Question

A solution containg 0.123 mol Na2C2O4 and 0.017 mol NaHC2O4 was treated with 23.50 mL of a 0.766 m HCl solution and diluted to 1.00 L. The solution pOH was found to be 9.31. Determie ka2 for oxalic acid or kb1 for the oxlatate ion.

Answer 1

moles of HCl added = 23.50 x 0.766 / 1000 = 0.018

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

0.123    0.018                  0.017    ----------------> initial

- 0.018       - 0.018    +0.018    --------------> Change

0.105       0    0.035 --------------> equilibrium

pOH = 9.31

pH = 14 - 9.31 = 4.69

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

4.69 = pKa2 + log [0.123 - 0.018/ 0.017 + 0.018]

pKa2 = 4.21

Ka2 = 6.12 x 10^-5

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

Kb1 = 1.63 x 10^-10