The solubility of bismuth iodide (BiI3) in water is 9.491 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi3+(aq) and I−(aq)
6.8e-8
The solubility of bismuth iodide (BiI3) in water is 11.091 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi3+(aq) and I−(aq)
To calculate the value of the solubility product (Ksp), you need to know the concentrations of the dissociated ions, Bi3+ and I-.
Given that the solubility of bismuth iodide (BiI3) in water is 9.491 x 10-3 M, we can assume that the concentration of both Bi3+ and I- is equal to this value, since BiI3 dissociates in a 1:3 ratio according to the balanced equation:
BiI3 (s) ⇌ Bi3+ (aq) + 3I- (aq)
Therefore, [Bi3+] = 9.491 x 10-3 M and [I-] = 3 × 9.491 x 10-3 M = 2.8473 x 10-2 M
Now, we can calculate the solubility product (Ksp) using the formula:
Ksp = [Bi3+][I-]^3
Ksp = (9.491 x 10-3 M) × (2.8473 x 10-2 M)^3
Ksp ≈ 6.156 x 10-8
Therefore, the value of the solubility product (Ksp) for bismuth iodide (BiI3) is approximately 6.156 x 10-8.