Why isn't auto ionisation of water considered when we find the pH during salt hydrolysis? $\ce{H+}$ concentration from the auto-ionisation of water is $10^{-7}$
If we have $\ce{HA}$ as weak acid and $\ce{BOH}$ as weak base having $\mathrm pK_\mathrm{a} = 3.2$ and $\mathrm pK_\mathrm{b} =3.4$ respectively, we are get a salt $\ce{AB}$.
When we find $\ce{H+}$ concentration for this salt using the formula,
$$\ce{H+} = \sqrt{\frac{K_\mathrm wK_\mathrm a}{K_\mathrm b}}$$
We get $\ce{H+}$ concentration in the solution to be $10^{-8}$.
In such a scenario, why don't we add $\ce{H+}$ concentration from the auto-ionisation of water to the $\ce{H+}$ concentration?
What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as
$$
\ce{A- + H+ &lt;=&gt; HA,H2O &lt;=&gt; H+ + OH-}\stackrel{\text{add}}{\Longrightarrow}\ce{A- + H2O &lt;=&gt; HA + OH-}.
$$
In a word, ionization has been taken into consideration in your calculation. There are multiple equilibria in the system, so you cannot simply add $10^{-7}$.