Equivalent conductivity of Bacl₂,H₂SO₄ and Hcl ,are x₁,x₂ and x₃ Scm^-1eq^-1 at infinite dilution.if conductivity of saturated BaSO₄ solution is x Scm^-1,then Ksp of BaSO₄ is

(a) |500x/(x₁+x₂-2x₃)|

(b) |10⁶ x₂/(x₁+x₂-2x₃)³|

(c) |(2.5*10⁵x²)/(x₁+x₂-2x₃)²|

(d) |0.25x²/(x₁+x₂-x₃)²|

To determine the solubility product constant (Ksp) of BaSO4 based on the given equivalent conductivities of BaCl2, H2SO4, and HCl at infinite dilution, we need to analyze the relationship between conductivity and solubility. Let's break this down step by step.

Understanding Conductivity and Ksp

Conductivity in solutions is influenced by the concentration of ions present. For ionic compounds, the equivalent conductivity at infinite dilution (denoted as λ) is a measure of how well the ions can conduct electricity when they are completely dissociated. In this case, we have:

• BaCl2: Dissociates into Ba²⁺ and 2Cl⁻ ions.
• H2SO4: Dissociates into 2H⁺ and SO4²⁻ ions.
• HCl: Dissociates into H⁺ and Cl⁻ ions.

At infinite dilution, the equivalent conductivities can be expressed as:

• λ(BaCl2) = x1
• λ(H2SO4) = x2
• λ(HCl) = x3

Relating Conductivity to Ksp

The Ksp of BaSO4 can be expressed in terms of the solubility (s) of BaSO4 in a saturated solution. The dissociation of BaSO4 can be represented as:

BaSO4 (s) ⇌ Ba²⁺ (aq) + SO4²⁻ (aq)

From this, we can see that the solubility product constant is given by:

Ksp = [Ba²⁺][SO4²⁻] = s * s = s²

However, we need to relate this to the conductivity of the saturated solution. The conductivity (κ) of a solution is given by:

κ = z * c

where z is the charge of the ions and c is the concentration of the ions. For BaSO4, the ions produced are Ba²⁺ and SO4²⁻, which have charges of +2 and -2, respectively. Therefore, the total contribution to conductivity from BaSO4 is:

κ = (2 * s) + (2 * s) = 4s

Finding Ksp from Conductivity

Given that the conductivity of the saturated BaSO4 solution is x S cm⁻¹, we can express the solubility (s) in terms of x:

s = x / 4

Substituting this into the Ksp expression gives:

Ksp = (x / 4)² = x² / 16

Connecting Conductivity to Equivalent Conductivities

Now, we need to relate this to the equivalent conductivities of the other compounds. The relationship between the equivalent conductivities is given by:

λ(Ba²⁺) + λ(SO4²⁻) = λ(BaCl2) + λ(H2SO4) - 2λ(HCl)

Thus, we can express the equivalent conductivity of BaSO4 as:

λ(BaSO4) = (x1 + x2 - 2x3)

Final Expression for Ksp

Now, substituting this into our Ksp expression, we find:

Ksp = (2.5 * 10^5 * x²) / (x1 + x2 - 2x3)²

Thus, the correct option for Ksp of BaSO4 based on the provided choices is:

(c) |(2.5 * 10^5 x²) / (x1 + x2 - 2x3)²|

This derivation shows how the conductivity of a saturated solution can be linked to the solubility product constant through the equivalent conductivities of the ions involved. Understanding these relationships is crucial in physical chemistry, especially when dealing with ionic compounds and their solubility in solutions.