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) 1.086 B) 48.89 C) 28.59 D) 97.78 E) 22.83

Question

Quizplus · Accepted Answer

The correct answer is B.The balanced chemical equation for the neutralization reaction is:H2C2O4 + 2NaOH → Na2C2O4 + 2H2OFrom the equation, we can see that 1 mole of H2C2O4 reacts with 2 moles of NaOH.The number of moles of NaOH used in the titration is:37.98 mL * (1 L / 1000 mL) * (0.2283 mol / L) = 0.00866 molTherefore, the number of moles of H2C2O4 in the sample is:0.00866 mol / 2 = 0.00433 molThe molar mass of H2C2O4 is 90.03 g/mol. Therefore, the mass of H2C2O4 in the sample is:0.00433 mol * 90.03 g/mol = 0.3896 gThe percent of H2C2O4 in the solid is:(0.3896 g / 0.7984 g) * 100% = 48.89%

Oxalic Acid Is a Diprotic Acid