The pH of 0.5 L of 1.0 M NaCl solution after electrolysis for 965 s...

The reaction involved is, $\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text{ Electrolysis }} \frac{1}{2} \mathrm{H}_2+\mathrm{Cl}_2+\mathrm{NaOH}$ Amount of NaCl present in 0.5 L of $1.0 \mathrm{M}, \mathrm{NaCl}=0.5$ mol Quantity of electricity passed $=965 \times 5$ coulombs. $\therefore 4825$ coulombs will decompose NaCl. $$\begin{aligned} & =\frac{4825}{96500} \mathrm{ mol} \\ & =0.05 \mathrm{ mol} \end{aligned}$$ NaOH formed in the solution will also be 0.05 mol. Volume of solution $=0.5 \mathrm{ L}\therefore$ Molarity of NaOH in the solution $$\begin{aligned} & =\frac{0.05 \mathrm{moles}}{0.5 \mathrm{ L}}=0.1 \mathrm{M} \\ & =10^{-1} \mathrm{M} \end{aligned}$$$$\begin{array}{rlrl} & \therefore & \mathrm{pOH} & =1 \\ \text{ or } & \mathrm{pH} & =14-1=13\end{array}$$