# 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?

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

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 $$10^{-7}$$.