If a functional group has a pKa of 3, then what is the (approximate) likelihood that this functional group will be protonated at pH 5?

Question

BlackTom AI · Accepted Answer

To assess protonation probability, use the Henderson–Hasselbalch framework for an acid HA with pKa = 3.
At pH 5, the pH–pKa difference is 5 − 3 = 2, so the ratio [A−]/[HA] = 10^(pH−pKa) = 10^2 = 100.
The fraction that is protonated (i.e., in the HA form) is [HA]/([HA] + [A−]) = 1/(1 + [A−]/[HA]) = 1/(1 + 100) ≈ 1/101 ≈ 0.0099, or about 1%.
Now let’s examine the other choices to see why they’re not correct:
- 10%: This would correspond to a smaller pH−pKa difference, around 1 (since [A−]/[HA] ≈ 10, giving ~1/11 ≈ 9%), not 2.
- 50%: This occurs when pH ≈ pKa, because then [A−] ≈ [HA], giving equal protonated and deprotonated forms; here pH is 2 units above pKa, far from equal.
- 90%: That would require pH to be well below pKa (more acidic than the pKa by about 1 unit or more) so that HA dominates; at pH 5 and pKa 3, HA is far less favored.
- 99%: This would correspond to pH ≪ pKa by about 2 units or more; again, our pH 5 is not that strongly acidic relative to pKa 3, so protonation is far from 99%.
In summary, with pH two units above the pKa, the protonated fraction is about 1%, aligning with the given correct option.

If a functional group has a pKa of 3, then what is the (approximate) likelihood that this functional group will be protonated at pH 5?

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