Homework Assignment 1995 - 4 (Answers)

Problem 2

First, plot the I.V. data on semi-log graph paper.

With the Cp0 = 6.7 mg/L and Cp6hr (from the line) = 2.6 mg/L the elimination rate constant can be calculated as 0.158 hr-1 and the volume of distribution as 14.9 L.

Plotting the oral data on semi-log paper gives:

Using the Method of Residual The Calculations

Time (hr) Cp (mg/L) Cp late (mg/L) Residual (mg/L)
0 0

0.25 2.28 17.3 15.0
0.5 4.33 16.7 12.4
0.75 5.67 16.1 10.4
1 6.89 15.6 8.7
1.5 8.81 14.5 5.7
2 9.31 13.5 4.2
3 9.68 11.7 2.0
4 10.25 10.1
6 7.51 7.6
9 4.98 5.0
12 3.25 3.2

Graphing this data with Cplate (green) and Residual (red) lines gives:

From the two lines kel = 0.144 hr-1 and ka = 0.740 hr-1 (assuming ka > kel)

Thus ka/kel = 5.14 and the method of residuals can be used.

The intercept 'A' value of 18 mg/L can be used to calculate V/F by rearranging equation 52:

V/F = Dose * ka / (A * (ka-kel)) = 300 x 0.740 / (18 x (0.740 - 0.144)) = 20.7 L

Since V = 14.9 L a value for F can be calculated as V/(V/F) = 14.9/20.7 = 0.720

Summary in Tabular Format
Parameter Value Units
kel (I.V.) 0.158 hr-1
V (I.V.) 14.9 L
kel (p.o.) 0.144 hr-1
ka (p.o.) 0.740 hr-1
F 0.720

Another Approach

Using the Wagner-Nelson method gave a ka value of 0.68 hr-1

The AUC(i.v.) was 42.1 mg.hr/L and the AUC(p.o.) was 100.9 mg.hr/L thus F = 0.80 after adjusting for dose.