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

Options

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))`

Answer

C

Explanation

`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))`