Solution:

To find the solubility product of BaSO4, we can use the relationship between conductivity and solubility product. The solubility product (Ksp) is a measure of the extent to which a compound dissociates in a solution.

Given:
Conductivity of the saturated solution of BaSO4 (σ) = 3.06 × 10^-6 ohm^-1cm^-1
Equivalent conductance (Λ) = 1.53 ohm^-1cm^2 eq^-1

1. Finding the molar conductivity (Λm):
The molar conductivity is the equivalent conductance divided by the concentration of the solution. In this case, we have a saturated solution, so the concentration of BaSO4 is equal to its solubility (S).

Λm = Λ / S

Since the concentration of BaSO4 is equal to its solubility, we can rewrite the equation as:

Λm = Λ / [BaSO4]

2. Finding the concentration of BaSO4:
The concentration of BaSO4 can be determined by finding the reciprocal of the molar conductivity:

[BaSO4] = 1 / Λm

3. Finding the solubility product (Ksp):
The solubility product (Ksp) is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. In the case of BaSO4, it dissociates into one Ba2+ ion and one SO42- ion:

BaSO4 ⇌ Ba2+ + SO42-

The solubility product expression can be written as:

Ksp = [Ba2+] * [SO42-]

Since BaSO4 dissociates into one Ba2+ ion and one SO42- ion, their concentrations are equal to the concentration of BaSO4:

Ksp = [BaSO4] * [BaSO4]

Substituting the concentration of BaSO4 from step 2, we have:

Ksp = ([BaSO4])^2

4. Calculating the solubility product:
Now, substituting the concentration of BaSO4 from step 2 into the solubility product expression, we get:

Ksp = (1 / Λm)^2

5. Substitute the given values:
Ksp = (1 / Λm)^2
= (1 / 1.53 ohm^-1cm^2 eq^-1)^2

Calculating the value, we find:

Ksp ≈ 0.446 eq^-2 cm^4 ohm^-2

Therefore, the solubility product of BaSO4 is approximately 0.446 eq^-2 cm^4 ohm^-2.