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) Potassium chloride $(KCl)$ is the salt of a strong acid $(HCl)$ and a strong base $(KOH)$. Hence,it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows:
$KCl_{(s)} \xrightarrow{H_2O} K_{(aq)}^{+} + Cl_{(aq)}^{-}$
In acidified and alkaline water,the ions do not react and remain as such.
Aluminium $(III)$ chloride $(AlCl_3)$ is the salt of a strong acid $(HCl)$ and a weak base $[Al(OH)_3]$. Hence,it undergoes hydrolysis in normal water:
$AlCl_{3(s)} + 3H_2O_{(l)} \to Al(OH)_{3(s)} + 3H^+_{(aq)} + 3Cl^-_{(aq)}$
In acidified water,$H^+$ ions suppress the hydrolysis of $Al^{3+}$,so $AlCl_3$ exists as $Al^{3+}_{(aq)}$ and $Cl^-_{(aq)}$ ions:
$AlCl_{3(s)} \xrightarrow{H^+_{(aq)}} Al^{3+}_{(aq)} + 3Cl^-_{(aq)}$
In alkaline water,the $Al(OH)_3$ formed reacts with $OH^-$ ions to form a soluble complex:
$Al(OH)_{3(s)} + OH^-_{(aq)} \to [Al(OH)_4]^-_{(aq)}$