Question
Derive an expression of Ostwald's dilution law for weak acid.
✓
Answer
Ostwald’s dilution law: It relates the dissociation constant of the weak acid $(K_a)$ with its degree of dissociation $(\alpha )$ and the concentration $(c).$
Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,
$CH _3 COOH \rightleftharpoons CH _3 COO ^{-}+ H ^{+}$
The dissociation constant of acetic acid is,
$K _{ a }=\frac{\left[ H ^{+}\right]\left[ CH _3 COO ^{-}\right]}{\left[ CH _3 COOH \right]}$
Substituting the equilibrium concentration in the equation
$K_a=\frac{(\alpha C)(\alpha C)}{(1-\alpha) C}$
$K_a=\frac{\alpha^2 C^2}{(1-\alpha) C}$
$K_a=\frac{\alpha^2 C}{(1-\alpha)}..........(1)$
We know that weak acid dissociates only to a very small extent compared to one, a is so small.
equation $(1)$ becomes,
$K_a = \alpha ^2 C$
$\alpha^2=\frac{K_a}{C}$
$\alpha=\sqrt{\frac{K_a}{C}}.........(2)$
Similarly, for a weak base,
$K_b = \alpha ^2 C$
$\alpha=\sqrt{\frac{K_b}{C}}........(3)$
The concentration of $H$ can be calculated using the $K_a$ value as below,
$[H^+] = \alpha C$
$\alpha=\frac{\left[ H ^{+}\right]}{C}$
Substituting a value in equation $(2),$
$\frac{\left[ H ^{+}\right]}{ C }=\sqrt{\frac{ K _{ a }}{ C }}$
$\left[ H ^{+}\right]=\sqrt{\frac{ K _{ a }}{ C }} \cdot C$
$\left[ H ^{+}\right]=\sqrt{\frac{ K _{ a } \cdot C ^2}{ C }}$
$\left[ H ^{+}\right]=\sqrt{ K _{ a } \cdot C }$
For weak base
$\left[ OH ^{-}\right]=\sqrt{ K _{ b } \cdot C }$
Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,
$CH _3 COOH \rightleftharpoons CH _3 COO ^{-}+ H ^{+}$
The dissociation constant of acetic acid is,
$K _{ a }=\frac{\left[ H ^{+}\right]\left[ CH _3 COO ^{-}\right]}{\left[ CH _3 COOH \right]}$
| $CH_3COOH$ | $H^+$ | $CH_3COO^−$ | |
| Initial number of moles | $4$ | $-$ | $-$ |
| Degree of dissociation of $CH_3COOH$ | $\alpha$ | $-$ | $-$ |
| Number of moles at equilibrium | $1 − \alpha$ | $\alpha$ | $\alpha$ |
| Equilibrium concentration | $(1 − \alpha ) C$ | $\alpha C$ | $\alpha C$ |
$K_a=\frac{(\alpha C)(\alpha C)}{(1-\alpha) C}$
$K_a=\frac{\alpha^2 C^2}{(1-\alpha) C}$
$K_a=\frac{\alpha^2 C}{(1-\alpha)}..........(1)$
We know that weak acid dissociates only to a very small extent compared to one, a is so small.
equation $(1)$ becomes,
$K_a = \alpha ^2 C$
$\alpha^2=\frac{K_a}{C}$
$\alpha=\sqrt{\frac{K_a}{C}}.........(2)$
Similarly, for a weak base,
$K_b = \alpha ^2 C$
$\alpha=\sqrt{\frac{K_b}{C}}........(3)$
The concentration of $H$ can be calculated using the $K_a$ value as below,
$[H^+] = \alpha C$
$\alpha=\frac{\left[ H ^{+}\right]}{C}$
Substituting a value in equation $(2),$
$\frac{\left[ H ^{+}\right]}{ C }=\sqrt{\frac{ K _{ a }}{ C }}$
$\left[ H ^{+}\right]=\sqrt{\frac{ K _{ a }}{ C }} \cdot C$
$\left[ H ^{+}\right]=\sqrt{\frac{ K _{ a } \cdot C ^2}{ C }}$
$\left[ H ^{+}\right]=\sqrt{ K _{ a } \cdot C }$
For weak base
$\left[ OH ^{-}\right]=\sqrt{ K _{ b } \cdot C }$
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