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

Explanation

Aluminium chloride, `AlCl_(3)`, is a salt of weak base, `Al(OH)_(3)` and strong acid, `HCl`.
a. In normal water: `AlCl_(3)` undergoes hydrolysis to form `Al(OH)_(3),H^(o+)` and `Cl^(ө)` ions.
`Al(OH)_(3_((s)))+3H_(2)O_(l)to Al(OH)_(3_((s)))+3H_(aq)^(o+)+3Cl_(aq)^(ө)`
In acidified water: `H^(o+)` ions react with `Al(OH)_(3)` to form `Al_(aq)^(3+)` ions and `H_(2)O`. Thus in acidic water, `AlCl_(3)` exists as `Al^(3+)` and `Cl_(aq)^(ө)` ions.
`AlCl_(3(s))overset("Acidfied water")to Al_(aq)^(3+)+3Cl_(aq)^(Θ)`
c. In alkaline water, `Al(OH)_(3)` reacts with `.^(Θ)OH` ion to form soluble tetrahydroxoluminate complex or metaaluminate ion.
`Al(OH)_(3(s))+.^(ө)OHto [Al(OH)_(4)]_(aq)^(ө)` + or `AlO_(2)^(ө)+2H_(2)O`
Potassium chloride, `KCl` is a salt of strong acid `(HCl)` and strong base `(KOH)`.
it does not undergo hydrolysis in normal water. It ionises to give `K_(aq)^(o+)` and `Cl_(aq)^(ө)` ions.
`KCl_(s)overset("water")to K_(aq)^(o+)+Cl_(aq)^(ө)` Since the aqueous solution of `KCl` is neutral, therefore, in acidified water as well as alkaline water, the ions do not react and remains as such.