3.38g / 90.035g/mol = 0.375mol
0.375mol / 0.500L = 0.751M

pKa1 = 1.25
pKa2 = 4.27

=> per aver pH = 2.40 è MEGLIO optare per la coppia H2C2O4 / HC2O4-

t.c.:
[H+] = Ka1*[H2C2O4]/[HC2O4-]

Dove:

H2C2O4 + KOH -----------> KHC2O4 + H2O
0.751_______X______________0_________/__
0.751-X_____0______________X_________/__

=> [H+] = 10^-2.40 = 5.60*10^-2 * (0.751-X) / X

=> X = 7.01*10^-2mol/L

7.01*10^-2mol/L * 500mL = 0.800mol/L * V

=> V = 43.81mL