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.

Answer 1

$KCl$ is a salt of strong acid $HCl$ and strong base $KOH$. Such salts are neutral in nature and do not undergo hydrolysis. In neutral water, acidic water and alkaline water, $KCl$ dissociates as:

$KCl\stackrel{Water}{\to }{K}^{+}+C{l}^{-}$
$AlC{l}_3$ is a salt of strong acid $HCl$ and weak base $Al(OH{)}_3$.

In normal water, it is hydrolyzed.

$AlC{l}_3+3H_{2}O\to Al(OH{)}_3+3H^{+}+3C{l}^{-}$
In acidic water, $H^{+}$ ions react with $Al(OH{)}_3$ (obtained from the hydrolysis of $AlC{l}_3$) forming water and $A{l}^{3+}$ ions. Hence it exists as $A{l}^{3+}+C{l}^{-}$.

$AlC{l}_3\underset{water}{\overset{acidified}{\to }}A{l}^{3+}+3C{l}^{-}$

In alkaline water, the $Al(OH{)}_3$ obtained from hydrolysis of $AlC{l}_3$ reacts with hydroxide ions of alkali. The reaction is:

$Al(OH{)}_3+O{H}^{-}\to [Al(OH{)}_4{]}^{-}$