Charge-State Distributions of Polyprotic Drugs

Why Charge-State Distributions Matter

A polyprotic drug can occupy several charge states in the same solution. Knowing which states are present, and in what proportions, is more informative than assigning the whole molecule a single label such as “ionized.” The fractions depend on the solution pH and on each sequential pKa. They can be calculated by extending the acid–base equilibrium logic used for monoprotic compounds.

The calculation is a speciation calculation: it partitions the total concentration of a specified polyprotic compound among its protonation states at equilibrium. It does not, by itself, predict solubility, permeability, or tissue distribution.

Sequential Equilibria and General Formula

For an acid with \(n\) ionizable protons, label the states in order of deprotonation. State 0 is fully protonated, state 1 has lost its first proton, and state \(j\) has lost \(j\) protons. Let \(pK_{a,j}\) be the pKa for the equilibrium from state \(j-1\) to state \(j\). Under the usual simple acid model, each step is a sequential equilibrium:

Here the displayed charges assume that the fully protonated acid scaffold is neutral and that each listed deprotonation adds one negative charge. If the compound has other ionizable groups or a different formal charge, the state labels must reflect its actual structure; the fraction calculation still follows the sequential equilibrium ratios.

For a base, use the same ordering by loss of protons from its fully protonated form. For example, the states of a triprotic base are \(\mathrm{H_3B^{3+}}\), \(\mathrm{H_2B^{2+}}\), \(\mathrm{HB^+}\), and \(\mathrm{B}\). The pKa values in this tutorial always refer to consecutive deprotonation steps in that order. They need not be written in ascending order: for a polyprotic base they are often descending.

For each step, the acid dissociation relation gives the ratio of the concentration in the more deprotonated state to that in the preceding state. Using \(pH=-\log_{10}[H^+]\) and \(pK_{a,j}=-\log_{10}K_{a,j}\) under the concentration approximation, that ratio is \(10^{pH-pK_{a,j}}\). Multiplying adjacent ratios gives the relative amount in every state compared with the fully protonated state.

Set the fully protonated state’s relative weight to \(w_0=1\), and set \(w_j=C_j/C_0\) for the other states. Dividing each weight by their sum converts relative amounts to fractions. This normalization is essential: the weights themselves are not fractional abundances.

Diprotic Compounds

For a diprotic acid \(\mathrm{H_2A}\), the states are \(\mathrm{H_2A}\), \(\mathrm{HA^-}\), and \(\mathrm{A^{2-}}\), with sequential pKas \(pK_{a,1}\) and \(pK_{a,2}\). Their weights are 1, \(10^{pH-pK_{a,1}}\), and \(10^{2pH-pK_{a,1}-pK_{a,2}}\), respectively. The same weights describe a diprotic base when its states are ordered from fully protonated to fully deprotonated; only the formal charges differ.

Triprotic Compounds

A triprotic compound has four sequential states. For an acid with a neutral fully protonated form, these are \(\mathrm{H_3A}\), \(\mathrm{H_2A^-}\), \(\mathrm{HA^{2-}}\), and \(\mathrm{A^{3-}}\). For a triprotic base, the corresponding states are conventionally written from the fully protonated cation to the neutral form. In either case, calculate each weight by adding the pH-minus-pKa contribution for every proton removed so far, then normalize all four weights.

Checks, Diagnostics, and Limitations

A short derivation explains why the normalized formula satisfies the basic checks. Since every weight is positive, each \(f_j=w_j/\sum_r w_r\) is positive and no fraction exceeds 1. Summing across all included states gives \(\sum_j f_j=\sum_jw_j/\sum_rw_r=1\). The adjacent ratio also remains intact after normalization:

This ratio provides a useful independent check. If \(pH=pK_{a,j}\), the two adjacent states have equal abundance. If pH is one unit above that pKa, the more deprotonated state is ten times as abundant as the preceding state. This does not guarantee that either state is the most abundant overall, because other sequential states also contribute.

Common failure modes include reversing the pKa order for a base, omitting one term from a cumulative exponent, and treating the unnormalized weights as fractions. Check the state sequence against the definitions of the pKas, verify every adjacent ratio, and confirm that normalized fractions sum to 1. For very large exponents, direct evaluation of powers of ten can overflow or underflow numerically; subtracting the largest log-weight before exponentiating preserves the normalized fractions.

This sequential, macroscopic-state calculation is appropriate when the stated pKas describe the transitions among the included states. If a molecule has multiple non-equivalent ionizable sites, several microscopic protonation arrangements can contribute to the same net charge. A single sequential set of macroscopic pKas can describe total charge-state fractions, but it does not necessarily identify the abundance of each individual microstate. Use a microstate-specific equilibrium model when the question requires those distinctions. The earlier material on ionization fractions for monoprotic compounds supplies the two-state case; the same adjacent-ratio reasoning extends here without assuming that only one charged state exists.