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Answer the following : Derive the relationship between the degree of dissociation and dissociation constant in weak electrolytes. - Chemistry

Answer in Brief

Answer the following :

Derive the relationship between the degree of dissociation and dissociation constant in weak electrolytes.

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Solution 1

i. Consider 1 mol of weak base BOH dissolved in V dm3 of solution. The base dissociates partially as

\[\ce{BOH_{(aq)} ⇌ B^+_{ (aq)} + OH^-_{ (aq)}}\]

ii. The base dissociation constant is:

`"K"_"b" = (["B"^+]["OH"^-])/[["BOH"]]`  ....(1)

iii. Let the fraction dissociated at equilibrium is α and that remains undissociated is (1 − α).

\[\ce{BOH_{(aq)}⇌ B^+_{ (aq)} + OH^-_{ (aq)}}\]
Amount present at equilibrium (1 - α) α α
Concentration at equilibrium `(1 - α)/"V"` `α/"V"`

`α/"V"`

iv. at equilibrium,

[BOH] = `(1 - α)/"V"` mol dm-3,

[B+] = [OH-] = `α/"V"` mol dm-3

v. Substituting these concentrations in equation (1),

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

vi. If c is the initial concentration of base 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= `(α^2"c")/(1 - α)`  ...(3)

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

Kb = `α^2/"V"` and K= α2c  

α = `sqrt(("K"_"b")/"c")` or α = `sqrt("K"_"b". "V")`  ...(4)

The degree of dissociation of a weak base is inversely proportional to square root of its concentration and is directly proportional to square root of volume of the solution containing 1 mol of weak base.

Solution 2

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

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

ii. The acid dissociation constant is given as:

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

iii. 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 the 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"`

iv. Thus, at equilibrium,

[HA] = `(1 - α)/"V"` mol dm-3,

[H+] = [A-] = `α/"V"` mol dm-3

v. Substituting these in equation (1),

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

vi. If c is the initial concentration of 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

Ka = `(α^2"c")/(1 - α)`  ...(3)

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

Ka = `α^2/"V"` and Ka = α2c  ...(4)

α = `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 or directly proportional to the square root of volume of the solution containing 1 mol of the weak acid.

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APPEARS IN

Balbharati Chemistry 12th Standard HSC Maharashtra State Board
Chapter 3 Ionic Equilibria
Exercise | Q 4.05 | Page 62
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