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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 more useful than potassium?

Answer
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Hint: salt of strong base and weak acid on hydrolysis produces more hydroxyl ions as compared to hydrogen ions as strong base dissociates completely whereas weak acid dissociates only to some extent. Alkali metals are highly electropositive. They are strong reducing agents. alkali metals react with water. Sodium is primarily found in blood plasma and interstitial fluids.

Complete answer:
-When sodium carbonate is added to water, on hydrolysis it produces carbonic acid and sodium hydroxide. Carbonic acid is a weak acid, it does not dissociate completely to produce hydrogen ions. Sodium hydroxide is a strong base so it dissociates completely to produce hydroxyl ions.so, as more hydroxyl ions are present, the solution becomes alkaline.
\[N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}+2NaOH\]
When aqueous solution is used for electrolysis of alkali metals, as hydrogen ions are more electropositive, they discharge at anode as hydrogen gas so alkali metal ions remain in solution, hence aqueous salt solution cannot be used for electrolysis of alkali metals.
When fused chlorides undergo electrolysis, cation moves to cathode, undergoes reduction and is deposited in solid form.
Anion chloride ion moves to anode, undergoes oxidation and chlorine gas is liberated.
-sodium ion is found in blood plasma and interstitial fluids.it helps in transmission of signals in nerves. All purposes for which potassium is used, sodium can be used as it is easily obtained.

Note:
When salt of strong acid and strong base undergo hydrolysis, the resultant solution is neutral. When salt of weak base and strong acid undergoes hydrolysis, the resultant solution is acidic. When aqueous solution of alkali salt undergoes electrolysis, hydrogen gas is liberated at cathode and chlorine gas is liberated at cathode and corresponding alkali hydroxide is formed.