Do you expect different products in solution when aluminium $(III)$ chloride and potassium chloride are treated separately with $(i)$ normal water,$(ii)$ acidified water,and $(iii)$ alkaline water? Write equations wherever necessary.

(N/A) $AlCl_{3}$ is a Lewis acid and undergoes hydrolysis in water.
$(i)$ In normal water: $AlCl_{3(s)} + 3H_{2}O_{(l)} \rightarrow Al(OH)_{3(s)} + 3H^{+}_{(aq)} + 3Cl^{-}_{(aq)}$
$(ii)$ In acidified water: $Al(OH)_{3}$ reacts with $H^{+}$ to form $Al^{3+}$ ions. The final solution contains $Al^{3+}_{(aq)}$ and $Cl^{-}_{(aq)}$ ions.
$(iii)$ In alkaline water: $AlCl_{3(s)} + 4OH^{-}_{(aq)} \rightarrow [Al(OH)_{4}]^{-}_{(aq)} + 3Cl^{-}_{(aq)}$
$KCl$ is a salt of a strong acid $(HCl)$ and a strong base $(KOH)$. It does not undergo hydrolysis.
$(i, ii, iii)$ In all three cases (normal,acidified,and alkaline water),$KCl$ simply dissociates into ions: $KCl_{(s)} \rightarrow K^{+}_{(aq)} + Cl^{-}_{(aq)}$.