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Titration Analysis of Aspirin Tablets

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Chemistry 12
12/Oct/2011
Titration- Analysis of Aspirin Tablets
Objective: Determine the percentage of aspirin (acetylsalicylic acid) present in two different commercial tablets by titrating the solution with a base. Also determine whether the aspirin is a strong or weak acid according to the Bronsted- Lowry and Lewis theories and deduce the formula of the acid- base reaction.
Independent Variable: The amount of base (NaOH) in moles that are needed to neutralize the solution.
Dependent Variable: Percentage of aspirin (acetylsalicylic acid) found in each tablet.
Materials:
* Balance * 2 aspirin samples from different brands * 50 cm3 conical flask * 10.00cm3 of 95% alcohol * 0.100 mol dm-3 sodium hydroxide * Phenolphthalein …show more content…

H804 | | |

Percentage of aspirin: 0.5207 g | X | 100% | / | 0.100 g | = | 520.65 % C9H804 |

Percent Uncertainty: 0.10 ml | x | 100% | / | 28.90 ml | = | -+ 0.35% | 0.001 g | x | 100% | / | 0.603 g | = | -+ 0.17 % |

-+ 0.35% | + | -+ 0.17 % | = | -+ 0.52% |

Trial 3
NaOH needed to neutralize: 27.90 ml NaOH | 1 L | 0.10 mol NaOH | = | 0.00279 mol NaOH -+0.11ml | | 1000 ml | 1 L NaOH | | | |

Grams of Aspirin: 0.00279 mol NaOH | 1 mol C9H8O4 | 180.157 g C9H804 | = | 0.5026 g C9H8O4 | | 1 mol NaOH | 1 mol C9H804 | | |

Percentage of aspirin: 0.5026 g | X | 100% | / | 0.100 g | = | 502.64 % C9H804 |

Percent Uncertainty: 0.10 ml | x | 100% | / | 27.90 ml | = | -+ 0.36% | 0.001 g | x | 100% | / | 0.599 g | = | -+ 0.17 % |

-+ 0.36% | + | -+ 0.17 % | = | -+ 0.53% |

Aspirin B
Trial 1
NaOH needed to neutralize: 5.30 ml NaOH | 1 L | 0.10 mol NaOH | = | 0.00053 mol NaOH -+0.11ml | | 1000 ml | 1 L NaOH | | | |

Grams of Aspirin: 0.00053 mol NaOH | 1 mol C9H8O4 | 180.157 g C9H804 | = | 0.0955 g C9H8O4 | | 1 mol NaOH | 1 mol C9H804 | | |

Percentage of aspirin: 0.0955 g | X | 100% | / | 0.100 g | = | 95.48 % C9H804 |

Percent Uncertainty: 0.10 ml | x | 100% | / | 5.30ml | = | -+ 1.89% | 0.001 g | x | 100% | / | 0.217 g | = | -+ 0.46 % |

-+ 1.89% | + | -+ 0.48 % | = | -+ 2.35 % | Trial 2
NaOH needed to neutralize: 5.60 ml NaOH | 1 L | 0.10 mol NaOH | = | 0.00056 mol NaOH -+0.11ml | | 1000 ml | 1 L NaOH | | | |

Grams of

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