A current of $9.65$ A is passed through $100$ ml $N a C l$ solution containing sufficient $N a C l$ for $100$ s. What will be the $p H$ of the resulting solution?
Assume $100 %$ current efficient and constancy of the volume of solution.

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Solution

$N a c l \to N a^{+} + c l^{-}$
$H_{2} O \to H^{+} + O H^{-}$
at anodes $c l^{-}$ will get oxidized
$2 c l^{-} \to c l_{2} + 2 e^{-}$
At cathode, $H^{+}$ will be reduced
$2 H^{+} 2 e^{-} \to H_{2} (g)$
$Q = I \times t = (9.65) (100) = 965$ coulombs
$965 C = 0.01 F$
$(∵ I F = 96500$ coulombs)
moles of Nacl = $\frac{M}{n}$
$\frac{m a s s}{E q . w t} = \frac{Q}{96500} = F$
$0.01 F = \frac{m a s s}{E q . w t}$
$[n_{m o l e s} = 0.01]$
morality $\frac{0.01}{100} \times 1000 = 0.1 M$
$[O H^{-}] = 0.1 H$
$P O H = 1$
$P H = 14 - 1 = 13.$
$∴$ PH of resulting solution $= 13 .$

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