Thu18November0755PM 6

A drug that is weakly acidic is injected intravenously into a patient. It has a pKa of 8.4.

What percentage of this drug is unionised in plasma at a normal physiological pH?

(Please select 1 option)

99

10

1 Incorrect answer selected

90 This is the correct answer

50

Explanation

There are two issues pertinent to Primary FRCA. The first is knowledge and understanding of the Henderson-Hasselbalch equation and secondly the mathematical concepts of logarithms and antilogarithms.

The pKa is defined as the pH at which drug exists as 50% ionised and 50% unionised forms.

The Henderson-Hasselbalch equation can be used to calculate the proportion of ionised to unionised form of a drug.

pH = pKa + log ([A-]/[HA])

or

pH = pKa + log [(salt)/(acid)]

or

pH = pKa + log ([ionised]/[unionised])

Therefore if the pKa - pH = 0, then 50% of drug is ionised and 50% is unionised.

In this example the drug has a pKa of 8.4 and the pH of blood is 7.4

7.4 = 8.4 + log ([ionised]/[unionised])

7.4 - 8.4 = log ([ionised]/[unionised])

log -1 = log ([ionised]/[unionised])

The antilog is simply the inverse log calculation. Namely, if you know the logarithm of a number, you can compute the value of the number itself by taking the antilog. The antilogarithm is defined as:

y = antilog x = 10x

Antilog to the base 10 of 0 = 1, -1 = 0.1, -2 = 0.01, -3 = 0.001 and, -4 = 0.0001.

[A-]/[HA] = 0.1

Assuming that we can apply the approximation [A-] << [HA} then this means the acid is 0.1 x 100% = 10% ionised

so the percentage of (non-ionized) acid will be 100% - 10% = 90%

Answer Statistics

1

7%

2

43%

3

5%

4

35%

5

12%

Times answered: 242