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Intramuscular Injection Inputs From Solutions and Suspensions

Learn how to represent intramuscular solutions and suspensions as PBPK inputs, and use their release and absorption rates to predict concentration–time profiles.

Intermediate 12 min read

What You'll Learn

  • Write mass-balance equations for solution and suspension depots.
  • Distinguish the rate of systemic input from the plasma concentration peak.
  • Calculate input and plasma peak times for first-order absorption.
  • Compare solution and suspension peak concentrations and AUC under explicit assumptions.
  • Recognize when a simple first-order release model may not represent the injection-site process.

Why the Formulation Changes the Input

An intramuscular dose does not necessarily enter the systemic circulation at the time it is injected. Drug in a solution is already in a form that can be absorbed, whereas drug in a suspension must first become available from the particulate depot. That additional step can delay and spread out systemic input. In a PBPK model, the distinction matters because the time course of input can change the plasma peak even when the total amount reaching the circulation is the same.

The useful comparison is between input rate and plasma concentration. A solution with first-order absorption has its greatest input rate immediately after injection, but its plasma concentration generally peaks later because absorption and elimination occur together. A suspension can have an input rate that first rises and then falls: release from the depot must build drug in the absorbable pool before systemic input reaches its maximum.

Definition: An intramuscular solution input represents drug available for absorption from the injection site, with no separate particulate-release step. A suspension input represents a solid depot that releases drug into a locally absorbable pool before absorption. In both cases, an input function is the rate at which drug enters the systemic circulation, usually expressed as mass per unit time.

Mass-Balance Models

Let \(D\) be the injected dose, \(S(t)\) the mass remaining in a suspension depot, and \(M(t)\) the mass in the local absorbable pool. Let \(k_r\) be the first-order suspension release rate constant and \(k_a\) the first-order absorption rate constant. Rate constants have units of inverse time. For the simple suspension model, the equations are

$$ \frac{dS}{dt}=-k_r S,\qquad \frac{dM}{dt}=k_rS-k_aM,\qquad S(0)=D,\quad M(0)=0. $$

Assume the fraction of absorbed drug that reaches the systemic circulation is \(F\), with \(0\leq F\leq1\). The systemic input rate is \(R_{\mathrm{in}}(t)=F k_aM(t)\). The solution input is the simpler case in which the whole dose is initially in the absorbable pool: \(M(0)=D\), with no solid depot or release step.

These models assume a well-mixed, lumped injection site; first-order transfer; constant rate parameters; no loss from the local depot except through the stated transfers; and a constant systemic availability fraction. For concentration predictions below, assume a one-compartment systemic model with volume \(V\) and first-order elimination rate constant \(k_{\mathrm{el}}\). These are simplifying assumptions for interpreting the input, not claims that every injection site behaves this way.

Solution Input and Its Plasma Profile

For a solution, the local available mass declines as \(M(t)=D e^{-k_at}\). Thus systemic input begins at \(F D k_a\) and declines exponentially. In the one-compartment systemic model, amount \(A(t)\) follows \(dA/dt=R_{\mathrm{in}}-k_{\mathrm{el}}A\), with \(A(0)=0\), and concentration is \(C(t)=A(t)/V\). When \(k_a\ne k_{\mathrm{el}}\), solving this balance gives

$$ R_{\mathrm{in,sol}}(t)=F D k_a e^{-k_at},\qquad C_{\mathrm{sol}}(t)=\frac{F D}{V}\frac{k_a}{k_a-k_{\mathrm{el}}} \left(e^{-k_{\mathrm{el}}t}-e^{-k_at}\right). $$

The concentration peak is not at time zero. Differentiating the concentration expression and setting the derivative equal to zero yields \(k_a e^{-k_at}=k_{\mathrm{el}}e^{-k_{\mathrm{el}}t}\). Therefore, for positive unequal rates,

$$ T_{\max,\mathrm{sol}}= \frac{\ln(k_a/k_{\mathrm{el}})}{k_a-k_{\mathrm{el}}}. $$

Worked Example: Solution Absorption and Plasma Peak

Consider a hypothetical intramuscular solution dose of \(D=100\ \mathrm{mg}\), with \(F=0.80\), \(V=20\ \mathrm{L}\), \(k_a=1.2\ \mathrm{h}^{-1}\), and \(k_{\mathrm{el}}=0.20\ \mathrm{h}^{-1}\). Assume the model conditions above hold.

The concentration peak time is

$$ T_{\max,\mathrm{sol}}= \frac{\ln(1.2/0.20)}{1.2-0.20} =\ln(6) =1.792\ \mathrm{h}. $$

At that time, \(e^{-0.20(1.792)}\approx0.6988\) and \(e^{-1.2(1.792)}\approx0.1164\). Substitution gives

$$ C_{\max}= \frac{(0.80)(100\ \mathrm{mg})}{20\ \mathrm{L}} \frac{1.2}{1.0}(0.6988-0.1164) =2.796\ \mathrm{mg/L}. $$

The total systemic amount is \(FD=80\ \mathrm{mg}\). Integrating the concentration profile, or using the one-compartment mass balance, gives \(AUC=FD/(V k_{\mathrm{el}})=80/(20\times0.20)=20.0\ \mathrm{mg\,h/L}\). The peak is delayed relative to injection even though the input rate is largest at \(t=0\).

Suspension Release Before Absorption

For the suspension equations, \(S(t)=D e^{-k_rt}\). Solving the equation for \(M(t)\), when \(k_a\ne k_r\), gives

$$ M(t)=D\frac{k_r}{k_a-k_r}\left(e^{-k_rt}-e^{-k_at}\right). $$

Multiplying by \(F k_a\) gives the systemic input function:

$$ R_{\mathrm{in,susp}}(t)= F D\frac{k_a k_r}{k_a-k_r} \left(e^{-k_rt}-e^{-k_at}\right). $$

The expression starts at zero because \(M(0)=0\). Its maximum occurs when its derivative is zero, or \(k_a e^{-k_at}=k_r e^{-k_rt}\). For positive unequal rates this gives

$$ T_{\max,\mathrm{input}}= \frac{\ln(k_a/k_r)}{k_a-k_r}. $$

This is the time of maximum systemic input, not necessarily the time of maximum plasma concentration. Elimination smooths and shifts the concentration profile, so its peak must be calculated from the systemic mass balance.

Worked Example: When Suspension Input Reaches Its Maximum

Suppose a hypothetical \(100\ \mathrm{mg}\) suspension dose has \(F=0.80\), \(k_r=0.15\ \mathrm{h}^{-1}\), and \(k_a=1.0\ \mathrm{h}^{-1}\). The model assumes first-order release and absorption without local loss.

The time of maximum systemic input is

$$ T_{\max,\mathrm{input}}= \frac{\ln(1.0/0.15)}{1.0-0.15} =\frac{\ln(6.667)}{0.85} =2.232\ \mathrm{h}. $$

The systemic input rate at this time is

$$ R_{\mathrm{in}}(2.232)= (0.80)(100\ \mathrm{mg})\frac{(1.0)(0.15)}{0.85} \left(e^{-0.15(2.232)}-e^{-1.0(2.232)}\right) \approx8.59\ \mathrm{mg/h}. $$

The input has risen from zero to this maximum and subsequently declines. By contrast, for a solution with the same dose, \(F\), and \(k_a=1.0\ \mathrm{h}^{-1}\), the input rate begins at \(80\ \mathrm{mg/h}\) and decreases from there. That comparison concerns input rates; it does not by itself establish the relative plasma peak times.

Comparing Concentrations and Total Exposure

A suspension concentration profile follows by passing its input through systemic elimination. Using the response to an exponential input, define \(g(q,t)=(e^{-qt}-e^{-k_{\mathrm{el}}t})/(k_{\mathrm{el}}-q)\). Then, for distinct \(k_r\), \(k_a\), and \(k_{\mathrm{el}}\),

$$ C_{\mathrm{susp}}(t)= \frac{F D}{V}\frac{k_a k_r}{k_a-k_r} \left[g(k_r,t)-g(k_a,t)\right]. $$

The plasma peak is the time at which this concentration reaches its maximum. In general, it has no simple two-rate formula like the solution peak; it can be found by differentiating the expression or numerically locating its maximum. A useful check is the mass balance: at a concentration peak, systemic input equals elimination, \(R_{\mathrm{in}}(t)=k_{\mathrm{el}}V C(t)\).

If all drug is eventually released and absorbed and \(F\) is constant, the total systemic amount is \(FD\) for either formulation. With linear elimination, the complete AUC is consequently \(FD/(V k_{\mathrm{el}})\). Release kinetics can therefore change peak height and timing without changing complete AUC under these assumptions.

Worked Example: Same Complete AUC, Different Peak

Compare a hypothetical solution and suspension, each with \(D=100\ \mathrm{mg}\), \(F=0.80\), \(V=20\ \mathrm{L}\), \(k_a=1.0\ \mathrm{h}^{-1}\), and \(k_{\mathrm{el}}=0.20\ \mathrm{h}^{-1}\). For the suspension, let \(k_r=0.15\ \mathrm{h}^{-1}\). Assume complete eventual absorption, no local loss, and linear elimination.

For the solution, the plasma peak time is

$$ T_{\max,\mathrm{sol}}= \frac{\ln(1.0/0.20)}{1.0-0.20} =\frac{\ln(5)}{0.80} =2.012\ \mathrm{h}. $$

Its peak concentration, using the solution profile, is

$$ C_{\max,\mathrm{sol}}= \frac{80}{20}\frac{1.0}{0.80} \left(e^{-0.20(2.012)}-e^{-1.0(2.012)}\right) \approx2.674\ \mathrm{mg/L}. $$

For the suspension, substituting the rates and dose into \(C_{\mathrm{susp}}(t)\) and locating its maximum gives \(T_{\max}\approx6.95\ \mathrm{h}\) and \(C_{\max}\approx1.242\ \mathrm{mg/L}\). The numerical peak can be checked using \(R_{\mathrm{in}}=k_{\mathrm{el}}VC\): near \(6.95\ \mathrm{h}\), both sides are approximately \(4.97\ \mathrm{mg/h}\). Both formulations have

$$ AUC=\frac{FD}{Vk_{\mathrm{el}}} =\frac{80\ \mathrm{mg}}{(20\ \mathrm{L})(0.20\ \mathrm{h}^{-1})} =20.0\ \mathrm{mg\,h/L}. $$

In this example, the modeled suspension produces a later, lower plasma peak but the same complete AUC. That conclusion depends on the stated assumptions; incomplete absorption, local loss, or nonlinear systemic disposition could change the exposure comparison.

Derivation and Model Checks

The suspension solution for \(M(t)\) can be justified directly with an integrating factor. Starting from \(dM/dt+k_aM=k_rD e^{-k_rt}\), multiply both sides by \(e^{k_at}\), integrate from zero to \(t\), and use \(M(0)=0\):

$$ e^{k_at}M(t)=k_rD\int_0^t e^{(k_a-k_r)u}\,du =k_rD\frac{e^{(k_a-k_r)t}-1}{k_a-k_r}. $$

Multiplying by \(e^{-k_at}\) gives \(M(t)=Dk_r(e^{-k_rt}-e^{-k_at})/(k_a-k_r)\), as stated. This form also provides checks: \(M(0)=0\), the units are mass, and \(M(t)\) remains nonnegative for either ordering of the two rate constants. The apparent singularity when \(k_a=k_r\) is removable; taking the limit gives \(M(t)=Dk_at e^{-k_at}\).

1
Check the initial condition.
A solution begins with \(D\) in the absorbable pool; the modeled suspension begins with \(D\) in the solid depot and \(M(0)=0\).
2
Check mass transfer.
For the suspension, release decreases \(S\) and increases \(M\); absorption decreases \(M\) and supplies systemic input. Do not count the same transfer twice.
3
Check long-run exposure.
If all \(D\) is transferred and absorbed, systemic input integrated over time is \(FD\). If the modeled input does not approach that amount, inspect the equations and any local-loss assumptions.

Diagnostics, Alternatives, and Limitations

A solution model is appropriate when the injected drug is treated as immediately available for absorption at the site. A two-stage suspension model is useful when a delayed, prolonged input is needed and a single first-order absorption step cannot reproduce the observed profile. A still simpler empirical alternative is to specify a measured or estimated input function directly, but that approach may not distinguish release from absorption and may be less informative when predicting a different formulation.

Several patterns should prompt a model check. A computed negative depot mass or negative input rate violates the intended mass balance. A suspension input that is nonzero at injection conflicts with the assumed initial condition \(M(0)=0\). A fitted suspension model that predicts an implausibly early input peak may have release and absorption rates poorly distinguished by the available profile. Finally, a predicted AUC inconsistent with \(FD/(Vk_{\mathrm{el}})\) requires explanation, such as incomplete input, local loss, a different availability fraction, or a mismatch in the systemic model.

A common mistake is to call \(T_{\max,\mathrm{input}}\) the plasma \(T_{\max}\). These are different quantities: input is a rate, while plasma concentration reflects the history of input minus elimination. The correction is to calculate or simulate \(C(t)\) and locate its maximum; the input-peak formula alone cannot provide that plasma peak. Another limitation is that a first-order release constant is a lumped approximation. It should not be interpreted as a universal physical mechanism or assumed to describe every formulation and injection site.

Key takeaway: A solution places drug directly into the absorbable pool; a suspension adds a depot-release step. That extra step can delay and reduce the plasma peak, while complete AUC remains \(FD/(Vk_{\mathrm{el}})\) only when all drug is absorbed and systemic disposition is linear.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. For a solution with \(k_a=0.8\ \mathrm{h}^{-1}\) and \(k_{\mathrm{el}}=0.2\ \mathrm{h}^{-1}\), calculate the plasma peak time.
  2. In the suspension model, what is the initial systemic input rate, and why?
  3. If \(D=60\ \mathrm{mg}\), \(F=0.75\), \(V=15\ \mathrm{L}\), and \(k_{\mathrm{el}}=0.25\ \mathrm{h}^{-1}\), find complete AUC under the linear, complete-absorption assumptions.
  4. Explain why a suspension's maximum input time need not equal its maximum plasma concentration time.

Solutions. (1) \(T_{\max}=\ln(0.8/0.2)/(0.8-0.2)=\ln(4)/0.6=2.310\ \mathrm{h}\). (2) It is zero because \(M(0)=0\), so \(R_{\mathrm{in}}(0)=Fk_aM(0)=0\). (3) \(AUC=FD/(Vk_{\mathrm{el}})=(0.75)(60)/(15\times0.25)=12.0\ \mathrm{mg\,h/L}\). (4) Plasma concentration accumulates and is eliminated over time; its peak occurs when systemic input equals elimination, which need not coincide with the time when input itself is greatest.