Question

What is the solubility (in g/L) of magnesium oxalate, MgC₂O₄, in 0.020 M sodium oxalate, Na₂C₂O₄? Solve the equation exactly. (K_sp of MgC₂O₄ is 8.5 x 10^-5; MW = 112 g/mol).

Answer 1

The chemical equation involved would be :

MgC2O4 <=======> Mg2+... +... C2O42-

Now, Ksp = [Mg+] [C2O42-]

Putting the respective values in above equation, we get

8.5 x 10^-5 = [Pb2+] [CrO4^2-]

use the [C2O4^2-] from the Na2C2O4= 0.020 M (because C2O4^2- from MgC2O4 is so small compared to the 0.020 M from the Na2C2O4.)

8.5 x 10^-5 = s x [0.020]

4.25 x 10^-3 M = s

Now, solubility in gL-1 = Molarity x molar mass of PbCrO4

= 4.25 x 10^-3 M x 112 g mol-1

= 0.476 g