Question

A weak acid, HA, has a pK, of 4.093. If a solution of this acid has a pH of 4.786, what percentage of the acid is not ionized? Assume all H+ in the solution came from the ionization of HA. percentage not ionized: | 60.7

Answer 1

HA & weak anid . So, it will displate only partially in water, HA (as) +20 (d) = H30®lags+ A@lags Filially: C I where, Ateatri cond) e ca initial conc. of ed Given, d=degree of Ionization of HA pH = 4.786 ✓ - hegn. [Uz ovaj = 4.786 + [H308 =10 Bu .4.786 =1.036816521x10-5 t c2 = 10636816521800-5 Nowo, ko = [H300J CART d.ca (HA) Ki-2 → Kq cond + ka = 1.63681654 x10 SZ Id Now, pka = 4.093 thegio Ke = 40093 - K= 10-4.093 = 8.0723503022105 So, 8072350302x105 - 1.636816521x10-52 + = 10636816591 41945 80072350302 Xigs

+ 5 -1 = 0.909768271 = 3 = 1.2.02.718 17 1202768271 FLE 0.831415346 So, degree of ionitation of HA , D E 01831415346 So, befcentage ionized a 2xloo ofo =0.231415346 X100% = 83.1415346 % → belcentage not ionized = (100m 23.14153461% = 16.85846531 ofe