Final answer:
56 milliliters of 0.100 M HClO₃ are needed to neutralize 40.0 mL of 0.140 M KOH, using the 1:1 stoichiometry of the acid-base neutralization reaction.
Step-by-step explanation:
The question is asking for the volume of 0.100 M HClO₃ needed to neutralize a known volume and concentration of KOH, which is a typical acid-base neutralization problem often encountered in high school chemistry classes. We'll use the stoichiometry of the reaction and the molarity concept to find the answer.
The balanced chemical equation for the reaction between hydrochloric acid (HClO₃) and potassium hydroxide (KOH) is:
HClO₃(aq) + KOH(aq) → KClO₃(aq) + H₂O(l)
From this equation, we see that the reaction ratio between HClO₃ and KOH is 1:1. Therefore, each mole of KOH will react with one mole of HClO₃.
First, we calculate the number of moles of KOH:
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moles of KOH = volume (in L) × molarity = 0.040 L × 0.140 M = 0.0056 moles
Since the ratio of HClO₃ to KOH is 1:1, we need the same amount of moles of HClO₃ to neutralize the KOH. Thus, we need 0.0056 moles of HClO₃.
Next, we find the volume of 0.100 M HClO₃ required:
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volume of HClO₃ = moles of HClO₃ / molarity of HClO₃ = 0.0056 moles / 0.100 M = 0.056 L
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volume of HClO₃ in milliliters = 0.056 L × 1000 mL/L = 56 mL
Therefore, 56 milliliters of 0.100 M HClO₃ are required to neutralize 40.0 mL of 0.140 M KOH.