How Much HCl Is Needed to Neutralize NaOH?

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To determine the volume of 0.100 M HCl needed to neutralize 25.0 mL of 0.350 M NaOH, one must first calculate the moles of NaOH present. The reaction requires an exact amount of hydrogen ions to react with the hydroxide ions, eliminating the need to identify a limiting reagent in this case. Since the volumes of the reactants are equal, the problem simplifies to finding the moles of NaOH and then calculating the corresponding volume of HCl required for neutralization. The conclusion reached is that 87.5 mL of HCl is needed for complete neutralization.
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What volume of a .100 M HCl is needed to neutralize 25.0 mL of a .350 Molarity Sodium Hydroxide?

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I don't understand why this reaction doesn't require you to determine the limiting reagent (there is none).

The book explicitly says "This problem requires the addition of just enough Hydrogen ions to react exactly with the OH- ions present."

Even with this description my mind is still a little fuzzy on how I can determine when a problem needs a limiting reagent to be found and when a problem doesn't need a limiting reagent found.

Please shed some light on this issue.
 
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oops, I know why now.


Since 25 mL is the only exact volume (excluding the molarities). You are finding the exact amount of H that will react with the exact amount of Hydroxides. If HCl had an amount of mL different than NaOH . . . then you would determine the LR.

:D
 
Looks like you got it. Do you still need help?

It's pretty easy. Just figure out how many moles of NaOH you have. Then figure out how many mL's HCl that would be.
 
Did you come to the right answer? 87.5 ml
 
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