Final answer:
The molar solubility of silver chromate in water is determined using its solubility product constant (Ksp). By setting up an equilibrium expression and solving for S, the molar solubility is found to be approximately 1.14 × 10-4 M.
Step-by-step explanation:
The molar solubility of silver chromate (Ag2CrO4) in water can be determined from its solubility product constant (Ksp). Given the solubility product Ksp of 2.76 × 10-12 for silver chromate, we can set up an equilibrium expression for its dissociation in water:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq)
Let the molar solubility of Ag2CrO4 be S moles per liter. In a saturated solution, the concentration of Ag+ will be 2S and that of CrO42- will be S. The solubility product expression is:
Ksp = [Ag+]2[CrO42-] = (2S)2×(S) = 4S3
Substituting the value of Ksp into the equation gives:
2.76 × 10-12 = 4S3
Solving for S, we find that the molar solubility of Ag2CrO4 is approximately 1.14 × 10-4 M.