Deriving the buffer formula

Essentially I want to derive the buffer formula: $\ce{pH}$ = $\mathrm{p}K\_\mathrm {a}$ + $\log$ $\left(\frac{\alpha}{1-\alpha} \right)$ to $\alpha$ = $\left(\frac{1}{10^{\mathrm{p}K\_\mathrm{a}-\ce{pH}}+1} \right)$
I'm going to include everything I had done trying to solve this here (and also show where I got stuck):


$$\ce{pH} = \mathrm{p}K\_\mathrm{a} + \log(\alpha) - \log(1-\alpha)$$


$$\ce{pH} - \mathrm{p}K\_\mathrm{a} + \log(1-\alpha) = \log(\alpha)$$


$$\alpha = 10^{\ce{pH}-\mathrm{p}K\_\mathrm{a}+\log(1-\alpha)}$$


$$\alpha = 10^{\ce{pH}} \times 10^{-\mathrm{p}K\_\mathrm{a}} \times 10^{\log(1-\alpha)}$$


$$\alpha = 10^{-\log\ce{[H+]}} \times 10^{\log(K\_\mathrm{a})} \times 10^{\log(1-\alpha)}$$


$$\alpha = \ce{[H+]}^{-1} \times K\_\mathrm{a} \times (1-\alpha)$$


$$K\_\mathrm{a} = \left(\frac{\alpha \times \ce{[H+]}}{(1-\alpha)}\right)$$


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.


Edit: Managed to complete. For those who are interested in how I had done it in the end:
$$\ce{pH}-\ce{pK\_\mathrm{a}} = \log\left(\frac{\alpha}{1-\alpha}\right)$$


$$10^{pH-pK\_\mathrm{a}} = \left(\frac{\alpha}{1-\alpha}\right)$$


$$10^{pH-pK\_\mathrm{a}} - (\alpha 10^{pH-pK\_a}) = \alpha$$


$$10^{pH-pK\_a} = \alpha + \alpha 10^{pH-pK\_\mathrm{a}}$$


$$10^{pH-pK\_a} = \alpha(1 + 10^{pH-pk\_a})$$


$$\left(\frac{10^{pH-pK\_a}}{1 + 10^{pH-pK\_a}}\right) = \alpha$$


divide numerator by itself = 1 and thus also divide denominator by the numerator.


$$\left(\frac{10^{pH-pK\_a}}{10^{pH-pK\_a}}\right)/\left(\frac{1+10^{pH-pK\_a}}{10^{pH-pK\_a}}\right) =$$ 


$$\left(\frac{1}{1+10^{pH-pK\_a}/10^{pH-pK\_a}}\right)$$


working out the denominator:


$$\left(\frac{1}{10^{pH-pK\_a}}\right) + \left(\frac{10^{pH-pK\_a}}{10^{pH-pK\_a}}\right) ->$$


$$\left(\frac{1}{10^{pH-pK\_a}}\right) +1$$


now work out the last fraction:


$$\left(\frac{1}{10^{pH-pK\_a}}\right) = \left(\frac{1}{10^{pH} \times 10^{-pK\_a}}\right) = (10^{pH})^{-1} \times (10^{-pK\_a})^{-1}$$
finally gives:


$$\alpha = \left(\frac{1}{10^{pK\_a-pH}+1}\right)$$




$$\ce{pH} = \mathrm{p}K\_\mathrm {a} + \log \left(\frac{\alpha}{1-\alpha} \right)$$
Multiply the equation by $-1$:


$$-\ce{pH} = -\mathrm{p}K\_\mathrm {a} + \log \left(\frac{1-\alpha} {\alpha}\right)$$
Rearrange the equation
$$-\ce{pH} +\mathrm{p}K\_\mathrm {a} = \log \left(\frac{1-\alpha} {\alpha}\right)$$


Apply the definition of $\log$


$$10^{-\ce{pH} +\mathrm{p}K\_\mathrm {a}} = \frac{1-\alpha} {\alpha}= \frac{1}{\alpha} -1$$
Rearrange the equation
$$1+ 10^{-\ce{pH} +\mathrm{p}K\_\mathrm {a}} = \frac{1}{\alpha} $$
So,
$$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K\_\mathrm {a}} } $$


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