Oxalic acid (H 2 C 2 O 4 ) is a substance found in Oxalis plant (sour grass) and

Question

Oxalic acid (H2C2O4) is a substance found in Oxalis plant (sour grass) and in spinach.   It is a strong chelating agent in that it can bind to metal ions such as Ca2+ and Fe2+and3+, thus in high amounts is poisonous. It is used as a non-chlorine bleaching agent and in cleaners.

            Calculate the pH and concentration of oxalate ion, [C2O42-], in a 0.02 M solution of oxalic acid (H2C2O4)

Once in water, H2C2O4 undergoes two ionizing equilibriums:

                                    H2C2O4(aq)     H+(aq)   +   HC2O4-(aq)                       Ka1 = 5.9 x 10-2

And

                                    HC2O4-(aq)    H+(aq) +   C2O42-(aq)                    Ka2 = 6.4 x 10-5

Explanation / Answer

the [H+] comes essentially only from Ka1

5.90 x 10–2 = [H+] [HC2O4] / [H2C2O4]

5.90 x 10–2 = [X] [X] / [0.02 - X]

5.90 x 10–2 [0.02 - X] = X2

0.00118 - 0.059X = X2

X2 + 0.059X - 0.00118 = 0

X = [HC2O4] = [H+] = 0.01578 Molar

pH = - log [H+] = -log[0.01578]

pH = 1.8

======================================...

now find the [C2O4] , when the [H+] & [HC2O4] where just found to be 0.01578 Molar,

6.4 x 10–5 = [H+] [C2O42-] / [HC2O4]

6.4 x 10–5 = [0.01578 ] [C2O42-] / [0.01578 ]

6.4 x 10–5 = [C2O42-]

your answer is
pH = 1.8 ; [C2O4 2–] = 6.3 x 10–5 M

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