The correct option is C 3.2×10−2 M
Given, [BOH=25 M, Kb=1×10−12, [HCl]=215 M, Kw=1×10−14 at 25o C
N1V1Base=N2V2Acid
2.5×25=215×V2
or V2(acid)=152=7.5 mL
BOH+HCl⇌BCl+H2O
millimoles of BOH=2.5×25=1 mmol
∴ salt BCl formed = 1 mmol
Total volume of solution =2.5+7.5=10 mL=10×10−3 L
∴ [BCl] in solution =1×10−310×10−3=0.1 M
For the salt of weak base and the strong acid
[H+]=√KwCKb=√10−14×0.110−12
[H+]=3.2×10−2M