Posted by **Kayleigh** on Tuesday, January 29, 2013 at 3:50pm.

Determine the maximum mass of calcium iodate that can be dissolved in a 10.0L jug of 3.25x10^-2 molar calcium sulphide solution.

- chemistry -
**Devron**, Tuesday, January 29, 2013 at 4:03pm
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

- chemistry -
**DrBob222**, Tuesday, January 29, 2013 at 4:57pm
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.

- chemistry -
**Devron**, Tuesday, January 29, 2013 at 5:01pm
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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