Intravenous Infusion
Previous rates of
administration were
instantaneous IV bolus
and
first order absorption.
As a rate (mg/hr)
a first order rate
constantly declines
even though a
constant
percentage of
what remains is
handled.
Intravenous Infusion
A first order rate (ka)
such as 0.693 hr-1,
will peak at about
2.6 hours when the
T½ = 4 hours.
6.25 mg/hr
25 mg/hr
100 mg/hr
If we deal with ka as
we would K, since
ka = 0.693,
then T½ka = 1 hour.
This means that following
a 400 mg dose:
Time Am’t Unabs
1
200
2
100
3
50
4
25
5
12.5
6
6.25
Intravenous Infusion
Zero-Order Rate
200 mg/hr
Tmax = 2 hours
6.25 mg/hr
25 mg/hr
100 mg/hr
A zero order rate
is constant (mg/hr).
With a 400 mg dose,
a rate of 200mg/hr
results in all drug
being delivered
in 2 hours
Time
hr
1
2
3
Am’t
Delivered
(mg)
200
200
0
The infusion rate
is identified as k0
with units of am’t/time.
Intravenous Infusion-continuous
Zero-Order Rate
4 mg/hr
Approaching maximum
concentration of 7.3 mg/L
If a drug is infused at a
constant rate (4 mg/hr)
for a long period of time,
concentrations will rise
and approach a constant
concentration (CSS).
At that concentration (CSS),
the rate of elimination
of the drug equals the
infusion rate.
IN = OUT
CSS = k0/VK
Intravenous Infusion-continuous
Zero-Order Rate
4 mg/hr
Approaching maximum
concentration of 7.3 mg/L
The rate at which
the steady-state
concentration
is approached is
proportional
to the half-life.
Number
Of
T½
Percent
of CSS
1
2
3
4
5
6
7
50.0
75.0
87.5
93.75
96.875
98.438
99.219
Intravenous Infusion
IV Intermittent Infusion
Dose = 400 mg
Infusion Time = 2 hours
Infusion Rate (k0) = 200 mg/hr
Many drugs are
administered by
intermittent
intravenous infusions,
where the time of the
infusion is short and
SS is not achieved.
15 min - -lactams
30 min – aminoglycosides
1 hr vancomycin
4 hour amphotericin.
For these drugs the
infusion rate (mg/hr) is:
k0 = dose / infusion time
Intravenous Infusion
During the infusion period
the concentrations increase.
The equation predicting
these concentrations is:
CSS = (k0/KV) *(1-e-kt)
IV Intermittent Infusion
Dose = 160 mg
Infusion Time = 4 hours
Infusion Rate (k0) = 40 mg/hr
Vol = 10 L; T½ = 2hr
CSS = 8.66 mg/L
This has rough similarity
to the oral dosing equation
Dose
Clearance
Intravenous Infusion
During the infusion period
the concentrations increase.
The equation predicting
these concentrations is:
CSS = (k0/KV) *(1-e-kt)
IV Intermittent Infusion
Dose = 160 mg
Infusion Time = 4 hours
Infusion Rate (k0) = 40 mg/hr
Vol = 10 L; T½ = 2hr
CSS = 8.66 mg/L
Here 160 mg is infused
over 4 hours to a patient
with a V of 10 L and a
half-life of 2 hours
The concentration at the end
of the infusion (4 hr) is:
CSS = (k0/KV) *(1-e-kt)
= (40/(0.3465 * 10))*(1--0.3465*4)
= (11.54)(1-0.25)
= 8.66 mg/L
Intravenous Infusion
During the infusion period
the concentrations increase.
The equation predicting
these concentrations is:
CSS = (k0/KV) *(1-e-kt)
IV Intermittent Infusion
Dose = 160 mg
Infusion Time = 4 hours
Infusion Rate (k0) = 40 mg/hr
Vol = 10 L; T½ = 2hr
CSS = 8.66 mg/L
Here 160 mg is infused
over 4 hours to a patient
with a V of 10 L and a
half-life of 2 hours
What is the Cl in this patient?
Cl = KV
= 0.3465 * 10
= 3.465 L/hr
If the infusion is continuous
CSS = (k0/KV) *(1-e-kt) = (k0/KV)
Cl can be obtained from k0 & CSS
Intravenous Infusion
As soon as the infusion is
complete, the concentrations
will begin to decline.
If distribution is instantaneous
This equation calculates the
concentration from the start of the
infusion to the end of the infusion
at time t
CSS = (k0/KV) *(1-e-kt)
This equation calculates the
concentration from the END of the
infusion on and so t is the duration of
infusion and t’ is the time from the end
of the infusion
CSS = [(k0/KV) *(1-e-kt)]e-Kt’
Intravenous Infusion
8
6
4
2
This equation calculates the
concentration from the END of the
infusion on and so t is the duration of
infusion and t’ is the time from the end
of the infusion
CSS = [(k0/KV) *(1-e-Kt)]e-Kt’
To calculate the concentration
at 6 hours from the Start of a
4 hour infusion, t will be 4 hours
and t’ is the time from the end
of the infusion to the time of
interest – 2 hours.
CSS= [(k0/KV) *(1-e-kt)]e-Kt’
= [(11.54)(1-0.25)]e-0.3465*2
= [8.66]*0.500
= 4.33 mg/L
Intravenous Infusion
8
6
4
2
This equation will calculate the
concentrations during the infusion
but requires that the time t’ is zero.
Example:
Calculate the concentration @ 2 hours
CSS = [(k0/KV) *(1-e-Kt)] e-Kt’
= (11.54)*(1-e-0.3465*2) e-0.3465*0
= [(11.54)(1-0.50)] x 1
= [11.54*0.500] x 1
= 5.77 mg/L
Intravenous Infusion
8
6
4
2
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Time
Conc
(hr)
(mg/L)
6
3.333
8
1.667
1. What loading dose would
achieve the end of infusion
concentration immediately?
2. Would a loading dose be
considered necessary for
this drug in this patient?
Intravenous Infusion
8
6
4
2
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Time
Conc
(hr)
(mg/L)
6
3.333
8
1.667
[ ] end of infusion requires K.
T½ = 2 hours; K = 0.3465 hr-1
C4 = 3.333 eKt
= 3.333 e(0.3465 * 2)
= 6.667 mg/L
Intravenous Infusion
88
66
44
22
Volume based
on this concentration
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Time
Conc
(hr)
(mg/L)
6
3.333
8
1.667
Calculate Volume:
C4 = [(k0/KV) *(1-e-Kt)]e-Kt’
V = [(k0/KC4) *(1-e-Kt)]e-Kt’
t’= 0; t=4; C4 = 6.667; k0=50 mg/hr
V = [(50/(0.3465*6.667) *(1-e-0.3465*4]
= 16.23 L
Intravenous Infusion
8
8
6
6
4
4
2
2
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Time
Conc
(hr)
(mg/L)
6
3.333
8
1.667
1. What loading dose would
achieve the end of infusion
concentration immediately?
Load = V * CSS
= 16.23 L * 6.667 mg/L
= 108.21 mg
Intravenous Infusion
8
8
6
6
4
4
2
2
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Time
Conc
(hr)
(mg/L)
6
3.333
8
1.667
1. What loading dose would
achieve the end of infusion
concentration immediately?
2. Would a loading dose be
considered necessary for
this drug in this patient?
Intravenous Infusion
8
8
6
6
4
4
2
2
Question
A patient is given an intravenous
infusion of 200 mg over 4 hours.
Six and 8 hrs after the infusion
began samples were taken and
the concentrations measured.
Would a loading dose be
considered necessary for
this drug in this patient?
How long does it take
to get to steady state?
3.3 (90%) – 5 (96.8) T½
With a T½ of 2 hours
…between 6.6 and 10 hours.
Bolus?? How rapid is response required?
Does a 108.2 mg bolus achieve CSS?
Intravenous Infusion
Question
If this patient receives a
continuous infusion (> 24 hr)
what is the steady-state
concentration?
Intravenous Infusion
Question
If this patient receives a
continuous infusion (> 24 hr)
what is the steady-state
concentration?
CSS = [(k0/KV) *(1-e-Kt)]
When t is large e-Kt becomes
Small and equation becomes
CSS = [(k0/KV)]
= 50 (0.3465 x 16.23)
= 8.89 mg/L
What bolus loading dose
would achieve this true
steady-state?
Intravenous Infusion
Question
If this patient receives a
continuous infusion (> 24 hr)
what is the steady-state
concentration?
CSS = [(k0/KV) *(1-e-Kt)]
When t is large e-Kt becomes
Small and equation becomes
CSS = [(k0/KV)]
= 50 (0.3465 x 16.23)
= 8.89 mg/L
Load = Css V
= (8.89 mg/L)(16.23 L)
= 144.3 mg
Intravenous Infusion
Question
If this patient receives a bolus of
144.3 mg and simultaneously
receives an intravenous infusion
of 50 mg/hr, what will the
concentration-time profile look
like? … what concentrations
are attained?
Intravenous Infusion
Question
If this patient receives a bolus of
144.3 mg and simultaneously
receives an intravenous infusion
of 50 mg/hr, what will the
concentration-time profile look
like? … what concentrations
are attained?
What would the initial
concentration of the bolus be?
C0 = Dose / V
= 144.3 mg / 16.23 L
= 8.89 mg/L
Intravenous Infusion
7
4
2
Question
If this patient receives a bolus of
144.3 mg and simultaneously
receives an intravenous infusion
of 50 mg/hr, what will the
concentration-time profile look
like? … what concentrations
are attained?
What would the concentration
-time profile for the bolus look like?
T½ =
Intravenous Infusion
7
4
2
Question
If this patient receives a bolus of
144.3 mg and simultaneously
receives an intravenous infusion
of 50 mg/hr, what will the
concentration-time profile look
like? … what concentrations
are attained?
What would the concentration
-time profile for the bolus look like?
T½ = 2 hours
Should the conc.–time profile
from the bolus and the simultaneous
infusion just be the sum
of both profiles?
Intravenous Infusion
7
4
2
Question
Time Bolus Infusion Total
(hr) (mg/L) (mg/L) (mg/L)
0.0
8.891
0.00
8.891
0.5
7.477
1.414 8.891
1.0
6.287
2.604 8.891
1.5
5.287
3.604 8.891
2.0
4.446
4.445 8.891
3.0
3.144
5.747 8.891
4.0
2.223
6.668 8.891
6.0
1.112
7.779 8.891
8.0
0.556
8.335 8.891
12.0
0.139
8.752 8.891
16.0
0.035
8.856 8.891
24.0
0.002
8.889 8.891
Intravenous Infusion
7
4
2
When bolus and infusion
are exactly matched
concentration starts
at true steady state
and remains
unchanged
Question
Time Bolus Infusion Total
(hr) (mg/L) (mg/L) (mg/L)
0.0
8.891
0.00
8.891
0.5
7.477
1.414 8.891
1.0
6.287
2.604 8.891
1.5
5.287
3.604 8.891
2.0
4.446
4.445 8.891
3.0
3.144
5.747 8.891
4.0
2.223
6.668 8.891
6.0
1.112
7.779 8.891
8.0
0.556
8.335 8.891
12.0
0.139
8.752 8.891
16.0
0.035
8.856 8.891
24.0
0.002
8.889 8.891
Intravenous Infusion
7
Question
4
2
What happens if the bolus
is larger or smaller?
Don’t let this be you on Exam Day.