A) A liter of solution saturated at 25degrees C. with Calcium Oxalate, CaC 2 O 4

Question

A) A liter of solution saturated at 25degrees C. with Calcium Oxalate, CaC2O4, Isevaporated to dryness, giving a .0061 gm residue ofCaC2O4. Calculate the conentrations of theIons, and the molar solubilityand the solubility product constant forthis salt at 25 degressC. Show Balanced Reaction. B) Answer the above question using asolution of .15M CaCl2 as the solvent instead ofpure water. This is what I did: CaC2O4 yeilds Ca+2 +C2O4-2 .0061/128.1=4.7x10-9 KSP= (4.7x10-5)2=2.27x10-9 Am I right or am I missing the Conc. of ions, molarsolubility, and solubility product constant? A) A liter of solution saturated at 25degrees C. with Calcium Oxalate, CaC2O4, Isevaporated to dryness, giving a .0061 gm residue ofCaC2O4. Calculate the conentrations of theIons, and the molar solubilityand the solubility product constant forthis salt at 25 degressC. Show Balanced Reaction. B) Answer the above question using asolution of .15M CaCl2 as the solvent instead ofpure water. B) Answer the above question using asolution of .15M CaCl2 as the solvent instead ofpure water. This is what I did: CaC2O4 yeilds Ca+2 +C2O4-2 .0061/128.1=4.7x10-9 KSP= (4.7x10-5)2=2.27x10-9 Am I right or am I missing the Conc. of ions, molarsolubility, and solubility product constant? Am I right or am I missing the Conc. of ions, molarsolubility, and solubility product constant?

Explanation / Answer

0.0061 grams *(1 mol/128.1 grams) = 4.76 e-5 M, the molarsolubility [Ca2+] = 4.76e-5 [C2O42-] = 4.76e-5 M Ksp = [Ca2+][C2O42-] = (4.76e-5)*(4.76e-5) = 2.27e-9 Side note: CaC2O4 actually has a solubility of 0.00067 g/L. You canconvert then calculate that the Ksp = ((0.00067 g/0.1 L)*(1/128))^2 = 2.74e-9, so you are close B) So there is an additional 0.15 M of Ca2+ [Ca2+] = 0.1500476 [C2O42-] = 4.76e-5 M Ksp = (0.1500476)*(4.76e-5) = 7.14 e-6

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