Question

Derive the equation which implies that the degree of dissociation of weak acid is inversely proportional to the square root of its concentration.

Answer 1

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

\[\ce{HA_{ (aq)} <=> H^+_{ (aq)} + A^-_{ (aq)}}\]

The acid dissociation constant is given as:

`"K"_"a" = (["H"^+]["A"^-])/["HA"]`    .....(1)

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

\[\ce{HA_{(aq)} <=> H^+_{ (aq)} + A^-_{ (aq)}}\]
Amount present at equilibrium (mol)	(1 - α)	α	α
Concentration at equilibrium (mol dm−3)	`(1 - α)/"V"`	`α/"V"`	`α/"V"`

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

[H⁺] = [A^–] = `α/"V"` mol dm⁻³

Substituting these in equation (1),

`"K"_"a" = ((alpha//"V")(alpha//"V"))/((1 - alpha)//"V") = alpha^2/((1 - alpha)"V")`    ....(2)

If c is the initial concentration of an acid 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" = (alpha^2"c")/(1 - alpha)`    ....(3)

For the weak acid HA, α is very small, or (1 − α) ≅ 1.

With this equation (2) and (3) becomes:

K_a = α²/V and k_a = α²c     ....(4)

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

The equation (5) implies that the degree of dissociation of a weak acid is inversely proportional to the square root of its concentration.