(a) The solubility of bismuth iodide in water is 9.091 x 10-3 M. Calculate the value of the solubility product. The dissolution products are and
solubility = (x)*((x*3))^3
x=9.091*10-3
solubility = 1.84*10^-7
u r the best!
5.043*10^-7
To calculate the value of the solubility product (Ksp) for bismuth iodide in water, we need to know the equation for the dissolution reaction and the stoichiometry of the reaction.
The equation for the dissolution of bismuth iodide (BiI3) in water can be represented as:
BiI3(s) ⇌ Bi3+(aq) + 3I-(aq)
From the balanced equation, we can see that one molecule of bismuth iodide dissociates into one bismuth ion (Bi3+) and three iodide ions (I-) in the solution.
Given that the solubility of bismuth iodide in water is 9.091 x 10^(-3) M, we can deduce that the concentration of the bismuth ion (Bi3+) in the solution is also 9.091 x 10^(-3) M.
However, the concentration of iodide ions (I-) is three times that of the bismuth ion concentration since three iodide ions are produced for each molecule of bismuth iodide that dissolves.
Therefore, the concentration of iodide ions (I-) is 3 * (9.091 x 10^(-3) M) = 2.727 x 10^(-2) M.
Now, we can calculate the solubility product (Ksp) by multiplying the concentration of the bismuth ion and the concentration of the iodide ions raised to the power of their stoichiometric coefficients.
Ksp = [Bi3+][I-]^3 = (9.091 x 10^(-3) M) * (2.727 x 10^(-2) M)^3
Calculating this expression will yield the value of the solubility product (Ksp) for bismuth iodide.