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.

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


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:

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

  • 2
    $\begingroup$ Hi and welcome to Chemistry.Stackexchange. Nice that you already used MathJax. However, for long equations it is preferrable to use one line of MathJax in the way I have just formatted your post. Further helpful hints: \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, 2015 at 19:11
  • $\begingroup$ Just to make sure that I understand you correctly: You want to derive $\alpha = \left(\frac{1}{10^{\mathrm{p}K_\mathrm{a}-\ce{pH}}+1} \right)$ from the starting formula $\ce{pH}$ = $\mathrm{p}K_\mathrm {a}$ + $\log$ $\left(\frac{\alpha}{1-\alpha} \right)$, right? $\endgroup$
    – Philipp
    Sep 29, 2015 at 22:25
  • $\begingroup$ Yes, that is correct. Ideally from the point where I got stuck. But if there's a faster/better way, I am open to that also. $\endgroup$
    – user21398
    Sep 29, 2015 at 23:21
  • 4
    $\begingroup$ Instead of deleting the question, please feel free to add the derivation as an answer. You are encouraged to answer your own question if you found out how to do it. That way, the question can serve future Stack Exchange users, too. $\endgroup$
    – Jan
    Sep 30, 2015 at 13:11

1 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, $$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} } $$


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