± Weak Base Calculations

The reaction is NX3(aq) + H2O(l) <>HNX3^+ (aq) + OH^- (aq)

a) If Kb for NX3 is 5.5×10−6, what is the pOH of a 0.175M aqueous solution of NX3?
b) If Kb for is 5.5×10−6, what is the percent ionization of a 0.325M aqueous solution of NX3?
c) If Kb for NX3 is 5.5×10−6 , what is pKathe for the following reaction?
HNX3 ^+ (aq) + H2O(l) <> NX3(aq) + H3O^+ (aq)

a) Substitute:
5.5×10^−6 = x^2 / 0.175;
Solving, x = [OH-] = 0.001 M, so pOH = 3 (pH = 11).

b) By the same technique as a), [OH-] = 0.0013 M, so the percent ionization is 100% * 0.0013 M / 0.325 M = 0.41 %.

c) For a weak acid / weak base conjugate pair,
KaKb = Kw. Therefore,
Ka = (1x10^-14) / (5.5x10^-6) = 1.8x10^-9.
pKa= -log(1.8*10^-9) = 8.74