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Question #1:

Css = ?

Css = Q / ke * Vd where:

Q = 150 mg/min

ke = 0.046/hr

and Vd = 40 L

Css = 150 mg/min /((0.046/60min)*40,000ml)

notice how the units are consistent: hr in ke changed to min to match Q (mg/min) and 40 L is represented as 40,000 ml

Answer: Css = 4.89 mg/ml

 

Question #2:

Css = ?

Css = Q / ke * Vd where:

Q = 150 mg/min

ke = 0.046/hr

and Vd

Css = 150 mg/min /((0.046/60min)*12,000ml)

notice how the units are consistent: hr in ke changed to min to match Q (mg/min) and 12 L is represented as 12,000ml

Answer: Css = 16.23 ug/ml

 

Question #3:

Css = ?

Css = Q / ke * Vd where:

Q = 300 mg/min

ke = 0.046/hr

and Vd 40 L

so

Css = 300 ug/min /((0.046/60min)*40,000ml)

notice how the units are consistent: hr in ke changed to min to match Q (mg/min) and 40 L is represented as 40,000ml

Answer: Css = 9.74 mg/ml

 

Question #4:

Css = ?

Css = Q / ke * Vd where:

Q = 450 m g/min

ke = 0.046/hr

and Vd 40 L>

so

Css = 450 mg/min /((0.046/60min)* 40,000ml)

notice how the units are consistent: hr in ke changed to min to match Q (mg/min) and 40 Lis represented as 40,000ml

Answer: Css = 14.61 mg/ml

 

Question #5:

f = 1 -e -ket

this equation allows calculation of the time (t) need to reach a new steady-state concentration following a constant infusion of drug; f is the fraction of the shift from one steady state to the other. The question here asks for a 50% or 0.5 shift.

so: f = 1 -e -ket

or 0.5 = 1 - e-(0.046/hr)t

or 0.5 = e(-0.046/hr)t

now take antilogs:

-0.693 = -0.046 t

-0.693/-0.046 = t

Answer: t = 15 hours; note that a drug will always reach 50% of Css in one t1/2

 

Question #6:

Q = ?

Css = Q / ke * Vd where:

Css = 25 ug/ml

Q = ? m g/min

ke = 0.046/hr

and Vd = 40 L

25 ug/ml = Q / ((0.046/60 min) * 40,000 ml)

Answer: Q = 760 ug/min

 

Question #7:

f = 1 -e -ket

this equation allows calculation of the time (t) need to reach a new steady-state concentration following a constant infusion of drug; f is the fraction of the shift from one steady state to the other. The question here asks for a 90% or 0.9 shift.

so: f = 1 -e -ket

or 0.9 = 1 - e-(0.046/hr)t

or 0.1 = e(-0.046/hr)t

now take antilogs:

-2.3 = -0.046 t

-2.3/-0.046 = t

Answer: t = 50 hours;