Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with (a) normal water, (b) acidified water and ( c) alkaline water? Write equations wherever necessary.

Explanation

KCl is the salt of a strong acid and a strong base. It does not undergo hydrolysis in normal water. It just dissociates to give `K^(+)(aq) and Cl^(-)(aq)` ion.
`KCl(s) overset("Water")to K^(+)(aq) + Cl^(-)(aq)`
Since the aqueous solution of KCl is neutral , therefore, in acidified water or in alkaline water,the ions do not react further and stay as such.
`AlCl_(3)` on the other hand, is a salt of a weak base `Al(OH)_(3)` and a strong acid HCl. Therefore, in normal water, it undregoes hydrolysis to form `Al(OH)_(3), H^(+) and Cl^(-)` ions.
In acidic water, the `H^(+)` ions react with `Al(OH)_(3)` to form `Al^(3+)(aq)` ions and `H_(2)O`. Thus, the acidic water, `AlCl_(3)` exists as `Al^(3+)` (aq) and `Cl^(-)` ions.
`AlCl_(3)(s) overset("Acidified water")to Al^(3+)(aq) + 3Cl^(-)(aq)`
In alkaline water, `Al(OH)_(3)` reacts to form soluble tetrahydroxoaluminate complex or meta-aluminate ion,
i.e., `Al(OH)_(3)(s) + OH^(-)(aq) to underset("Tetrahydroxoaluminate")([Al(OH)_(4)]^(-)(aq) ) ` or ` underset("meta-aluminate ion")(AlO_(2)^(-)(aq)) + 2H_(2)O(l)`
The complete equation may be written as :
`AlCl_(3)(s) overset("Alkaline water")to underset(AlO_(2)^(-)(aq) + 2H_(2)O(l) + 3Cl^(-)(aq))underset({:darr OH^(-):}" ")(Al[OH]_(4)^(-)(aq)+3Cl^(-)(aq))`