To create a buffer mixture, I understand that mixing an equal concentration of a weak acid and it's salt (conjugate base) will form an acidic buffer.

However, why is it that when $\pu{25 cm^3, 9.100 M}$ $\ce{CH3COOH}$ and $\pu{50 cm^3, 0.100 M}$ $\ce{CH3COONa}$ can also form a buffer solution?

Is there any specific rule for the creation of buffer solution from weak acid and it's salt?

  • $\begingroup$ You need excess weak acid. Not an equal amount of both of them. $\endgroup$
    – user60221
    Commented Apr 9, 2018 at 13:08
  • $\begingroup$ Welcome to ChemistrySE. Please feel free to take a tour of our site to know how our site works and what's on-topic here. Hope you get a satisfactory answer! $\endgroup$ Commented Apr 9, 2018 at 14:10

2 Answers 2


There is no such rule that you need equal concentrations of weak acid and salt of conjugate base to make a buffer. (In this case acidic buffer)

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a $\mathrm{pH}$ of $\mathrm{4.76}$ because $\mathrm{pK_a}$ of acetic acid is $\mathrm{4.76}$. It doesn’t matter what the concentrations were, as long as they were the same, the $\mathrm{pH}$ remains at $\mathrm{4.76}$.

You can change the pH of the buffer solution by changing the ratio of salt to acid, or by taking a different acid and one of its salts.

Note: The $\mathrm{pH}$ of a buffer is calculated using Henderson-Haselbalch equation, which is, $$\mathrm{pH = pK_a + log\frac{[Conjugate Base]}{[Acid]} }$$

Also, as indicated by @VicNeal in the comments, the buffer solution has highest buffer capacity at equal concentrations of the constituents, meaning its more effective at resisting the change in $\mathrm{pH}$.


The relationship between $\mathrm{pK_a}$ and $\mathrm{pH}$ of a weak acid, $\ce {HA}$, is mathematically represented by Henderson-Hasselbach equation where $\ce {A-}$ is its conjugate base (there is a equation for a weak base as well). $$\mathrm{pH = pK_a + log\frac{[\ce {A-}]}{[\ce {HA}]}}$$ Thus, the $\mathrm{pH}$ of a buffer of weak acid can be calculated using above equation, which gives close approximate value. As @MollyCooL pointed out, the highest buffer capacity is in effect when $\mathrm{pH}$ of the buffer is equal to $\mathrm{pK_a}$ of the acid, meaning, $\mathrm{log\frac{[\ce {A-}]}{[\ce {HA}]}= pH - pK_a =0}$. Hence, $\mathrm{\frac{[\ce {A-}]}{[\ce {HA}]}= 1}$ or $\mathrm{[\ce {A-}]=[\ce {HA}]}$. At this $\mathrm{pH}$, the effective buffer capacity is $\mathrm{pK_a \pm 1}$. Because of this reason, it is common practice with biochemists that they made bufer solutions at or about $\mathrm{pK_a}$ of the relavent acid or base. For example, if they need buffer of $\mathrm{pH}\space 4.5$, they use acetic acid/sodium acetate (acetic acid: $\mathrm{pK_a} = 4.76$), but use $\ce {KH2PO4/K2HPO4}$ system to make $\mathrm{pH}\space 7.0$ buffer (phosphoric acid: $\mathrm{pK_{a2}} = 7.2$). To understand this concept, see Plot 1 below:

enter image description here This plot illustrates various protonation states of acetic acid as the pH changes. When $\mathrm{pH}$ is below $\mathrm{pK_a}$ of the acid ($\mathrm{pH \lt 3.5}$), most of the acid has been protonated. When $\mathrm{pH}$ is significantly above $\mathrm{pK_a}$ ($\mathrm{pH \gt 6.0}$), most of the acid is deprotonated. If the $\mathrm{pH}$ close of equal to the $\mathrm{pK_a}$, the acid is about $\mathrm 50\%$ protonated and $\mathrm 50\%$ deprotonated. That means, acetic acid is not suitable for the situations where sought $\mathrm{pH}$ is below $\mathrm 3.5$ or above $\mathrm 6.0$.

Now, let's do little quick and dirty calculations for your buffer solution:

$$\pu {Final acetic acid concentration = 9.100 \space M\times\frac{0.025 \space L}{0.075 \space L} = 3.033 \space M}$$ $$\pu {Final acetate concentration = 0.100 \space M\times\frac{0.050 \space L}{0.075 \space L} = 0.067 \space M}$$ Now, apply these two values to Henderson-Hasselbach equation: $$\mathrm{pH = 4.76 + log\frac{[\ce {0.067}]}{[\ce {3.033}]}= 4.76-1.66=3.1}$$ Thus, the buffer system selected is not good choice for $\mathrm {pH} \space 3.1$ buffer. ascorbic acid (ascorbic acid: $\mathrm{pK_{a1}} = 4.10$) or phosphoric acid (phosphoric acid: $\mathrm{pK_{a1}} = 2.15$) would have been a better choice. Nonetheless, citric acid (citric acid: $\mathrm{pK_{a1}} = 3.15$) is the ideal one for this $\mathrm {pH}$.

The plot is from: https://bio.libretexts.org/LibreTexts/University_of_California_Davis/BIS_2A%3A_Introductory_Biology_(Easlon)/Readings/06.1%3A_pKa

  • $\begingroup$ Your image wasn't working. Is this the plot you were referring to . Images are recommend to be uploaded to stack.imgur since many times the image is deleted on the source site which renders your answer/question obsolete. $\endgroup$ Commented Apr 12, 2018 at 2:29

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