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The degree of dissociation of 0.1 M weak...

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. __________ ?

Text Solution

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The correct Answer is:
2

` 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`
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