A solubility product problem: BaSO4 has Ksp = 1.1×10^-10. If [Ba2+] = 1.0×10^-4 M, what [SO4^2-] would cause precipitation?

A solubility product problem: BaSO4 has Ksp = 1.1×10^-10. If [Ba2+] = 1.0×10^-4 M, what [SO4^2-] would cause precipitation?

• A. 1.0×10^-6 M
• B. 1.1×10^-6 M
• C. 1.1×10^-4 M
• D. 1.0×10^-4 M

Answer: (B)

The precipitation of a sparingly soluble salt starts when the ionic product [Ba2+][SO4^2-] exceeds the solubility product Ksp. At the point where precipitation just begins, the product equals Ksp. So the sulfate concentration that will cause precipitation, given a fixed barium concentration, is [SO4^2-] = Ksp / [Ba2+]. Plugging in the numbers: [SO4^2-] = (1.1 × 10^-10) / (1.0 × 10^-4) = 1.1 × 10^-6 M. At this concentration, the system is at the edge of precipitation; any higher sulfate concentration would push the product above Ksp and BaSO4 would start to precipitate. Therefore the required sulfate concentration is 1.1 × 10^-6 M.

The precipitation of a sparingly soluble salt starts when the ionic product [Ba2+][SO4^2-] exceeds the solubility product Ksp. At the point where precipitation just begins, the product equals Ksp. So the sulfate concentration that will cause precipitation, given a fixed barium concentration, is [SO4^2-] = Ksp / [Ba2+].

Plugging in the numbers: [SO4^2-] = (1.1 × 10^-10) / (1.0 × 10^-4) = 1.1 × 10^-6 M. At this concentration, the system is at the edge of precipitation; any higher sulfate concentration would push the product above Ksp and BaSO4 would start to precipitate. Therefore the required sulfate concentration is 1.1 × 10^-6 M.