400 ml of 1M NaCl is electrolysed using 0.04F (inert electrodes). After electrolysis pH of the solution left will be

Electrolysis of aq.NaCl gives hydrogen gas at cathode and oxygen gas at anode, the electrolyte solution contains NaOH after electrolysis.
Number of equivalents of NaOH formed =0.04

$\mathrm{Normality} \left(\mathrm{N}\right)\hspace{0.17em}= \frac{0.04\times 1000}{400}=0.1\mathrm{N}$

[OH^-]= 0.1N

pOH= -log [OH^-]

pOH= -log (0.1) 

pOH =1 

pH+pOH = 14 

pH =14-1 

pH= 13