Question

If pKa of a weak electrolyte is 4.2, how much percentage of it will be ionized at pH of 5.2?

• A. 50%
• B. $>$50%
• C. $<$50%
• D. Cannot be predicted

Hint:

If pH $>$ pKa, the drug is more ionized; if pH $<$ pKa, it's less ionized.

Answer 1

The Correct Option is B

Using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \Rightarrow 5.2 = 4.2 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \Rightarrow \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 1 \Rightarrow \frac{[\text{A}^-]}{[\text{HA}]} = 10 \] This implies that $>$90% of the drug exists in ionized form. Hence, more than 50% is ionized.