Intravenous Infusion Previous rates of administration were instantaneous IV bolus and first order absorption. As a rate (mg/hr) a first order rate constantly

Intravenous InfusionIntravenous Infusion

Previous rates of administration were

instantaneous IV bolusinstantaneous IV bolus and

first order absorption.first order absorption.

As a rate (mg/hr) a first order rate

constantly declineseven though a

constant percentage of

what remains is handled.


A first order rate (ka) such as 0.693 hr-1, will peak at about 2.6 hours when the

T½ = 4 hours.

If we deal with ka as we would K, since

ka = 0.693, then T½ka = 1 hour.

This means that followinga 400 mg dose:

Time Am’t Unabs 1 200 2 100 3 50 4 25 5 12.5 6 6.25

25 mg/hr

100 mg/hr

6.25 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 Am’t Delivered

hr (mg) 1 200 2 200 3 0

25 mg/hr

100 mg/hr

6.25 mg/hr

Zero-Order Rate200 mg/hr

Tmax = 2 hours

The infusion rate is identified as k0

with units of am’t/time.

Intravenous Infusion-continuousIntravenous Infusion-continuous

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


Approaching maximum concentration of 7.3 mg/L

Intravenous Infusion-continuousIntravenous Infusion-continuous

Number Percent Of of CSS

T½

1 50.0 2 75.0 3 87.5 4 93.75 5 96.875 6 98.438 7 99.219


Approaching maximum concentration of 7.3 mg/L

The rate at which the steady-state concentration

is approached is proportional

to the half-life.


Many drugs are administered by

intermittent intravenous infusions, where the time of the infusion is short and SS is not achieved.

15 min - -lactams30 min – aminoglycosides

1 hr vancomycin 4 hour amphotericin.

For these drugs the infusion rate (mg/hr) is:

k0 = dose / infusion time

IV Intermittent InfusionDose = 400 mg

Infusion Time = 2 hoursInfusion Rate (k0) = 200 mg/hr


During the infusion period the concentrations increase.

The equation predicting these concentrations is:CSS = (k0/KV) *(1-e-kt)

This has rough similarity to the oral dosing equation



Vol = 10 L; T½ = 2hrCSS = 8.66 mg/L

Dose Clearance




Here 160 mg is infused over 4 hours to a patient with a V of 10 L and a

half-life of 2 hours



Vol = 10 L; T½ = 2hrCSS = 8.66 mg/L

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




Here 160 mg is infused over 4 hours to a patient with a V of 10 L and a

half-life of 2 hours



Vol = 10 L; T½ = 2hrCSS = 8.66 mg/L

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


As soon as the infusion is complete, the concentrations will begin to decline. If distribution is instantaneousThis 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’


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 ofinterest – 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

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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

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QuestionA patient is given an intravenousinfusion 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?

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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

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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

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Volume based on this concentration8

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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

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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?

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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) T½With 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?

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QuestionIf 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 becomesCSS = [(k0/KV)]

= 50 (0.3465 x 16.23)= 8.89 mg/L

What bolus loading dose would achieve this true

steady-state?



CSS = [(k0/KV) *(1-e-Kt)]When t is large e-Kt becomes Small and equation becomesCSS = [(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


QuestionIf 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



What would the concentration-time profile for the bolus look like?

T½ =

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What would the concentration-time profile for the bolus look like?

T½ = 2 hours

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Should the conc.–time profilefrom the bolus and the simultaneous

infusion just be the sum of both profiles?


QuestionTime 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 8.89112.0 0.139 8.752 8.89116.0 0.035 8.856 8.89124.0 0.002 8.889 8.891

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QuestionTime 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 8.89112.0 0.139 8.752 8.89116.0 0.035 8.856 8.89124.0 0.002 8.889 8.891

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2When bolus and infusion

are exactly matched concentration starts at true steady state

and remains unchanged


Question

What happens if the bolusis larger or smaller?

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Don’t let this be you on Exam Day.

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Intravenous Infusion Previous rates of administration were instantaneous IV bolus and first order absorption. As a rate (mg/hr) a first order rate constantly