From Dissociation Equilibrium to Species Fractions
For an ionizable drug, pH and pKa determine the relative amounts of its protonated and deprotonated forms at equilibrium. In a PBPK model, this distinction matters whenever a property or process depends on which species is present. The key calculation is not just the Henderson–Hasselbalch ratio: it is converting that ratio into fractions of the total concentration of the two forms.
This tutorial considers one monoprotic acid or one monoprotic base at a time. A monoprotic compound has one relevant proton-dissociation equilibrium, so the calculation involves two species. For a base, pKa conventionally means the pKa of its conjugate acid. The fractions below refer to these two forms only; they do not describe additional protonation states or other chemical species.
Here, square brackets denote species concentrations in the same compartment and on the same concentration basis. The fractions are dimensionless, lie between zero and one inclusive, and sum to one for the two-species system. The acid form \(HA\) and base form \(B\) are neutral; \(A^-\) and \(BH^+\) are charged.
Deriving the Fraction Equations
For the acid dissociation \(HA \rightleftharpoons H^+ + A^-\), the concentration-form dissociation constant is \(K_a=[H^+][A^-]/[HA]\). Rearranging gives \([A^-]/[HA]=K_a/[H^+]\). Using \(pH=-\log_{10}[H^+]\) and \(pK_a=-\log_{10}K_a\) in the concentration-based form yields the Henderson–Hasselbalch ratio:
Let \(r=[A^-]/[HA]\). Since the total concentration of the two forms is \([HA]+[A^-]\), substitution of \([A^-]=r[HA]\) gives \(f_{A^-}=r/(1+r)\) and \(f_{HA}=1/(1+r)\). Inserting the acid ratio produces both fractions:
For a base, the equilibrium is the dissociation of its conjugate acid \(BH^+\). Its concentration-form constant is \(K_a=[H^+][B]/[BH^+]\), so rearrangement gives \([B]/[BH^+]=K_a/[H^+]=10^{pH-pK_a}\). In this case let \(r=[B]/[BH^+]\). Dividing each species by the total \([BH^+]+[B]\) gives the neutral and ionized fractions:
These equations have the same ratio \(10^{pH-pK_a}\), but its numerator and denominator refer to different species in the acid and base cases. For an acid, increasing pH increases the deprotonated, ionized fraction. For a base, increasing pH increases the unprotonated, neutral fraction.
Worked Calculations
Worked Example: Ionized Fraction of a Weak Acid
Suppose a hypothetical monoprotic acid has \(pK_a=4.80\) and is in a compartment at \(pH=6.50\). Assume the two-species equilibrium and concentration-based equation are appropriate. The deprotonated-to-protonated ratio is:
Thus the ionized fraction is the ratio divided by one plus the ratio, while the neutral fraction is its complement:
The hypothetical acid is therefore about 98.04% ionized and 1.956% neutral under these conditions. The values sum to \(0.9804+0.01956=0.99996\); the small shortfall from one is due to rounding.
Worked Example: Ionized Fraction of a Weak Base
Suppose a hypothetical base has a conjugate-acid \(pK_a=8.20\), at \(pH=7.40\). Its neutral-to-ionized ratio is:
The ionized fraction is one divided by one plus this ratio. Using the unrounded power in the calculation gives:
So the hypothetical base is about 86.32% ionized and 13.68% neutral. The ionized fraction is high because the pH is below the conjugate-acid pKa. The fractions sum to one apart from display rounding.
Worked Example: Fractions Across pH for an Acid and a Base
Consider a hypothetical weak acid with \(pK_a=6.00\) and a hypothetical weak base whose conjugate acid has \(pK_a=8.00\). Calculate the fractions at \(pH=5.00,\ 7.00,\) and \(9.00\), assuming the two-species concentration equations apply. For the acid, the ratio \([A^-]/[HA]\) is \(10^{pH-6.00}\). At the three pH values this ratio is \(0.1000,\ 10.00,\) and \(1000\), respectively. For the base, \([B]/[BH^+]=10^{pH-8.00}\), giving ratios \(0.001000,\ 0.1000,\) and \(10.00\).
| pH | Acid: \(f_{A^-}\) | Acid: \(f_{HA}\) | Base: \(f_{BH^+}\) | Base: \(f_B\) |
|---|---|---|---|---|
| 5.00 | 0.09091 | 0.9091 | 0.9990 | 0.0009990 |
| 7.00 | 0.9091 | 0.09091 | 0.9091 | 0.09091 |
| 9.00 | 0.9990 | 0.0009990 | 0.09091 | 0.9091 |
For example, at \(pH=7.00\), the acid ratio is \(10.00\), so \(f_{A^-}=10/(1+10)=0.9091\). At the same pH, the base ratio is \(0.1000\), so \(f_{BH^+}=1/(1+0.1000)=0.9091\), with \(f_B=0.1000/(1+0.1000)=0.09091\). Each row has a pair of fractions that sums to one for each compound. The table shows that moving upward in pH favors \(A^-\) for the acid but favors \(B\) for the base.
Checks, Diagnostics, and Alternatives
A useful algebraic check is that the two fractions are complementary. For the acid, for example, their sum is:
The same identity holds for the base fractions. At \(pH=pK_a\), the relevant ratio is one, so both forms are present at equal concentration and each fraction is \(0.5\). Well below the pKa, a weak acid is mostly \(HA\), while a weak base is mostly \(BH^+\). Well above the pKa, the acid is mostly \(A^-\) and the base mostly \(B\). These checks are useful for detecting sign errors in the exponent.
The equations calculate fractions, not absolute concentrations. If the total concentration of the two forms is known, each concentration can be found by multiplying that total by its fraction. For example, with a total acid concentration \(C_T=[HA]+[A^-]\), \([A^-]=f_{A^-}C_T\) and \([HA]=f_{HA}C_T\). No total concentration is needed to calculate the fractions themselves.
The direct alternative is to solve the equilibrium and total-concentration equations together. For an ideal two-species system, this produces the same result as the fraction equations above; the ratio-to-fraction derivation is simply a compact way to do that algebra. Where nonideal behavior is important, activity-based equilibrium calculations or experimentally determined conditional pKa values may be more appropriate. These approaches address the medium’s effect on the equilibrium rather than changing the basic two-species fraction calculation.
| Check | Expected result | What a failure may indicate |
|---|---|---|
| Fraction sum | Ionized plus neutral fraction equals one, within rounding | A mismatched denominator, arithmetic error, or omitted species |
| At \(pH=pK_a\) | Each of the two fractions equals 0.5 | A sign error or use of the wrong pKa |
| Change with increasing pH | Acid ionization increases; base ionization decreases | The acid and base formulas may have been interchanged |
| Extreme pH relative to pKa | Fractions approach zero or one in the expected direction | Check the ratio definition and whether pH and pKa use consistent conventions |
Why the Assumptions and Scope Matter
The two-species assumption is essential. If additional protonation states contribute, a fraction calculated from a single monoprotic equilibrium is not the fraction of the total across all species. Likewise, a pH value from one compartment should not automatically be used to describe a different compartment if their pH values differ. The calculation gives an equilibrium distribution at the specified pH; it does not establish that equilibrium is reached in every biological situation.
In PBPK applications, pH and pKa are therefore inputs that should be checked for chemical identity, medium, and units or conventions before calculated fractions are used. The equations are not a substitute for deciding whether the two-form model is adequate. They also do not by themselves determine solubility, permeability, or tissue partitioning; those require additional relationships and assumptions.
Check Your Understanding
Answer each question using the two-species assumptions stated above. Solutions follow the questions.
- A hypothetical acid has \(pK_a=5.00\). At \(pH=5.00\), what are its ionized and neutral fractions?
- A hypothetical acid has \(pK_a=6.00\) at \(pH=4.00\). Calculate its ionized fraction and state which form predominates.
- A hypothetical base has conjugate-acid \(pK_a=7.00\) at \(pH=8.00\). Calculate its neutral and ionized fractions.
- For a monoprotic acid, what should happen to the ionized fraction as pH increases while pKa stays fixed?
- A calculation gives ionized and neutral fractions of 0.72 and 0.41. Name a check that fails and give a plausible cause.
Solutions. (1) Since \(pH=pK_a\), the ratio is \(10^0=1\), so each fraction is \(1/(1+1)=0.5000\). (2) The ratio \([A^-]/[HA]=10^{4.00-6.00}=0.01000\), so \(f_{A^-}=0.01000/(1.01000)=0.009901\); neutral \(HA\) predominates. (3) The base ratio \([B]/[BH^+]=10^{8.00-7.00}=10.00\). Thus \(f_B=10/11=0.9091\) and \(f_{BH^+}=1/11=0.09091\). (4) The ionized acid fraction increases, since \(10^{pH-pK_a}\) increases with pH and \(r/(1+r)\) increases as \(r\) increases. (5) The fractions sum to \(1.13\), not one. A likely cause is inconsistent denominators, an arithmetic or transcription error, or treating an omitted species as if the system had only two forms.