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
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
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)
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
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
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.
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 😓
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.
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.