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For a 'C'M concentarted solution of a we...

For a 'C'`M` concentarted solution of a weak electrolyte `A_(x)B_(y)alpha`(degree of dissociation) is

A

`alpha=sqrt(K_(eq)/(C(x+y)))`

B

`alpha=sqrt(K_(eq)/(C(xy)))`

C

`alpha=(K_(eq)//C^(z+y-1) X^(x)y^(y))^(1/(x+y))`

D

`alpha=(K_(eq)/C_(xy))`

Text Solution

Verified by Experts

The correct Answer is:
C
`A_(x)B_(y) to xA^(y+) + yB^(x-)`
t = 0 C 0 0
`t=t_(eq)C(1-alpha) xCalpha yCalpha`
`K_(eq)=((xCalpha)^(x)(yCalpha)^(y))/(C(1-alpha)) "where" 1-alpha approx 1 implies alpha=(K_(eq)/(C^(x+y-1)x^(x)y^(y)))^(1/(x+y))`
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