Final answer:
To neutralize 50.0 ml of 0.500 M KOH with 0.345 M HCl, one would require 72.46 ml of the HCl solution.
Step-by-step explanation:
To calculate the volume of 0.345 M HCl needed to neutralize 50.0 ml of 0.500 M KOH, we use the concept of mole-to-mole ratio from the balanced chemical equation:
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
From this equation, we see that the reaction ratio is 1:1. Thus, 1 mole of HCl is required to neutralize 1 mole of KOH. First, we calculate the moles of KOH in 50.0 ml:
Moles of KOH = 0.500 M × 0.050 L = 0.025 moles KOH
Since the mole ratio is 1:1, we will also need 0.025 moles of HCl to neutralize it. Now, we calculate the required volume of 0.345 M HCl to get 0.025 moles:
Volume of HCl = Moles of HCl ÷ Molarity of HCl
Volume of HCl = 0.025 moles ÷ 0.345 M
Volume of HCl = 0.07246 L, or 72.46 ml
Therefore, 72.46 ml of 0.345 M HCl is needed to neutralize the sample of KOH.