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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.

user275257
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1 Answers1

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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 $\alpha$

$\alpha\cdot 10^{pK_a-pH }=1-\alpha$

$\alpha+\alpha\cdot 10^{pK_a-pH}=1$

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

$\alpha=\frac{1}{1+ 10^{pK_a-pH}}$

callculus42
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  • I think this might be easier to follow if you leave out the mysterious “multiply by $-1$” step. Solving for $a$ then gives $\frac{10^{pH-pK_a}}{1+10^{pH-pK_a}}$, then you divide the numerator and denominator by the numerator to yield the desired form. – amd Sep 29 '15 at 18:37
  • This was my first idea. But after the divison it is not getting easier, in my opinion. And many ways lead to rome. – callculus42 Sep 29 '15 at 18:57
  • Certainly, there are many ways to derive the answer. However, if you didn’t know the final result you were driving at ahead of time, what would motivate you to multiply by $-1$ in the middle of your calculation? To me, it seems more natural to solve for $a$ first and then see if the resulting expression can be simplified. – amd Sep 29 '15 at 19:06
  • But the final result has been posted. That´s why I calculated it straighforward. – callculus42 Sep 29 '15 at 19:20