Intravenous infusion
Presented by:
Mona Ahmed Elhabak, Ph.D
Lecturer of Pharmaceutics and Industrial Pharmacy, faculty of Pharmacy, ACU.
Introduction
IV infusion plasma concentration versus time curve
- No drug was present in the body at zero time.
- Drug level rises from zero drug concentration
- Gradually becomes constant when a plateau or steadystate drug concentration is reached.
 At steady state, the rate of drug leaving the body is
equal to the rate of drug (infusion rate) entering the
body.
 At steady state, the rate of change in the plasma drug
concentration:
dCp/dt = 0
IV infusion
One-Compartment Model Drugs
 The pharmacokinetics of a drug given by constant IV infusion follows a zero-order input process in which
the drug is infused directly into the systemic blood circulation.
dDB = R
R: rate of drug input (zero order)
dt
 For most drugs, elimination of drug from the plasma is a first-order process.
dDB = K DB
K: elimination rate constant (first order)
dt
DB: amount of drug in the body
 Therefore, in this one-compartment model, the infused drug follows zero-order input and first-order
output.
 The change in the amount of drug in the body at any time (dDB/dt) during the infusion is the rate of input
minus the rate of output and is represented by the following equation:
dDB = R - K DB
dt
IV infusion
One-Compartment Model Drugs
dDB = R - K DB
dt
eq.1
By dividing by Vd
dCp = R - K Cp
dt
Vd
By integration of eq.1
Cp =
R (1- e-kt )
Vd K
eq.2
N.B: Cp=DB/Vd
Steady-State Drug Concentration (CSS)
At steady state: the rate of infusion equals the rate of elimination.
Therefore, the rate of change in the plasma drug concentration is equal to zero.
dCp = 0
dt
dCp = R
dt
Vd
- K Cp = 0
R = K Cp
Vd
CSS =
R =
Vd K
R
Cl
eq.3
Whenever the infusion stops either at steady state (A) or before reaching steady state
(B), the drug concentration declines according to first-order kinetics with the slope
of the elimination curve equal to -kel/2.303
Time Needed to Reach CSS
Cp =
R (1- e-kt )
Vd K
CSS = R
eq.3
Vd K
eq.2
Cp = CSS (1- e-kt )
eq. 4
 The time to reach steady-state could be determined by knowing the time to reach half the steady-state
which can be derived:
Cp = CSS (1- e-kt ) = CSS /2
(1- e-kt ) =
1/2
(- e-kt ) = ½ - 1
 It must be noticed that the time to reach half the steady-state :
- has the same value for the elimination half-life
- is dependent on the elimination process not the infusion
t= 0.693/K
Time Needed to Reach CSS
 The time to reach 99% steady-state:
Cp = CSS (1- e-kt ) = 99%Css
(1- e-kt ) = 99%
Ln 0.01= - Kt
t= Ln 0.01/- K = 4.69 / K
t= 4.69 / (0.693/t1/2) = 6.65 t1/2
Time Needed to Reach CSS
 Mathematically, the time to reach true steady-state drug concentration, CSS, would take an infinite time.
 The time required to reach the steady-state drug concentration in the plasma is dependent on the
elimination rate constant of the drug.
 After one half-life the plasma drug concentration is 50% the steady state concentration (plateau) value,
after 2 half-lives, the plasma drug concentration is 75% of the steady state value.
 95% of the steady state is reached in 4.32 ≈ 5 half lives ,
 99% of the steady state is reached in 6.65 half lives.
Time Needed to Reach CSS
Calculate the time for a drug whose t1/2 is 6 hours to reach at least 95% of the steadystate plasma drug concentration :
Time to reach 95% Css = 5t1/2 = 5 x 6 hours = 30 hours.
Situations sometimes demand that the plateau be reached rapidly.
?
Increase in infusion rate
Will not shorten the time to reach the steady state
drug concentration.
The time to reach steady state is the same.
A higher steady-state drug concentration
will be obtained
IV infusion
Loading Dose Plus IV Infusion: One-Compartment Model
 The loading dose (bolus dose is used to obtain the desired concentration as rapidly as
possible.
 At the start of an infusion, A bolus dose equal to the amount desired in the body at
plateau is given and the infusion rate is adjusted to maintain the therapeutic level.
Introduction
Loading Dose Plus IV Infusion: One-Compartment Model
 Remember if a bolus dose of a certain drug is given at the same time an IV infusion of the
same drug is infused, then the total drug concentration in the blood is the sum of the 2
resulting concentrations.
 Therefore, if an IV loading dose is given,
followed by an IV infusion, steady- state
plasma drug concentrations are obtained
immediately and maintained.
Introduction
Loading Dose Plus IV Infusion: One-Compartment Model
Bolus:
C1
= Cpº e-kelt
Don’t forget:
=
DL
Vd
DB = Vd Cp
Infusion:
C2
=
R
Vd kel
(1- e-kelt)
e-kelt
Introduction
Loading Dose Plus IV Infusion: One-Compartment Model
DL = CSS Vd
DL =
R
kel
CSS =
R
Vd kel
Introduction
Problems
1. An anesthetic agent was administered by IV infusion at a rate of 2 mg/hr. The drug has an
elimination rate constant of 0.1 hr –1, and a volume of distribution (one compartment)
equals to 10 L. What loading dose is recommended if the physician wants the drug level to
reach 2μg/mL immediately?
Solution
Introduction
Problems
2. A drug was given to a patient by simultaneous administration of a loading dose of 10 mg and
an infusion rate of 2 mg/hr (the drug has a t½ of 3 hr and a volume of distribution of 10 L)?
What is the concentration of a drug following 6 hr of administration?
Solution:
Introduction
Problems
N.B: To determine the concentration of the drug in the body after infusion has been stopped,
first the final concentration of drug at the end of the infusion is to be calculated and considered
as C0. Then, use the IV bolus dose equation (C = C0 e–kt) for calculations for any further point in
time.
3. A patient was infused for 6 hr with a drug (k = 0.01 hr– 1; VD = 10 L) at a rate of 2
mg/hr. What is the concentration of the drug in the body 2 hr after cessation of the
infusion?
Solution: