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:
$$\alpha = \left(\frac{1}{10^{pK_a-pH}+1}\right)$$
\log
will give an upright logarithm, and chemical stuff such as $\ce{[H+]}$ can be formatted much easier with the\ce
command:$\ce{[H+]}$
. Further helpful MathJax stuff can be found in the sandbox (Takes a while to fully load.) $\endgroup$ – Jan Sep 29 '15 at 19:11