pH of 400 mL of 1 M solution of NaCl after passing 0.4 F of electricity is $\qquad$\

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Professor PC Thomas Classes · Answer

$\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \xrightarrow{	ext { Electrolysis }} \mathrm{NaOH}+\underset{	ext { (cathode) }}{\mathrm{H}_{2}}+\underset{	ext { (anode) }}{\mathrm{Cl}_{2}}$ No. of moles of NaCl taken $=$ $$ \frac{1}{1000} 	imes 400=0.4 \mathrm{~mol} $$ $\mathrm{Na}^{+}+\overline{\mathrm{e}} ightarrow \mathrm{Na} \xrightarrow{\mathrm{H}_{2} \mathrm{O}} \mathrm{NaOH}+\frac{1}{2} \mathrm{H}_{2}$\ IF $ightarrow 1$ mole NaOH\ $0.04 \mathrm{~F} ightarrow 0.04$ mole NaOH in 400 mL solution.\ $	herefore$ Molar con. of $\mathrm{NaOH}=\frac{0.04}{400} 	imes 1000=0.1$\ $[\mathrm{OH}]=10^{-1}$ ie $\left[\mathrm{H}^{+}ight]=10^{-13}$\ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]=-\log 10^{-13}=13$\