The Ksp of Ag₂CrO₄ = 1.1 × 10⁻¹⁰, what is the value of solubility, S?

The Ksp of Ag₂CrO₄ = 1.1 × 10⁻¹⁰.

The dissolution of Ag₂CrO₄ is represented by the equation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)

Let S be the molar solubility of Ag₂CrO₄. Then:
[Ag⁺] = 2S
[CrO₄²⁻] = S

The Ksp expression is:
Ksp = [Ag⁺]²[CrO₄²⁻] = (2S)²(S) = 4S³

Substituting the given Ksp value:
1.1 × 10⁻¹⁰ = 4S³

Solving for S:
S³ = (1.1 × 10⁻¹⁰) / 4
S³ = 2.75 × 10⁻¹¹
S = ³√(2.75 × 10⁻¹¹)
S ≈ 3.02 × 10⁻⁴ M

Therefore, the value of solubility, S, is approximately 3.02 × 10⁻⁴ M.