# Calculating volumes of acid/base required to make buffer using pH and pKa

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?

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