The correct option is A 0.01
Given, Ka=10−5, [NaX]=0.1 M
According to question, NaX is a mixture of weak acid and strong base.
Therefore X−− only undergoes hydrolysis
at eqb. X−C−Ch+H2OKa⇌HXCh+OH−Ch
'h' is hydrolysis constant and 'C' is concentration
Kh=[HX][OH−][X−][H2O]
we know,
⇒Kh=KwKa=10−1410−5
⇒Kh=10−9
we know
h=√KhC=√10−90.1=√10−8
h=10−4
⇒degree of hydrolysis (h)=10−4
percentage degree of hydrolysis=10−4×100=0.01