$HCl+NaOH\to NaCl+H_{2}O$

pH = $-log\left[H^{+}\right]$, $moles=concentration\times volume$

Explanation: For HCl, pH = 2

pH = $-log\left[H^{+}\right]$

2 = $-log\left[H^{+}\right]$

$\left[H^{+}\right]$ = ${10}^{-2}$ = 0.01M

$moles=concentration\times volume$ = 0.01 x 200 =2

Similarly, for NaOH pH =12 then pH + pOH = 14

12 + pOH = 14

pOH = 14-12 = 2

pOH = $-log\left[O{H}^{-}\right]$

2 = $-log\left[O{H}^{-}\right]$

$\left[O{H}^{-}\right]={10}^{-2}$ = 2

moles = 0.01 x 300 = 3

Total volume = 200 + 300 = 500ml

Out of 3 moles of NaOH, 2 moles of NaOH will be neutralized with 2 moles of HCl (3-2=1).

the $\left[O{H}^{-}\right]$ concentration remaining = $\frac{1}{500}$ = $2\times {10}^{-3}$M

pOH = $-log\left[O{H}^{-}\right]$ = $-log[2\times {10}^{-3}]$= 2.70

Then pH + pOH = 14

pH = 14 - 2.70 = 11.3.