Question

A solution containing 0.1238 mol Na2C2O4 and 0.0176 mol NaHC2O4 was treated with 23.58 mL of...

A solution containing 0.1238 mol Na2C2O4 and 0.0176 mol NaHC2O4 was treated with 23.58 mL of a 0.766 M HCl solution and diluted to 1.00 L. the solution pOH was fond to be 9.32. determine Ka2 for oxalic acid or Kb1 for the oxalate ion.

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Answer #1

moles of HCl added = 23.58 x 0.766 / 1000 = 0.01806

C2O42-    +    HCl   ----------------> HC2O4-  

0.1238       0.01806                    0.0176    ----------------> initial

- 0.01806    - 0.01806    +0.01806    --------------> Change

0.14186    0 0.03566 --------------> equilibrium

pOH = 9.32

pH = 14 - 9.32 = 4.68

pH = pKa2 + log [salt - C / acid + C]

4.68 = pKa2 + log [0.1238 - 0.01806/ 0.0176 + 0.01806]

pKa2 = 4.2079

Ka2 = 6.20 x 10^-5

Kb1 = 10^1-14 / 6.12 x 10^-5

Kb1 = 1.61 x 10^-10

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