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

To determine the different products formed when aluminum (III) chloride (AlCl₃) and potassium chloride (KCl) are treated with normal water, acidified water, and alkaline water, we can analyze each case step by step.

### Step 1: Reaction of Aluminum (III) Chloride with Normal Water
When aluminum (III) chloride is dissolved in normal water, it undergoes hydrolysis:
\[ \text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl} \]
**Products:** Aluminum hydroxide (Al(OH)₃) and hydrochloric acid (HCl).
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