Answer:-
i. Sodium carbonate (\(Na_2CO_3\)) is a salt of weak acid \(H_2CO_3\) and strong base \(NaOH\). When dissolved in water, it dissociates completely.
\[\ce{Na2CO3_{(aq)} -> 2Na^+_{(aq)} + CO3^{2-}_{(aq)}}\]
ii. The \(Na^+\) ions of salt have no tendency to react with \(OH^-\) ions of water since the possible product of the reaction is \(NaOH\), a strong electrolyte.
iii. On the other hand, the reaction of \(CO_3^{2-}\) ions of salt with the \(H_3O^+\) ions from water produces unionized \(H_2CO_3\).
\[\ce{CO3^{2-}_{(aq)} + 2H2O_{(l)} <-> H2CO3_{(aq)} + 2OH^-_{(aq)}}\]
Thus, the hydrolytic equilibrium for \(Na_2CO_3\) is,
\[\ce{Na2CO3_{(aq)} + H2O_{(l)} <-> H2CO3_{(aq)} + 2Na^+_{(aq)} + 2OH^-_{(aq)}}\]
iv. As a result of excess \(OH^-\) ions produced, the resulting solution of \(Na_2CO_3\) is alkaline.
v. Similarly, ammonium chloride (\(NH_4Cl\)) is the salt of strong acid \(HCl\) and weak base \(NH_4OH\). When \(NH_4Cl\) is dissolved in water, it dissociates completely as,
\[\ce{NH4Cl_{(aq)} -> NH4^+_{(aq)} + Cl^-_{(aq)}}\]
vi. \(Cl^-\) ions of salt have no tendency to react with water because the possible product \(HCl\) is a strong electrolyte.
vii. The reaction of \(NH_4^+\) ions with \(OH^-\) ions forms unionized \(NH_4OH\). The hydrolytic equilibrium for \(NH_4Cl\) is then written as,
\[\ce{NH4^+_{(aq)} + 2H2O_{(l)} <-> NH4OH_{(aq)} + H3O+_{(aq)}}\]
viii. Due to the presence of an excess of \(H_3O^+\) ions, the resulting solution of \(NH_4Cl\) is acidic.