Question

The $pH$ of $0.5\text{L}$ of $1.0\text{M NaCl}$ after the electrolysis for $965\text{s}$ using $5.0\text{A}$ current is

Answer 1

$\text{At cathode}2H_{2}O+2e^{-}\to H_2+2O{H}^{-}\text{At anode}2C{l}^{-}\to C{l}_2+2e^{-}\text{Moles of}O{H}^{-}\text{formed = zit}=\frac{1\times 5\times 965}{96500}=0.05\text{mol}\left[O{H}^{-}\right]=\frac{0.05}{0.5}=1\times {10}^{-1}{\text{mol L}}^{-1}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.1\right)=1\therefore pH=13$