If a functional group has a pKa of 9, then what is the deprotonated:protonated ratio for this functional group at pH 4?

Question

BlackTom AI · Accepted Answer

We start by identifying what the question is asking: the deprotonated to protonated ratio for a functional group with pKa 9 at pH 4.
First, apply the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the deprotonated form and [HA] is the protonated form.
Rearranging for the ratio gives: log([A-]/[HA]) = pH − pKa = 4 − 9 = −5.
Exponentiating both sides yields [A-]/[HA] = 10^(-5) = 1/100,000, which means the deprotonated form is 1 part for every 100,000 parts of the protonated form.
Option-by-option analysis:
- Option "1:100,000": This matches the calculated ratio of deprotonated to protonated forms (1 part deprotonated per 100,000 parts protonated). Correct.
- Option "100,000:1": This would imply [A-]/[HA] = 100,000, which corresponds to pH far above pKa, not 4; incorrect.
- Option "1:10,000": This would be a ratio of 1 to 10,000 (1e-4), which corresponds to a pH of pKa − 4 (not 4) and is not what we calculated; incorrect.
- Option "10,000:1": This would be 1e4, implying pH much higher than pKa; incorrect.
- Option "None of the others is correct": Since the first option does match the calculation, this statement is false.
Thus, the correct interpretation is that the deprotonated:protonated ratio at pH 4 for a pKa of 9 is 1:100,000, aligning with the Henderson-Hasselbalch result.

If a functional group has a pKa of 9, then what is the deprotonated:protonated ratio for this functional group at pH 4?

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