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

pH = 4.09

Step-by-step 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