Question

Question 10.27 State as to why

(a) a solution of Na2CO3 is alkaline ?

(b) alkali metals are prepared by electrolysis of their fused chlorides ?

(c) sodium is found to be more useful than potassium ?

Class XI The s-Block Elements Page 306

Answer 1

(a) Na2CO3 is the salt of weak acid (H2CO3) and strong base (NaOH) therefore it undergoes Hydrolysis to produce strong base, NaOH and hence, its aqueous solution is alkaline in nature.
Na2CO3 + H2O --->2NaOH(strong base) + H2CO3(weak acid)

(b) there are some important reason :-alkali metals are strong reducing agents ,hence can't be extracted by reduction of their oxides and other compounds.

Being highly positive in nature it is not possible to displace them from their slat solutions by any other element.

Alkali metals also can't be obtained by the electrolysis of the aqueous solution of their salt because of H2 is librated at cathode instead of alkali metal.
That's why alkali metals are prepared by electrolysis of their fused chloride,
NaCl --->Na+ + Cl- {fusion}
during electrolysis,
at cathode,
2Na+ + 2e- --->2Na
at anode,
2Cl- ---> Cl2 + 2e-


(c) sodium is found to be more useful than potassium as it is highly reactive but not as reactive as potassium . sodium is used
(i) as a coolant reactor.
(ii) in the manufacturing of tetraethyl lead-an anti knock additive for petrol.
4C2H5Cl + 4Na-pb ---> (C2H5)4Pb + 3Pb + 4NaCl

(iii) in sodium vapor discharge lamps.
(iv) as laboratory reagents for organic analysis.