E5 ATQ

ETERMINATION OF ACETYLSALICYLIC ACID IN QUANTITATIVE DETERMINATION  ASPIRIN T ABLETS BY B ACK -T -TITRATION D. J. Q. Y TAC TAC HEMICAL E NGINEERING NGINEERING , C OLLEGE OLLEGE OF E NGINEERING NGINEERING DEPARTMENT OF C HEMICAL U NIVERSITY  , DILIMAN  , QUEZON C ITY   , P HILIPPINES NIVERSITY OF THE P HILIPPINES  HILIPPINES  ITY  HILIPPINES DATE PERFORMED: 27 SEP 2014 INSTRUCTOR S NAME: ERICKA ITANG ’

 ANSWERS TO QUESTIONS 1. Discuss the use of a more dilute NaOH solution for the standardization of NaOH. This was done to make the neutralization reaction fast and allow the solution to change color as soon as the titrant was added. This also helps minimize deviations in calculated molarity of NaOH since its volume is in the denominator in calculating for the molarity [1]. 2. Discuss the rationale behind the dilution and aliquoting of the aspirin sample. This is to obtain less amount of HCl to be used for titration and to conserve chemicals and minimize chemical waste. 3. Discuss the principles behind the use of back titration in the analysis of aspirin tablet. Back-titration was used because the reaction of acetylsalicylic acid in aspirin and NaOH is slow. This is better than the usual titration since endpoint is easier to locate. 4. Discuss the importance of simmering the tablet in the NaOH solution for 20 minutes. This was done to allow complete hydrolysis of aspirin to acetic and salicylic acids since aspirin easily hydrolizes in basic solutions and this could cause it to breakdown. It is also only sparingly soluble in water[1]. After hydrolysis, neutralization of NaOH comes next. Simmering the tablet also helps speed up the reaction. 5. Discuss the possible sources of errors and their effect on calculated parameters. Errors could come from failing to add distilled water to aspirin during simmering. This causes the sample to dry up and would therefore affect in calculating for the %acetylsalicylic acid. Other sources of errors could come from improper conduct of titration such as not using white background when observing the color which could cause the solutions to be overtitrated and not removing the bubbles in the burette during titration. REFERENCES [1] Skoog, D. A., West, D. M., Holler, F. J., Crouch, S. R. Fundamentals of Analytical Chemistry 9 th ed. Cengage Learning. 2013. Page 1 of 3

CALCULATIONS

  =180.17    =204.23  Standardization of 0.050M NaOH Solution Diluted NaOH Reaction: KHC8H4O4 (aq) + NaOH (aq)  KNaC8H4O4 (aq) + H2O (aq) mol NaOH = mol KHP

0.204. 201323 / )0.995  (  %  =  = = 0.0354 M NaOH 0.0277  Standard NaOH

 = ,   = 0.0354  = 0.7 08 M NaOH %ASA Average Molarity of HCl = 0.04695724696M Total mmol NaOH added to aspirin =

0.0050.711  = 3.56  NaOH

Excess mmol reacted with HCl =

 = 2.5 1  NaOH 0.047010.7   

mmol ASA =

.−. = 0.522 mmol ASA 

180. 1 6    1   0. 5 22   × × 1000   1    ×100%=62.53%  % = 0.1504 

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Statistical Analysis

=  = 0.03540.0357 =0.0356 2   

Range = Highest Value-Lowest Value = 0.0357-0.0354 = 0.000317

Standard Deviation (s) =

−  =  .−.+.−. =  − −

0.000224

 ×1000= . ×1000= 6.2 9 ppt  . s =0.051 ± .. =0.036±0.002 Confidence Limits(at 95% confidence level) =  ± √  √  Relative Standard Deviation (RSD) =

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