A current strength of `96.5 A` is passed for `10s` through `1L` of a solution of `0.1 M` aqueous `CuSO_(4)`. Calculate the `pH` of the solution.

Explanation

Aqueous solution of `CuSO_(4)` on electrolysis gives gives `Cu` at cathode and `O_(2)` at anode and `H^(o+)` ions in the solution.
Eq of current `=(96.5Axx10s)/(96500C)=0.01Eq=10^(-2)Eq`
`:. 10^(-2)Eq` of `Cu=10^(-2)`Eq of `O_(2)=10^(-2)`Eq of `H^(o+)=10^(-2)F`
`:. [H^(o+)]=(10^(-2)Eq)/("Volume of solution in L")=(10^(-2)Eq)/(1L)`
`=10^(-2)N `or `M`
`:. pH-log (10^(-2))=2`