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

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.