Ca(IO3)2(s) ⇌ Ca^(2+)(aq) + 2IO3^(-)(aq)
First, you poured 30 mL of 0.100 M Ca^(2+) solution over a 2.000 g sample of solid Ca(IO3)2 and allowed the mixture to sit for 30 minutes. Some of the solid dissolved, causing the calcium ion concentration to increase by a certain quantity and the iodate concentration to increase by two times that certain quantity. Next, you isolate the filtrate through gravity filtration. What do you expect the iodate concentration to be in the filtered solution? How would the iodate concentration differ if only a 0.500 g sample of the solid was used? What if the solvent was water instead?
Chemistry - DrBob222, Sunday, February 7, 2010 at 12:13am
This is a solubility product problem with Ca^+2 as a common ion.
Ca(IO3)2 ==> Ca^+2 + 2IO3^-
Ksp = (Ca^+2)(IO3^-)^2
If we call the solubility of Ca(IO3)2 = X, then (Ca^+) = X in the equilibrated solution and (IO3^-) = 2X.
Now plug those values into Ksp.
(0.1 + X)(2X)^2 = Ksp.
Solve for X and multiply by 2 to find IO3^-.
Next, no difference if 0.5 g sample of Ca(IO3)2 used. Finally, your last question doesn't make any sense? The solvent IS water isn't it? instead of what? It dissolves in water.