1.

In an experiment of 0.04F was passed through 400 ml of 1 M solution of NaCl. What would be the pH of the solution after the electrolysis?

Answer»

8
10
13
6

SOLUTION :`NACL+H_(2)Ooverset("Electrolysis")toNaOH+underset("At anode")((1)/(2)H_(2))+underset("At cathode")((1)/(2)Cl_(2))`
400 ml of 1 M NaCl solution contains NaCl
`=(1)/(1000)xx100mol=0.4mol`
As `Na^(+)+e^(-)toNa overset(H_(2)O)toNaOH+(1)/(2)H_(2)`
1 F produces 1 MOL of NaOH
`therefore0.04F` produces NaOH=0.04 mol
Thus, 400 ml of the solution now contain 0.04 mol of NaOH
`therefore`Molar concentration of NaOH SOL.
`=(0.04)/(400)xx1000=0.1M`
`[OH^(-)]=0.1M[H^(+)]=10^(-13)MthereforepH=13`


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