The correct option is B 13.3
mmol of HCl=20×1=20
mmol of NaOH added=30×1=30
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)Initial: 20 30 0 0Final: 0 10 20 20
After titration total HCl will react and excess NaOH is present in the solution.
mmol of NaOH remaing =10
Final concentration=Moles of NaOH remaining Total Volume
Total Volume =30+20=50 mL
[NaOH]final=1050 M=0.2 MpOH=−log[OH−]pOH=−log[0.2]pOH=1−log(2)=0.7pH=14−0.7=13.3