The equilibrium dissociation of the weak acid can berepresented as: HA (aq) + H2O (l) H3O+(aq) + A- (aq) [HA]initial= 0.080 M Percent ionization = 2.3 % Percent ionization = [H3O+] /[HA]initial * 100 2.3 % = ([H3O+]/0.080 M) *100 [H3O+] = 0.00184 M Ka= [H3O+][A-] / [HA] HA (aq) + H2O (l) H3O+ (aq) + A- (aq) Initial 0.080M - 0 0 Change (0.080 -0.00184) +0.00184 + 0.00184 Equilibrium 0.07816M 0.00184 M 0.00184M Ka = [H3O+][A-] /[HA] =( 0.00184 )2 / (0.07816) = 4.3 * 10-5 pKa = -logKa = - log(4.3 *10-5) =4.37 pKa = -logKa = - log(4.3 *10-5) =4.37 Solution The equilibrium dissociation of the weak acid can berepresented as: HA (aq) + H2O (l) H3O+(aq) + A- (aq) [HA]initial= 0.080 M Percent ionization = 2.3 % Percent ionization = [H3O+] /[HA]initial * 100 2.3 % = ([H3O+]/0.080 M) *100 [H3O+] = 0.00184 M Ka= [H3O+][A-] / [HA] HA (aq) + H2O (l) H3O+ (aq) + A- (aq) Initial 0.080M - 0 0 Change (0.080 -0.00184) +0.00184 + 0.00184 Equilibrium 0.07816M 0.00184 M 0.00184M Ka = [H3O+][A-] /[HA] =( 0.00184 )2 / (0.07816) = 4.3 * 10-5 pKa = -logKa = - log(4.3 *10-5) =4.37 pKa = -logKa = - log(4.3 *10-5) =4.37.