Question:
At pKa = pH, which of the following is true for a drug?
At pKa = pH, which of the following is true for a drug?
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When pH equals pKa, the log of the ratio is zero, so the split is even.
Updated On: Jun 24, 2026
- Concentration of drug is 50% ionic and 50% non-ionic
- Absorption of drug is 50% ionic and 50% ionic
- Concentration of drug is 75% ionic and 25% non-ionic
- Concentration of drug is 25% ionic and 75% non-ionic
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The Correct Option is A
Solution and Explanation
Step 1: The Henderson-Hasselbalch equation links pH, pKa and the ratio of ionized to unionized drug. For a weak acid: \(\text{pH} = \text{pKa} + \log\dfrac{[\text{ionized}]}{[\text{unionized}]}\).
Step 2: When pH equals pKa, the equation gives \(\log\dfrac{[\text{ionized}]}{[\text{unionized}]} = 0\), which means the ratio is 1, since \(\log 1 = 0\).
Step 3: A ratio of 1 means the ionized and unionized forms are present in equal amounts, that is 50% ionic and 50% non-ionic. Hence option A is correct.
Step 2: When pH equals pKa, the equation gives \(\log\dfrac{[\text{ionized}]}{[\text{unionized}]} = 0\), which means the ratio is 1, since \(\log 1 = 0\).
Step 3: A ratio of 1 means the ionized and unionized forms are present in equal amounts, that is 50% ionic and 50% non-ionic. Hence option A is correct.
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