) The solubility of bismuth iodide (BiI3) in water is 5.891 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi3+(aq) and I−(aq)
its still little bit confuse, could you explain the deatail?
5.043*10^-7
To calculate the value of the solubility product (Ksp) for bismuth iodide (BiI3), we need to use the balanced chemical equation for its dissolution:
BiI3(s) ⇌ Bi3+(aq) + 3I^-(aq)
The solubility of BiI3 in water is given as 5.891 x 10^-3 M, which represents the concentration of Bi3+ ions in the equilibrium mixture.
Since BiI3 dissociates into one Bi3+ ion and three I^- ions, the concentration of I^- in the equilibrium mixture would be 3 times the concentration of Bi3+. Therefore, the concentration of I^- would be 3 * 5.891 x 10^-3 M = 1.7673 x 10^-2 M.
Now we can write the expression for the solubility product (Ksp):
Ksp = [Bi3+][I^-]^3
Substituting the calculated concentrations:
Ksp = (5.891 x 10^-3 M)(1.7673 x 10^-2 M)^3
Calculating the value of Ksp:
Ksp = 5.891 x 10^-3 M * 1.7673 x 10^-2 M * 1.7673 x 10^-2 M * 1.7673 x 10^-2 M
Ksp = 8.917 x 10^-10 (rounded to three significant figures)
Therefore, the value of the solubility product (Ksp) for bismuth iodide is 8.917 x 10^-10.