Q. how to calculate isoelectric point with 3 pKa’s

Q: How to handle equal or very close pKvalues?

If two relevant \mathrm{p}K_a values are equal or within <1 pH unit, average them as usual. If they are identical, the pI equals that common value. Very close values broaden the buffering region, but pI remains the arithmetic mean.

Q: Can I use Henderson Hasselbalch to find pI?

Yes. Compute fractional protonation for each ionizable group with \alph= \frac{1}{1 + 10^{\mathrm{p}K_- \mathrm{pH}}} . Sum charges over all groups. Solve for pH that yields net charge zero, which will equal the average of the two pKvalues surrounding the neutral species.

Q: Example calculation for \mathrm{p}K_{a1}=2.0, \mathrm{p}K_{a2}=9.0, \mathrm{p}K_{a3}=10.5 ?

Identify neutral species between \mathrm{p}K_{a2} and \mathrm{p}K_{a3} if the midpoint charge sequence fits. Then \mathrm{pI} = \frac{9.0 + 10.5}{2} = 9.75 .

Q: How to confirm pI by computing net charge versus pH?

Calculate each group's fraction protonated with Henderson Hasselbalch at candidate pH values. Multiply by group charges and sum. Adjust pH until total charge equals zero. The root should match the average of the two pKvalues that flank the neutral species.

Answer

Order the pK_a values as \( \mathrm{p}K_{a1} \le \mathrm{p}K_{a2} \le \mathrm{p}K_{a3}\). Identify which two pK_a values flank the neutral species. For a triprotic molecule that changes charge +1 to 0 to -1 as protons are removed, the isoelectric point is the average of the two pK_a values on either side of the neutral form.

Formula: \[
\mathrm{pI} \;=\; \frac{\mathrm{p}K_{a(i)} \;+\; \mathrm{p}K_{a(j)}}{2},
\]

where i and j are the pK_a values for the +1 to 0 and 0 to -1 transitions respectively.

If the neutral form lies between \(\mathrm{p}K_{a1}\) and \(\mathrm{p}K_{a2}\) then \(\mathrm{pI} = \frac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2}\).

Detailed Explanation

Isoelectric point, or pI, is the pH at which a molecule has net zero charge. For a molecule with three ionizable groups (three pKa values), find which two pKa values bracket the neutral (net zero) species and take their average.

Step 1. List the three pKa values in ascending order: \( \mathrm{p}K_{a1} \lt \mathrm{p}K_{a2} \lt \mathrm{p}K_{a3} \).

Step 2. Determine which protonation state has net charge zero. For typical amino acids with three pKa values this falls into one of two patterns.

Step 3. Identify the two pKa values that flank the neutral species. The isoelectric point is the average of those two pKa values. In formula form:

\[ \mathrm{pI} \;=\; \frac{\mathrm{p}K_{a,\text{lower}} \;+\; \mathrm{p}K_{a,\text{upper}}}{2} \]

Notes on common cases. If the side chain is acidic (example: aspartic or glutamic acid), the neutral (zwitterionic) form lies between the two lowest pKa values, so use those two. If the side chain is basic (example: lysine or arginine), the neutral form lies between the two highest pKa values, so use those two.

Example 1. Aspartic acid (acidic side chain). Using representative pKa values \( \mathrm{p}K_{a1}=2.10 \), \( \mathrm{p}K_{a2}=3.86 \), \( \mathrm{p}K_{a3}=9.82 \). The neutral species lies between \( \mathrm{p}K_{a1} \) and \( \mathrm{p}K_{a2} \). Compute

\[ \mathrm{pI} \;=\; \frac{2.10 \;+\; 3.86}{2} \;=\; \frac{5.96}{2} \;=\; 2.98 \]

Example 2. Lysine (basic side chain). Using representative pKa values \( \mathrm{p}K_{a1}=2.18 \), \( \mathrm{p}K_{a2}=8.95 \), \( \mathrm{p}K_{a3}=10.53 \). The neutral species lies between \( \mathrm{p}K_{a2} \) and \( \mathrm{p}K_{a3} \). Compute

\[ \mathrm{pI} \;=\; \frac{8.95 \;+\; 10.53}{2} \;=\; \frac{19.48}{2} \;=\; 9.74 \]

Summary. Identify which two pKa values flank the neutral form, then compute the average of those two pKa values using the formula above. This gives the isoelectric point for a molecule with three pKa values.

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Chemistry FAQs

What is the isoelectric point, pI, in three pKsystem?

The pI is the pH where net charge is zero. For triprotic molecule, identify the two successive deprotonation equilibrithat flank the neutral species. Average their dissociation constants. Use \( \mathrm{pI} = \frac{\mathrm{p}K_{a\_low} + \mathrm{p}K_{a\_high}}{2} \).

How do I choose which two pKvalues to average?

Find the species distribution by charge as protons are removed. The neutral species occurs between the protonation states that differ by one proton. Average the two pKvalues that correspond to those two protonation steps, surrounding the neutral form.

What if the side chain is ionizable, giving three pKvalues?

Determine the sequence of charges as pH increases. If the neutral form lies between \(\mathrm{p}K_{a1}\) and \(\mathrm{p}K_{a2}\), average those. If the neutral form is between \(\mathrm{p}K_{a2}\) and \(\mathrm{p}K_{a3}\), average those. The side chain shifts which pair you use.

How to handle equal or very close pKvalues?

If two relevant \(\mathrm{p}K_a\) values are equal or within <1 pH unit, average them as usual. If they are identical, the pI equals that common value. Very close values broaden the buffering region, but pI remains the arithmetic mean.

Can I use Henderson Hasselbalch to find pI?

Yes. Compute fractional protonation for each ionizable group with \( \alph= \frac{1}{1 + 10^{\mathrm{p}K_- \mathrm{pH}}} \). Sum charges over all groups. Solve for pH that yields net charge zero, which will equal the average of the two pKvalues surrounding the neutral species.

Example calculation for \(\mathrm{p}K_{a1}=2.0\), \(\mathrm{p}K_{a2}=9.0\), \(\mathrm{p}K_{a3}=10.5\) ?

Identify neutral species between \(\mathrm{p}K_{a2}\) and \(\mathrm{p}K_{a3}\) if the midpoint charge sequence fits. Then \( \mathrm{pI} = \frac{9.0 + 10.5}{2} = 9.75 \).

How to confirm pI by computing net charge versus pH?