Question

The oxalate ion has the formula C2O42- and is often abbreviated "ox". A metal ("M") oxalate has the formula M2(ox)3 and the value of its solubility product constant Ksp is 1.8 10-24. A) Calculate the concentration of oxalate ion in a saturated solution of M2(ox)3, neglecting activity effects and any acid-base or complexation reactions that might compete with the solubility equilibrium.

_____?________ mol/L C₂O₄^2-

B) Calculate the concentration of oxalate ion in a saturated solution of M₂(ox)₃ that is also 0.054 mol/L in M(NO₃)₃, neglecting activity effects and any acid-base or complexation reactions that might compete with the solubility equilibrium.

Answer 1

Given that, Ksp of M₂(ox)₃ = 1.8 * 10^-24

Let S be the solubility.

Now Solubility reaction is:

M2(ox)3 (s)		2M3+	3 ox2-
		2s	3s

Ksp = [M³⁺]²[ox^-]³ = (2s)²(3s)³ = 1.8 * 10^-24

108 S⁵ = 1.8 * 10^-24

S⁵ = 1.67 * 10^-26

S = 6.99 *10^-6 M

Thus, Concentration of Oxalate ion = 3S = 3* 6.99 *10^-6 = 2.096 *10^-5 mol/L or 2.10 * 10^-5 mol/L

b) Now we have 0.054 mol/L of M(NO₃)₃ in solution also Which dissociates to give 0.054 M of M³⁺

Thus, Now, total amount of M³⁺ in solution = 2S + 0.054. But S <<<< 0.054. So, here we can neglect S

and [M³⁺] = 0.054

Ksp =[M³⁺]²[ox^-]³ = (0.054)²(3s)³ = 1.8 * 10^-24

27S³ = 6.17 *10^-22

S³ = 2.29 *10^-23

S = 2.84 *10^-8 mol/L

Hence, the concentration of oxalate ion = 3S = 3*2.84 *10^-8 mol/L = 8.51 * 10^-8 mol/L