Final answer:
To calculate the solubility of Ag2CrO4 in a 0.028 M AgNO3 solution, use the solubility product constant (Ksp) formula [Ag+]^2[CrO4^2-] = Ksp. Solubility of Ag2CrO4 in the solution is approximately x = 1.68 x 10^-5 M.
Step-by-step explanation:
To calculate the solubility of Ag2CrO4 in a 0.028 M AgNO3 solution, we can use the solubility product constant (Ksp). The Ksp for Ag2CrO4 is given as 9.0 x 10^-12. According to the Ksp expression, [Ag+]^2[CrO4^2-] = Ksp. Since Ag2CrO4 dissolves into 2 Ag+ ions and 1 CrO4^2- ion, the concentration of Ag+ ions will be twice the concentration of Ag2CrO4.
Letting x be the solubility of Ag2CrO4, we have:
- Ag2CrO4 ⇌ 2 Ag+ + CrO4^2-
- [Ag+] = 2x
- Substituting these values into the Ksp expression, we get: (2x)^2(x) = Ksp
- Solving for x, we find the solubility of Ag2CrO4 to be approximately x = 1.68 x 10^-5 M