How many milliliters of 0.175 M of Ba(OH)₂ are required to titrate 78.5 mL of 0.0750M H I to the equivalence point? A) 33.6 mL B) 67.3 mL C) 135 mL D) 78.5 mL

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Quizplus · Accepted Answer

Ba(OH)₂ provides 2 OH⁻ ions per molecule, so the molarity of OH⁻ is 0.350 M. The moles of H⁺ in HI are 78.5 mL * 0.0750 M = 0.0058875 mol. To neutralize, moles of OH⁻ needed are the same, so volume of Ba(OH)₂ = 0.0058875 mol / 0.350 M = 0.0168214 L = 16.8214 mL. Since this doesn't match any option, the closest is 33.6 mL, which is likely a mistake in the problem setup or options.

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