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?