A 1.00-L solution saturated at 25°C with calcium oxalate (CaC2O4) contains 0.0061 g of CaC2O4. The solubility constant (Ksp) for this salt is [CaC2O4=128.09 g/mol]

Answer 1

0.0061g/128.09 = 4.76E-5
.........CaC2O4 ==> Ca^2+ + C2O4^2-
equil....4.76E-5..4.76E-5...4.76E-5

Ksp = (Ca^2+)(C2O4^2-)
Substitute from the ICE chart and solve for Ka.