Concept understanding — Ionisation of Weak Acids and Ostwald's Dilution Law
A weak acid HA is only partially dissociated: HA+H2O⇌H3O++A−, with dissociation constant Ka=[HA][H+][A−] (writing H3O+ as H+ for convenience); the analogous weak-base constant is Kb=[BOH][B+][OH−].
Ostwald's dilution law relates Ka to the degree of dissociation α (the fraction of moles that dissociate at equilibrium) and the concentration C. Working through acetic acid, CH3COOH⇌H++CH3COO−, with equilibrium concentrations (1−α)C, αC, αC: Ka=(1−α)C(αC)2=1−αα2C. Since a weak acid dissociates only slightly, α≪1 so (1−α)≈1, giving the simplified law Ka≈α2C, i.e. α=Ka/C -- as dilution increases (C falls), α rises, which is Ostwald's dilution law in words: dilution increases the degree of dissociation of a …
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CBSE 2023Set ANNUAL5 marks
Q.(a) Derive an expression for Ostwald's Dilution Law.
OR
(b)
(i) Why aniline does not undergo Friedel Craft's reaction ?
(ii) How nylon-2-nylon-6 is prepared ?
›Reveal solutionSolution
(a) Ostwald's dilution law derives Ka≈Cα2 for a weak electrolyte, showing α∝1/C (dissociation rises on dilution). (b) Aniline's lone pair is tied up by AlCl3, killing Friedel-Crafts reactivity; nylon-2-nylon-6 is a glycine/6-aminohexanoic-acid copolymer.
(a) Derivation of Ostwald's Dilution Law: Consider a weak, binary electrolyte AB ionising in solution: AB⇌A++B− Let the initial concentration be C (mol/L) and the degree of dissociation be α. At equilibrium: [AB]=C(1−α), [A+]=Cα, [B−]=Cα. The dissociation (ionisation) constant is: Ka=[AB][A+][B−]=C(1−α)(Cα)(Cα)=1−αCα2 For a weak electrolyte, α≪1, so (1−α)≈1, giving the approximate form: Ka≈Cα2⇒α≈CKa=KaV (using C=1/V, where V is the volume containing 1 mole). This is Ostwald's dilution law: since Ka is constant at a given temperature, α increases as C decreases (i.e. on dilution) — the degree of dissociation of a weak electrolyte increases with dilution.
OR (b)(i) Why aniline does not undergo Friedel-Crafts reaction: Friedel-Crafts alkylation/acylation requires a Lewis acid catalyst, typically anhydrous AlCl3. Aniline's -NH2 group carries a lone pair of electrons on nitrogen, which is a good Lewis base; this lone pair immediately reacts with (donates to) the Lewis acid catalyst AlCl3 to form a salt-like complex, C6H5−N+H2−AlCl3−. Once complexed this way, the nitrogen becomes positively charged, strongly withdrawing electron density from the ring by induction, and the aromatic ring becomes deactivated and effectively meta-directing — the exact opposite of what's needed for electrophilic Friedel-Crafts substitution. Because the required Lewis acid catalyst is consumed/deactivated by the amine itself (rather than activating an electrophile), the Friedel-Crafts reaction fails for aniline.