What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)₂(aq) solution A) 55.34 mL B) 11.07 mL C) 16.14 mL D) 32.29 mL E) 47.10 mL

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Quizplus · Accepted Answer

The balanced chemical equation for the reaction between HCl and Ba(OH)2 is: HCl + Ba(OH)2 → BaCl2 + 2H2O. This shows that 2 moles of HCl are needed to neutralize 1 mole of Ba(OH)2. Using the molarity and volume of Ba(OH)2, we find the moles of Ba(OH)2: (0.2350 M)(0.02355 L) = 0.00553825 moles. Since the mole ratio of HCl to Ba(OH)2 is 2:1, we need 2 * 0.00553825 = 0.0110765 moles of HCl. Using the molarity of HCl to find the volume needed: Volume = moles / molarity = 0.0110765 moles / 0.3428 M = 0.03229 L, which is 32.29 mL.

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