[slayer]

Kb process of HCO3-
Equation #1
HCO3- (aq) + H2O (l) ------> OH- (aq) + H2CO3 (aq) Kb= ??

Ka process of HCO3-
Equation #2
HCO3- (aq) + H2O (l) ------> H3O+ (aq) +CO3 2- (aq) Ka= 4.7E-11

If the Kb value of HCO3- is not given, you can use the Ka value of it's conjugate acid, H2CO3 (4.5E-7) to determine Kb of HCO3- in [Kw = Kb*Ka] === [Kb = Kw/Ka] where Kw= 1.0E-14

Conceptually, why are we able to use the Ka process of it's conjugate acid again? Why can't we just use the Ka value from equation #2 to determine Kb of HCO3-?

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=polyprotic-dissociation-constants

Write all reactions (acid/base dissociations) for the processes and assign numbers, you'll see why.