The degree of dissociation of 0.1 M weak acid is 10^(-2) and the degree of dissociation of the same acid in 0.025M concentration is '*', then find 100x. __________ ?

The degree of dissociation of 0.1 M weak acid is 10^(-2) and the degre

1.	The degree of dissociation of 0.1 M weak acid is 10^(-2) and the degree of dissociation of the same acid in 0.025M concentration is '*', then find 100x. __________ ?
Answer» Solution :` K_a =CALPHA ^(2)=10 ^(-1)(10 ^(-2) )^(2)= 10 ^(-5) ` ` X= sqrt( (K_a)/(C)) =sqrt((10 ^(_5))/(25 xx 10 ^(-3))) rArr x = (10^(_1))/(5)` ` therefore 100 x =(0.1)/(5) xx 100 =2`