# chemistry

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a) A liter of a solution saturated at 25degree with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061gm residue of CaC2O4. calculate the concentrations of the ions, and molar solubility and the solubility product constant for this salt at 25 degree C.

b)answer the above question, using a solution of 0.15M CaCl2 as the solvent instead of pure water.

So you have 0.0061 g CaC2O4 per liter of solution.
How many mols is that?
Isn't that also the molarity since it is in 1 L?
Now write the solubility equation.
CaC2O4 ==> Ca^+2 + C2O4^=
Ksp = (Ca^+)(C2O4^=) .
Plug in the values for the molarity and calculate Ksp. Post you work if you get stuck.
For part b, same thing EXCEPT you have more Ca^+ than in the saturated solution of water alone.

• chemistry -

I know you're in Alpert's class

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