Why Milk-to-Plasma Partitioning Needs More Than a pH Comparison
A prediction of drug concentration in milk can begin with a maternal unbound plasma concentration, but the two concentrations are not automatically equal. A drug’s ionization may differ between plasma and milk because their pH values differ. In addition, drug bound to milk protein or fat contributes to a measured total milk concentration even though it is not part of the unbound aqueous concentration. Separating these effects makes the prediction interpretable and helps identify when a simple partition model is unsuitable.
The central equilibrium assumption is that the neutral drug species can move passively between plasma water and milk water until their chemical activities are equal. Under an approximation of similar activity coefficients, this means their neutral concentrations are equal. Ionized drug does not have to be equal across the two fluids: different pH values can change the amount ionized on each side. Binding to milk components then relates the unbound aqueous concentration to the total concentration measured in milk.
Here, “unbound” means not bound to macromolecules or milk fat; it does not mean “uncharged.” Unbound plasma concentration includes both unbound ionized and unbound neutral drug. That distinction matters because the pH correction translates between the total unbound concentrations in the two aqueous compartments.
Deriving the pH-Dependent Partition Ratio
For a monoprotic weak base, write the equilibrium as \(B+H^+\rightleftharpoons BH^+\). The Henderson–Hasselbalch relationship gives the ratio of ionized to neutral drug as \([BH^+]/[B]=10^{pKa-pH}\). Thus, the total unbound aqueous concentration is the neutral concentration multiplied by an ionization factor. For a monoprotic weak acid, \(HA\rightleftharpoons H^++A^-\), and \([A^-]/[HA]=10^{pH-pKa}\). The corresponding factor follows from that ratio.
At passive equilibrium, take the neutral concentrations in plasma and milk to be equal. Dividing the milk unbound concentration by the plasma unbound concentration therefore gives the ratio of their ionization factors. Denote this pH-dependent ratio by \(R_{\mathrm{ion}}\). The milk binding correction is separate: by definition, \(C_{m,tot}=C_{m,u}/f_{u,m}\).
The last equation follows by dividing the predicted total milk concentration by \(C_{p,tot}\). Since \(C_{p,u}=f_{u,p}C_{p,tot}\), the ratio is \(C_{p,u}R_{\mathrm{ion}}/(f_{u,m}C_{p,tot})=(f_{u,p}/f_{u,m})R_{\mathrm{ion}}\). The pH ratio alone is not generally the conventional total \(M/P\): that ratio also depends on binding in both plasma and milk.
Worked Predictions
Worked Example: Weak-Base Ion Trapping and Milk Binding
Suppose a hypothetical weak base has \(pKa=8.0\), maternal plasma pH is 7.4, and milk pH is 7.0. Let \(C_{p,u}=0.200\) mg/L, \(f_{u,m}=0.80\), and \(f_{u,p}=0.10\). Predict total milk concentration and the total \(M/P\).
For a weak base, the pH ratio is \(R_{\mathrm{ion}}=[1+10^{8.0-7.0}]/[1+10^{8.0-7.4}]\). Evaluating without rounding the intermediate powers gives \(R_{\mathrm{ion}}=11/[1+10^{0.6}]=11/4.9810717\approx2.208360\).
The predicted total milk concentration is \(C_{m,tot}=0.200\times2.208360\times(1/0.80)=0.552090\) mg/L, or 0.5521 mg/L to four significant figures. Total plasma concentration is \(C_{p,tot}=C_{p,u}/f_{u,p}=0.200/0.10=2.00\) mg/L. Therefore \(M/P=0.552090/2.00=0.276045\), or 0.2760.
The value above one for \(R_{\mathrm{ion}}\) reflects greater total unbound drug in milk than plasma under the specified pH conditions. The milk binding correction increases total milk concentration by a factor of \(1/0.80=1.25\). The total \(M/P\) is below one because the total plasma concentration is much larger than the unbound plasma input.
Worked Example: A Weak Acid With Lower Milk Concentration
For a hypothetical weak acid, let \(pKa=4.5\), plasma pH be 7.4, milk pH be 7.0, \(C_{p,u}=0.550\) mg/L, \(f_{u,m}=1.00\), and \(f_{u,p}=0.44\). Assume no measurable milk binding in this example. Find total milk concentration and total \(M/P\).
For a weak acid, \(R_{\mathrm{ion}}=[1+10^{7.0-4.5}]/[1+10^{7.4-4.5}]\). Using unrounded powers, \(R_{\mathrm{ion}}=317.227766/795.328235\approx0.39886396\).
Thus \(C_{m,tot}=0.550\times0.39886396/1.00=0.2193752\) mg/L, which rounds to 0.2194 mg/L. The total plasma concentration is \(0.550/0.44=1.25\) mg/L, so \(M/P=0.2193752/1.25=0.1755002\), or 0.1755.
The result is consistent with the pH calculation: for this weak acid, the higher plasma pH increases the ionized fraction relative to milk, so the total unbound milk concentration is lower. This conclusion depends on the assumed equilibrium and the stated pH values; it is not a general claim that all acidic drugs have low milk concentrations.
Worked Example: Milk Binding Can Raise Total Concentration Without Ion Trapping
Consider a hypothetical neutral drug with \(C_{p,u}=0.120\) mg/L, \(f_{u,p}=0.60\), and \(f_{u,m}=0.25\). For a neutral drug, take \(R_{\mathrm{ion}}=1\). Predict total milk concentration and total \(M/P\).
The milk prediction is \(C_{m,tot}=0.120\times1\times(1/0.25)=0.480\) mg/L. Total plasma concentration is \(0.120/0.60=0.200\) mg/L. Therefore \(M/P=0.480/0.200=2.40\).
This result does not imply pH-driven ion trapping: the pH factor is one. Rather, the assumed milk unbound fraction is lower than the plasma unbound fraction, so milk binding raises the total milk concentration relative to the unbound aqueous concentration. The prediction depends on the validity of the milk binding estimate.
Choosing and Interpreting the Binding Inputs
If a prediction starts from measured \(C_{p,u}\), plasma binding is not needed to calculate \(C_{m,tot}\) when the target is total milk concentration: use \(C_{p,u}R_{\mathrm{ion}}/f_{u,m}\). Plasma binding is needed to calculate the conventional total \(M/P\), because its denominator is total plasma concentration. Alternatively, if total plasma concentration is the input, first obtain \(C_{p,u}=f_{u,p}C_{p,tot}\), then apply the milk prediction.
The milk unbound fraction is a practical summary of binding in the milk matrix. It can include association with milk proteins and fat, provided it was measured or estimated for a matrix and concentration range relevant to the prediction. If a model instead applies an explicit lipid-partition or fat-binding factor, that factor should replace or be combined consistently with the binding description; applying both to account for the same association would double-count it.
Milk composition and pH are not fixed across all people or sampling times. If the intended prediction concerns a specific milk sample, use a pH and binding estimate appropriate to that sample when possible. A model based on one pH or one milk unbound fraction is a conditional prediction, not a universal drug property. For drugs with multiple ionizable groups, the monoprotic expressions are insufficient; the unbound concentration must instead be calculated from the relevant species fractions.
Diagnostics, Alternatives, and Limitations
A useful diagnostic is to inspect each component separately: the pH ratio, the milk binding correction, and—if reporting total \(M/P\)—the plasma binding correction. A large predicted \(M/P\) driven by a weak-base pH ratio should be checked against the drug’s \(pKa\), the pH values, and the assumption of passive equilibrium. A large prediction driven mainly by a very small \(f_{u,m}\) calls for scrutiny of the binding estimate, its units, and whether it reflects the same milk matrix as the target concentration.
The equilibrium expression is an approximation, not a complete time-dependent lactation model. Milk is produced and removed, concentrations change over time, and sampling may occur before equilibrium is reached. Transporters, local metabolism, non-equilibrium exchange, or departures from equal neutral-species activity can all cause observed concentrations to differ from the prediction. A measured milk-to-plasma ratio can help evaluate the model, but a single paired measurement may not identify which assumption caused a discrepancy.
The pH-based calculation is useful when the drug’s ionization and passive distribution are plausible and the required inputs are available. Alternatives include using an empirically measured milk-to-plasma ratio or a more mechanistic compartment model with explicit exchange and binding processes. An empirical ratio may summarize several mechanisms but may not transfer to a different pH, sampling time, or population. A more detailed model can represent additional mechanisms, but it requires data to support those extra parameters. In all cases, the predicted total milk concentration is a concentration estimate; translating it into infant exposure is a separate calculation.
Confirm whether the input is unbound or total plasma concentration and whether the target is unbound or total milk concentration.
Use the weak-base or weak-acid expression only when the monoprotic approximation is appropriate, with pH and \(pKa\) on the same scale.
For total milk concentration from \(C_{p,u}\), divide by \(f_{u,m}\). Include \(f_{u,p}\) when converting to total plasma concentration or total \(M/P\).
Assess equilibrium, passive transfer, pH relevance, and the binding estimate. Carry full precision through intermediate steps and round the reported result at the end.
Common Mistake
A frequent error is to treat the pH ratio as the total \(M/P\), or to multiply a total plasma concentration by that ratio and then apply a milk binding correction. The pH ratio relates the two unbound aqueous concentrations under the equilibrium assumptions; it does not by itself include plasma or milk binding. Starting with total plasma concentration requires conversion to unbound concentration first. Starting with \(C_{p,u}\) requires only the pH ratio and the milk binding correction to predict total milk concentration.
Check Your Understanding
Use the assumptions stated in each question, and retain unrounded intermediate values until the final result.
- A neutral drug has \(C_{p,u}=0.080\) mg/L and \(f_{u,m}=0.50\). What is the predicted total milk concentration if passive equilibrium holds?
- For a weak base, milk pH is lower than plasma pH. Which way does \(R_{\mathrm{ion}}\) tend to move, and what assumption underlies that prediction?
- A drug has \(C_{p,u}=0.30\) mg/L, \(f_{u,p}=0.30\), \(f_{u,m}=0.60\), and \(R_{\mathrm{ion}}=1.2\). Calculate total milk concentration and total \(M/P\).
- Why is the pH-only ratio not generally the conventional total \(M/P\)?
- Name one condition under which the simple equilibrium prediction may not match a measured milk concentration.
Solutions. (1) For a neutral drug \(R_{\mathrm{ion}}=1\), so \(C_{m,tot}=0.080/0.50=0.160\) mg/L. (2) Lower milk pH increases the weak base’s ionized-to-neutral ratio in milk, so \(R_{\mathrm{ion}}\) tends to exceed one; this prediction assumes passive equilibrium of the neutral species and the stated pH values. (3) \(C_{m,tot}=0.30\times1.2/0.60=0.600\) mg/L. Total plasma concentration is \(0.30/0.30=1.00\) mg/L, so \(M/P=0.600/1.00=0.600\). (4) The pH-only ratio relates unbound aqueous concentrations, whereas total \(M/P\) also includes plasma and milk binding. (5) Examples include active transport, failure to reach equilibrium, or a pH value that does not represent the sampled milk.