Question

Consider a weak base like, NH3 in water:
(i) Write the base ionization constant, Kb expression.
(ii) Write the equation to calculate the pKb of this base given above.
(iii) Using the above base dissociation process, show that Kb x Ka = Kw for this
base and its conjugate acid.

Answer 1

(i) The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows:

HA(aq)+H2O(l)⇌H3O+(aq)+A-(aq)

As we noted earlier, the concentration of water is essentially constant for all reactions in aqueous solution, so [H2O] in Equation can be incorporated into a new quantity, the acid ionization constant ( Ka ), also called the acid dissociation constant:

Answer 2

Answer:

NH3 + H2O => NH4^+ + OH^-

i) Kb = ([NH4^+][OH^-])/[NH3]

ii) pKb = -logKb

iii) pKa + pKb = -logKw

you know that : pKa = -log Ka  and pKb = -logKb

therefore = logKa + logKb = logKw

               =log(Ka + Kb) = logKw

divide both sides by log and you get:

Ka + Kb = Kw

     

Explanation: