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)

Answer 1

.........BiI3 ==> Bi^3+ + 3I^-
Equil..solid.10.291E-3..3*10.291E-3

Substitute into Ksp expression and solve.

Answer 2

6.8e-8

Answer 3

6.8e-8 is oncorrect!