Question

Part A If Kb for NX3 is 3.5×10⁻⁶, what is the pOH of a 0.175 M aqueous solution of  NX3?

Part B If Kb for NX3 is 3.5×10⁻⁶, what is the percent ionization of a  0.325 M aqueous solution of  NX3?

Part C If Kb for NX3 is 3.5×10⁻⁶ , what is the the pKa for the following reaction?

HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq)

Express your answer numerically to two decimal places.

Answer 1

Part A)

NX3 --> N+ + X3-

X3- + H2O <==> HX3 + OH-

Kb = 3.5 x 10^-6 = [HX3][OH-]/[X3-]

3.5 x 10^-6 = x^2/0.175

x = [OH-] = 7.83 x 10^-4 M

pOH = -log[OH-] = 3.106

Part B)

[X3-] = 0.325 M

So,

3.5 x 10^-6 = x^2/0.325

x = [OH-] = 1.07 x 10^-3 M

percent ionization = [OH-] x 100/[X3-]

                             = 1.07 x 10^-3 x 100/0.325

                             = 0.328%

Part C)

Ka = Kw/Kb

      = 1 x 10^-14/3.5 x 10^-6

      = 2.86 x 10^-9

pKa = -log[Ka]

       = 8.54