The solubility of bismuth iodide (BiI3) in water is 10.291 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi3+(aq) and I−(aq)

.........BiI3 ==> Bi^3+ + 3I^-

Equil..solid.10.291E-3..3*10.291E-3

Substitute into Ksp expression and solve.

6.8e-8

6.8e-8 is oncorrect!

To calculate the value of the solubility product (Ksp), we need to know the concentration of the dissolved ions in the solution. Since the solubility of bismuth iodide (BiI3) is given, we can use this information to determine the concentration of the dissolved ions.

BiI3 dissociates into Bi^3+ and 3 I^- ions in water, so the stoichiometry is 1:3. Therefore, the concentration of Bi^3+ in the solution is the same as the solubility, which is 10.291 x 10^-3 M.

To find the concentration of I^- ions, we multiply the concentration of Bi^3+ by the stoichiometric coefficient of I^- in the dissociation reaction, which is 3. Therefore, the concentration of I^- ions is 3 times the solubility: 3 x 10.291 x 10^-3 M = 30.873 x 10^-3 M.

Now we have the concentrations of the dissolved ions: [Bi^3+] = 10.291 x 10^-3 M and [I^-] = 30.873 x 10^-3 M.

The solubility product (Ksp) is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. Therefore, the Ksp can be calculated as follows:

Ksp = [Bi^3+][I^-]^3
= (10.291 x 10^-3 M)(30.873 x 10^-3 M)^3

Now, we can simplify this expression and calculate the numerical value of the solubility product.