SME
Biotransport
Pharmacokinetics
Use this for “500 mg is injected into a 50 L volume and cleared at 5 L/h. What is the level 12 h later, and what is the half-life?”
Chart
C(t) = 10e−0.1t
A log y-axis: an exponential draws as a straight line.
This is an example. Type your own values to start a new problem.
C₀ (mg/L)
=
D (mg)
V (L)
k (h⁻¹)
=
CL (L/h)
V (L)
C (mg/L)
=
C₀ (mg/L)
e
−
h⁻¹
×
h
D is the dose injected and V the volume of distribution, the volume the drug seems to spread through.
t½
Half-life
calculated
AUC
Area under the curve
calculated · mg·h/L
Formulas
C₀
=
D
V
10
=
500
50
k
=
CL
V
0.1
=
5
50
t
1
2
=
ln(2)
k
6.931
=
ln(2)
0.1
C
=
C₀
×
e
−
k
×
t
3.012
=
10
×
e
−0.1 × 12
AUC
=
D
CL
100
=
500
5
Assumptions
•
The body is one well-mixed compartment, and the whole dose enters it at once (an IV bolus).
•
Elimination is first order: dC/dt = −kC, so the same share of the drug leaves each hour.
•
Clearance CL is the volume of blood cleared of drug each hour, so k = CL ÷ V.
Step-by-step
WE KNOW
D = 500 mg, V = 50 L, CL = 5 L/h, t = 12 h
FIND
starting concentration (C₀), elimination rate constant (k), half-life (t½), concentration at t (C), area under the curve (AUC)
Step 1 · Find starting concentration (C₀)
Use C₀ = D ÷ V
The dose spreads through the whole volume at once.
C₀
=
500
50
C₀
=
10
mg/L
Step 2 · Find elimination rate constant (k)
Use k = CL ÷ V
Clearance over volume is the share cleared each hour.
k
=
5
50
k
=
0.1
h⁻¹
Step 3 · Find half-life (t½)
Use t½ = ln(2) ÷ k
Half
is
left
when
e
−
kt
=
1
2,
so
kt
=
ln
2.
t
1
2
=
ln(2)
0.1
t
1
2
=
0.6931
0.1
t
1
2
=
6.931
h
Step 4 · Find concentration at t (C)
Use C = C₀ × e−k × t
The
level
falls
by
the
factor
e
−
kt
as
the
drug
is
cleared.
C
=
10
×
e
−0.1 × 12
C
=
10
×
e
−1.2
C
=
10
×
0.3012
C
=
3.012
mg/L
Step 5 · Find area under the curve (AUC)
Use AUC = D ÷ CL
All of the dose is cleared in the end, so the area is the dose over the clearance.
AUC
=
500
5
AUC = 100 mg·h/L
Check
Put the numbers back in. Both sides should match.
10
=
500
50
✓
0.1
=
5
50
✓
6.931
=
ln(2)
0.1
✓
3.012
=
10
×
e
−0.1 × 12
✓
100
=
500
5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Repeated doses and the steady state
Use this for “A 500 mg dose every 8 h, clearance 5 L/h. What is the average steady-state level?”
A constant infusion
Use this for “What infusion rate holds 10 mg/L when clearance is 5 L/h, and how long until the level is 90% of the way there?”
An oral dose: peak time and peak level
Use this for “A 500 mg tablet is absorbed at kₐ = 1 h⁻¹ into 50 L and cleared at k = 0.1 h⁻¹. When does the level peak, and how high?”