Question

Ionization percentage of a drug when pH of the medium = pKa of drug is:

• A. 100%
• B. 50%
• C. 0%
• D. 1%

Hint:

The $pK_a$ of a drug is formally defined as the specific $pH$ value at which the drug is exactly $50\%$ ionized and $50\%$ unionized.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Most drugs are either weak acids or weak bases and exist in solution in both ionized (charged) and unionized (uncharged) forms.
The degree of ionization depends on the $pH$ of the medium and the dissociation constant ($pK_a$) of the drug.
Key Formula or Approach:
The relationship is described by the Henderson-Hasselbalch equation:
- For a weak acid: \[ pH = pK_a + \log\left(\frac{[\text{Ionized}]}{[\text{Unionized}]}\right) \] - For a weak base: \[ pH = pK_a + \log\left(\frac{[\text{Unionized}]}{[\text{Ionized}]}\right) \]

Step 2: Detailed Explanation:
Let us analyze the situation when the $pH$ of the medium is exactly equal to the $pK_a$ of the drug: \[ pH = pK_a \] Substituting this into the Henderson-Hasselbalch equation: \[ pK_a = pK_a + \log\left(\frac{[\text{Ionized}]}{[\text{Unionized}]}\right) \] \[ 0 = \log\left(\frac{[\text{Ionized}]}{[\text{Unionized}]}\right) \] Taking the antilog of both sides: \[ 10^0 = \frac{[\text{Ionized}]}{[\text{Unionized}]} \implies 1 = \frac{[\text{Ionized}]}{[\text{Unionized}]} \] \[ [\text{Ionized}] = [\text{Unionized}] \] This mathematical equality means that exactly half of the drug molecules are in the ionized state and the other half are in the unionized state.
Therefore, the percentage of ionization is exactly $50\%$.

Step 3: Final Answer:
The ionization percentage is $50\%$, which corresponds to Option (B).