Question

Part A: If Kb for NX3 is 9.0×10−6, what is the pOH of a 0.175 M aqueous solution of NX3? Express your answer numerically.

Part B: If Kb for NX3 is 9.0×10⁻⁶, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Part C: If Kb for NX3 is 9.0×10⁻⁶ , what is the the pKa for the following reaction?

HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq)

Answer 1

A) for weak bases

pOH = 1/2[pKb - logC]

pKb = -log Kb = - log [9.0 x 10^-6] = 5.04

pOH = 1/2 [5.04 - log0.175]

pOH = 1/2 [5.797]

pOH = 2.898

pH = 14 - 2.898

pH = 11.1

B) for weak bases

kb = C² / 1-

9.0 x10^-6   = 0.325² /1-

9.0 x 10^-6  - 9.0 x 10^-6  - 0.325²  = 0

  = 0.0052

% ionization = 100 x   = 100 x 0.0052 = 0.52

C) Ka x Kb = Kw

Ka x 9.0 x 10^-6  = 1.0 x 10^-14

Ka = 1.0 x10^-14 / 9.0 x 10^-6

Ka = 1.0 x 10^-9

pKa = -logKa

pKa = - log [1.0 x 10^-9]

pKa = 9.0