Statement-1 : When K₂CrO₄ is added in a solution containing Ba₂+,Ca₂+ ions and dilute acid, only BaCrO₄ is precipitated 
Statement-2 : The Ksp of CaCrO₄ is higher than BaCrO₄

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Detailed Explanation:

We are given the following information:

• Ksp of PbBr₂ = 8 × 10^-5
• The salt is 80% dissociated.

First, let's consider the dissociation of PbBr₂:

PbBr₂(s) ⇌ Pb²⁺(aq) + 2Br^-(aq)

Let the solubility of PbBr₂ be 'S'. The dissociation of PbBr₂ will give: - The concentration of Pb²⁺ ions will be 0.80S (since 80% dissociates). - The concentration of Br^- ions will be 2 × 0.80S = 1.6S (as two Br^- ions are produced for each PbBr₂ unit dissociated). The solubility product (Ksp) is given by:

Ksp = [Pb²⁺] × [Br^-]²

Substituting the values for the concentrations:

Ksp = (0.80S) × (1.6S)²

We know Ksp = 8 × 10^-5, so we can solve for 'S':

8 × 10^-5 = (0.80S) × (1.6S)²

After solving the equation for 'S', we get the solubility of PbBr₂. The solubility is given by:

S = (10^-4) / (1.6 × 1.6)^1/3

Thus, the solubility of PbBr₂ is calculated using the given values of Ksp and the degree of dissociation.

Concept Behind the Solution:

This question involves the concept of solubility product (Ksp) and the degree of dissociation. The solubility product (Ksp) helps in calculating the solubility of salts in solution. The degree of dissociation tells us how much of the salt dissociates in water, which directly affects the concentrations of the ions involved in the equilibrium.

By applying the Ksp expression, we can solve for the solubility by substituting the concentrations of ions that are governed by the degree of dissociation.