How to rationalise the drop in pH when excess aluminium nitrate is added into an aqueous ammonium solution? Consider the flowing reaction 
  $$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$
  Explain &ldquo;when excess powdered aluminium nitrate is added to the reaction mixture and the $\mathrm{pH}$ value of the solution is lowered.&rdquo;


There are 3 questions that confuse me: 


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?
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? 
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"
$$\ce{[Al(H2O)6]^{3+} + H2O&lt;=&gt; [Al(H2O)5(OH^-)]^{2+} +H3O+}$$

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