Q.i. Define degree of dissociation. Derive Ostwald's dilution law for the $\mathrm{CH_3COOH}$.

Step 1. Definition. Degree of dissociation, $\alpha$, is the fraction of the total moles of an electrolyte that has dissociated into its ions at equilibrium: $\alpha=\dfrac{\text{moles dissociated}}{\text{total moles}}$ (Eq. 3.1); percent dissociation $=\alpha\times100$ (Eq. 3.2).

Step 2. Set up the equilibrium. For CH3COOH, $CH_3COOH(aq)\rightleftharpoons CH_3COO^-(aq)+H^+(aq)$. Start with 1 mol of CH3COOH dissolved in V dm3 of solution. At equilibrium, a fraction $\alpha$ has dissociated, so the amounts present are: $(1-\alpha)$ mol CH3COOH, $\alpha$ mol CH3COO-, $\alpha$ mol H+.

Step 3. Convert to concentrations. Dividing each amount by the volume V: $[CH_3COOH]=\dfrac{1-\alpha}{V}$, $[CH_3COO^-]=[H^+]=\dfrac{\alpha}{V}$ mol dm-3.

Step 4. Substitute into Ka. $K_a=\dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]}=\dfrac{(\alpha/V)(\alpha/V)}{(1-\alpha)/V}=\dfrac{\alpha^2}{(1-\alpha)V}$ (Eq. 3.5).

Step 5. Introduce concentration c. With $c=1/V$ (the initial molar concentration), this becomes $K_a=\dfrac{\alpha^2c}{1-\alpha}$ (Eq. 3.6) -- the exact form of Ostwald's dilution law for CH3COOH.

Step 6. Simplify for a weak acid. Since CH3COOH is weak, $\alpha$ is small, so $(1-\alpha)\approx1$, and $K_a\approx\alpha^2c$ (Eq. 3.7), giving $\alpha\approx\sqrt{K_a/c}$ (Eq. 3.8) -- showing $\alpha$ is inversely proportional to $\sqrt{c}$ (or directly proportional to $\sqrt{V}$).

Define degree of dissociation, then re-derive Ostwald's law step by step for CH3COOH specifically, following the general derivation of section 3.4.2.

• Skipping the ICE-style equilibrium-amount setup and jumping straight to the final formula, losing the derivation the question asks for.
• Forgetting to state the small-alpha approximation explicitly as a SEPARATE, simplified step from the exact Eq. 3.6 result.