At what pH is a functional group expected to be ~90% deprotonated if its pKa is 4?

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Question: At what pH is a functional group expected to be ~90% deprotonated if its pKa is 4?

Answer options: ["5", "2", "3", "4", "6"]

Now let's analyze each option step by step.

Option 1: 5
- This corresponds to pH = pKa + 1. According to the Henderson–Hasselbalch equation, for a weak acid HA with conjugate base A−, the ratio [A−]/[HA] = 10^(pH − pKa).
- With pH = 5 and pKa = 4, the ratio is 10^(1) = 10, meaning there are 10 parts deprotonated for every 1 part protonated. Therefore, the fraction deprotonated is 10/(10 + 1) ≈ 0.909, i.e., about 90%. This matches the target of ~90% deprotonated.

Option 2: 2
- Here pH − pKa = 2 − 4 = −2. The ratio [A−]/[HA] = 10^(−2) = 0.01, meaning 1 part deprotonated per 99 parts protonated. The fraction deprotonated is 0.01/1.01 ≈ 1%, far from 90%.

Option 3: 3
- pH − pKa = 3 − 4 = −1, so [A−]/[HA] = 10^(−1) = 0.1. The deprotonated fraction is 0.1/1.1 ≈ 9%, again far from 90%.

Option 4: 4
- Here pH equals pKa, so [A−]/[HA] = 10^(0) = 1. This yields a 1:1 ratio, meaning 50% deprotonated. That is not ~90%.

Option 5: 6
- pH − pKa = 6 − 4 = 2, so [A−]/[HA] = 10^2 = 100. The deprotonated fraction is 100/101 ≈ 99%, which is more than ~90%.

Summary of reasoning:
- The target is ~90% deprotonated, which occurs when the ratio [A−]/[HA] is about 9:1, i.e., pH ≈ pKa + 1.
- Among the options, pH 5 yields ~90% deprotonation, consistent with the Henderson–Hasselbalch relationship.
- pH 2 gives ~1% deprotonation (too low), pH 3 gives ~9% (still too low), pH 4 gives 50% (not close to 90%), and pH 6 gives ~99% (too high if specifically ~90%).

At what pH is a functional group expected to be ~90% deprotonated if its pKa is 4?

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