Evaluate the solubility of calcium oxalate [CaC_2O_4] in water. The oxalate ion

Question

Evaluate the solubility of calcium oxalate [CaC_2O_4] in water. The oxalate ion acts as weak base, reacting with water to form HC_2O_2^-. This ion again acts as a weak base, reacting with water once again to form oxalic acid, H_2C_2O_4. Using the systematic treatment of equilibrium approach, calculate the H_2C_2O_4, HC_2O_4^-, C_2O_4^-2, Ca^2+, H^+ and OH^- concentrations if the pH is held constant at 6.0. You MUST show all of your work to receive credit. Use additional pages, if necessary. Using the GoalSeek function in Excel, calculate the actual pH of the system if the pH is not held constant. Upload your spreadsheet showing this calculation to Blackboard.

Explanation / Answer

I'll help you with part a). To do this we just need the pKb values for oxalate ion.
pKb1 = 9.734 ----> Kb1 = 1.84x10-10
pKb2 = 12.75 ----> Kb2 = 1.78x10-13

With the pH value we can calculate the H+ and OH- concentration:
[H+] = 10-6 = 1x10-6 M
[OH-] = 1x10-14 / 1x10-6 = 1x10-8 M

Now the equilibrium concentrations and reactions:
r: C2O42- + H2O <--------> HC2O4- + OH-   Kb2 = 1.78x10-13
i: Y 0 0
e: Y-1x10-8   1x10-8   1x10-8

1.78x10-13 = (1x10-8)2 / Y
Y = (1x10-8)2/1.78x10-13
Y = [C2O42-] = [Ca2+] = 5.62x10-4 M

For the other hydrolisis:
r: HC2O4- + H2O <---------> H2C2O4 + OH- Kb1 = 1.84x10-10
i: 1x10-8 0 1x10-8
e: 1x10-8 - Z Z 1x10-8+Z

1.84x10-10 = (1x10-8+Z)Z / (1x10-8 - Z)
1.84x10-10 = 1x10-8Z/1x10-8
Z = [H2C2O4] = 1.84x10-10 M

Hope this helps

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