Question:easy
At pKa = pH -
At pKa = pH -
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Put pH = pKa into Henderson-Hasselbalch; log of 1 is zero.
Updated On: Jun 23, 2026
- Conc. of drug is 50% ionic and 50% non-ionic
- Absorption of drug is 50% ionic and 50% ionic
- Conc. of drug is 75% ionic and 25% non-ionic
- Conc. of drug is 25% ionic and 75% non-ionic
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The Correct Option is A
Solution and Explanation
Step 1: The pKa is, by definition, the pH value at which an ionisable drug exists half in its charged form and half uncharged.
Step 2: Write the Henderson-Hasselbalch equation for a weak acid: $pH = pKa + \log\frac{[A^-]}{[HA]}$.
Step 3: Substituting the condition $pH = pKa$ removes the pKa term, leaving $\log\frac{[A^-]}{[HA]} = 0$.
Step 4: Since $\log 1 = 0$, the ratio $\frac{[A^-]}{[HA]} = 1$, meaning ionised and unionised concentrations are equal - a 50:50 split.
Step 5: Hence at pKa = pH the drug concentration is 50% ionic and 50% non-ionic; the 75/25 and 25/75 options are wrong.\[\boxed{50\% \text{ ionic and } 50\% \text{ non-ionic}}\]
Step 2: Write the Henderson-Hasselbalch equation for a weak acid: $pH = pKa + \log\frac{[A^-]}{[HA]}$.
Step 3: Substituting the condition $pH = pKa$ removes the pKa term, leaving $\log\frac{[A^-]}{[HA]} = 0$.
Step 4: Since $\log 1 = 0$, the ratio $\frac{[A^-]}{[HA]} = 1$, meaning ionised and unionised concentrations are equal - a 50:50 split.
Step 5: Hence at pKa = pH the drug concentration is 50% ionic and 50% non-ionic; the 75/25 and 25/75 options are wrong.\[\boxed{50\% \text{ ionic and } 50\% \text{ non-ionic}}\]
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