Question

At 298 K, the limiting molar conductivity of a weak mono basic acid is 4 × 10² S cm² mol⁻¹. At 298 K, for an aqueous solution of the acid, the degree of dissociation is ‘α’ and the molar conductivity is y × 10² S cm² mol⁻¹. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 10² S cm² mol⁻¹.

1. The value of ‘α’ is ______.
2. The value of ‘y’ is ______.

Answer 1

1. The value of ‘α’ is 0.216.
2. The value of ‘y’ is 0.863.

Explanation:

Given: Limiting molar conductivity at 298 K,

`Lambda_m^circ` = 4 × 10² S cm² mol⁻¹

Initial molar conductivity of solution,
Λ_m = y × 10² S cm² mol⁻¹

Degree of dissociation (α) = `Lambda_m/Lambda_m^circ` 

= `(y xx 10^2)/(4 xx 10^2)`

= `y/4`

After 20 times dilution, molar conductivity becomes:

Λ'_m = 3y × 10² S cm² mol⁻¹

Now degree of dissociation:

α' = `(Lambda'_m)/(Lambda_m^circ)`

= `(3y xx 10^2)/(4 xx 10^2)`

= `(3y)/4`

By using ostwald’s dilution law for weak acid:

`K_alpha = (c alpha^2)/(1 - alpha)`

Let original concentration be ccc, then after 20 × dilution it becomes `c/20`.

So,

`K_alpha = (c alpha^2)/(1 - alpha)`

= `((c/20) alpha^2)/(1 - alpha')`

Now substitute α = `y/4` and

α′ = `(3y)/4`

`(c (y^2/16))/(1 - y/4) = ((c/20) * (9 y^2)/16)/(1 - (3y)/4)`

Cancel `(cy^2)/16` from both sides.

`1/(1 - y/4) = (9/20)/(1 - (3y)/4)`

Now simplify:

LHS: `1/(1 - y/4)`

= `1/((4 - y)/4)`

= `4/(4 - y)`

RHS: `(9/20)/(1 - (3y/4)`

= `9/20 xx 1/((4 - 3y)/4)`

= `9/20 xx 4/(4 - 3y)`

= `36/(20(4 - 3y))`

So,

`4/(4 - y) = 36/(20(4 - 3y))`

⇒ 4 × 20(4 − 3y) = 36(4 − y)

⇒ 80(4 − 3y) = 36(4 − y)

⇒ 320 − 240y = 144 − 36y

⇒ 320 − 144 = 240y − 36y

⇒ 176 = 204y

⇒ y = `176/204`

⇒ y = `44/51`

⇒ y = 0.8627

Now `alpha = y/4`

= `0.8627/4`

α = 0.216