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**Advantages:**

i) The absorption and elimination processes can be quite similar and still accurate determinations of ka can be made.

ii) The absorption process doesn't have to be first order. This method can be used to investigate the absorption process. I have used this type of method to investigate data obtained after IM administration and found that two absorption steps maybe appropriate. Possibly a fast step from drug in solution and a slower step from drug precipitated at the injection site.

**Disadvantages:**

i) The major disadvantage of this method is that you need to know the elimination rate constant, from data collected following intravenous administration.

Theory: The working equations can be derived from the mass balance equation:-

**Equation 18.3.1 Mass Balance Equation**

or

**Equation 18.3.2 Mass Balance Equation**

Differentiating each term with respect to time gives:-

**Equation 18.3.3 Differentiated Equation**

or

**Equation 18.3.4 Rate of Change of Amount Absorbed**

or

Integrating gives:-

**Equation 18.3.5 Amount Absorbed versus Time**

or

**Equation 18.3.6 Amount Absorbed divided by Volume versus Time**

Taking this to infinity where Cp equals 0

**Equation 18.3.7 Maximum Amount Absorbed divided by Volume of Distribution**

Finally (A_{max} - A), the amount remaining to be absorbed can also be expressed as the amount remaining in the GI, Xg

**Equation 18.3.8 Amount Remaining to be Absorbed**

We can use this equation to look at the absorption process. If absorption is first order

or

Thus a plot of ln (A_{max} - A) versus time will give a straight line for first order absorption with a slope = -ka

**kel (from IV data) = 0.2 hr ^{-1}**

Time(hr) | PlasmaConcentration (mg/L) | Column3 ΔAUC | Column4 AUC | Column 5kel * AUC | A/V[Col2 + Col5] | (A_{max } - A)/V |

0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 4.9 |

1.0 | 1.2 | 0.6 | 0.6 | 0.12 | 1.32 | 3.58 |

2.0 | 1.8 | 1.5 | 2.1 | 0.42 | 2.22 | 2.68 |

3.0 | 2.1 | 1.95 | 4.05 | 0.81 | 2.91 | 1.99 |

4.0 | 2.2 | 2.15 | 6.2 | 1.24 | 3.44 | 1.46 |

5.0 | 2.2 | 2.2 | 8.4 | 1.68 | 3.88 | 1.02 |

6.0 | 2.0 | 2.1 | 10.5 | 2.1 | 4.1 | 0.8 |

8.0 | 1.7 | 3.7 | 14.2 | 2.84 | 4.54 | 0.36 |

10.0 | 1.3 | 3.0 | 17.2 | 3.44 | 4.74 | 0.16 |

12.0 | 1.0 | 2.3 | 19.5 | 3.9 | 4.9 | - |

∞ | 0.0 | 5.0 | 24.5 | 4.9 | 4.9 | - |

The data (A_{max}-A)/V versus time can be plotted on semi-log and linear graph paper.

**Figure 18.3.1 Semi-log plot of (A _{max}-A)/V versus Time**

**Figure 18.3.2 Linear plot of (A _{max}-A)/V versus Time**

Plotting (A_{max}-A)/V versus time produces a straight line on semi-log graph paper and a curved line on linear graph paper. From the slope of the line on the semi-log graph paper ka can be calculated to be 0.306 hr^{-1}.

- Wagner, J and Nelson, E. 1964
*J. Pharm. Sci.*,**53**, 1392

Copyright 2001-3 David W. A. Bourne (david@boomer.org)

This file was last modified: Wednesday 26 May 2010 at 07:51 PM