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(C) The electrolysis of aqueous $NaCl$ solution is represented by the reaction: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.Accord

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$13$

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(C) The electrolysis of aqueous $NaCl$ solution is represented by the reaction: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.According to Faraday's law,the number of moles of $NaOH$ produced is equal to the number of Faradays passed,which is $0.04 \ mol$.The concentration of $NaOH$ formed is $M = \frac{n}{V(L)} = \frac{0.04 \ mol}{0.4 \ L} = 0.1 \ M$.Since $NaOH$ is a strong base,$[OH^-] = 0.1 \ M = 10^{-1} \ M$.$pOH = -\log[OH^-] = -\log(10^{-1}) = 1$.Using the relation $pH + pOH = 14$,we get $pH = 14 - 1 = 13$.

In an experiment,$0.04 \ F$ was passed through $400 \ mL$ of a $1 \ M$ solution of $NaCl$. What would be the $pH$ of the solution after the electrolysis? (Change in volume will be negligible)

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