Question

Oxalic acid (a diprotic acid MW=90.0 g/mole) is sometimes used instead of KHP for the standardization of NaOH. If a 0.145 gram sample of a oxalic acid is titrated to a phenolphthalein endpoint with 23.75 mL of the NaOH solution, calculate the molar concentration of the base. (Hint: What is the balanced equation for NaOH and a diprotic acid?) SHOW ALL WORK

Answer 1

m(H2C2O4) = 0.145 g

M(H2C2O4) = 90 g/mole

V(NaOH) = 23.75 mL

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

n(H2C2O4) = m/M = 0.145 g / 90 g/mole = 0.0016 mol

n(NaOH) = 2·n(H2C2O4) = 2·0.0016 = 0.0032 mol

C(NaOH) = n/V = 0.0032 mol / 0.02375 L = 0.135 mol/L = 0.135 M