Question

B. If Kb for NX3 is 1.5×10−6, what is the percent ionization of a 0.325 M aqueous solution of NX3? percent ionization = to three sig figs

C. If Kb for NX3 is 1.5×10−6 , what is the the pKa for the following reaction? HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq) to 2 decimal places

Answer 1

C) We know that KaxKb=Kw

3 Iintial o325 1 0.325. 0 225

The KB for the reaction
NX3 + H2O <---------> OH- + HNX3
Kb = [OH-] [HNX3+] / [NX3] = 1.5 x 10^-6
pure liquids like H2O are not written in Kb or Ka expression
and you have to find Ka for this the reaction can be written as

HNX3+ + H2O <-----------> NX3 + H3O
Ka = [NX3] [H3O+] / [HNX3+] = ?
now multiply Ka and Kb
Ka X Kb = [NX3] [H3O+] / [HNX3+] X [OH-] [HNX3+] / [NX3] = [H3O+] [OH-]
We know that [H3O+] [OH-] = Kw = 10^-14
so Ka X Kb = Kw=1x10^-14
Ka = 10^-14/Kb = 10^-14 / (1.5 x 10^-6) = 6.6 x 10^-9

pKa=-log(Ka)=-log(6.6x10^-9)=9-log(6.6)

pKa=8.18.