I am a "virtual student'' in my school without a teacher and I need some urgent help with a question not covered in my textbook or other assignments.

 

The question is: what is the solubility in g/L of AgI in a 0.040 M solution of MgI2 Ksp for AgI is 8.3 x 10 (to the power of -17) I have no idea where to start on this one and any help is greatly appreciated

Solubility constants are a lot like acid constants. For a compound AaBb..., Ksp = [A]^a*^b*... for each species, where the smaller case letters are the subscripts of the species. You are given the solubility of AgI as 8.3 x 10^-17.

 

To begin with, you have a 0.04M solution of MgI2. In 1 liter, this would mean there is 0.040 mols MgI2 in the solution. This completely ionizes into 0.04mols Mg2+ and 0.080 moles of I-.

 

Now, take Ksp = 8.3 x 10^-17 = [Ag] (the coefficients are one because subscripts of Ag and I are both 1)

 

Make an ICE table

 

AgI --><-- Ag+ + I-

 

I | | 0 | 0.08

C | | X | X

E | | X | 0.08 + x

 

AgI is not included in the ice table because it is a solid

 

Let x denote the amount of moles of AgI that can be used. AgI completed ionizes into X moles Ag+ and X moles I-. HOWEVER, we already have 0.08 moles of I- from the MgI2. So we then have Ksp = [ x ]*[0.08+x]

 

Essentially, since x is extremely small compared to 0.08, you'll get Ksp is approximately 0.08x

 

Once you find x you should be able to convert that into grams. Then since you used one liter it would become g/L

 

If you need more help, research "the common ion effect"

Thanks so much for the help! It's great to have people out there willing to lend a helping hand, or keyboard LOL