Find the percent ionization of a 0.250 M solution of HC2H3O2.?

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