Correct Answer - B
`pH of HCI =2` , `:. [HCI]=10^(-2)M`
`pH of NaOH = 12` , `:. [NaOH]= 10^(-2)M`
`{:(,HCI+,NaOHrarr,NaCI+,H_(2)O),("Meq. before reaction", 200xx10^(-2),300xx10^(-2), ,),(, =2,=3,0,0),("Meq. after reaction", 0,1,2,2):}`
`:. [OH^(-)]` left from `NaOH=(1)/(500)=2xx10^(-3)M`
`:. pOH= -log [OH^(-)]= -log 2x10^(-3)`
`:. pOH=2.6989 and pH = 11.3010`