Posted by Kayleigh on Tuesday, January 29, 2013 at 3:50pm.
Hopefully, Dr. Bob222 corrects this if this is wrong.
The maximum mass of calcium iodate that can dissolve in a 10L solution of calcium sulfide can not exceed the number of moles of calcium sulfide in the 10L solution. Therefore, solve for the number of moles in the solution.
3.25x10^-2 M=moles/L, solve for moles
moles=Molarity * Volume (L)=(10.0L)*(3.25x10^-2 M)= moles of calcium sulfide
moles of calcium sulfide = moles of calcium iodate
moles of calcium iodate * 389.88 g/mol =maximum mass of calcium iodate that can dissolve in a 10L solution of calcium sulfide
I would do it this way. This is a Ksp problem with a common ion, Ca^2+ from CaS.
Ca(IO3)2 ==> Ca^2+ + 2IO3^-
...s..........s........2s
Ksp = (Ca^2+)(I03^-)^2
Substitute Ksp --look it up in a text or the web.
(Ca^2+) = s from Ca(IO3)2 + 0.00325M from CaS
(IO3^-) = 2s.
Substitute Ca^2+ and IO3^- and solve for s = solubility Ca(IO3)2 in M = mols/L.
Convert to g/L. g = mols/L x molar mass Ca(IO3)2.
Then multiply by 10 to find grams in 10 L.
You can get some of this information from Devron's post.
I thought about doing it that way, but the Ksp wasn't provided in the initial post, so I thought the way I initially did it would work, but I still wasn't that sure.
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