## Examples of Calculations

Equations (13) to (15) can be used to solve three types of problems involving first-order processes. These types of problems are illustrated in the following examples:

TYPE 1: Given the rate constant or half-life and the initial concentration, calculate the concentration at some time in the future.

Example. A drug solution containing 500 units/mL has a half-life of 10 days. What will the concentration be in seven days?

k = 0693 = ^691 = 0.069jday t1=2 10 day ' y ln C = kt C

^ = antiln (0.483) = e0483 = 1.62 C = 308 units jmL

TYPE 2: Given the half-life or rate constant and the initial concentration, calculate the time required to reach a specified lower concentration.

Example. A drug solution has a half-life of 21 days. How long will it take for the potency to drop to 90% of the initial potency?

0:693

100%

TYPE 3: Given an initial concentration and the concentration after a specified elapsed time, calculate the rate constant or half-life.

Example. A drug solution has an initial potency of 125 mg/5 mL. After one month in a refrigerator, the potency is found to be 100 mg/5 mL. What is the half-life of the drug solution under these storage conditions?

125 mg/5 mL

For each type of problem the following assumptions are made: (i) The process follows first-order kinetics, at least over the time interval and concentration range involved in the calculations and (ii) all time and concentration values are accurate.

The latter assumption is particularly critical in solving problems such as type 3, where a rate constant is being calculated. It would be unwise to rely on only two time points to calculate such an important value. Normally, duplicate or triplicate assays would be performed at six or more time points throughout as much of the reaction as possible. The resulting mean assay values and standard deviation values would be plotted on semilogarithmic graph paper and a straight line carefully fitted to the data points. The half-life could then be determined using equation (14).

Example. A solution of an ester in pH 9.5 buffer (25°C) was assayed in triplicate several times over a 20-hour period. The data obtained are presented in Table 2. The results were plotted on semilogarithmic graph paper as shown in Figure 3. Calculate the pseudo-first-order rate constant for the hydrolysis of the ester at pH 9.5 (25°C).

By fitting a straight line ("best-fit" line) through the data points in Figure 3 (this can be done by eye using a transparent straight edge) and extrapolating to t = 0, the intercept C0 is found to be 3.13 mg/mL. The half-life is the time at which the concentration equals 1.57 mg/mL (Co/2), and this is found by interpolation to be 2.4 hours. The value of k can be given by

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