A colleague found an interesting article on the development of the H-H Equation. Inside this linked pdf, you will find a link to the article as well as same brief comments on the H-H Equation.
The Henderson-Hasselbalch Equation is: (1) a rearrangement of the K_{a} expression followed by (2) the use of negative logarithms.
It is:
[A¯] pH = pK_{a} + log ––––– [HA]
The alternate form starts from the ionization equation for a generic base called B:
B + H_{2}O ⇌ HB^{+} + OH¯
By the way, here is an example of the above generic equation, using ammonia:
NH_{3} + H_{2}O ⇌ NH_{4}^{+} + OH¯
The B simply represents the entire base and HB^{+} represents the substances with an additional H^{+}.
Next, we write the K_{b} expression for this reaction:
[HB^{+}] [OH¯] K_{b} = –––––––––– [B]
Next, we isolate the [OH¯] on the left-hand side of the equation:
[B] [OH¯] = K_{b} · ––––– [HB^{+}]
We negative log each of the three terms of the above equation to give pOH, pKb and we flip the third term so as to make it an addition, not a subtraction. Why make it an addition? Not sure, probably because that was the way Henderson & Hasselbalch did it. I really don't know.
Here is the alternate form of the Henderson-Hasselbalch Equation, expressed in terms of pOH and pK_{b}:
[HB^{+}] pOH = pK_{b} + log ––––– [B]
Using the words base and acid gives us this:
[acid] pOH = pK_{b} + log ––––– [base]