1.

For a compound solution of a weak electrolyte A_(x)B_(y) of concentration 'C', the degree of dissociation alpha is given as

Answer»

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

Solution :`{:(A_(x)B_(y),HARR,xA^(y+),+,yB^(x-),),(C,,0,,0,"(Initially)"),(C(1-alpha),,Cxalpha,,CY alpha,"(At equilibrium)"):}`
Where `alpha` = DEGREE of dissociation.
`:. K_(eq) = (((Cx alpha)^(x)(Cy alpha)^(y))/(C(1-alpha)))` For concentrated solution of WEAK electrolyte, `alpha` is very small. Therefore, `(1-alpha) ~~ 1`.
`:. alpha = ((K_(eq))/(C^(x+y-1).X^(x).y^(y)))^((1)/(x+y))`.


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