The correct option is D 223
Given, [BOH]=0.1N, pH=6.6, Kb=6.25×10−8,V=100 ml
Let the mass of HCl added be x mg
BOH+HCl⇌BCl+H2OMillimoles 10 x36.5 0 0at eqb 10−x36.5 0 x36.5
This constitutes basic buffer.
For basic buffer, the [OH−] is given by Henderson equation
pOH=−log Kb+log[BCl][BOH]
[OH−]=Kb[BOH][BCl]..........(1)
we know, pH+pOH=14
6.6+pOH=14
pOH=7.4
−log[OH−]=7.4
log[OH−]=−7.4
[OH−]=3.98×10−8
now substitue values in equation (1),
3.98×10−8=6.25×10−8×(365−x36.5)(36.5x)
(365−x36.5)(36.5x)=0.6368
(365−xx)=0.6368
On solving, x = 223.92
∴ Weight of HCl required = 223 mg.