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8.6 At equilibrium, HA => H* +A~ [H*] = [A>] = (0.135) (0.040 M) = 5.40 x 1073 [HA] = (1 ~ 0.135) (0.040 M) = 3.46 x 10-? M The dissociation constant, K,, of the acid is therefore x, = EEDA) _ 640x107)" ig 8.8 The dissociation constant of a monoprotic acid at 298 K is 1,47 x 1073, Calculate the degree of dissociation by (a) assuming ideal behavior and (b) using a mean activity coefficient y, = 0.93. The concentration of the acid is 0,010 M. Let a be the degree of dissociation of the monoprotic acid. The corresponding concentrations of all species are HA = Ht + A Initial 0.010 0 o Mm At equilibrium 0.010 (1 —«) 0.0100 0.0100 M +] AW 2 ® k= 147 x 197 = FETA) _ _ 0.0100)" {HA} 0.010 (1 —a@) 1.0 x 10~4a? + 1.47 x 10a — 1.47 x 10-5 =0 a = 0.32 Therefore, assuming ideal behavior, the acid is 32% dissociated, (b) x, — duds. _ [Ht], [AT] “aa (HA] yaa Since HA is an uncharged species and the solution is dilute, Yaa iS approximately 1. Purthermore, v4 ¥_ = 2. The K, expression becomes k= 147 x 19-3 _ LEAT 2 _ @.010a)? 0.93? ed = UMA Se _ (0.010a)" (0.93)* (HA] 0.010 (1 — ee) 8.65 x 10-Sa? + 1.47 x 10-5a — 1.47 x 10-5 =0 o=0.34 Therefore, accounting for non-ideality, the acid is 34% dissociated. 
