The solubility product of Ag ₂ CrO ₄ is 4 10⁻¹² . What is the amount of Ag ₂ CrO ₄ in gram present in 100 ml its saturated solution? (molar mass of Ag ₂ CrO ₄ = 331.73 g/mol).

Step-by-step solution

Let the solubility of Ag ₂ CrO ₄ be s mol/L. The dissociation reaction is: Ag ₂ CrO ₄ 2 Ag ^+ + CrO ₄²⁻ The solubility product K_ sp is given by: K_ sp = [ Ag ^+]^2 [ CrO ₄²⁻] = (2s)^2(s) = 4s^3 Given K_ sp = 4 10⁻¹² , we have: 4s^3 = 4 10⁻¹² s^3 = 10⁻¹² s = 10⁻⁴ mol/L The number of moles of Ag ₂ CrO ₄ in 100 ml ( 0.1 L) of saturated solution is: Moles = s V = 10⁻⁴ 0.1 = 10⁻⁵ mol The amount of Ag ₂ CrO ₄ in grams is: Mass = Moles Molar mass Mass = 10⁻⁵ 331.73 = 3.3173 10⁻³ g Answer: 3.3173 10⁻³ g