Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H₂C₂O₄)in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization. A)48.89 B)97.78 C)28.59 D)1.086 E)22.83

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Quizplus · Accepted Answer

Oxalic acid is diprotic, meaning it can donate two protons. The reaction with NaOH is: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O. Moles of NaOH = 0.03798 L * 0.2283 mol/L = 0.008673 moles. Since it takes 2 moles of NaOH to neutralize 1 mole of oxalic acid, moles of oxalic acid = 0.008673 / 2 = 0.0043365 moles. Molar mass of oxalic acid = 90.03 g/mol. Mass of oxalic acid = 0.0043365 moles * 90.03 g/mol = 0.3904 g. Percent = (0.3904 g / 0.7984 g) * 100 = 48.89%.

Oxalic Acid Is a Diprotic Acid