• # question_answer 32) State as to why: (a) a solution of $N{{a}_{2}}C{{O}_{3}}$is alkaline? (b) alkali metals are prepared by electrolysis of their fused chlorides? (c) sodium is found to be more useful than potassium?

(a) $N{{a}_{2}}C{{O}_{3}}$is a salt of a strong base, sodium hydroxide and a weak acid, carbonic acid, therefore it undergoes hydrolysis, i.e., $N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}O\rightleftharpoons \underset{Strong}{\mathop{2NaOH}}\,+\underset{Weak}{\mathop{{{H}_{2}}C{{O}_{3}}}}\,$ $O{{H}^{-}}$ion concentration increases. Hence the solution is alkaline. (b) The discharge potentials of alkali metals are much higher than hydrogen. Therefore, when aqueous solution of alkali halide is electrolysed, hydrogen instead of alkali metal is discharged at cathode. Hence, to prepare an alkali metal, electrolysis of its fused chloride is carried out. (c) Sodium is relatively more abundant and cheaper than potassium. It shows all the properties of potassium. It is comparatively less reactive and its reaction can be better controlled.