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Question

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.


Solution

Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in natrue and does not undergo hydrolysis in normal water. It dissociates into ions as follows:
KCl(s)water−−K+(aq)+Cl(aq)
In acidfied and alkaline water, the ions do not react and remain as such.
Aluminium (III) chloride is the salt of the strong acid (HCl) and weak base [Al(OH)3].
Hence, it undergoes hydrolysis in normal water.
AlCl3(s)+3H2O(l)Normal−−−WaterAl(OH)3(s)+3H+(aq)+3Cl(aq)
In acidified water, H+ ions react with Al(OH)3 forming water and giving Al3+ ions. Hence, in acidified water, AlCl3 will exist as Al3+(aq) and Cl(aq)
AlCl3(s)Acidified−−−−−waterAl3+(aq)+3Cl(aq) ions.
In alkaline water, the following reaction takes place:
Al(OH)3(s)+OH(aq)[Al(OH)4](aq)+2H2O(l)

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