I'm not sure whether I should post this on the chemistry stack exchange or the mathematics since it mainly consists of mathematical knowledge but also has some chemistry terms in it. Note: I am not that good at math but I gave it a try.
Essentially I want to derive the buffer formula: $pH$ = $pK_a$ + $log$ $\left(\frac{\alpha}{1-\alpha} \right)$ to $\alpha$ = $\left(\frac{1}{10^{pK_a-pH}+1} \right)$ I'm going to include everything I had done trying to solve this here (and also show where I got stuck):
$pH$ = $pK_a$ + $log(\alpha)$ $-$ $log(1-\alpha)$
$pH$ $-$ $pK_a$ + $log(1-\alpha)$ = $log(\alpha)$
$\alpha$ = $10^{pH-pK_a+log(1-\alpha)}$
$\alpha$ = $10^{pH}$ $\times$ $10^{-pK_a}$ $\times$ $10^{log(1-\alpha)}$
$\alpha$ = $10^{-log[H^+]}$ $\times$ $10^{log(K_a)}$ $\times$ $10^{log(1-\alpha)}$
$\alpha$ = $[H^+]^{-1}$ $\times$ $K_a$ $\times$ $(1-\alpha)$
$K_a$ = $\left(\frac{\alpha \times [H^+]}{(1-\alpha)}\right)$
So at this point I got stuck and have no clue how to get rid of an $\alpha$ to complete the derivation. Appreciate any help.
You can do it straightforward.
$pH = pK_a + log \left(\frac{\alpha}{1-\alpha} \right)$
$pH - pK_a = log \left(\frac{\alpha}{1-\alpha} \right)$
Multiplying by (-1)
$ pK_a-pH = -log \left(\frac{\alpha}{1-\alpha} \right)$
$ pK_a-pH = log \left(\frac{1-\alpha}{\alpha} \right)$
Take both sides of the equation as a exponent of 10.
$10^{pK_a-pH }=\frac{1-\alpha}{\alpha}$
Multiplying both sides by $\alpha$
$\alpha\cdot 10^{pK_a-pH }=1-\alpha$
$\alpha+\alpha\cdot 10^{pK_a-pH}=1$
$\alpha\left(1+ 10^{pK_a-pH }\right)=1$
$\alpha=\frac{1}{1+ 10^{pK_a-pH}}$