Derived the equation pH = pKa + Log [base] for a weak acid and its conjugate base, [acid]
You didn't copy all of the H-H equation but you meant
pH = pKa + log (b/a)
For an acid, HA, it is
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Solve for (H^+).
(H^+) = Ka*(HA)/(A^-) but that is
(H^+) = Ka*(acid)/(base) or
(H^+) = Ka*(a/b)
Multiply by -1
-(H^+) = -Ka*(a/b) then take log
-log(H^+) = -log Ka -log (a/b)
But -log(H^+) and -log Ka is pH and pKa respectively.
pH = pKa - log(a/b). USUALLY those - signs are replaced with + signs and we can do that by reversing the quotient to
pH = pKa + log (b/a). I'm not sure why you posted this, you can find it in almost any chemistry book, or on the web. Of course this way I do all the thinking and you don't do any thinking. You might like to know that when I was a senior in college I couldn't derive my way out of a wet paper bag but with years of experience I learned to do it. Good luck.