Someone derive the Henderson-Hasselbalch equation for me?

Acid dissociation equation:

[HA] > [H+] + [A-]    

Ka = [H+][A-] / [HA]

take the negative log of both sides

-log[Ka] = -log{ [H+][A-] / [HA] } = -log[H+] + log { [A-] / [HA] }

pKa = pH + log{ [A-] / [HA] }

Calculating logs without a calculator is not as difficult as you might expect it to be. First, you should be familiar with the following log rules:

log (xy) = log x + log y
log (x/y) = log x - log y
log (x)^y = y(log x)

(keep this in mind for now)

You can use these rules to derive the Henderson-Hasselbalch equation from the Ka expression for an acid:

HA -> A- + H+

Ka = [A-][H+]/[HA]

Taking the -log of both sides gives us:

-log Ka = -log {[A-][H+]/[HA]}

The left side is the definition of pKa:

pKa = -log {[A-][H+]/[HA]}

The right side can be re-written using the log rules:

pKa = -log [A-] - log [H+] - (-log [HA])

The second right side term is the definition of pH:

pKa = -log [A-] + pH + log [HA]

Solving for pH:

pH = pKa + log [A-] - log [HA]

Using the log rules in reverse gives us the equation as it is normally written:

pH = pKa + log [A-]/[HA]