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Topic: Titration Tartaric Acid Problem  (Read 6746 times)

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Offline miden24

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Titration Tartaric Acid Problem
« on: November 29, 2014, 10:27:34 PM »
A winemaker wishes to determine the concentration of tartaric acid (H2C2H4O6) in his wine. He therefore titrates a 20.00 mL sample of his wine with 11.70 mL of 0.0485 M of sodium hydroxide. What is the molarity of the tartaric acid in the wine?

(0.0485 M * 0.0117 L) = 5.67 * 10^-4 moles of NaOH

5.67 * 10^-4 moles of NaOH * (1 mole H2C4H4O6/ 2 moles of NaOH) = 2.84 * 10^-4 moles of tartaric acid

2.84 * 10^-4 moles of tartaric acid / 0.0117 L = 0.0243 M of tartaric acid?

Is my answer right?

Offline Borek

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Re: Titration Tartaric Acid Problem
« Reply #1 on: November 30, 2014, 04:31:25 AM »
Don't use rounded down intermediates in your calculations.

But that's not the main problem with your solution. You were right up to the number of moles of tartaric acid (.284 mmol is a correct answer), yet the concentration you calculated is wrong.
« Last Edit: November 30, 2014, 05:26:40 AM by Borek »
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Offline cseil

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Re: Titration Tartaric Acid Problem
« Reply #2 on: November 30, 2014, 01:13:10 PM »
A winemaker wishes to determine the concentration of tartaric acid (H2C2H4O6) in his wine. He therefore titrates a 20.00 mL sample of his wine with 11.70 mL of 0.0485 M of sodium hydroxide. What is the molarity of the tartaric acid in the wine?

(0.0485 M * 0.0117 L) = 5.67 * 10^-4 moles of NaOH

5.67 * 10^-4 moles of NaOH * (1 mole H2C4H4O6/ 2 moles of NaOH) = 2.84 * 10^-4 moles of tartaric acid

2.84 * 10^-4 moles of tartaric acid / 0.0117 L = 0.0243 M of tartaric acid?

Is my answer right?

When the equivalent point is reached, tartaric acid has reacted completely with NaOH. What does that mean?
« Last Edit: November 30, 2014, 02:46:27 PM by cseil »

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