Question

Write equilibrium reactions for the following weak bases in water and then write the Kb expression in terms of concentrations of the weak base, its conjugate acid, and the conjugate base of water, OH-. Remember to ignore the alkali metal ion if present as they are only spectator ions in aqueous solutions and have nothing to do with acid-base chemistry here.

NH3 equilibrium Kb reaction____________________________________ Kb:

KClO equilibrium Kb reaction____________________________________ Kb:

LiC2H3O2 equilibrium Kb reaction____________________________________ Kb:

CsNO2 equilibrium Kb reaction____________________________________ Kb:

NaCHO2 equilibrium Kb reaction____________________________________ Kb:

KC6H5O equilibrium Kb reaction Kb:

Answer 1

NH3(aq) + H2O(l) <=======> NH4+(aq) + OH-(aq) ; Kb = [NH4(aq)][OH-(aq)] / [NH3(aq)]

ClO-(aq) + H2O(l) <=======> HClO(aq) + OH-(aq) ; Kb = [HClO(aq)][OH-(aq)] / [ClO-(aq)]

C2H3O2-(aq) + H2O(l) <======> C2H4O2(aq) + OH-(aq) ; Kb = [C2H4O2(aq)][OH-(aq)] / [ C2H3O2-(aq)]

NO2-(aq) + H2O(l) <=======> HNO2(aq) + OH-(aq) ; Kb = [HNO2(aq)][OH-(aq)] / [NO2-(aq)]

CHO2-(aq) + H2O(l) <=======> CH2O2(aq) + OH-(aq) ; Kb = [CH2O2(aq)][OH-(aq)] / [CHO2-(aq)]

C6H5O-(aq) + H2O(l) <========> C6H5OH(aq) + OH-(aq) ; Kb = [C6H5OH(aq)][OH-(aq)] / [C6H5O-(aq)]