Asked by kaken

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 Iāˆ’(aq)
Answered by DrBob222
I worked this above for Bi2S3. Same procedure.
Answered by Anonymous

27(3.691*10^-3)^4
Answered by rex
5.043*10^-7
Answered by kaken
can you explain me how to do it please?
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