Help with Common ion effect!!!!

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Shingo123

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1) How many moles of AgIO3 (Ksp=3.1 x 10^-8) will dissolve in one liter of a 10^-5M soln of NaIO3?

the answer is ([3.1x10^-8]^1/2) - (10^-5).

How'd they get that?


Am I on the right track with thinking, the added ions of IO3, causes a shift to the left of the equation. Thus it would decrease both the [Ag+] and [IO3] by 10^-5?

AgIO3 -> Ag+ + IO3


[IO3] = [Ag+]= x-10^-5


3.1x10^-8 = x2

X= [3.1x10^-8]^1/2

Thus the solubility of AgIO3 = Ag = IO3, and are all decreased by 10^-5.

Is this right?


I would really appreciate it if someone could go through this.

Thanks!
 
Last edited:
guess no one can explain this one.

im getting 1.6 x10^-4.

the molar solubility decreases because when you increase the [IO3], the reaction will be driven back to the left. So when that happens, more of the solid will form (AgIO3 is insoluble)
 
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