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

Mass of vial: 6.0134g

Mass of vial and Aspirin: 7.5142g

Mass of Aspirin: 1.5008g

Resulting Aspirin solution made up to: 250.0cm3

Titration I: Standardization of 0.1M sodium hydroxide solution

Titrant (in burette): Potassium Hydrogen Phthalate

Titrate (in conical flask): 25.0cm3 of Sodium Hydroxide

Indicator used: Phenolphthalein

Colour of indicator changed from: Purple to Colourless

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

Average volume: 9.95 cm3

Calculation:

CH3COOC6H4COOH(aq) + 2NaOH(aq) → CH3COONa(aq) + HOC6H4COONa(aq) + H2O(l)

Number of mole of HK(C8H4O4) = (0.1005)(9.95/1000)

                           = 0.000999975

Number of mole of 25.0 cm3 NaOH(left) = Number of mole of HK(C8H4O4)

                                = 0.000999975

Number of mole of NaOH(left) = (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.