You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Ostwald Dilution Law

If α is the d...

Question

If $α$ is the degree of ionization, $C$ the concentration of a weak electrolyte and $K_{a}$ the acid ionization constant, then the correct relationship between $α$ , C and $K_{a}$ is

$\alpha$ ² = √(K_a xC).

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\alpha$ ² = √(C / K_a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\alpha$ = √(K_a / C).

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\alpha$ = √(C / K_a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$\alpha$ = √(K_a / C).

According to the Ostwald's dilution formula $α^{2} - \frac{K_{a} (1 - α)}{C}$ . But for weak

electrolytes $α$ is very small. So that $(1 - α$ ) can be taken as 1. So that $α = \sqrt{\frac{K_{a}}{C}}$ .

Suggest Corrections

Similar questions

If $α$ is the degree of ionization, C the concentration of a weak electrolyte and $K_{a}$ , the acid ionization constant, then the correct relationship between $α, C$ and $K_{a}$ is:

Q. Consider a weak monobasic acid

H A

having a concentration C with degree of dissociation,

α << 1.

. The pH of this weak acid in terms of acid dissociation constant

K_{a}

and concentration

C

is :

The degree of dissociation of weak electrolyte is inversely proportional to the square root of concentration. It is called Ostwald's dilution law.

α = \sqrt{\frac{K_{a}}{c}}

, as the temperature increases, degree of dissociation will increase

\frac{α_{1}}{α_{2}} = \sqrt{\frac{K_{a_{1}}}{K_{a_{2}}}}

, if concentration is same

\frac{α_{1}}{α_{2}} = \sqrt{\frac{c_{2}}{c_{1}}}

, if acid is same

a_{1}

and

a_{2}

are in ratio of

1 : 2, K_{a_{1}} = 2 \times 10^{- 4}

. What will be

K_{a_{2}}

The degree of dissociation of weak electrolytes is inversely proportional to the square root of concentration. It is called Ostwald's dilution law.

α = \sqrt{\frac{K_{a}}{c}}

As the temperature increases, degree of dissociation will increase

\frac{α_{1}}{α_{2}} = \sqrt{\frac{K_{a_{1}}}{K_{a_{2}}}}

If concentration is same

\frac{α_{1}}{α_{2}} = \sqrt{\frac{c_{2}}{c_{1}}}

If acid is same

0.01 M

C H_{3} C O O O H

has

4.24 %

degree of dissociation, the degree of dissociation of

0.1 M

C H_{3} C O O O H

will be

Q. What is the percent ionization

(α)

of a

0.01 M

H A

solution?
(Given that

K_{a} = 10^{- 4}

)

Join BYJU'S Learning Program

If α is the degree of ionization, C the concentration of a weak electrolyte and Ka the acid ionization constant, then the correct relationship between α, C and Ka is

The correct option is C = √(Ka / C). According to the Ostwald's dilution formula α2 − Ka(1 − α)C. But for weak electrolytes α is very small. So that (1 − α) can be taken as 1. So that α = √KaC.

If $α$ is the degree of ionization, $C$ the concentration of a weak electrolyte and $K_{a}$ the acid ionization constant, then the correct relationship between $α$ , C and $K_{a}$ is

The correct option is C
$\alpha$ = √(K_a / C).

According to the Ostwald's dilution formula $α^{2} - \frac{K_{a} (1 - α)}{C}$ . But for weak

electrolytes $α$ is very small. So that $(1 - α$ ) can be taken as 1. So that $α = \sqrt{\frac{K_{a}}{C}}$ .