Cracking the SAT Chemistry Subject Test

For NH 3 (aq):  KbNH 3  =   [NH 4 +][OH−]/[NH 3 ]

Rearranging each so that [NH 4 +] is by itself on the left yields

For NH 4 +(aq): [NH 4 +]    =   [NH 3 ][H+]/pKaNH 4 +

For NH 3 (aq):  [NH 4 +]    =   KbNH 3  [NH 3 ]/[OH−]

Now these can be set equal to one another as follows:

[NH 3 ][H+]/pKaNH 4 +   =   KbNH 3  [NH 3 ]/[OH−]

The [NH 3 ]’s cancel out, and then grouping the K’s and concentration,

respectively, leaves

(pKaNH 4 +)(KbNH 3 )    =   [OH−][H+]

Since we’ve already seen that [OH−] × [H+] = Kw = 10−14 M 2 at 25°C for any

aqueous solution, then

(pKaNH 4 +)(KbNH 3 )    =   [OH−][H+]   =   Kw  =   10−14 M     2 , or

pKaNH 4 +   +   pKbNH 3     =   14

In other words:

The sum of  the pKa and pKb of  a   conjugate   pair    of  a   weak    acid    and

weak    base    must    always  be  equal   to  14  at  25°C.

If the acid-base properties of one member of a conjugate pair are already known,
then the acid-base properties of the other can be inferred using the conjugate
rules. There are four conjugate rules that cover all of the possible combinations
of strong/weak acids/bases.

1. 
   

        The conjugate   acid    of  a   strong  base    is  neutral.