Question

A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pH when 25.00 mL of oxalic acid solution is titrated with 35.00 mL of NaOH

Answer 1

Answer:

pH = 4.09

Explanation:

molarity of oxalic acid in the solution

         = 0.1 x 25 / (25 + 35)

         = 0.0417 M

molarity of NaOH in the solution

          = 0.1 x 35 / (25 +35)

         = 0.0583 M

H2C2O4 + NaOH -------------------> NaHC2O4 + H2O

0.0417          0.0583                            0                0

0                      0.0166                         0.0417

now second acid -base titration  

NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O

0.0417             0.0166                              0                0

0.0251                 0                                  0.0166         ---

now

pH = pKa2 + log [Na2C2O4 / NaHC2O4]

pH = 4.27 + log (0.0166 / 0.0251)

pH = 4.09