Question

Part A: If Kb for NX3 is 9.5×10−6, what is the pOH of a 0.175 M aqueous solution of NX3?

Part B: If Kb for NX3 is 9.5×10⁻⁶, what is the percent ionization of a  0.325 M aqueous solution of  NX3?

Part C: If Kb for NX3 is 9.5×10⁻⁶ , what is the the pKa for the following reaction?

HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq)

Answer 1

A)

NX3 + H2O <-> NH3H+ + OH-

Kb = [NH3H+][OH-]/[NX3]

in equilibrium

[NH3H+]= x

[OH-] = x

[NX3] = 0.174-x

9.5*10^-6 = x*x/(0.175-x)

x = [OH-] = 0.0012846

pOH = -log(OH) = -log(0.0012846) = 2.8912

B)

repeat for 0.325 M of NX3

Kb = [NH3H+][OH-]/[NX3]

in equilibrium

[NH3H+]= x

[OH-] = x

[NX3] = 0.175-x

9.5*10^-6 = x*x/(0.325-x)

x = [OH-] = 0.0017523

for % ionization

% ion = [NXH+]/[NX3] * 100% = 0.0017523/0.325*100 = 0.539 %

C)

for pKa:

pKa + pKb = 14

pKa = 14 - pKb

pKb = -log(Kb) = -log(9.5*10^-6) = 5.022

so

pKa = 14-5.022

pKa = 8.978