**Answer the following :**

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

#### Solution 1

i. Consider 1 mol of weak base BOH dissolved in V dm^{3} 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_{b }= `(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 dm^{3} mol^{–1} then c = 1/V. Replacing 1/V in equation (2) by c,

we get

K_{b }= `(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:

K_{b} = `α^2/"V"` and K_{b }= α^{2}c

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 dm^{3} 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 |
`(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_{a }= `(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 dm^{3} 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 HA, α is very small, or (1 − α) ≅ 1. With this equation (2) and (3) becomes:

K_{a} = `α^2/"V"` and K_{a }= α^{2}c ...(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.