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posted by  displaced_ca_girl on 4/4/2008 10:06:38 AM  |  status: Closed  

Solubility

Course Textbook Chapter Problem
General Chemistry Principles of General Chemistry, Silberberg N/A N/A
Question Details:
What is the solubility of Ca(IO3)2 (Ksp = 7.1 x 10-7) in pure water?
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posted by Mr. Bean on 4/4/2008 10:43:46 AM  |  status: Live
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displaced_ca_girl's comment:
"Brilliant! Thank you! I know exactly what I was doing wrong now. "
Response Details:
Hello!

well remember Ksp is a relative measure of solubility and can only be used to compare solubililties of the same ratios as if you wanted to compare

CuS with NaCl  since they both dissociate with a 1:1 ratio you can compare there solubilities with ksp's

but solubility is always x !     Ksp is only the relative measure of solubility

so we have a salt Ca(IO3)2<==>Ca+ 2(IO3)     as you can see this is a 1:2 Ionization ratio
                                                           
                the ksp= [Ca]*[IO3]^2

                                 ksp= (x)*(2*x)^2

                                so we get ksp= 4x^3       so now we can solve for solubility which is  x
7.1*10^-7=4x^3

so x= ((7.1*10^-7)/(4))^(1/3) this gives us solubility= 5.6*10^-3M
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