Question

Derive an expression for Ostwald’s dilution law.

Answer 1

Ostwald’s dilution law: It relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).

Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,

$$\mathrm{CH}{\phantom{A}}_3\mathrm{COOH}\stackrel{-\rightharpoonup }{\leftharpoondown -}\mathrm{CH}{\phantom{A}}_3\mathrm{COO}{\phantom{A}}^{-}+\mathrm{H}{\phantom{A}}^{+}$$

The dissociation constant of acetic acid is,

K_a = ${\displaystyle \frac{\left[{\text{H}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}}$

	CH3COOH	H+	CH3COO−
Initial number of moles	1	-	-
Degree of dissociation of CH3COOH	α	-	-
Number of moles at equilibrium	1 − α	α	α
Equilibrium concentration	(1 − α) C	α C	α C

Substituting the equilibrium concentration in the equation

K_a = ${\displaystyle \frac{(\alpha \text{C})(\alpha \text{C})}{(1-\alpha ) \text{C}}}$

K_a = ${\displaystyle \frac{{\alpha }^2 {\text{C}}^2}{(1-\alpha ) \text{C}}}$

K_a = ${\displaystyle \frac{{\alpha }^2 \text{C}}{(1-\alpha )}}$ ........................(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 = α² C

α² = ${\displaystyle \frac{{\text{K}}_{\text{a}}}{\text{C}}}$

α = ${\displaystyle \sqrt{\frac{{\text{K}}_{\text{a}}}{\text{C}}}}$ ...............(2)

Similarly, for a weak base,

K_b = α² C

α = ${\displaystyle \sqrt{\frac{{\text{K}}_{\text{b}}}{\text{C}}}}$ ...............(3)

The concentration of H can be calculated using the K_a value as below,

[H⁺] = α C

α = ${\displaystyle \frac{\left[{\text{H}}^{+}\right]}{\text{C}}}$

Substituting a value in equation (2),

${\displaystyle \frac{\left[{\text{H}}^{+}\right]}{\text{C}}=\sqrt{\frac{{\text{K}}_{\text{a}}}{\text{C}}}}$

[H⁺] = ${\displaystyle \sqrt{\frac{{\text{K}}_{\text{a}}}{\text{C}}}.\text{C}}$

[H⁺] = ${\displaystyle \sqrt{\frac{{\text{K}}_{\text{a}}.{\text{C}}^2}{\text{C}}}}$

[H⁺] = ${\displaystyle \sqrt{{\text{K}}_{\text{a}}.\text{C}}}$

For weak base

[OH⁻] = ${\displaystyle \sqrt{{\text{K}}_{\text{b}}.\text{C}}}$