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What is the pH of a 0.39 M solution of sodium oxalate, Na2C2O4? Acid ionization constants for...

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Chemistry

What is the pH of a 0.39 M solution of sodium oxalate,Na2C2O4? Acid ionization constantsfor oxalic acid are Ka1 = 5.6 x 10-2and Ka2= 5.4 x 10-5.

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3.9 Ratings (582 Votes)

C2O4^2- + H2O   -----------------------> HC2O4- +   OH-

   0.39                                                         0            0

0.39 - x                                                        x            x

Kb2 = [HC2O4-][OH-] / [C2O4^2-]

Kw / Ka2 = x^2 / (0.39 - x)

1.0 x 10^-14 / 5.4 x 10-5.= x^2 / (0.39 - x)

1.85 x 10^-10 = x^2 / (0.39 - x)

x = 8.5 x 10^-6

[OH-] = 8.5 x 10^-6 M

pOH = -log[OH-] = -log (8.5 x 10^-6 )

         = 5.07

pH + pOH = 14

pH = 14 - 5.07 = 8.9

pH = 8.93
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