10g fairly concentrated solution of `CuSO_(4)` is electrolyzed using `0.01F` of electricity. Calculate:
(a) The weight of resulting solution
(b) Equivalents of acid or alkali in the solution.

Answer

Final weight `=9.6g,0.01Eq` of acid.

Explanation

`Cu^(2+)+H_(2)OtoCu(s)+2H^(+)+(1)/(2)O_(2)`.
Moles of electricity `=0.01`
`therefore` moles of `Cu^(2+)` consumed `=(0.01)/(2)`
and moles of `O_(2)` liberated `=(0.01)/(4)`
`therefore` mass of solution decreased `=((0.01)/(2)xx63.5)+((0.01)/(2)xx32)=0.3175+0.08=0.3975`.
`therefore` mass of resulting solution `=10-0.3975=9.6gm`
Equivalent of acid formed `=0.01`