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The degree of dissociation of weak elect...

The degree of dissociation of weak electrolyde is inversely proportional to the square root fo concentration. It is called Ostwald's dilution law.
`alpha = sqrt((K_(a))/(c))` As the tempertaure increases, degree of dissociation will increase.
`(alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2))))` if concentration is same.
`(alpha_(1))/(alpha_(2)) = sqrt((c_(2))/(c_(1)))` if acid is same.
`0.01M CH_(3)COOH` has `4.24%` degree of dissociation, the degree of dissociation of `0.1M CH_(3)COOH` will be

A

0.0133

B

0.0424

C

0.0524

D

0.0033

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The correct Answer is:
A
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The degree of dissociation of weak electrolyde is inversely proportional to the square root fo concentration. It is called Ostwald's dilution law. alpha = sqrt((K_(a))/(c)) As the tempertaure increases, degree of dissociation will increase. (alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2)))) if concentration is same. (alpha_(1))/(alpha_(2)) = sqrt((c_(2))/(c_(1))) if acid is same. pH of 0.005 M HCOOH [K_(a) = 2 xx 10^(-4)] is equal to

The degree of dissociation of weak electrolyde is inversely proportional to the square root fo concentration. It is called Ostwald's dilution law. alpha = sqrt((K_(a))/(c)) As the tempertaure increases, degree of dissociation will increase. (alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2)))) if concentration is same. (alpha_(1))/(alpha_(2)) = sqrt((c_(2))/(c_(1))) if acid is same. a_(1) and a_(2) are in ratio of 1:2, K_(a_(1)) = 2xx10^(-4) . What will be K_(a_(2)) ?

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