Question

Read the information given below and answer the questions that follow.

Kohlrausch observed an interesting pattern between the values of molar conductance at infinite dilution (`Λ_m^∞`) for strong electrolytes. It was observed that different pairs of electrolytes having a common cation or anion had almost same difference of `Λ_m^∞`. On the basis of his observation, he postulated a law known as Kohlrausch’s Law of Independent Migration of ions. The values of molar conductivities at infinite dilution for some cations and anions are as follows: Ion `bb(λ_m^∞ (S  cm^2  mol^(−1)))` Ba2+ 127.2 Cl− 76.3 Ca2+ 119.0 \[\ce{SO^{2-}_4}\] 160.0

1. State the Kohlrausch’s Law of Independent Migration of ions.
2. Calculate the molar conductance at infinite dilution `Λ_m^∞` for BaCl₂.
3. Arrange the values of `Λ_m^∞`, for CaSO₄ and BaCl₂ in increasing order.

Answer 1

(i) Kohlrausch’s Law states that at infinite dilution, where dissociation of an electrolyte is complete, each ion makes a definite contribution to the total molar conductivity of the electrolyte, independent of the nature of the other ion with which it is associated. Mathematically, it is expressed as:

`Λ_m^∞ = v_+ λ_+^∞ + v_- λ_-^∞`

(ii) \[\ce{BaCl2 -> Ba^2+ + 2Cl^-}\]

`λ_m^∞(Ba^(2+)) = 127.2` S cm² mol⁻¹

`λ_m^∞(Cl^-) = 76.3` S cm² mol⁻¹

`Λ_m^∞(BaCl_2) = λ_m^∞(Ba^(2+)) + 2 xx λ_m^∞(Cl^-)`

= 127.2 + (2 × 76.3)

= 127.2 + 152.6

= 279.8 S cm² mol⁻¹

(iii) \[\ce{CaSO4 -> Ca^2+ + SO^2-_4}\]

`λ_m^∞(Ca^(2+)) = 119.0` S cm² mol⁻¹

`λ_m^∞(SO_4^(2-)) = 160.0` S cm² mol⁻¹

`Λ_m^∞(CaSO_4) = λ_m^∞(Ca^(2+)) + λ_m^∞(SO_4^(2-))`

= 119.0 + 160.0

= 279.0 S cm² mol⁻¹

For CaSO₄ and BaCl₂ in increasing order.

`Λ_m^∞(CaSO_4) < Λ_m^∞(BaCl_2)`

279.0 < 279.8

The molar conductance at infinite dilution for BaCl₂ is 279.8 S cm² mol⁻¹, which is slightly higher than the 279.0 S cm² mol⁻¹ calculated for CaSO₄.