# How to create a buffer solution from NaOH and CH₃COOH?

In my chemistry book, they say that an approach to making a buffer solution of $\mathrm{pH}=5.09$ is by adding an appropriate amount of strong base $(0.052\ \mathrm{mol}\ \ce{NaOH})$ to $0.300\ \mathrm{l}$ of $0.025\ \mathrm{mol/l}\ \ce{CH3COOH}$. But, I haven’t been able to figure out how they got those numbers. If I am not mistaken the relevant equation is,

$$\ce{CH3COOH +OH- <=>CH3COO^- + H_2O}$$

where the weak acid and conjugate base are respectively $\ce{CH3COOH}$ and $\ce{CH3COO^-}$. My concern is that this equation above goes nearly to completion $(K=1.8\times10^9)$ so I don’t see how this can even be a buffer solution. If not, how do they get the values? I would appreciate any help.

• So, I am missing one more equation $\ce{CH3COOH +H2O <=> H3O+ +CH3COO-}$, where the concentration of $\ce{CH3COOH}$ is defined by the excess. After this its like any other buffer, we use the $K_a$ of the equation above and $[\ce{H3O+}]=K_a \frac{[\ce{CH3OOH}]_{\text{excess}}}{[\ce{CH3COO-}]}$ – user29568 Jul 12 '14 at 18:00