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“11g of ethyl ethanoate were mixed with 18cm^3 of 1.00moldm^-3 HCl(aq) in a flask and allowed to stand at a constant temperature until equilibrium was reached.
CH3CO2C2H5(l) + H2O(l) <=> (equilibrium) CH3CO2H(l) + C2H5OH(l)
Ethyl ethanoate + water <=> ethanoic acid + ethanol
When titrated with 1.00moldm^-3 NaOH(aq), 106cm^3 were required for neautralisation. Assume that 18cm^3 of 1.00moldm^-3 HCl(aq) contain 18g water.
What is the value of Kc?
If its difficult to reply by comment it would be great if you could email me.. Thank you in advance :)
CH3CO2C2H5(l) + H2O(l) <=> (equilibrium) CH3CO2H(l) + C2H5OH(l)
Ethyl ethanoate + water <=> ethanoic acid + ethanol
When titrated with 1.00moldm^-3 NaOH(aq), 106cm^3 were required for neautralisation. Assume that 18cm^3 of 1.00moldm^-3 HCl(aq) contain 18g water.
What is the value of Kc?
If its difficult to reply by comment it would be great if you could email me.. Thank you in advance :)
8 years ago
Chemistry Question asked by Marya
Add a CommentUse the NaOH to work out the concentration of ethanoic acid formed. Then use the molar ratio to work out the mole of the ester, acid and alcohol. Use that to work out Kc.
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Accepted Answer
Moles NaOH used = 1.00 x 106/1000 = 0.106 mol
Moles total acid neutralised = 0.106 mol as H+:OH- is 1:1
Moles HCl used = 1.00 x 18/1000 = 0.0180 mol
Moles CH3COOH at eqm = 0.106 - 0.0180 = 0.0880 mol
Hence moles C2H5OH at eqm = 0.0880 mol
Moles CH3COOC2H5 taken = 11/88 = 0.125 mol
Moles CH3COOC2H5 reacted = 0.0880 mol
Moles CH3COOC2H5 at eqm = 0.125 - 0.0880 = 0.037mol
Moles H2O at eqm = 1.00 - 0.0880 = 0.912 mol (18g H2O = 1.00 mol)
Kc = [CH3COOH] [C2H5OH] / [CH3COOC2H5] [H2O]
As the concentration terms in the Kc expression are balanced moles can be used instead of mol dm-3 as the volume terms will cancel each other out
Kc = 0.0880 x 0.0880 / 0.037 x 0.912 = 0.229
Moles total acid neutralised = 0.106 mol as H+:OH- is 1:1
Moles HCl used = 1.00 x 18/1000 = 0.0180 mol
Moles CH3COOH at eqm = 0.106 - 0.0180 = 0.0880 mol
Hence moles C2H5OH at eqm = 0.0880 mol
Moles CH3COOC2H5 taken = 11/88 = 0.125 mol
Moles CH3COOC2H5 reacted = 0.0880 mol
Moles CH3COOC2H5 at eqm = 0.125 - 0.0880 = 0.037mol
Moles H2O at eqm = 1.00 - 0.0880 = 0.912 mol (18g H2O = 1.00 mol)
Kc = [CH3COOH] [C2H5OH] / [CH3COOC2H5] [H2O]
As the concentration terms in the Kc expression are balanced moles can be used instead of mol dm-3 as the volume terms will cancel each other out
Kc = 0.0880 x 0.0880 / 0.037 x 0.912 = 0.229
Answered by [Deleted Member]
+ 3 
Thank you so much, I finally get it!!! :D
