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

Options

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

Answer

C

Explanation

`{:(,A_(x)B_(y),hArr,xA^(y+),+,yB^(x-)),("Initial",c,,0,,0),("At eqm.",c(1-alpha),,x c alpha,,ycalpha):}`
`K_(eq)=((xc alpha)^(x)(ycalpha)^(y))/(c )`
`=((x c alpha)^(x)(y c alpha)^(y))/(c )`
Taking `1- alpha~=1`
or `K_(eq)=(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)x^(x+y-1))]^((1)/(x+y))`