For the given electrolyte `A_(x)B_(y)`. The degree of dissociation `'alpha'` can be given as:

Options

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

Answer

C

Explanation

`{:(A_(x)B_(y),hArr,xA^(y+),+,yB^(x-),),(C,,0,,0,"(Initially)"),(C(1-alpha),,Cxalpha,,Cyalpha,"(At equilibrium)"):}`
Where `alpha = ` degree of dissociation.
` :. K_(eq)= (((Cxalpha)^(x)(Cyalpha)^(y))/(C(1-alpha)))` For concentrated solution of weak electrolyte,  is very small. Therefore , `(1-alpha) approx `
` :. alpha =((K_(eq))/(C^(x+y-1).x^(x).y^(y)))^(1/(x+y))` .