Final answer:
25.00 mL of 0.200 M HCl will completely react with 20 mL of 0.250 M KOH in a process known as acid-base neutralization.
Step-by-step explanation:
This question is all about acid-base neutralization reactions.
HCl
(hydrochloric acid) and
KOH
(potassium hydroxide) are an acid and a base respectively. When they react together, they neutralize each other and produce water and a salt (in this case, potassium chloride) in a process often referred to as titration. We must first calculate the amount of moles of HCl since it's a limiting reactant in the reaction by using the formula (Molarity = moles/Volume(L)). Therefore, moles of HCl = 0.200 M *0.025 L = 0.005 mol. Because the stoichiometric coefficient for HCl and KOH is 1:1 in the balanced acid-base neutralization equation, we know that the moles of KOH required will also be 0.005 mol. The amount of KOH solution required can then be calculated using the formula Volume(L) = moles/Molarity. So Volume(L) of KOH = 0.005 mol / 0.250 M = 0.02 L or 20 mL of the KOH solution.
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