For a 'C'M concentarted solution of a we...

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

`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)`

Text Solution

Verified by Experts

The correct Answer is:

`{:(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))`.

Topper's Solved these Questions

IONIC EQUILIBRIUM
ERRORLESS |Exercise Ordinary Thinking (Acids and Bases)|154 Videos
IONIC EQUILIBRIUM
ERRORLESS |Exercise Ordinary Thinking (Common ion effect, Isohydric solutions, Solubility product, Ionic product of water and Salt hydrolysis )|137 Videos
HYDROCARBON
ERRORLESS |Exercise JEE SECTION (Matrix Match type questions)|6 Videos
NITROGEN CONTAINING COMPOUNDS
ERRORLESS |Exercise JEE (ADVANCED) 2018 MORE THAN ONE CHOICE CORRECT ANSWER|1 Videos

For a concentrated solution of a weak electrolyte A_x,B_y, the degree of dissociation is given as

`alpha=sqrt(K_(eq)//C(x+y))`

`alpha=sqrt(K_(eq)C//(xy))`

`alpha=(K_(eq)//C^(x+y) x^(x) y^y)^(1//(x+y))`

`alpha=sqrt(K_(eq)//xyC)`

For the weak electrolyte , their degree of dissociation increase

One increasing dilution

On decreasing diluation

On increasing pressure

None of these

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

(a) `alpha = (i-1)/((x+y-1)`

(b) `i = (1-alpha)+x alpha+ y alpha`

( c) `alpha = (1-i)/((1-x-y))`

(d) Either of these