The pH of 0.5 L of a 1.0 M $N a C l$ solution after electrolysis for 965 seconds using 5.0 A current (100% efficiency) is:

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12.7

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1.3

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Solution

The correct option is B 13
Charge $q = i \times t = 5.0 \times 965 C = 4825 C = \frac{4825}{96500} F = 0.05 F$
$= 0.05 e q v o f O H^{-} = 0.05 m o l e s o f O H^{-}$
$[O H^{-}]$ = $\frac{0.05 moles}{0.5 L} = 0.1 M$
pOH=1 $\Rightarrow$ pH=13

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