Question

Calculate the equilibrium constant for the following reaction of a weak acid and base HClO(aq) + NH3(aq) NH4+(aq) + ClO-(aq) given Ka (HClO) = 2.8 × 10-8, Kb (NH3) = 1.8 × 10-5 and Kw = 1.0 × 10-14.

Answer 1

The given reaction is

HClO(aq) + NH3(aq) = NH4+(aq) + ClO-(aq)

We broke up the reactions in subreactations

Dissociation of weak acid

HClO = H+ + ClO-

Weak Acid (HClO) dissociation constant

Ka = [H+][ClO-]/[HClO]

Dissociation of base

NH3 + H2O = NH4+ + OH-

Base dissociation constant

Kb = [NH4+][OH-] / [NH3]

Dissociation of water

H2O = H+ + OH-

Kw = [H+][OH-]

Overall reaction

HClO + NH3 = NH4+ + ClO-

Equilibrium constant expression of the reaction

Kc = [NH4+][ClO-] / [NH3][HClO]

Kc = ([NH4+]/[NH3]) x ([ClO-]/[HClO])

From Kb, Ka and Kw expressions

Kc = (Kb/[OH-]) x (Ka/[H+])

= ( Ka x Kb) / ([H+][OH-])

= (Ka x Kb) / Kw

=(2.8 x 10^-8) x (1.8 x 10^-5) / (1 x 10^-14)

= 50.40