The solubility of barium chromate, bacro4, is 2.81 × 10−3 g/l. calculate the solubility product of this compound.

Solubility product constants are values to describe the saturation of ionic compounds with low solubility. A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound. It is calculated from the product of the ion concentration in the solution. For barium chromate, the dissociation would  be as follows:

BaCrO4 = Ba^2+ + (CrO4)^2-

So, the expression for the solubility product would be:

Ksp = [Ba^2+] [(CrO4)^2-]

we let x = [BaCrO4] = [Ba2+] = [(CrO4)2-] = 2.81x10^-3 g/L ( 1 mol / 253.35 g ) = 1.11x10^-5

Ksp = x(x)
Ksp= x^2
Ksp = (1.11x10^-5)^2
Ksp = 1.23x10^-10

The Ksp of Barium chromate at that same temperature for the solubility would be 1.23x10^-10.

The answer is letter c