Question

Part A

If Kb for NX3 is 4.0×10⁻⁶, what is the pOH of a 0.175 M aqueous solution of NX3?

Part B

If Kb for NX3 is 4.0×10⁻⁶, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Express your answer numerically to three significant figures.

Part C

If Kb for NX3 is 4.0×10⁻⁶ , 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

                   NX3 + H2O --------> HNX3^+ + OH-

       I         0.175                          0              0

       C       -x                                 +x             +x

       E      0.175-x                         +x              +x

         Kb    = [HNX3^+][OH-]/[NX3]

       4*10^-6 = x*x/0.175-x

4*10^-6*(0.175-x) = x^2

    x   = 0.000834

[OH-] = x = 0.000834M

POH = -log[OH-]

         = -log0.000834   = 3.08

                NX3 + H2O --------> HNX3^+ + OH-

       I         0.325                          0              0

       C       -x                                 +x             +x

       E      0.325-x                         +x              +x

         Kb    = [HNX3^+][OH-]/[NX3]

       4*10^-6 = x*x/0.325-x

4*10^-6*(0.325-x) = x^2

    x   = 0.001138

[OH-] = x = 0.001138M

Percentage of ionisation =   [OH-]*100/[Base]

                                          = 0.001138*100/0.325   = 0.35%

part-C

HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+

   Ka = Kw/Kb

         = 1*10^-14/4*10^-6

        = 2.5*10^-9

PKa = -logKa

        = -log2.5*10^-9

      = -log0.0000000025

      = 8.6020