Answer 1

i.

Consider an equilibrium of weak acid CH₃COOH that exists in solution partly as the undissociated species CH₃COOH and partly H⁺ and CH₃COO^– ions. Then

\[\ce{CH3COOH_{(aq)} ⇌ H^+_{ (aq)} + CH3COO^-_{ (aq)}}\]

ii.

The acid dissociation constant is given as:

`"K"_"a" = (["H"^+]["CH"_3"COO"^-])/[["CH"_3"COOH"]]`  ....(1)

iii.

Suppose 1 mol of acid CH₃COOH is initially present in volume V dm³ of the solution. At equilibrium, the fraction dissociated would be α, where α is the degree of dissociation of the acid. The fraction of an acid that remains undissociated would be (1 - α).

\[\ce{CH3COOH_{(aq)}⇌ H^+_{ (aq)} + CH3COO^-_{ (aq)}}\]
The amount present at equilibrium (mol)	(1 - α)	α	α
Concentration at equilibrium (mol dm−3)	`(1 - α)/"V"`	`α/"V"`	`α/"V"`

iv.

Thus, at equilibrium [CH₃COOH] = `(1 - α)/"V"` mol dm^-3,

[H⁺] = [CH₃COO^-] = `α/"V"` mol dm^-3

v.

Substituting these in equation (1),

K_a = `(α / "V" α / "V") /((1 - α) / "V") = α^2/(1 - α  "V")`  ...(2)

vi.

If c is the initial concentration of CH₃COOH in mol dm^–3 and V is the volume in dm³ mol^–1 then c = 1/V. Replacing 1/V in equation (2) by c,

we get

K_a = `(α^2"c")/(1 - α)`  ...(3)

vii.

For the weak acid CH₃COOH, α is very small, or (1 − α) ≅ 1. With this equation (2) and (3) becomes:

K_a = `α^2/"V"` and K_a = α²c  ...(4)

α = `sqrt("K"_"a")/"c"` or α = `sqrt("K"_"a". "V")`  ...(5)

Equation (5) implies that the degree of dissociation of a weak acid (CH₃COOH) is inversely proportional to the square root of its concentration or directly proportional to the square root of the volume of the solution containing 1 mol of the weak acid.