determine drug absorption. pKa of lidocaine is 7.9

determine drug absorption. pKa of lidocaine is 7.9

 

 

The pKa of lidocaine is 7.9. Which fraction of the drug will be in the lipid soluble form in the extracellular fluids (pH = 7.4)?

A. 10%

B. 24%

C. 50%

D. 76%

E. 90%

 

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percent ionized formula

 

 

 

100 / [1 + 10^((x) (pH - pKa))]

where x = -1 if acid drug or 1 if basic drug

 

 

100 / [1 + 10^((-1) (7.4 - 7.9))]

= 24.03%

 

 

Answer: B. 24%

 

 

 

 

HA(aq) ⇌ H^+(aq) + A^-(aq)
( HA ≡ 리도카인 )
( A^- ≡ 이온화된 리도카인 )

 

Henderson-Hasselbalch 식
( 식 설명 https://ywpop.tistory.com/1926 )

 

pH = pKa + log([A^-]/[HA])
log([A^-]/[HA]) = pH – pKa = 7.4 – 7.9 = -0.5
[A^-]/[HA] = 10^(-0.5) = 0.316228
[A^-] : [HA] = 0.316228 : 1

 

[A^-] / ([HA]+[A^-]) × 100
= 0.316228 / (1+0.316228) × 100 = 24.03% 

 

 

 

 

예제 1) The pH of the stomach is 2.5. The pKa of sodium pentothal is 7.4 and it is acidic. If a patient is given sodium pentothal orally instead of IV, will it put the patient to sleep?
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100 / [1 + 10^((-1) (2.5 – 7.4))] = 0.001259%
---> None of the sodium pentothal is ionized in the stomach. Therefore the patient would absorb 100% of this medication and it would put him to sleep.

 

 

예제 2) A basic drug with a pKa of 7.8 is a known teratogen. If given IV to a pregnant woman whose blood pH is 7.4, will this drug cross the placenta and effect the baby?
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100 / [1 + 10^((1) (7.4 – 7.8))] = 71.53%
---> 72% of the drug is ionized which means 28% of the drug is unionized and will pass through the placenta to effect the baby.