calculate the % dissociation of a 0.750M acetic acid solution and 0.50 M sodium acetate solution.

Question

I got the % for acetic acid to be 0.48% but cant finish the question.

Answer 1

Short cut...
________
[H+]= √Ka[Acid]
_____________
= √1.75E-5[0.750] M
= 0.0036M
%Ionization = ([H+]/[HOAc])100%
=[(0.0036)/(0.750)]100% = 0.48% is correct for pure water, but NaOAc provides a common ion that reduces the %ionization. Use the Ice table
HOAc <=> H + OAc
Ceq 0.75 X .50

Ka = [H][OAc]/[HOAc] = [(x)(0.50)]/(0.750) = 1.75E-5
Solve for x = 2.625E-5 M = [H]
%Izn = [(2.625E-5)/(0.750)]100%
=0.0035%