Aqueous solution of  ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is subjected to electrolysis using inert electrodes. Then the true statement is

$\mathrm{aq}{\mathrm{Na}}_2{\mathrm{SO}}_4$is a salt of SA – SB 

$\therefore {\mathrm{p}}^{\mathrm{H}}=7$

Due to electrolysis:

 At Cathode

Discharge potential of  ${\mathrm{H}}_2\mathrm{O}$is lesser than ${\mathrm{Na}}^{+}$

$\therefore 2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

At Anode

Discharge potential of  ${\mathrm{H}}_2\mathrm{O}$ is lesser than ${\mathrm{SO}}_{{ }^4}^{-2}$

$2{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}+{\mathrm{O}}_2$

          Left over solution is concentrated ${\mathrm{Na}}_2{\mathrm{SO}}_4$ . Composition doesn’t change but concentration increases.  ${\mathrm{p}}^{\mathrm{H}}$ of a salt of SA-SB always is 7 irrespective of its concentration.