# Why is it advised to use neutral ferric chloride solution while performing confirmatory test for acetate ions? [duplicate]

The reaction of $$\ce{FeCl3}$$ with $$\ce{CH3COO-}$$ is

$$\ce{3Fe^{+3} +6CH3COO- +H2O<=> [Fe3(OH)2(CH3COO)6]^{+} +2H^{+}}$$

Recently I read that a solution of $$\ce{FeCl3}$$ hydrolyses very quickly to form $$\ce{Fe(OH)3 +3H^{+}}$$ as described by the following equilibrium:

$$\ce{FeCl3(aq) +3H2O <=> Fe(OH)3(s) +3H^{+}} \tag{1}\label{eqn:1}$$

with hydrolysis constant $$K_\mathrm{h}$$. I found a $$K_\mathrm{sp}$$ value of $$\ce{Fe(OH)3}$$ of $$\pu{4E-38}$$, so the equilibrium

$$\ce{Fe(OH)3 <=>Fe^{+3} +3OH-}$$

has $$K_\mathrm{sp} =\pu{4E-38}$$ and after a bit of rearrangement I got $$K_\mathrm{h}$$ of $$(\ref{eqn:1})$$ as $$\frac{K_\mathrm w^{3}}{K\mathrm{sp}}$$.

Hence at $$\pu{25 ^\circ C}$$, since the ionic product of water is $$10^{-14}$$, calculation gives $$K_\mathrm{h}= \pu{2.5E-5}$$. If we maintain neutral pH (ie. $$10^{-7}$$) we then get $$\alpha$$ (dissociation constant) for $$(\ref{eqn:1})$$ as almost 1.0.

How is this possible? If this is true then how can $$\ce{Fe^3+}$$ ions be in solution for complex formation even if $$\ce{CH3COO-}$$ ion is not hydrolysed to acetic acid because of such low concentration of $$\ce{H^+}$$ in solution?