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

`AlCl_3` is a salt of a weak base `Al(OH)_3` and a strong acid `HCl`. Therefore , in normal water, it undergoes hydrolysis as :
`AlCl_3 (s) + 3H_2 O (l) to Al(OH)_3 (s) + 3H^+ (aq) + 3Cl^(-) (aq)`
In acidic water, `H^+` ions react with `Al(OH)_3` to form `Al^(3+)` ions and water . Thus, in acidic water `AlCl_3` exists as `Al^(3+)(aq)` and `Cl^(-)` ions.
`AlCl_3(s) underset("water")overset("acidified")to Al^(3+)(aq) + 3Cl^(-)(aq)`
In alkaline water, `Al(OH)_3` reacts to form tetra hydroxoaluminate complex or metaaluminate ion :
`Al(OH)_3(s) + OH^(-)(aq) to underset("Tetrahydroxo aluminate ion")([Al(OH)_4]^(-)` or `underset("Metaaluminate ion")(AlO_2^(-)+ 2H_2 O(l))`
The complete reaction may be written as :
`AlCl_3(s) underset("water")overset("alkaline")to {:([Al(OH)_4]^(-)(aq) + 3Cl^(-) (aq)),(darr OH^(-)),(AlO_2^(-) + 2H_2 O(l) + 3Cl^(-)(aq)):}`
Potassium chloride is a salt of a strong acid and a strong base. It does not undergo hydrolysis in normal water. It dissociates to `K^(+) (aq) and Cl^(-)(aq)` ions as :
`KCl(s) overset("water")to K^(+)(aq) `
Therefore, the aqueous solution of KCl is neutral and hence the ions donot react in acidified or alkaline water.