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Oxalic acid , HOOCCOOH (aq) , is a diprotic acid with Ka1= 0.056 and Ka2=1.5x10-4. Determine...

Oxalic acid , HOOCCOOH (aq) , is a diprotic acid with Ka1= 0.056 and Ka2=1.5x10-4. Determine the pH of a 0.12 M oxalic acid solution.

HOOCCOOH(aq) = HOOCCOO-(aq) +

H+(aq) HOOCCOO-(aq) = -OOCCOO-(aq) + H+(aq)

Expert Solution

H2C2O4 ---------------------> HC2O4- + H+

0.12 0 0

0.12 - x x x

Ka1 = x^2 / 0.12 - x

0.056 = x^2 / 0.12 - x

x = 0.0586

[H+] = 0.0586 M

HC2O4- ------------------> C2O42- + H+

0.0586 0 0.0586

0.0586 - x x 0.0586 + x

Ka2 = x * (0.0586 + x) / (0.0596 - x)

1.5 x 10^-4 = 0.0586 x + x^2 / 0.0586 - x

8.79 x 10^-6 - 1.5 x 10^-4 x = 0.0586 x + x^2

x^2 + 0.05875x - 8.79 x 10^-6 = 0

x = 1.49 x 10^-4

[H+] = 1.49 x 10^-4 M

total [H+] = 0.0586 + 1.49 x 10^-4 = 0.0588 M

pH = -log [H+] = -log (0.0588)

pH = 1.23

queen_honey_blossom answered 1 year ago

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