0.1 mole of acetic acid has been dissolved per dm3 of the solution, the percentage ionization of acetic acid will be

Explanation

A, B and D are incorrect because these are not correct percentage ionization valuesC is correct since The solution is given as CH3COOH dissociates to a small extent:CH3COOH ↔ CH3COO- + H+Calculate [H+] from the Ka equation.Ka = [H+] [[CH3COO-] / [CH3COOH]Because [H+] =[CH3COO-] and dissociation is very small, we can write the equation asKa = [H+]² / 0.1Ka = 1.8*10-51.8*10-5 = [H+]² / 0.1[H+]² = (1.8*10-5)* 0.1[H+]² = 1.8*10-6[H+] = 0.0013M% dissociation = 0.0013/0.1*100 = 1.34%