# Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?

Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base. e.g. $$\ce{CH3COOH + H2O <=>[K_\rm{a}] CH3COO- + H3O+}\tag1$$ $$\ce{CH3COO- + H2O <=>[K_\rm{h}] CH3COOH + OH-}\tag2$$

Now, $\displaystyle{K_\rm{a} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$ and $\displaystyle{K_\mathrm{h} = \frac{K_\rm{w}}{K_\mathrm{a}}}$.

Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.

It is also mentioned in a book that

Assume that the extent of protonation of acetate ions and the deprotonation of acetic acid molecules is so small that the concentrations of both species are nearly the same as their initial values.

Why is this so? If $K_\rm{a}$ is low, then $K_\rm{h}$ should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the $\ce{CH3COO-}$ coming through the acid dissociation but how can we neglect the hydrolysis of the $\ce{CH3COO-}$ coming from the salt?

Edit $-$

My calculation for equilibrium concentrations.

$\ce{[CH3COOH]_\mathrm{eq}} =$ initially added acid $-$ deprotonated acid $+$ hydrolysed salt

$\ce{[CH3COO-]_\mathrm{eq}} =$ initially added salt $-$ hydrolysed salt $+$ deprotonated acid

$\ce{[H3O+]}_{\mathrm{eq}} =$ deprotonated acid $-$ hydrolysed salt

Let the concentration of

initially added acid $= c_1$

initially added salt $= c_2$

deprotonated acid at eq. $= x$

hydrolysed salt at eq. $= y$

Thus, $$K_{\mathrm{a}} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(x - y)(c_2 - y + x)}{(c_1 - x + y)}$$

Now, assuming $x << c_1, \, c_2$,

$$K_{\mathrm{a}} = \frac{(- y)(c_2 - y)}{(c_1 + y)}$$

• You may take it into account and compare the results. – Ivan Neretin Sep 21 '17 at 5:10
• @IvanNeretin Please see my edited question. – Apoorv Potnis Sep 21 '17 at 5:54
• You said before that you neglect $y$, not $x$. – Ivan Neretin Sep 21 '17 at 6:15
• @IvanNeretin I said that I understand why deprotonation is neglected, i.e. why $x$ is neglected. I don't understand why the hydrolysis, i.e. $y$ is neglected. – Apoorv Potnis Sep 21 '17 at 6:21
• Hydrolysis = cleavage of chem. bonds by addition of water (en.wikipedia.org/wiki/Hydrolysis). Where is there a hydrolysis here? How can acetic acid be hydrolyzed? – logical x 2 Sep 21 '17 at 10:40

$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,

$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$

So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:

$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$

where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.

Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M} \cdot [\ce{H+}]}{\pu{1.0 M}}$$

Therefore, we have:

$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$

If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of

$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$

and after addition of acetate, we have

$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$

Meaning that the solution got more basic after the addition of acetate, as excpected.

• Hopefully that provides a good starting point for understanding. – logical x 2 Sep 21 '17 at 12:35
• @deuxexmachina After you added 1.0 M acetate ions, how did you take equilibrium concentration to be 1.0 M? How is initial concentration equal to equilibrium concentration? – Apoorv Potnis Sep 24 '17 at 4:51
• @ApoorvPotnis Yeah, I figured that this point wasn't really clear yet. So first of all, you can neglect that the acid CH3COOH will dissociate to CH3COO-, as by the above argument, it only dissociates to a very little extent. However, your concern is the other way around, and you argue that CH3COO-Na+ could be protonated to form CH3COOH + NaOH. But: look at the pKa of NaOH and CH3OO-Na: NaOH is a much stronger base! Therefore, you can assume that $\mathrm{[NaCH3COO] = [NaCH3COO]_0}$. – logical x 2 Sep 24 '17 at 7:38
• @ApoorvPotnis You can also find a similar description here: study.com/academy/lesson/… for a similar explanation – logical x 2 Sep 24 '17 at 7:40
• Can you add your comment about $\ce{NaOH}$ being a stronger base into the answer so that I could accept it? – Apoorv Potnis Nov 6 '17 at 12:50