Final answer:
The solubility of Ag₂CrO₄ in grams per liter can be calculated using the solubility product constant (Ksp) and the stoichiometry of the compound. The Ksp value of Ag₂CrO₄ is 9.0×10-12. The solubility of Ag₂CrO₄ is approximately 0.311 grams per liter.
Step-by-step explanation:
The solubility of Ag₂CrO₄ in grams per liter can be calculated using the solubility product constant (Ksp) and the stoichiometry of the compound. The Ksp value of Ag₂CrO₄ is 9.0×10-12. In a saturated solution of Ag₂CrO₄, the concentration of silver ions ([Ag+]) and chromate ions ([CrO²¯]) will be equal.
Let's assume that x is the concentration of Ag+ and CrO²¯ ions in the saturated solution. Therefore, [Ag+] = [CrO²¯] = x. Since Ag₂CrO₄ dissociates into 2 Ag+ ions and 1 CrO²¯ ion, the Ksp expression becomes:
Ksp = [Ag+]²[CrO²¯] = (x)²(x) = x³
Substituting the Ksp value into the equation and solving for x, we get:
x³ = 9.0×10-12
x ≈ 9.4×10-4 M
The solubility of Ag₂CrO₄ in grams per liter can now be calculated by multiplying the molar solubility (9.4×10-4 M) by the molar mass of Ag₂CrO₄ (331.74 g/mol):
Solubility = (9.4×10-4 M) * (331.74 g/mol) = 0.311 g/L