For a fairly concentrated solution of a weak electrolyte `A _(y) , B _(y)` the degree of dissociation is given by

Options

A. `alpha = sqrt((K_(eq)xy)/(C ))`
B. `alpha = sqrt((K_(eq)C)/(x y))`
C. `alpha = sqrt((K_(eq)C)/((x+y)))`
D. `alpha = ((K_(eq))/(C^(x+y-1)x^(x)y^(y)))^(1//(x+y))`

Answer

D

Explanation

The reaction can be expressed as
`{:(,A_(x)B_(y),hArr,xA^(y+),+,yB^(x-)),("Initial concentration",C,,0,,0),("Conc. At equilibrium",C(1-alpha),,xC alpha,,yC alpha):}`
`K_(eq) = ((x C alpha)^(x)(y C alpha)^(y))/(C(1-alpha)) = ((x C alpha)^(x)(yC alpha)^(y))/(C )` (taking `1-alpha = 1`) = `(x^(x)y^(y)C^(x+y)alpha^(x+y))/(C ) = x^(x )y^(y) C^(x+y-1)alpha^(x+y)`
`alpha = ((K_(eq))/(x^(x)y^(y)C^(x+y-1)))^(1//(x+y))`