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Lelu [443]
3 years ago
5

Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity. If an aspirin tablet weighing 0.615 g is titr

ated with standardized 0.1121 M KOH, the endpoint is reached after 20.52 mL of KOH have been added. What is the percentage of aspirin in the tablet
Chemistry
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

67.4 % of C₉H₈O₄

Explanation:

To make titrations problems we know, that in the endpoint:

mmoles of acid = mmoles of base

mmoles = M . volume so:

mmoles of acid = 20.52 mL . 0.1121 M

mmoles of acid = mg of acid / PM (mg /mmoles)

Let's determine the PM of aspirin:

12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol

mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol

mass (mg) = 414.4 mg

We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g

We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %

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Explanation:

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Learn more about Gibbs free energy here:

brainly.com/question/13765848

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