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$\ce{H+}$ 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?

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  • $\begingroup$ Nobody likes complex formulas. If all factors are considered, it can easily go out of hand. Formulation of few complications is quite easy. $\endgroup$
    – Poutnik
    Sep 1, 2020 at 12:58

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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

$$ \ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\Longrightarrow}\ce{A- + H2O <=> HA + OH-}. $$

In a word, ionization has been taken into consideration in your calculation. There are multiple equilibria in the system, so you cannot simply add $10^{-7}$.

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    $\begingroup$ A more complex formula has to be considered, if the salt total/initial concentration is comparable with [H+] or [OH-]. In such cases, pH is not a simple function of just the respective pKx. $\endgroup$
    – Poutnik
    Sep 1, 2020 at 11:44

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