ACT Science Practice Question #1765 (Hard (570)) | Test Citadel
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ACT Science Difficulty: Hard (570)

Digital ACT Science Practice Question #1765

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At 12.5 mL of added NaOH (the half-equivalence point), the pH equals 4.76. This value represents the: Experiment Context (Potentiometric Titration of Weak Acid): A 25.0 mL sample of 0.10 M acetic acid (CH3COOH) was titrated with 0.10 M NaOH. The pH was monitored using a glass electrode: Table 4: - 0.0 mL NaOH: pH = 2.87 - 6.25 mL NaOH: pH = 4.28 - 12.5 mL NaOH: pH = 4.76 - 20.0 mL NaOH: pH = 5.35 - 25.0 mL NaOH: pH = 8.72 - 30.0 mL NaOH: pH = 11.96
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At half-equivalence point, pH = pKa.

Masterclass Solution & Distractor Trap Analysis

At half-equivalence, [HA] = [A-], so pH = pKa by the Henderson-Hasselbalch equation....

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

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