Question:
If the pKa of phenobarbitone is 3.4 what would be pH 4.4:
If the pKa of phenobarbitone is 3.4 what would be pH 4.4:
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When $pH = pKa + 1$, a weak acid is 90% ionized.
Updated On: May 25, 2026
- 1.0
- 0.01
- 0.5
- 0.90
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The Henderson-Hasselbalch equation relates pH, pKa, and the ratio of ionized to unionized drug forms.
Step 2: Meaning
For a weak acid like phenobarbitone: $pH = pKa + \log(\frac{Ionized}{Unionized})$.
Step 3: Analysis
$4.4 = 3.4 + \log(\frac{I}{U}) \implies 1.0 = \log(\frac{I}{U}) \implies \frac{I}{U} = 10^1 = 10$. The fraction ionized is $\frac{I}{I+U} = \frac{10}{10+1} = \frac{10}{11} \approx 0.909$.
Step 4: Conclusion
The extent of ionization (or ratio) corresponds to 0.90. Final Answer: (D)
The Henderson-Hasselbalch equation relates pH, pKa, and the ratio of ionized to unionized drug forms.
Step 2: Meaning
For a weak acid like phenobarbitone: $pH = pKa + \log(\frac{Ionized}{Unionized})$.
Step 3: Analysis
$4.4 = 3.4 + \log(\frac{I}{U}) \implies 1.0 = \log(\frac{I}{U}) \implies \frac{I}{U} = 10^1 = 10$. The fraction ionized is $\frac{I}{I+U} = \frac{10}{10+1} = \frac{10}{11} \approx 0.909$.
Step 4: Conclusion
The extent of ionization (or ratio) corresponds to 0.90. Final Answer: (D)
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