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**Advantages:**

i) The absorption and elimination processes can be quite similar and accurate determinations of ka can still 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. The method can provide very useful information about the absorption processes with different dosage forms.

**Disadvantages:**

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

ii) The required calculations are more complex.

**Equation 9.3.1 Mass Balance Equation**

or

**Equation 9.3.2 Mass Balance Equation**

Differentiating each term with respect to time gives:-

**Equation 9.3.3 Differentiated Equation**

or

**Equation 9.3.4 Rate of Change of Amount Absorbed**

or

Integrating gives:-

**Equation 9.3.5 Amount Absorbed versus Time**

or

**Equation 9.3.6 Amount Absorbed divided by Volume versus Time**

Taking this to infinity where Cp equals 0

**Equation 9.3.7 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 9.3.8 Amount Remaining to be Absorbed**

We can use this equation to look at the absorption process. If, and only if, absorption is a single first order process.

or

Thus a plot of ln (A_{max} - A) *versus* time will give a straight line for first order absorption with a slope(ln) = -ka. Note that linear or other types of plots may reveal other absorption behavior. For example a straight line on linear graph paper might suggest that absorption follows zero order kinetics, such as with an infusion step or an attempted mimicking of zero order absorption with a patch or another slow release delivery device.

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

**Figure 9.3.2 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. This would support the assumption that absorption can be described as a single first process. The first-order absorption rate constant, ka, can be calculated to be 0.306 hr^{-1} from the slope of the line on the semi-log graph paper.

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

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