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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.

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Try considering reaction stoichiometry. You know the extent of the reaction. Now consider stoichiometry.

EDIT: If these are really the numbers your textbook gave you, then the textbook is wrong; you cannot create a buffer system with these numbers specifically. I suspect that you didn't transcribe the numbers from your textbook correctly. In any case the general idea, however, is correct - you can indeed form a buffer solution using sodium hydroxide and a solution of acetic acid.

All you have to do to realize this is to know what a buffer is - a buffer has significant concentrations of both an acid and its conjugate base (or base and its conjugate acid) - and consider reaction stoichiometry.

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  • $\begingroup$ If the reaction goes to completion then what would happen to the buffer when you add a base, since there is no weak acid left. Can you please be more concise? I copied the answers verbatim, if they are wrong can you suggest correct answers. $\endgroup$
    – user29568
    Commented Jul 12, 2014 at 17:43
  • $\begingroup$ "what would happen to the buffer when you add a base, since there is no weak acid left." Do you always run out of acid when you add base? What if you add less strong base than you have acid? That was my point. Consider reaction stoichiometry. $\endgroup$
    – Dissenter
    Commented Jul 12, 2014 at 17:45
  • $\begingroup$ So, the amount of OH-(~NaOH) should be half that of the weak acid? $\endgroup$
    – user29568
    Commented Jul 12, 2014 at 17:51
  • $\begingroup$ If you want a one to one acid to conjugate base concentration buffer, yes. Not all buffers must have such a ratio. $\endgroup$
    – Dissenter
    Commented Jul 12, 2014 at 17:51
  • $\begingroup$ 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 $[\ce{H3O+}]=K_a \frac{[\ce{CH3OOH}]_{\text{excess}}}{[\ce{CH3COO-}]}$ $\endgroup$
    – user29568
    Commented Jul 12, 2014 at 18:00

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