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The pH of 0.5L of 1.0"M NaCl" after the ...

The pH of `0.5L` of `1.0"M NaCl"` after the electrolysis for 965 s using `5.0` A current (`100%` efficiency ) is :

A

`1.00`

B

`13.00`

C

`12.70`

D

`1.30`

Text Solution

Verified by Experts

The correct Answer is:
C
Mole of `OH^(-)` formed = `(1xx5xx965)/(96500)=0.05`
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