Calculate the pH of 0.5 of 1.0 M NaCl solution after electrolysis when a current of 5.0 ampere is passed for 965 seconds.

Explanation

`NaCl(aq)+H_(2)Ooverset("Electrolysis")rarr(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)+NaOH(aq)`
Amount of NaCl present in 0.5 L 1 M solution =0.5 mol
Quantity of charge passed (Q)`=(5A)xx(965 s)`=4825 As =4825 C
Now, 96500 C of charge decompose NaCl=1 mol.
4825 C of charge decompose NaCl `=((4825 C))/((96500 C))xx1 "mol"=0.05 "mol"`
Amount of NaOH formed in solution =0.05 mol
Volume of solution =0.5 L
Molarity of NaOH solution `=((0.05 "mol"))/((0.5 L))=0.1 " mol " L^(-1)=10^(-1) M`
pOH=1 or pH =14-1=13.