[13%] Calcium oxalate, CaC2O4, has a Ksp of 2.7 × 10-9. Oxalic acid, H2C2O4, has two ionizable hydro — Solubility Equilibria and Acid-Base Chemistry Chemistry Question
Problem Context
[13%] Calcium oxalate, CaC2O4, has a Ksp of 2.7 × 10-9. Oxalic acid, H2C2O4, has two ionizable hydrogens with pKa1 = 1.27 and pKa2 = 4.28.
Draw a Lewis structure for oxalate ion, C2O4 2-, including all bonds, lone pairs, and formal charges.
Calculate the molar solubility of calcium oxalate in pure water.
Model Answer
If the molar solubility is S, then [Ca2+] = [C2O4 2-] = S
Ksp = [Ca2+][C2O4 2-] = S^2 = 2.7 × 10^-9
S = 5.2 × 10^-5 mol L^-1
Calculate the molar solubility of calcium oxalate in a 0.100 M CaCl2 solution.
Model Answer
Ksp = [Ca2+][C2O4 2-] = [0.100][C2O4 2-] = 2.7 × 10^-9
[C2O4 2-] = 2.7 × 10^-8 mol L^-1
Since all the oxalate comes from the dissolving calcium oxalate, this is the solubility.
A 0.100 mol sample of solid CaC2O4 is suspended in 1.00 L of water and HCl(g) is bubbled through the solution until all of the solid just dissolves. What is the pH of the final homogeneous solution? You may assume the final volume of solution is 1.00 L.
Model Answer
Since all the calcium oxalate has dissolved, [Ca2+] = 0.100 M. Since it has just dissolved, the Ksp expression is valid, and [C2O4 2-] = 2.7 × 10^-8 mol L^-1 as in part b. The total amount of oxalate, [C2O4 2-] + [HC2O4 -] + [H2C2O4], must add up to 0.100 M. Since so little of it is C2O4 2-, we will assume that almost all of it is doubly protonated, so [H2C2O4] ≈ 0.100 M. We can relate the concentration of C2O4 2- to that of H2C2O4 by combining the two Ka equilibria:
H2C2O4 ⇌ HC2O4 - + H+ (Keq = Ka1 = 0.054)
HC2O4 - ⇌ C2O4 2- + H+ (Keq = Ka2 = 5.2 × 10^-5)
H2C2O4 ⇌ C2O4 2- + 2 H+ (Keq = Ka1Ka2 = 2.8 × 10^-6)
[C2O4 2-][H+]^2 / [H2C2O4] = (2.7 × 10^-8)[H+]^2 / 0.100 = 2.8 × 10^-6
[H+] = 3.2 M
pH = –log10[H+] = –0.51
At this pH, about 1 part in 60 of the oxalate is in the form of HC2O4 -, so our assumption about the concentration of H2C2O4 is indeed valid.
How many moles of HCl(g) are added in part d?
Model Answer
To achieve this concentration of H+, one would need to add 3.2 mol HCl, plus 0.2 mol more to protonate the 0.1 mol of C2O4 2-, for a total of 3.4 mol.
Will the molar solubility of CaC2O4 in 0.1 M NaHC2O4 be significantly greater than, significantly less than, or within 10% of the molar solubility of calcium oxalate in pure water? Justify your answer.
Model Answer
Addition of HC2O4 - cannot increase the solubility of calcium oxalate, since it cannot decrease the oxalate concentration by protonating it (since that would just replace the oxalate with another mole of oxalate!). It might be able to decrease the solubility by the common ion effect. To see if this is significant, we can calculate the concentration of C2O4 2- in the 0.100 M HC2O4 - solution. The major way this is formed is by deprotonation of HC2O4 - by itself:
HC2O4 - + H+ ⇌ H2C2O4 (Keq = 1/Ka1 = 18.6)
HC2O4 - ⇌ C2O4 2- + H+ (Keq = Ka2 = 5.2 × 10^-5)
2 HC2O4 - ⇌ C2O4 2- + H2C2O4 (Keq = Ka2/Ka1 = 9.8 × 10^-4)
[C2O4 2-][H2C2O4] / [HC2O4 -]^2 = ^2 / [0.100 - 2x]^2 = 9.8 × 10^-4
/ [0.100 - 2x] = 0.031
x = 2.9 × 10^-3 = [C2O4 2-]
Since this is 56 times larger than the oxalate concentration when CaC2O4 is dissolved in pure water (see part b), the common ion effect will significantly decrease the solubility of calcium oxalate in the NaHC2O4 solution compared to pure water.