What volume of 0.378 M hydroiodic acid (HI) is required to neutralize 21.6 mL of 0.520 M barium hydroxide, Ba(OH)₂?

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Quizplus · Accepted Answer

Barium hydroxide, Ba(OH)2, provides 2 OH- ions per formula unit. The neutralization reaction is: 2 HI + Ba(OH)2 → BaI2 + 2 H2O. Moles of Ba(OH)2 = 0.520 M * 0.0216 L = 0.011232 mol. Moles of HI needed = 2 * 0.011232 mol = 0.022464 mol. Volume of HI = 0.022464 mol / 0.378 M = 0.0594 L = 59.4 mL.

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