Question

Part A

If Kb for NX3 is 4.5×10⁻⁶, what is the pOH of a 0.175 M aqueous solution of NX3?

Express your answer numerically.

Part B

If Kb for NX3 is 4.5×10⁻⁶, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Express your answer numerically to three significant figures.

Part B If Kb for NX3 is 4.5×10−6, 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 4.5×10−6 , 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 + H2O ---------> HNX3⁺ + OH^-

I     0.175                         0             0

C       -x                            +x            +x

E       0.175-x                     +x           +x

   Kb   = [HNX3⁺][OH^-]/[NX3]

4.5*10^-6   = x*x/0.175-x

4.5*10^-6 *(0.175-x) = x^2

     x   = 0.000885

[OH-] = x= 0.000885M

POH =-log[OH-]

        = -log0.000885

        = 3.053

part-B

    NX3 + H2O ---------> HNX3⁺ + OH^-

I     0.325                         0             0

C       -x                            +x            +x

E       0.325-x                     +x           +x

   Kb   = [HNX3⁺][OH^-]/[NX3]

4.5*10^-6   = x*x/0.325-x

4.5*10^-6 *(0.325-x) = x^2

     x   = 0.0012

[OH-] = x= 0.0012M

percent ionisation   = Final [OH-] *100/conc of base

                                = 0.0012*100/0.325 =0.37%

part-C

Ka   = Kw/Kb

        = 1*10^-14/4.5*10^-6    = 2.23*10^-9