The standard curve that was generated from the first experiment, in determining the partition coefficient, shows an increased absorbance reading as the concentration increases and this is shown in Figure 1. The R2 value generated from the trend-line of the points plotted is 0.9808, therefore giving us an accurate equation of y = 1.392x – 0.0486 to determine the concentration of acetaminophen in the aqueous phase at pH 6 and 12. Plugging in the absorbance value of 1.225, at pH 12, for y in the equation gives us a value of 0.927 mg/ml for x. To determine the partition coefficient at pH of 6, 0.927 mg/ml was subtracted from 1 mg/ml and then divided by 0.927 mg/ml to give final value of 0.0787. This equation works because the 1-octanol and the…show more content… The R2 value of the graph was 0.9998 and this provided for an accurate use of the equation y = 719.8x to calculate the concentration of plasma in the different samples. Figure 4, shows all the routes of administration superimposed on one another for comparative reasons. PO showed a decrease in plasma concentration at the 15 min mark interval while IP showed a rapid increase in plasma concentration at the 15 min mark interval. IP concentration levels then decreased at the 30 min mark interval and then it slightly shot back up at the 60 min mark and stabilized. IV had the greatest plasma concentration over time, followed by PO and then IP. The area under the cure (AUC) for each route of administration was determined by cutting out the graphs and weighing them. The AUC was 606.0 mg for IV, 440.1 mg for the PO, and 237.1 mg for the IP. The bioavailability of the drug through IV is accepted as 100%. The bioavailability of the drug through PO administration is (440.1 mg/606.0 mg) x 100 = 72.6%. The bioavailability of the drug administered through 1- administration is (237.1 mg/606.1 mg) x 100 = 39.1%. The initial plasma concentration of acetaminophen in the rabbits was determined to be 316.02 µg/ml by taking the log of the IV graph, as shown in Figure 5, and then using the equation of that graph, y = 316.02e-0.024x,