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
