In: Chemistry
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.
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