Question

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.

Answer 1

Potassium chloride (KCl) is the salt of a strong acid (HCl) and 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) overset("water")to H_(aq)^(+)+Cl_(aq)^(+)`
In acidified and alkaline water, the lons do not react and remain as such Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base `[Al(OH)_(3)]` Hence, It undergoes hydrolysis in normal water.
`AlCl_(3_(g))+3H_(2)O_(l) underset("Water")overset("Normal")to Al(OH)_(3_(s))+3H_(aq)^(2)+3Cl_(aq)^(-)`
In acidified water, `H^(+)` ions react with `Al(OH)_(3)` forming water and giving `Al^(3+)` ions. Hence in acidified water, `AlCl_(3)` will exists as `Al^(3+) and Cl_(aq)^(-)` ions.
`AlCl_(3_(g)) underset("Water")overset("Acidified")to Al_(aq)^(3+)+3Cl_(aq)^(-)`
In alkaline water, the following reaction takes place.
`Al(OH)_(3_(s))+ubrace(OH_(aq)^(-)) to [Al(OH)_(4)]_(aq)^(-)+2H_(2)O_(l)`