HA is a weak acid. Which equilibrium corresponds to the equilibrium constant Kb for A- ?

Question

A) A- (aq) H3O (aq) <==> HA (aq) H2O (l)

B) HA (aq) (aq) <==> H2O (l) H (aq)

C) A- (aq) H2O (l) <==> HA (aq) OH- (aq)

D)H A (aq) H2O (l) <==>H2A (aq) OH- (aq)

E) A- (aq) OH- (aq) HOA2- (aq)

Answer 1

The correct answer is C):

The expression of Ka for the weak acid is:



HA --&gt; H+ + A-     :     Ka = [H+][A-]/[HA]

Now, with C) we have:

A- + H2O --&gt; HA + OH-     :     Kb = [HA][OH-]/[A-]

Given this, if we multiply Kb by one ([H+]/[H+]), we can prove this is the correct answer as follows:

Kb = [HA][OH-]/[A-] * [H+]/[H+] = [HA]Kw/[H+][A-] = Kw/Ka

Rearrange to get:

Kw = Ka * Kb

Excellent!