posted by wendy on .
this is my work, would you please check my work, thank you!
a liter of a solution saturated at 225 degree C with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061 gm residue of CaC2O4. calculate the concentrations of the ions, and the molar solubility and solubility product constant for this salt at 25 degree C.
(o.oo61mg)(1/1000mg x 128g) = 4.8E-8M
ksp= (4.8E-8)(4.8E-8)= 2.3E-15
answer the above question, using a solution of 0.15M CaCl2 as the solvent instead of pure water.
ksp= 2.3E-15= (x)(0.15+x)
= 2.3E-15= 0.15x
x= 1.5E-14 mol/L CaC2O4
I think the problem states that the solubility is 0.0061 gram instead of milligrams. Therefore your factor of 1/1000 is not needed. Otherwise the problem looks ok.
Part B is ok, also, with the procedure, but you will need to change Ksp.