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The pH of 0.5 L of 1.0 M NaCl after the electrolysis for 965 Sec using 5.0 A current (100% efficiency), is

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1.00

13.00

12.70

1.30

answer is B.

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Detailed Solution

q= i x t = 5 x 965 coulombs = 4825 coulombs = 0.05F

0.05 equivalent of OH - = 0.05 moles of OH-

Molarity = 0.05/0.5=0.1M

P[OH] = 1

PH=13

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