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)
4 answers
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?