i is correct at 0.045 mols NaOH.
ii. The NaOH neutralized all of the HCl (remember the catalyst doesn't change in concentration) which is 0.005 mols. This means that 0.045-0.005 = 0.040 mols of the NaOH reacted with whatever CH3COOH was left after equilibrium was attained.
iii. Your equation is correct.
iv.See my response to ii.
b.
........CH3COOH + EtOH ==> CH3COOEt + H2O
I.............0.1..............0.1...............0....................0
C.......- 0.040.........-0.040..........+0.040........+0.040
E............?.................? ? ?
Plug the E line into the Kc expression and solve for Kc.
Ethanoic acid can be reacted with alcohols to form esters, an equilibrium mixture being formed.
CH3CO2H + ROH <--> CH3CO2R + H2O
The reaction is usually carried out in the presence of an acid catalyst.
In an experiment to determine Kc a student placed together in a conical flask 0.10 mol of ethanoic acid, 0.10 of an alcohol ROH, and 0.005 mol of hydrogen chloride catalyst.
The flask was sealed and kept at 25°C for seven days.
After this time, the student titrated all of the contents of the flask with 2.00 mol/dm3 NaOH using phenolphthalein indicator.
At the end point, 22.5 cm3 of NaOH had been used.
a) i) Calculate the amount, in moles of NaOH used in the titration.
ii) What amount, in moles, of this NaOH reacted with the hydrogen chloride?
iii) Write a balanced equation for the reaction between ethanoic acid and NaOH.
iv) Hence calculate the amount, in moles, of NaOH that reacted with the ethanoic acid.
b) i) Use your results from a) to calculate the amount, in moles, of ethanoic acid present at equilibrium.
Sorry for the huge chunk of boring text, but here's what I got:
i) moles = concentration x volume, which is 2 x 22.5/1000 = 0.045 mol
ii) Ehhh, not too sure about this one, some help would be appreciated :)
iii) CH3COOH + NaOH --> CH3CO2Na + H2O
iv) Not too sure 'bout this one either
b) i) Can't work this one out
Thanks in advance for your time and help ^_^
1 answer
1 answer