The $pH$ of a solution formed by mixing $40mL$ of $0.1M$ $HCl$ with $10mL$ of $0.45M$ of $NaOH$ is:

The correct option is A $12$
Meq. of $HCl=40\times 0.1=4$; Meq. of $NaOH=0.45\times 10=4.5$

$HCl+NaOH⟶NaCl+H_{2}O$

$H^{+}$ and $O{H}^{-}$ neutralise each other.

$\therefore$ Meq. of $NaOH$ left $=4.5-4=0.5$

Total volume (ml) $=50ml$

$\therefore$ $\left[O{H}^{-}\right]=\frac{0.5}{50}=1\times {10}^{-2}$

pOH = $-log\left[O{H}^{-}\right]$

$=-log\left[{10}^{-2}\right]$

$=2$

$\therefore$ $pOH=2$ and $pH=12$