The correct option is A 0.95
mmol of HCl=50×0.2=10
mmol of NaOH added=20×0.1=2.0
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)Initial: 10 2.0 0 0Final: 8 0 2.0 2.0
After titration the remaining HCl=8 mmol .
Final Concentration =Moles of HCl remaining Total Volume
Total Volume =50+20=70 mL
[HCl]=870 M=0.11 MpH=−log[H+]pH=−log[0.11]pH=2−log(11)=0.95