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:
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?