1.

State as to why : (a) a solution of Na_(2)CO_(3) is alkaline ? (b) alkali metals are prepared by electrolysis of their fused chlorides ? (c) Sodium is found to be more useful than potassium ?

Answer»

Solution :(a)`Na_(2)CO_(3)` is a salt of a weak acid, carbonic acid `(H_(2)CO_(3))` and a STRONG base , sodium hydroxide `(NaOH)` therefore , it undergoes hydrolysis to undergoes hydrolysis to produce strong base NaOH and hence its aqeous solution is ALKALINE in nature .
`Na_(2)CO_(3) (s) + H_(2)O(l)to underset("(Strong base)") (2NaOH (aq)) + underset("(Weak acid)") (H_(2)CO_(3) ) (aq)`
(b) Since the discharge potential of alkali metals is much HIGHER than that of hydrogen , therefore , when the aqueous solution of any alkali metal chloride is subjected to electrolysis . `H_(2)` instead of the alkali metal is produced at the cathode . Therefore , to prepare alkali metals , electrolysis of their fused chlorides is carried out .
(c) Sodium ions are found primarily in the blood plasma and in the interstitial fluid which surrounds the cells while potassium ions are primarily present within the cell fluids . Sodium ions helps in the transmission of nerve signals , in regulating the flow of water across cell membranes and in the transport of SUGARS and amino acids into the cells . Thus , sodium is found to be more useful than potassium .


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