Anonymous -
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I am really stuck with this question - I just don't know how to write the equation? I have not been given the formula for the acid and cannot work out the product (other than water). Once I have the equation I will be okay with it.

Thanks in advance

25.0 cm3 of ethanedioic acid required 22.5 cm3 of 0.100 mol dm−3 potassium hydroxide solution for complete neutralisation. The concentration of ethanedioic acid is

A 0.0225 mol dm−3
B 0.0560 mol dm−3
C 0.0900 mol dm−3
D 0.0450 mol dm−3
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charco
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(Original post by Anonymous -)
I am really stuck with this question - I just don't know how to write the equation? I have not been given the formula for the acid and cannot work out the product (other than water). Once I have the equation I will be okay with it.

Thanks in advance

25.0 cm3 of ethanedioic acid required 22.5 cm3 of 0.100 mol dm−3 potassium hydroxide solution for complete neutralisation. The concentration of ethanedioic acid is

A 0.0225 mol dm−3
B 0.0560 mol dm−3
C 0.0900 mol dm−3
D 0.0450 mol dm−3
Ethanoic acid

Organic acid, "eth" means two carbon atoms; "anoic acid" means a COOH group (it includes one of the carbon atoms)

CH3COOH - the last hydrogen is the acidic one that reacts with the OH- from the KOH

CH3COOH + KOH ==> CH3COOK + H2O
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(Original post by charco)
Ethanoic acid

Organic acid, "eth" means two carbon atoms; "anoic acid" means a COOH group (it includes one of the carbon atoms)

CH3COOH - the last hydrogen is the acidic one that reacts with the OH- from the KOH

CH3COOH + KOH ==> CH3COOK + H2O
Hm, okay thank you for the help

I'm just a bit confused as it says ethanedioic acid rather than ethanoic acid
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df1
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wait but the original question says "ethaneDIoic acid" so surely there should be 2 COOH groups.Submit reply
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(Original post by df1)
wait but the original question says "ethaneDIoic acid" so surely there should be 2 COOH groups.Submit reply
That's where I'm a bit confused - does that then mean it is C2H2O4 ?
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df1
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I would form my equation as: HOOCCOOH + 2KOH ==> KOOCCOOK + 2H2O
Then, mol = conc x vol, so mol. of KOH = 0.1 x (22.5÷1000)= 9/4000
Ratio of KOH to HOOCCOOH is 2:1, so there are 9/8000 mol of HOOCCOOH.
Conc = mol / vol, so conc. of HOOCCOOH is (9÷8000)/(25÷1000)= 0.045 moldm-3 so the answer is D
(every time i divided by 1000 was to convert cm3 into dm3)
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(Original post by Anonymous -)
That's where I'm a bit confused - does that then mean it is C2H2O4 ?Submit reply

it would be a carbon chain of 2 carbons in length. Each carbon would be a COOH carbon. Basically 2 COOH in a row. I've sent how I would do the question now, referring to ethanedioic acid as HOOCCOOH
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(Original post by df1)
it would be a carbon chain of 2 carbons in length. Each carbon would be a COOH carbon. Basically 2 COOH in a row. I've sent how I would do the question now, referring to ethanedioic acid as HOOCCOOH
Thank you so much for the help
I just went through the workings that you sent and it makes total sense, so thank you again - I really appreciate it
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Thank you so much for the help
I just went through the workings that you sent and it makes total sense, so thank you again - I really appreciate it
No problem! That is a bit of a trickier titration question tbh because it combines organic chemistry too, so it'd probably be on paper 2 (the synoptic paper). If you're year 12 now the question should seem easier once you've covered more organic chemistry.
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(Original post by df1)
No problem! That is a bit of a trickier titration question tbh because it combines organic chemistry too, so it'd probably be on paper 2 (the synoptic paper). If you're year 12 now the question should seem easier once you've covered more organic chemistry.
Ah yeah, I'm in year 12 at the moment so hopefully it will all seem easier eventually - it's definitely a struggle currently 😓
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Ah yeah, I'm in year 12 at the moment so hopefully it will all seem easier eventually - it's definitely a struggle currently 😓
Yeah that's a really hard question for year 12 tbh so you've done well! Forming the product (neutralisation of carboxylic acids) is taught in year 13 anyways! But basically the +ve ion from the base (K+) replaces the +ve ion in the acid (the H in the COOH group) to form a salt.
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(Original post by df1)
Yeah that's a really hard question for year 12 tbh so you've done well! Forming the product (neutralisation of carboxylic acids) is taught in year 13 anyways! But basically the +ve ion from the base (K+) replaces the +ve ion in the acid (the H in the COOH group) to form a salt.
Thank you so much for explaining it to me, it means a lot!

The amount of substance side of Chemistry is definitely something that I'm struggling with at the moment, but hopefully I'll improve.
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Thank you so much for explaining it to me, it means a lot!

The amount of substance side of Chemistry is definitely something that I'm struggling with at the moment, but hopefully I'll improve.
I'm sure you will, just keep doing the practise questions and you'll be fine!
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I'm sure you will, just keep doing the practise questions and you'll be fine!
I'll try my best
I'm doing a Google Quiz at the moment, and they're quite complicated questions (some of them) - especially the last one
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