$2 NaCl +2 H _{2} O \xrightarrow{\text { electrolysis }}H _{2}+ Cl _{2}+2 NaOH$
Weight of $NaCl$ present in $0.5\, L =0.5\, mol$
Charge $-965 \times 5-4825\, C$
$\therefore $ Number of moles decomposed
$=\frac{1 \times 4825}{96500}=0.05\, mol$
$\therefore $ Number of moles of $NaOH$ formed is also $0.05$.
$\therefore $ Molarity $=\frac{0.05 \times 1000}{500}=0.1$
$pOH =1, pH =13$