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

Question

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 $x << c_1, \, c_2$,

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

Answer 1

$\mathrm{p}K_\mathrm{a} = 4.75$, 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.