pH where molecule has net zero charge. For amino acids and proteins, pI is the pH at which positive and negative charges balance, so the molecule does not migrate in an electric field.

How do I calculate pI for an amino acid with only two pKvalues?

For simple amino acids use the average of the two relevant pKvalues. \( \mathrm{pI} = \frac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2} \). For example, glycine uses the pKof the carboxyl and the amino groups.

How do I calculate pI for amino acids with charged side chains?

Identify the two ionizations that flank the neutral zwitterion. For acidic side chains average the two lowest pKvalues. For basic side chains average the two highest pKvalues. That average gives the pI.

How do I calculate pI for peptides or proteins?

Sum the fractional charges of all ionizable groups as function of pH using Henderson-Hasselbalch, then find the pH where net charge equals zero. Use numerical search or software for many residues.

How do I get the fractional protonation for an acid or base group?

For an acidic group, fraction deprotonated is \( \frac{1}{1 + 10^{\mathrm{p}K_- \mathrm{pH}}} \). For basic group, fraction protonated is \( \frac{1}{1 + 10^{\mathrm{pH} - \mathrm{p}K_a}} \).

Which equation links pH and pKfor conjugate acid base pair?

Use Henderson-Hasselbalch. \( \mathrm{pH} = \mathrm{p}K_+ \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \). It relates pH to the ratio of deprotonated and protonated forms.

Where do I find pKvalues for residues?

Use biochemical tables, primary literature, protein databases, or computational predictors. Experimental titration of the molecule gives the most accurate pKvalues for that environment.

What factors shift pI from textbook values?

Local environment, neighboring charges, solvent accessibility, ionic strength, temperature, and post translational modifications can change pK values, and therefore shift pI. Experimental measurement or advanced calculation accounts for these effects.

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