Question
Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent. Such reactions can be generically symbolized as NX3(aq) + H2O(I) HNX3+(aq) + OH-(aq) where NX3 is the base and HNX3+ is the conjugate acid. The equilibrium-constant expression for this reaction is Kb = [HNX3+][OH-] / [NX3] where Kb is the base ionization constant. The extent of ionization, and thus the strength of the base, increases as the values of Kb increases. If Kb for NX3 is 4.0 x 10-6, what is the pOH of a 0.175 M aqueous solution of NX3? If Kb for NX3 is 4.0 x 10-6, what is the percent ionization of a 0.325 M aqueous solution of NX3? If Kb for NX3 is 4.0 x 10-6, what is the the pKa for the following reaction? HNX3+(aq) + H2O(I) NX3(aq) + H3O+(aq)
Explanation / Answer
x2 / 0.175 = 4 * 10-6
x = 8.4 * 10-4 = [OH-] ----- pOH = 3.077
x2 / 0.325 = 4 * 10-6
x= 0.00114
% ionization = 0.00114 / 0.325 * 100% = 0.35% ionization
Ka of the conjugate acid = Kw/Kb = 10-14 / (4*10-6) = 2.5 *10-8