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.

A current strength of `96.5 A` is passed for `10s` through `1L` of a s

1.	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.
Answer» 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`

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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.

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