Dose Number and Dissolution-Limited Release

Why Capacity and Speed Must Be Considered Separately

A dose may be small enough to dissolve completely at equilibrium in a given fluid volume and still dissolve too slowly to matter during the time available for absorption. Conversely, a dose can exceed that volume’s equilibrium capacity while its soluble portion is released quickly. These are different questions: dose number describes capacity, whereas a dissolution timescale describes how quickly the system approaches its capacity.

In PBPK absorption models, confusing the two can lead to an incorrect explanation of low or delayed absorption. Dose number is a useful screening measure, but it is not a direct prediction of the amount absorbed or of how quickly drug enters solution. The dissolution rate also depends on factors such as interfacial mass transfer and particle surface area, treated in the earlier material on dissolution rate and particle size.

Definition: The dose number, $Do$, is the administered drug mass divided by the mass that could be dissolved at equilibrium in a specified dissolution volume: $$ Do=\frac{D}{C_s V} $$ Here $D$ is dose mass, $C_s$ is the equilibrium solubility in the relevant medium, and $V$ is the volume available for dissolution. $C_sV$ has units of mass, so $Do$ is dimensionless. The dose and solubility must be expressed on the same drug-mass basis.

The volume in this definition is the volume chosen to represent the fluid available for dissolution, not automatically a standard laboratory vessel volume or the total body-water volume. In a gastrointestinal model, the relevant volume may vary by region and time. Likewise, $C_s$ should represent the specified medium and conditions; a solubility measured in a different medium may not be an appropriate input.

What Dose Number Says at Equilibrium

If the entire dose is present and dissolved in volume $V$, its nominal concentration would be $D/V$. Dividing by equilibrium solubility gives

\frac{D/V}{C_s}=\frac{D}{C_sV}=Do.

Thus dose number is the dose-loading concentration relative to equilibrium solubility under the specified conditions. If $Do<1$, the dose is below the volume’s equilibrium capacity. If $Do>1$, the dose exceeds that capacity, so some solid must remain at equilibrium under the assumptions of the calculation. At $Do=1$, the entire dose would just equal the nominal capacity.

This conclusion follows from a simple mass balance. At equilibrium, the dissolved mass cannot exceed $C_sV$. If $D>C_sV$, the undissolved mass is at least $D-C_sV$; if $D\le C_sV$, capacity alone does not require any residual solid. In the idealized case that the only limit is this capacity, the maximum fraction of the dose that can be dissolved is

f_{\mathrm{diss,max}}=\min\left(1,\frac{C_sV}{D}\right) =\min\left(1,\frac{1}{Do}\right).

This is an equilibrium capacity statement, not a statement about absorption. It assumes the specified solubility and volume remain applicable and does not include precipitation, fluid-volume changes, drug loss, or removal of dissolved drug from the volume.

Worked Example: A Dose Below Equilibrium Capacity

Suppose a hypothetical 100 mg dose is placed in 250 mL of medium with equilibrium solubility $0.80$ mg/mL. Calculate dose number and interpret its capacity implication.

The saturation capacity is

C_sV=(0.80\ \mathrm{mg/mL})(250\ \mathrm{mL})=200\ \mathrm{mg}.

Therefore,

Do=\frac{100\ \mathrm{mg}}{200\ \mathrm{mg}}=0.500.

The dose is half the stated equilibrium capacity. Under the stated conditions, capacity does not require any of the 100 mg to remain undissolved at equilibrium. This calculation does not show that the dose dissolves quickly; that requires kinetic information.

A Dissolution Timescale

A simple interfacial mass-transfer model provides a way to express dissolution speed. Let $M_{\mathrm{diss}}$ be dissolved mass, $A$ the dissolving surface area, $k$ the interfacial mass-transfer coefficient, and $C$ the dissolved concentration. For a well-mixed volume with constant $V$, constant $A$, and constant $C_s$, write

\frac{dM_{\mathrm{diss}}}{dt}=kA(C_s-C), \qquad C=\frac{M_{\mathrm{diss}}}{V}.

This model assumes dissolution is driven by the difference between equilibrium solubility at the interface and the bulk concentration. It neglects changes in area, volume, solubility, and mass-transfer coefficient during the interval considered. Because $k$ has units of length per time, $kA$ has units of volume per time, and multiplying by concentration gives mass per time.

With fixed $V$, $A$, $k$, and $C_s$, substitution of $M_{\mathrm{diss}}=VC$ gives a linear differential equation. Its solution describes how concentration approaches $C_s$ while solid remains available:

V\frac{dC}{dt}=kA(C_s-C), \qquad \frac{dC}{dt}=\frac{C_s-C}{\tau_d}, \qquad \tau_d=\frac{V}{kA}.

The quantity $\tau_d$ is a characteristic dissolution timescale for this fixed-area model. For an initial concentration $C_0$, integration gives

C(t)=C_s-(C_s-C_0)e^{-t/\tau_d}.

For example, when $C_0=0$, the model reaches about 63.2% of $C_s$ after one $\tau_d$, since $1-e^{-1}\approx0.632$. It does not imply that every real particle system has a single constant timescale. Particle area can shrink or change as particles dissolve, and $k$, $C_s$, or $V$ may vary. In those cases, $\tau_d$ is best understood as a local or representative timescale, not a complete release prediction.

Worked Example: Capacity Is Adequate but Dissolution Is Slow

Use the hypothetical 100 mg dose, 250 mL volume, and $C_s=0.80$ mg/mL above. Suppose a simplified constant-area model has $k=0.020$ cm/min and $A=12$ cm². Compare the dissolution timescale with a hypothetical 30-minute availability period.

First convert the volume: $250$ mL equals $250$ cm³. The characteristic timescale is

\tau_d=\frac{250\ \mathrm{cm^3}}{(0.020\ \mathrm{cm/min})(12\ \mathrm{cm^2})} =\frac{250}{0.240}\ \mathrm{min} =1.04\times10^3\ \mathrm{min}.

This is approximately 17.4 hours, much longer than 30 minutes. Under the model, starting from $C_0=0$, the dissolved concentration at 30 minutes is

C(30)=0.80\left(1-e^{-30/1042}\right) \approx0.0227\ \mathrm{mg/mL}.

Multiplying by volume gives approximately $(0.0227\ \mathrm{mg/mL})(250\ \mathrm{mL})=5.68$ mg dissolved at 30 minutes. The dose number remains $0.500$: equilibrium capacity is ample, yet this illustrative kinetic model predicts slow release over the stated period. The numerical prediction depends on the fixed-area assumptions and should not be treated as a universal estimate for a real formulation.

Combining Dose Number with Time Available

A practical screening comparison is between $\tau_d$ and a relevant time available for dissolution, denoted here by $\tau_{\mathrm{avail}}$. This may represent a specified residence or observation interval in a simplified analysis. It is not a universal constant: the appropriate interval depends on the modeled process and location.

If $\tau_d$ is much smaller than $\tau_{\mathrm{avail}}$, dissolution may be rapid relative to that interval, subject to the assumptions of the kinetic model.
If $\tau_d$ is comparable to or larger than $\tau_{\mathrm{avail}}$, release may remain kinetically limited during the interval.
Dose number still determines whether equilibrium capacity is sufficient for the entire dose. A short timescale does not increase that capacity.

There is no universal threshold for “rapid.” The comparison is contextual and should account for uncertainty in surface area, $k$, solubility, and the chosen availability time. It also does not by itself establish how much drug is absorbed, because absorption, transit, and other processes affect the amount that remains in the dissolution environment.

Worked Example: The Same Dose Number, Different Release Speeds

Consider two hypothetical formulations, each with $D=200$ mg, $V=100$ mL, and $C_s=1.0$ mg/mL. Both have the same capacity and dose number:

C_sV=(1.0\ \mathrm{mg/mL})(100\ \mathrm{mL})=100\ \mathrm{mg}, \qquad Do=\frac{200\ \mathrm{mg}}{100\ \mathrm{mg}}=2.00.

Each dose is twice the equilibrium capacity; under the assumptions used for this calculation, at most 100 mg can be dissolved at equilibrium, leaving at least 100 mg undissolved.

Suppose formulation A has $k=0.010$ cm/min and reference area $A=5$ cm². Since $100$ mL equals $100$ cm³,

\tau_{d,A}=\frac{100\ \mathrm{cm^3}}{(0.010\ \mathrm{cm/min})(5\ \mathrm{cm^2})} =2000\ \mathrm{min}.

Suppose formulation B has $k=0.050$ cm/min and reference area $A=100$ cm². Then

\tau_{d,B}=\frac{100\ \mathrm{cm^3}}{(0.050\ \mathrm{cm/min})(100\ \mathrm{cm^2})} =20.0\ \mathrm{min}.

The dose numbers are identical, but the estimated timescales differ by a factor of 100. This illustrates why dose number cannot rank dissolution speed: the kinetic inputs differ. These $\tau_d$ values are illustrative constant-area estimates, not complete predictions of how either formulation behaves as its particles dissolve.

Assumptions, Diagnostics, and Alternatives

Before interpreting a dose number, check that the inputs describe the same situation. In particular, dose and solubility must use consistent mass units and the same chemical basis, such as active moiety rather than a mixture of active-moiety and salt masses. Confirm that volume is the intended dissolution volume and that solubility corresponds to the medium and conditions being modeled. A unit check should leave $Do$ dimensionless and $V/(kA)$ in units of time.

The most important conceptual diagnostic is to ask whether the conclusion concerns equilibrium capacity or release speed. Dose number alone cannot demonstrate rapid dissolution, persistent non-sink conditions, or a particular absorbed fraction. A kinetic timescale requires estimates of $k$ and $A$; if area is poorly characterized or changes substantially, a single $\tau_d$ may be misleading. For a fuller analysis, compare the simple estimates with a time-varying dissolution model and with the available regional or system-level data, while checking mass balance.

These measures complement rather than replace more detailed approaches. Dose number is an equilibrium capacity screen and is easy to compute, but it compresses a dynamic system into a single ratio. A characteristic timescale adds a kinetic comparison, but its simple form assumes fixed properties. A dynamic dissolution model can represent changing area, concentration, or volume, at the cost of additional parameters and assumptions. The appropriate level of detail depends on the question and on what the available data can support.

Common pitfall: “$Do<1$, so the dose will dissolve quickly.” This reasoning mistakes equilibrium capacity for dissolution kinetics. $Do<1$ means only that the stated volume has enough equilibrium capacity for the dose under the specified assumptions. Dissolution may still be slow if interfacial transfer or available particle area is small relative to the time available.

Using the Measures in a PBPK Model

For a model, calculate dose number using inputs that match the relevant location and conditions, then report the assumptions alongside the result. Use it to flag whether equilibrium capacity might constrain complete dissolution, not as a stand-alone absorption prediction. Separately estimate or model dissolution kinetics and compare their timescale with the time the drug is available in that location. If a model predicts complete dissolution despite $Do>1$, identify what additional mechanism or changing condition supports that result; if it predicts slow release despite $Do<1$, examine its kinetic inputs rather than revising the equilibrium capacity interpretation.

This separation is especially useful when comparing formulations. A change in dose, solubility, or relevant volume changes dose number. A change in particle area or interfacial transfer can alter dissolution timescale without changing dose number. Reporting both prevents a formulation’s capacity and its release speed from being conflated.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A hypothetical 150 mg dose is placed in 300 mL of medium with $C_s=0.75$ mg/mL. Calculate dose number and state the maximum equilibrium dissolved fraction.
In one sentence, explain why $Do=0.40$ does not establish that most of the dose dissolves within 10 minutes.
A fixed-area model has $V=200$ cm³, $k=0.025$ cm/min, and $A=40$ cm². Calculate $\tau_d$.
Two formulations have the same dose, solubility, and volume, but one has a larger reference surface area. What changes directly in the fixed-property timescale expression, and what remains unchanged?

Solutions. 1. The capacity is $(0.75\ \mathrm{mg/mL})(300\ \mathrm{mL})=225$ mg, so $Do=150/225=0.667$. Since $Do<1$, the maximum equilibrium dissolved fraction is 1, or 100%, under the stated assumptions. 2. Dose number describes equilibrium capacity, not speed; the dissolution rate and time available are also needed to assess dissolution within 10 minutes. 3. $\tau_d=200/[(0.025)(40)]=200/1.00=200$ min. 4. A larger $A$ decreases $\tau_d=V/(kA)$, all else equal; dose number is unchanged because dose, solubility, and volume have not changed.