How would we use our $\mathrm{p}K_{\mathrm{a}}$ of the acid component of the buffer to calculate $\large \frac{[\ce{A-}]}{[\ce{HA}]}$?

I know $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$$ rearranging gives me $$\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}$$ if for example the $\mathrm{pH}=4.5$, $\mathrm{p}K_{\mathrm{a}}=3.74$,

Is it correct to say $$\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{19}{250}$$ since $$ \begin{align} \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}F\\ \implies\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=4.5- 3.74\\ \implies\frac{[\ce{A-}]}{[\ce{HA}]}&=\frac{\operatorname{e}^{0.76}}{10} \end{align} $$ if $\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{V_\text{b}}{V_\text{a}}= \dfrac{19}{250}$ how would I use this to find the volumes of acid and base needed to make my buffer?


1 Answer 1


First of all, note that:

$$\log_{10}\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)=4.5- 3.74$$ $$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)\neq\dfrac{e^{0.76}}{10}$$


$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)= 10^{0.76}$$

This said, the volume will depend on the concentration of the acid and of the salt (if it's a solution) that you are provided with. This ratio will give you the relative concentrations required to get a buffer of the pH 4.5.


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