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junior college 1 | H2 Maths
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This is a Chemistry Question
Date Posted: 6 years ago
Views: 589
Nah, it's a Math question. Always treat A-Level Chemistry questions like PSLE/O-Level Math, just with chemical equations thrown in.
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This question tests on stoichiometry ratios, in which NaOH is used to completely neutralised by NaOH. Since oxalic acid is a dicarboxylic acid (acid having 2 COOH) groups, the stoichiometry ratio of NaOH: Oxalic Acid is 2:1. Since we can calculate immediately the amount of NaOH used to neutralise both acids and the amount of ethanoic acid present, we can calculate the amount of NaOH used to neutralise ethanoic acid and hence the amount used to neutralise oxalic acid. From there, further use stoichiometry to calculate the amount of oxalic acid present and hence its concentration. Hope this helps!
Date Posted:
6 years ago
Thank you!
Thank you!!
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First find amount of NaOH, and from there you know the amount of OH- ions.
OH- and H+ react in a 1-to-1 ratio to form water, so amount of H+ ions = amount of OH- ions.
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They gave [CH3COOH], so from there we can find the amount of CH3COOH, and the amount of H+ ions contributed by it in the reaction.
The remaining H+ ions must come from (COOH)2.
Remember that 1 (COOH)2 = 2H+, therefore amount of (COOH)2 needs to be halved.
Lastly, find [(COOH)2].
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Feels like doing a problem sum isn't it?
OH- and H+ react in a 1-to-1 ratio to form water, so amount of H+ ions = amount of OH- ions.
---
They gave [CH3COOH], so from there we can find the amount of CH3COOH, and the amount of H+ ions contributed by it in the reaction.
The remaining H+ ions must come from (COOH)2.
Remember that 1 (COOH)2 = 2H+, therefore amount of (COOH)2 needs to be halved.
Lastly, find [(COOH)2].
---
Feels like doing a problem sum isn't it?
Date Posted:
6 years ago
Bleh, 3 minutes too slow.
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