How to rationalise the drop in pH when excess aluminium nitrate is added into an aqueous ammonium solution?


> 
> Consider the flowing reaction 
>  $$\ce{NH3 + H2O <=> NH4+ + OH-}$$
>  Explain “when excess powdered aluminium nitrate is added to the reaction mixture and the $\mathrm{pH}$ value of the solution is lowered.”
> 
> 
> 


There are 3 questions that confuse me: 


1. When $\ce{OH-}$ is removed by aluminium(III) ions, $\ce{NH4+}$ is produced, which is acidic. In that case, is the decrease in $\mathrm{pH}$ due to the formation of $\ce{NH4+}$ ions or because of the removal of $\ce{OH-}$ ions?
2. When $\ce{OH-}$ is removed, according to $K\_\mathrm{c} = [\ce{H+}][\ce{OH-}]$, the $\ce{H+}$ ions will increase since $K\_\mathrm{c}$ is a constant value. Hence, the pH decreases. Is is a correct explanation to the drop in $\mathrm{pH}$ when $\ce{OH-}$ ions are removed?
3. Does addition of excess aluminium nitrate mean that the equilibrium reaction will go to complete and there is no any ammonia at the end?




As *andselisk* has already pointed out in a comment, the hexaquo complex of aluminium is a known "cation acid"
$$\ce{[Al(H2O)6]^{3+} + H2O<=> [Al(H2O)5(OH^-)]^{2+} +H3O+}$$


A practical application is the use of solid $\ce{KAl(SO4)2·12 H2O}$ (*alaun*) for blood clotting during wet shave.


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