The solubility of bismuth iodide in water is 4.291 x 10-3 M. Calculate the value of the solubility product. The dissolution products are and

9.153724183*10^-9

how do we solve this?

To calculate the solubility product (Ksp) of a compound, you need to know the concentrations of the dissolved ions. In this case, the compound is bismuth iodide (BiI3) and it dissociates into bismuth ions (Bi3+) and iodide ions (I-).

Given that the solubility of bismuth iodide in water is 4.291 x 10-3 M, it means that the concentration of both Bi3+ and I- ions is equal to this value.

So, [Bi3+] = 4.291 x 10-3 M and [I-] = 4.291 x 10-3 M

The solubility product expression for bismuth iodide can be written as:

Ksp = [Bi3+][I-]^3

Substituting the concentrations, we get:

Ksp = (4.291 x 10-3)^1 * (4.291 x 10-3)^3

Ksp = 4.291 x 10-3 * (4.291 x 10-3)^3

Ksp = 4.291 x 10-3 * 4.291^3 x 10-3^3

Ksp ≈ 1.065 x 10-11

Therefore, the value of the solubility product (Ksp) for bismuth iodide is approximately 1.065 x 10-11.