The pH of the solution obtained by mixing 800 mL of 0.05 N NaOH and 200 mL of 0.1 N HCl.

Moles of NaOH in mixture = 800 x 0.05 m moles = 40 m moles

            m moles of HCI = 200 x 0.1 = 20 m moles    

            20 m moles of HCI is neutralized by 

            20 m moles of NaOH

            The remaining 20 m moles are present in (800 + 200) litre of solution

Conc. of NaOH= $\frac{20}{1000}$moles per litre = $2\times {10}^{-2}\mathrm{M}$

$\mathrm{pOH}=-log(2\times {10}^{-2})$ = 2 – 0.3010 = 1.699 

pH = 14 – pOH = 14 – 1.699 = 12.3010