Question

7. (3 points) A 0.613 M HX acidic solution has a pH of 2.07. a. Find the percent ionization of the acid in this solution. b. Calculate the Ka value of HX. 8. (3 points) Calculate the pH of a 0.76 M solution of CH3NH2, Kb = 4.4 x 10".

Answer 1

PH=2.07 (H30t) = 10 P14 = 10- 2:07 CHYJ = 0.613 M = 0 ka =? x= degree of icnisation = 8.5114x153M MX cool.3 + H2O() = H30tcom+ xiag.) ; ka initical cx ca at eam c-cx [H3O+] = CX = 8:5114x10 3 a) percent ionization - X x 100% - CA X = 8:5114x153 0:613 V = 0.0138847 8. Siluxio 3 0:613 - XICO -x100 -) • iconization = 1.39% my (6) "Ka= - Co2 (1-x) a [H3O+] Tx-] = ka= (4x) (CW) - ka = (Hx] ka = 0.613 (0.01388)? I - 0.0 1388 ka = 1.20x104m) one

# [сHдон,)= С. 76H : к, = 4-ух бЧ Pks 5 - log (4.4 x 6") јD Рk= 3- 356S PH = 14+ [{ C Pky -Lloys)] = -({{3 3ss - Дуо чо] | P» = 12-26 ЈАН 1