Deriving the buffer formula

Question

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:

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

Answer 1

$$\ce{pH} = \mathrm{p}K_\mathrm {a} + \log \left(\frac{\alpha}{1-\alpha} \right)$$ 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, $$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} } $$