The solubility of bismuth iodide (BiI3) in water is 3.691 x 10-3 M. Calculate the value of the solubility product. The dissolution products are Bi3+(aq) and I−(aq)
I worked this above for Bi2S3. Same procedure.
27(3.691*10^-3)^4
5.043*10^-7
can you explain me how to do it please?
To calculate the value of the solubility product (Ksp) for bismuth iodide (BiI3), we'll need to use the balanced equation for the dissociation of the compound in water.
The balanced equation for the dissolution of BiI3 in water is:
BiI3(s) ⇌ Bi3+(aq) + 3I−(aq)
The given information states that the solubility of BiI3 in water is 3.691 x 10-3 M. This means that at equilibrium, the concentration of Bi3+ ions and I- ions in the solution is 3.691 x 10-3 M.
Now, let's assume that x is the molar solubility of BiI3. Since BiI3 dissociates completely into Bi3+ and I- ions, the concentration of Bi3+ ions will be x M, and the concentration of I- ions will be 3x M (as per the balanced equation).
Therefore, we can write the expression for the solubility product (Ksp) as follows:
Ksp = [Bi3+] * [I-]^3
Substituting the known values:
Ksp = (x) * (3x)^3
Ksp = 27x^4
Now, we need to determine the value of x. From the given information, we know that the solubility of BiI3 is 3.691 x 10-3 M. Since the concentration of Bi3+ ions is equal to x, we can write:
x = 3.691 x 10-3 M
Now, we can substitute the value of x into the expression for Ksp:
Ksp = 27 * (3.691 x 10-3)^4
Calculating this expression will give you the value of the solubility product (Ksp) for bismuth iodide (BiI3) in water.