The Stability Testing of Pharmaceuticals - Lab Report Example

Lab report on Stability Testing of Pharmaceuticals Introduction The quality of pharmaceuticals relies on the range of conditions provided for storage. The problem therefore arises on a better way of determining the influence of various conditions. The conditions include humidity and temperature. The biological activity of most drugs is dependent on their chemical integrity within the formulated dosage form (Huynh-Ba, 2009, 897). In other words, drugs should not undergo any chemical change upon storage within their specified shelf-life. Stability testing provides essential data on how the quality of a pharmaceutical varies over time under the influence of different environmental factors such as temperature, humidity and photostability. This enables recommended storage conditions, re-testing intervals and shelf-lives to be established (Hadzija & Ma, 2013, 46). The major purpose for this experiment is to explore the rate of hydrolysis of susceptible drugs. This will involve evaluating the acid hydrolysis of the ester methyl acetate to explore whether the hydrolysis follows a first order kinetic. In which case methyl acetate functions as an ester given the high rate of hydrolysis associated with the compound. This is unlike drugs which may take much longer to give out results (Huynh-Ba, 2010). Consequently, Methyl acetate is used as a model for a drug because the principle process is same in both cases. Method Apparatus: Conical flasks, burette, pipette, stop watch, water bath. Chemicals: Methyl acetate, 0.5 M of HCl, 0.1 M of NaOH and phenolphthalein indicator. Procedure: Four thermostatic water baths were set up at 30°C, 35°C, 40°C and 45°C respectively. 2 M of 50 ml HCL was then obtained and placed in a clean 250 ml of conical flask. 45 ml water was then added and properly mixed with the 2 M of 50 ml HCL. The flask was then placed in the specified water bath and left for 10 minutes to reach equilibrium. The 5 ml methyl acetate was pipetted into the flask followed by rapid mixing to avoid interference by external environmental factors. The stop watch was then started to record time for titration. 5 ml of the mixture was carried using a clean pipette and then placed in a conical flask having 50 ml deionized water. The content was titrated immediately and accurately with 0.2 NaOH. In this case, phenolphthalein was used to serve as an end-point indicator. The above procedure was repeated for the rest of time intervals: 10, 20, 30, 40 and 50 min using clean flasks containing 50 ml deionised water. Results and calculation Figure 1: showing titration results at 35 NaOH volume Time (min)   0 10 20 30 40 50 first 50 50 50 50 50 50 final 24.7 22.3 20.8 18.9 18.4 16.7 Amount required for titration 25.3 27.7 29.2 31.1 31.6 33.3 Fig 2: showing titration results at 300 NaOH volume Time (min)   0 10 20 30 40 50 first 0 0 0 0 0 0 final 25 26.5 29.1 29.7 30.5 31 Amount required for titration 25 26.5 29.1 29.7 30.5 31 Figure 3: showing titration results at 400 NaOH volume Time (min)   0 10 20 30 40 50 first 1 1 1 1 1 1 final 25 30.1 30.4 29.5 30.5 30.9 Amount required for titration 24 29.1 29.4 28.5 29.5 29.9 Figure 4: Showing titration results at 400 NaOH volume Time (min)   0 10 20 30 40 50 first 5.1 0.35 10.3 6.6 3.95 6.75 final 28.3 27.5 39.5 37.6 35.6 39.7 Amount required for titration 23.2 27.15 29.2 31 31.65 32.95 The equation used for determining the concentration of acid during titration is as follows:\ NaOH + HCL → NaCL + H2O We know that one mole of HCL reacts with exactly one mole of NaOH Sample calculations at 400 Amount required for titration = 29.1 ml x 10-3 = 0.0291 L V for HCl = 50 ml x 10-3 = 0.05 L Concentration = 0.2 M alculate the moles HCl(aq)       n(NaOH) = c(NaOH)) x V(NaOH)) = 0.2 x 0.0291 = 5.82 x 10-3 moles  From the balanced chemical equation find the stoichiometric (mole) ratio of acid to base: n(HCl) : n(NaOH)         1 : 1 moles NaOH HCl : NaOH is 1:1 So n(NaOH) = n(HCl) = 5.82 x 10-3 moles at the equivalence point Calculate concentration of NaOH: c(HCl) = n(HCl) ÷ V(HCl)     n(HCl) = 5.82 x 10-3 mol       V(HCl) = 0.05 L c(HCl) = 5.82 x 10-3 ÷ 50.0 x 10-3 = 0.1164 M or 0.1164 mol/L For acetic acid HC2H3O2(aq) + NaOH(aq) ---> NaC2H3O2(aq) + H2O(ℓ) The key is that there is a one-to-one molar ratio between the acetic acid and the sodium hydroxide Since it takes exactly the same amount of base to neutralize a given amount of acid, either strong or weak acid: We’ll have 5.82 x 10-3 ÷ 5.0 x 10-3 = 1.164 M First-order hydrolysis rate constants (k) Moles acetic acid: 2.00 g / 60.0516 g/mol = 0.0333047 mol Hence initial concentration = 0.0333047/5.0 x 10-3 = 6.66 M InKt In = -k*10 K = 0.1744 Half life t1/2 = t * ln(2)/ln(N0/Nt) 10 * In(2)* In = 1.2 Figure 6: Graph of log k against 1/T Discussion and conclusion The apparent first order rate constant for the reaction was determined using the integrated first order rate equation ln [CH3CO2CH3] = –kt + ln [CH3CO2CH3] From which it can be seen that a plot of log k versus 1/T should yield astraight line with slope allowin for calculation of activation energy. The data from the experiment along with the transformed data for determining the apparent first order rate constant k are found in table 1.These data were used to construct a plot of Log K versus time (see Figure 6) which according to equation above has a linear relationship giving out the activation energy. The slope of the best straight line through the data was calculated using linear regression to be –0.1744 min-1 with correlation coefficient of –0.9999. This means that the apparent first order rate constant k for the reaction is 0.1744 min-1. Further, the results also confirm the concept of half-life associated with hydrolysis of methyl acetate. After calculation, the half life of methyl acetate was found to be 1.2 min. This can also be confirmed from the graph which gives a straight line for the reduction of methyl acetate in the reaction. The line can be drawn on the graph to give a rate constant after a certain time between 2.6 and 1.3. The experiment was successful in achieving the major objective of the lab. Through plotting a graph of log k against 1/T it has proved that the rate of hydrolysis of many susceptible drugs is likely to follow first order kinetics as shown by the trend witnessed by methyl acetate. The graph gives a straight line, which acts as a model for determining the rate constant of hydrolysis of methyl acetate at any given temperature. It shows that the hydrolysis rate is proportional to the concentration of only one of the reactants or products of the reaction hence first order. The rate and order of the reaction depends only on the concentration of methyl acetate (Upadhya, 2006, 34). In summary, from the plot of ln [Methyl Acetate] versus time (see Figure 1) which according to equation (3) has a slope equal to –k’. The hydrolysis of methyl acetate was found to behave as a first order reaction in 2 M HCl with an apparent first order rate constant k of 0. 1744 min-1. Bibliography Huynh-Ba, K. (2009). Handbook of stability testing in pharmaceutical development: Regulations, methodologies, and best practices. New York: Springer. Ma, J. K. H., & Hadzija, B. (2013). Basic physical pharmacy. Burlington, MA: Jones & Bartlett Learning. Upadhyay, S. K. (2006). Chemical kinetics and reaction dynamics. New York: Springer. Huynh-Ba, K., & American Association of Pharmaceutical Scientists. (2010). Pharmaceutical stability testing to support global markets. New York: Springer. Read More