ACT Science Practice Question #1804 (Hard (540)) | Test Citadel
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ACT Science Difficulty: Hard (540)

Digital ACT Science Practice Question #1804

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The second equivalence point occurs at exactly 40.0 mL of NaOH. This indicates that complete neutralization of a diprotic acid requires: Experiment Context (Dual Equivalence Points of Oxalic Acid): A 20.0 mL sample of 0.10 M oxalic acid (H2C2O4, diprotic) was titrated with 0.10 M NaOH. The reaction proceeds in two distinct deprotonation stages: Table 17: - 0.0 mL NaOH: pH = 1.30 - 10.0 mL NaOH: pH = 1.25 (Half-equivalence 1, pKa1) - 20.0 mL NaOH: pH = 2.85 (First Equivalence Point) - 30.0 mL NaOH: pH = 4.27 (Half-equivalence 2, pKa2) - 40.0 mL NaOH: pH = 8.40 (Second Equivalence Point) - 50.0 mL NaOH: pH = 12.10 (Excess Base)
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Diprotic stoichiometry: 2 moles of base per mole of acid.

Masterclass Solution & Distractor Trap Analysis

Oxalic acid has two moles of ionizable H+ per mole of acid, requiring 2 equivalents of NaOH (20 mL + 20 mL = 40 mL)....

Distractor Analysis: Trap choice eliminates careless test-takers who confuse roots with coordinates...

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