2 Answers

CH3 COOH <======> H+ + CH3COO – Make an ICE table :
** I : [ CH3COOH ] = 0.250 mol/L ; [H+] = [ CH3COO ] = 0
** C ; [ CH3COOH] = x mol/L ; [ H+] = [ CH3COO ] = x mol / L
** E : [ CH3COOH] = 0.250 – x ; [ H+] = [ CH3COO ] = x mol / L
k 1.8 10^5 = x^2 / ( 0.250 – x ) ; x is the amount of CH3COOH is ionized. Since k is small so ( 0.250 – x ) = x, then solve for x. The percent ionization = (x/ 0.250 ) 100 %

x=2.12E^3
% ionization = 0.85