• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Objective: 1)To standardize the given unknown sodium hydroxide solution 2)To analyze the commercial aspirin tablet by back titration method

Extracts from this document...

Introduction

Course Code: 13554/Y1 Student Name: Chow Man Chung Lam Kwok Kei Lee Long Sing Chan Ka Chun Date performed: 28th September, 2010 Experiment 3: Acid base titration II Objective: 1) To standardize the given unknown sodium hydroxide solution 2) To analyze the commercial aspirin tablet by "back titration" method Apparatus & equipments used: 1) Burette, 50mL capacity 2) Bulb pipette, 25mL capacity 3) Volumetric flask, 250mL capacity 4) Mortar and pestle Chemicals used: 1) Standard Potassium Hydrogen Phthalate Solution, 0.1M 2) Sodium Hydroxide Solution, 1M 3) Commercial Aspirin Tablet 4) Phenol Red 5) Phenolphthalein Indicator Solution Results and Data Treatment: Weighting Data: Mass of vial: 6.0134g Mass of vial and Aspirin: 7.5142g Mass of ...read more.

Middle

0.00 0.00 25.00 0.00 Volume of titrant used (cm3) 35.20 24.85 24.80 24.90 Average volume: 24.85 cm3 Calculation: HK(C8H4O4)(aq) + NaOH(aq) --> NaK(C8H4O4)(aq) + H2O(l) Number of mole of HK(C8H4O4) = (0.1005)(24.85/1000) = 0.002497 ? HK(C8H4O4) : NaOH = 1 : 1 ? Number of mole of NaOH = Number of mole of HK(C8H4O4) = 0.002497 Molarity of NaOH(dil) = (0.002497) / (25/1000) = 0.09988M ~ 0.10M Molarity of NaOH = (0.10)(250/25) = 1.00M Titration II: Determination of the commercial Aspirin tablet Titrant (in burette): Potassium Hydrogen Phthalate Titrate (in Conical flask): 25.0cm3 of Aspirin and Sodium Hydroxide Mixture Solution Indicator used: Phenol Red Colour of indicator changed from: Purple to Yellow Titration No. 1(trial) ...read more.

Conclusion

= (0.000999975)(250/25) = 0.00999975 Number of mole of NaOH(original) = (1)(25/1000) = 0.025 Number of mole of NaOH(reaction) = 0.025-0.00999975 = 0.015 ? CH3COOC6H4COOH : NaOH = 1 : 2 ? Number of mole of CH3COOC6H4COOH = (Number of mole of NaOH) / 2 = 0.015 / 2 = 0.0075 Mass of CH3COOC6H4COOH = [(9)(12.0107)+(8)(1.00794)+(4)(15.9994)](0.0075) = (180.15742)(0.0075) = 1.3512 Percentage of CH3COOC6H4COOH in a Aspirin tablet = (1.3512 / 1.5008)(100%) = 90.03% Conclusion: In titration I, we found out given sodium hydroxide solution was ~1.00M and is fine with the given molarity. In titration II, we use the determined sodium hydroxide solution to react the 2-acetoxybenzoic acid which contain in 1.5g Aspirin tablet powder and found out it has 90.03% of 2-acetoxybenzoic acid in a Aspirin tablet by calculation. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Estimating the % purity of marble by back titration method

    qualitative and subjective instruction and also the indicator on turning pink indicate a slight alkalinity. Hence, the actual neutralization point was left behind 1-2 drops back. 3. Unevenly distributed impurities: CaCO3 is found naturally as marble in the earth. Thus, being a natural stone the impurity that it will contain

  2. Preparation of a standard solution of oxalic acid and using it to standardize a ...

    in the solute (NaOH). This is calculated through the mole ratio which in this case was 1:2, and dividing it by the volume of oxalic acid used in dm3. Weaknesses and limitations Too much phenolphthalein indicator may have been put into the solution.

  1. Objective: 1.To study the preparation of 1-bromobutane from 1-butanol by an SN2 reaction ...

    The flask was swirled to mix the content and a few anti-bumping granules were added. 7. The flask was fitted with the apparatus for reflux on a hot plat with an oil bath; a condenser was placed with tubing connected on top of the round-bottomed flask as shown in figure

  2. Objective: 1)To practice the procedure for preparing a standard solution 2)To perform ...

    1(trial) 2 3 4 Final burette reading (cm3) 23.00 23.30 23.10 23.15 Initial burette reading (cm3) 0.00 0.00 0.00 0.00 Volume of titrant used (cm3) 23.00 23.30 23.10 23.15 Average volume: 23.18 cm3 Calculation: Na2CO3 (s) + 2HCl --> 2NaCl(s) + H2O(l) + CO2(g) Number of mole of Na2CO3(aq) = 1.302g / 105.8089gmol-1 = 0.0123 ?

  1. The purpose of this Titration Analysis of ASA experiment was to use titration analysis ...

    are added to KHP(aq) , it is now possible for us to calculate the concentration of NaOH(aq) using molar ratio, but before we go any further, we need to sort out our uncertainties clearly! - Uncertainties Used For Calculations: - Concentration of KHP: (aq)

  2. Biodiesel Investigation - How the concentration of Potassium Hydroxide solution would affect the yield ...

    done before and allow the stirrer to create a reaction between the substances. Keep making qualitative observations and recording changes in temperature from the reaction and record the mass of the final solutions once stirring. 1. Finally, once all the products have been reacted, place them in centrifugal tubes and insert the tubes into the centrifuge.

  1. Chemistry Titration Acid Base Lab

    Unfortunately, in some cases the bubbles remain near the stopcock. If this is the case, it would be best to use a suction method approach. This involves one to partially open the stopcock allowing the contents of the burette to discharge in the beaker.

  2. The purpose of this experiment is to determine the concentration of a solution of ...

    + Na+ OH- (aq) →Na + A- (aq)+ H2O 1 mole of acid reacts with 1 mole of NaOH solution. Hence 0.02488 moles of potassium hydrogen phthalate will react with 0.02488 moles of NaOH solution. Concentration of solution = Moles of solution/Volume of solution =0.02488/0.02554 = 0.097450 M ±

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work