# 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?

• Note that there is also hydrolysis of $\ce{Al(NO3)3}$ (when excess is added), which also significantly drops pH. – andselisk Dec 24 '17 at 7:16

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.