The strength of elctrolytes is expressed...

The strength of elctrolytes is expressed in terms of degree of dissociation `alpha` For strong electrolyte `alpha` is close to one and for weak electrolytes `alpha` is quite small. According to Ostwald Dilution Law `alpha=sqrt((K)/(C))`
For an acid `[H^(+)] = sqrt(K_(a)C)`
For a base `[OH^(-)] =sqrt(K_(b)C`
The relative strengths of acids or bases can be compared in terms of the square roots of their `K_(a) " or " K_(b)` values.
A monoprotic acid in 0.1 M solution ionises to 0.001 % . Its ionisation constant is :

`1.0 xx 11^(-3)`

` 1.0 xx 10^(-6)`

` 1.0 xx 10^(-8)`

`1.0 xx 10^(-11)`

Verified by Experts

The correct Answer is:

`K_(a) =Calpha^(2) =0.1 xx (1xx 10^(-5))^(2) =10^(-11)`

EQUILIBRIUM
DINESH PUBLICATION|Exercise Comprehesion 10|1 Videos
EQUILIBRIUM
DINESH PUBLICATION|Exercise Comprehesion 11|1 Videos
EQUILIBRIUM
DINESH PUBLICATION|Exercise Comprehesion 8|1 Videos
ENVIRONMENTAL CHEMISTRY
DINESH PUBLICATION|Exercise Integer Type Questions|4 Videos
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS
DINESH PUBLICATION|Exercise UNIT TEST - 34|1 Videos

The strength of elctrolytes is expressed in terms of degree of dissociation alpha For strong electrolyte alpha is close to one and for weak electrolytes alpha is quite small. According to Ostwald Dilution Law alpha=sqrt((K)/(C)) For an acid [H^(+)] = sqrt(K_(a)C) For a base [OH^(-)] =sqrt(K_(b)C The relative strengths of acids or bases can be compared in terms of the square roots of their K_(a) " or " K_(b) values. At infinite dilution , the percentage ionisation of both strong and weak electrolytes is :

`1%`

`20%`

`50%`

`100%`

`A lt B lt C`

`C lt A lt B `

`B lt C lt A `

`B lt A ~~ C`

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

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

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

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