A 0.351 L solution of 1.71 M oxalic acid (Ka1 = 6.5e-2 and Ka2 = 6.1e-5)...

A 0.351 L solution of 1.71 M oxalic acid (Ka1 = 6.5e-2 and Ka2 = 6.1e-5) is titrated with 1.35 M KOH. What will the pH of the solution be when 0.7425 L of the KOH has been added?

Expert Solution

molles of oxalic acid = 0.351 x 1.71 = 0.600 mol

moles of KOH = 1.35 x 0.7425 = 1.00 mol

H2C2O4 + KOH --------------> HC2O4- + H2O

0.6 1.0 0 0

0 0.4 0.6

HC2O4- + KOH ---------------> C2O42- + H2O

0.6 0.4 0 0

0.2 0 0.4 -

pH = pKa2 + log [salt / acid]

= 4.22 + log (0.4 / 0.2)

pH = 4.52

queen_honey_blossom answered 1 year ago

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