When $50mL$ of $0.1M$ $NaOH$ is mixed with $50mL$ of $0.05M$ ${CH}_{3}COOH$ solution, pH becomes:

The correct option is A $12.39$

$50$ ml of $0.05M$ $NaOH$ neutralises $50$ ml of $0.05M$ ${CH}_{3}COOH$ to form ${CH}_{3}COONa$

the concentration of excess $NaOH$ will be $=\frac{0.05\times 50}{100}=0.025M$

Now the $pH$ of this excess $NaOH$ will surpass the $pH$ obtained from hydrolysis of ${CH}_{3}COONa$.

$\therefore$ $pH=14-pOH=14-(-log0.025)=12.39$