Question

help! will rate!

A 0.25 M NaA solution has a pH of 9.25 at 25°C. Calculate the pK value of HA.

Answer 1

ANSWEX! bien that, [Nan] = [] = 0.25M pH=9.95 Hode, we are given the conjugate base of weak acid Hm. A + H₂O ---) OH + HA 0 0 | I 0 E 0.25 -x 0.25-X +x + x tx +X pol +PH - 14 pol -14-PH : 514-9,25 = 4.75 So at equilibrium, we can calculate Tort) Room ke hion Рен. Оe Know, poh --109 ton-] 4075=-109 [otr]. 109 [OH-] = -4.75 Aribing from both side [OH-] = 10 4075

= 1784105m So at ewlibrium, [ok-3= Y = 1.78X10-50% At equilibrium, .:. [^] = 0.25-1.78X16-57 = 0,25 [HA] = X =17881-5 go, bage- ionization constant, Kb = [HA] [CH] / [A-] Kg = 1078x10-5 MX): 78 x 165m 10:25 Kb = 1:26x109 Now, we can calculate Ka forom the Ho. KW = kaxkb Kw = 1.07.1664/1.288469 Kwa 70 94X106 we know, Pka=-log ka Pka =-1097,918106 PRO = 5.10 the, pha value of HA is 15.107 - X