How do I work out the molarity and equilibrium constant using moles of reactants?

Question

The equation is C2H5OH(aq) + CH3CO2H(aq) ? CH3CO2C2H5(aq) + H2O(l)

 

Moles of ethanol: 0.0747

Moles of ethanoic acid: 0.0112

 

amount of ethanol used in experiment: 4cm³

amount of ethanoic acid used: 1cm³

 

(titration with NaOH value if it helps: 18.00 --> if the titration value isn't used in the calculation then why was a titration necessary at all?)

 

I'm very confused and would appreciate some help!

Answer 1

Hello Dana,

 

I would say the titration value is essential because with that, you can calculate the number of moles of acetic acid present at equilibrium. Two questions: The titration value is 18.00...what? Numbers without units are meaningless. Second, your balanced equation shows aqueous solutions, but no volume is specified. If that is so, you can only come up with a pseudo-K using the moles, and not a true K which would require the molarity.

 

Once you have the amount of acetic acid present at equilibrium, you can simply use the moles at equilibrium to figure out how many moles of ethanol and ethyl acetate are also present, and then determine the value of K since the volumes divide out (in this particular situation!).

 

Jim Scripko