$50mL$ of $0.2MHCl$ is titrated against the $0.1MNaOH$ solution. What will be the $pH$ of the solution when $20mLNaOH$ is added ?

The correct option is A 0.95
mmol of $HCl=50\times 0.2=10$
mmol of $NaOH\text{added}=20\times 0.1=2.0$

$HCl\left(aq\right)+NaOH\left(aq\right)\to NaCl\left(aq\right)+H_{2}O\left(l\right)Initial:102.000Final:802.02.0$

After titration the remaining $HCl=8mmol$ .
$\text{Final Concentration}=\frac{\text{Moles of HCl remaining}}{\text{Total Volume}}$
Total Volume $=50+20=70mL$
$\left[HCl\right]=\frac{8}{70}M=0.11MpH=-\log \left[H^{+}\right]pH=-\log \left[0.11\right]pH=2-\log \left(11\right)=0.95$