Ionisation of Weak Acid and Base Drugs Using pH and pKa
Most drugs are weak acids or weak bases. Whether such a drug is mostly charged or mostly neutral depends on one thing: the pH of the fluid it is sitting in, compared with the drug’s pKa. This matters because, in the simple textbook picture, the neutral form crosses fatty cell membranes easily, while the charged form does not. The Henderson–Hasselbalch equation lets you calculate the split between the two forms. This is a good example of how a formula you already know from buffers connects to pharmaceutical and biological chemistry, and it makes a neat numerical for postgraduate entrance exams.
The formulas
pH = pKa + log ([A−] ÷ [HA])
ionised : unionised = 10(pH − pKa)
Weak base B (ionises to BH+, with pKa of BH+):
pH = pKa + log ([B] ÷ [BH+])
ionised : unionised = 10(pKa − pH)
Fraction ionised = r ÷ (1 + r), where r is the ionised : unionised ratio.
Two quick rules follow. When pH = pKa, the drug is exactly 50 % ionised, because r = 1. And every one pH unit away from the pKa changes the ratio by a factor of ten. For a base we always quote the pKa of its conjugate acid BH+; mixing this up with pKb is a common error.
Typical textbook pH values are about 1–3 for stomach fluid, about 6–7.5 in the small intestine, and about 7.4 for blood plasma. These are approximate and vary between people and with food, so use the pH values given in your question. The drugs below are hypothetical; the pKa values are chosen for practice, not taken from any real product.
Worked examples
Example 1: a weak acid drug in stomach and intestine. A drug HA has pKa = 3.50. What fraction is ionised at pH 2.0 and at pH 6.5?
At pH 2.0: r = 10(2.0 − 3.5) = 10−1.5 = 0.0316. Fraction ionised = 0.0316 ÷ 1.0316 = 0.0307, i.e. 3.07 % ionised, 96.9 % neutral.
At pH 6.5: r = 10(6.5 − 3.5) = 103 = 1000. Fraction ionised = 1000 ÷ 1001 = 0.9990, i.e. 99.9 % ionised.
Check: at pH 2.0 the pH is 1.5 units below the pKa, so the acid should be mostly in the HA form — it is. At pH 6.5 it is 3 units above, so it should be almost fully A− — it is.
Example 2: a weak base drug. A base B has a conjugate-acid pKa of 9.00. Find the percentage ionised in blood plasma (pH 7.4) and in stomach fluid (pH 2.0).
At pH 7.4: r = 10(9.0 − 7.4) = 101.6 = 39.81. Fraction ionised = 39.81 ÷ 40.81 = 0.9755, i.e. 97.5 % ionised.
At pH 2.0: r = 10(9.0 − 2.0) = 107. Fraction ionised = 107 ÷ (107 + 1) ≈ 99.99999 %, effectively all ionised.
Check: a base is more protonated in acid, so it must be more ionised at pH 2 than at pH 7.4. Correct.
Example 3: pH needed for 99 % ionisation. At what pH is the acid of Example 1 (pKa = 3.50) 99 % ionised?
The ratio r = 99 ÷ 1 = 99. Then pH = pKa + log 99 = 3.50 + 1.996 = 5.50.
Check: pH 5.5 is almost exactly two units above the pKa, and two units mean a ratio of about 100 : 1, i.e. 99 %.
Example 4: the ion-trapping idea. Assume only the neutral form crosses a membrane, so its concentration is equal on both sides. For the acid of Example 1 (pKa = 3.50) compare the total drug concentration in plasma (pH 7.4) with that in stomach fluid (pH 2.0).
Total = [HA] × (1 + 10(pH − pKa)).
Plasma: 1 + 103.9 = 1 + 7943.3 = 7944.3.
Stomach: 1 + 10−1.5 = 1.0316.
Ratio plasma : stomach = 7944.3 ÷ 1.0316 = about 7700 : 1.
So a weak acid collects on the more basic side of a membrane, and a weak base on the more acidic side. For the base in Example 2 the same method gives (107 + 1) ÷ (1 + 39.81) = about 2.5 × 105 times more total drug in stomach fluid than in plasma. This is a simplified model: real absorption also depends on blood flow, surface area, transporters and protein binding, so treat the numbers as an illustration of the principle only.
Connection to partition coefficients
The octanol–water partition coefficient P describes the neutral form. Because the charged form stays mostly in water, the effective distribution coefficient at a given pH is smaller: D = P ÷ (1 + 10(pH − pKa)) for a weak acid. The two ideas together explain why the same drug can be well absorbed in one part of the gut and poorly in another.
Common mistakes
- Using the base formula for an acid (or the reverse). The sign of (pH − pKa) flips between them.
- Using pKb of the base directly in the equation. Convert first: pKa of BH+ = 14.00 − pKb at 25 °C.
- Calling r the fraction. The fraction ionised is r ÷ (1 + r), not r.
- Forgetting that every pH unit is a factor of ten, not a factor of one.
- Stating that the ionised form never crosses a membrane. It crosses much more slowly, but not zero.
Exam relevance
| Topic | Where the idea is used |
|---|---|
| Ionic equilibrium and buffers | Henderson–Hasselbalch, pKa, fraction ionised |
| Physical chemistry applications | Distribution of a weak electrolyte between two phases |
| Biochemistry and pharmacy-linked questions | Absorption, ion trapping and pH-dependent solubility |
Syllabus content differs between IIT-JAM, GATE, CSIR-NET and CUET-PG, so check each exam’s current official notification.
Check the ionised-to-neutral ratio with the Henderson–Hasselbalch tool: enter the pKa and the pH and compare with your own working.
Open the Henderson–Hasselbalch Buffer Calculator →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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