Neutralizing HCl with NaOH: Calculating Volume Needed

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Discussion Overview

The discussion revolves around a homework problem involving the neutralization of hydrochloric acid (HCl) with sodium hydroxide (NaOH) through titration. Participants explore the calculations needed to determine the volume of NaOH required to neutralize a given volume and concentration of HCl, as well as related concepts such as molarity and conversions between grams and moles.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant seeks help with a titration problem involving 100 ml of 2.5 M HCl and .25 M NaOH, expressing confusion about the calculations and terminology.
  • Another participant mentions that HCl and NaOH neutralize at a 1:1 ratio and suggests calculating the moles of HCl to find the required moles of NaOH.
  • There is a discussion about converting grams to moles using molecular weight, with participants noting the importance of understanding molarity.
  • A participant expresses uncertainty about how to approach titration problems and requests step-by-step guidance.
  • One participant provides a calculation example involving converting grams of KOH to moles and then determining the volume of 6.0 M HCl needed for neutralization.
  • Another participant confirms the calculation of the volume of HCl needed based on the moles of KOH, indicating a collaborative effort to clarify the problem-solving process.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating the required volume of NaOH for neutralization based on molarity and the 1:1 neutralization ratio. However, there is no consensus on the exact steps or clarity on how to approach the problem, as some participants express confusion and seek further clarification.

Contextual Notes

Some participants demonstrate uncertainty regarding the application of concepts like molarity and conversions, indicating a potential gap in foundational knowledge that may affect their understanding of titration calculations.

Who May Find This Useful

This discussion may be useful for students learning about acid-base titrations, molarity, and stoichiometry, particularly those seeking assistance with homework problems in chemistry.

amd123
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Help with TITRATIONS URGENT!

Homework Statement


You need to neutralize 100 ml of 2.5 M HCl. You plan on using .25 M NaOH. How many liters of NaOH will you need to do this?

Homework Equations


What does the above mean?

The Attempt at a Solution


I could not attempt as I didn't attend the lesson due to absence. Could someone give me the 411 on how to do this using factor label and mole conversions. I googled this and got complex things that looked like physics,graphs and calculus jargon which I'm not on a high enough level of understanding to comprehend... :P
 
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Edit: Bad science, corrected in my next post.
 
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I don't know what that means, could you explain how I would apply this to something more complex such as "How many particles of HCl would be neutralized by 200 ml of .2 M NaOH?"
 
Also if I have to use grams can I still use that equation? I just figured out what M and V meant, lol how silly of me...
 
[tex]M= \frac {number of moles}{Volume}[/tex]

You can calculate the number of moles of HCl, calculate how many moles of NaOH you need to neutralise this amount, then calculate the volume of NaOH you need.
 
Whoops, okay, got off track on my last post, let's forget that. I was approaching the problem as a dilution rather than a titration, which just happened to work out almost correctly by coincidence.

This problem is fairly simple; HCl and NaOH neutralize at a ratio of 1:1. You have a 100mL of 2.5M HCl. That's a tenth of a Liter, thus you have 0.25M HCl that needs to be neutralized. So you need to add 0.25M of NaOH to neutralize it.
 
umm what if I have grams of something?
 
Not hard to convert grams to moles using the molecular weight.
 
I still have no idea on how to approach these... could you give me steps on how to solve a titration, I just asked a bunch of people in my class and they don't know.
 
  • #10
You've been given the tools on how to solve it, mate. Molarity = # of moles divided by the volume of the solution. With this, calculate the number of moles of HCl, with this number, calculate how many moles of NaOH you need to neutralise it if it's a 1:1 ratio, with this number, calculate how many litres of NaOH you need if you're given the molarity and you calculated the moles of NaOH needed for titration.
 
  • #11
Okay, I'll try it just the way you said with something else because I already did that problem...
https://www.physicsforums.com/latex_images/16/1663699-0.png

15 grams (KOH) times 1 Mol KOH divided by 56.01=.27 mols

Now that I converted that to mols how do i find what volume of 6.0 M HCl would neutralize that? I have no idea... :(
 
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  • #12
6.0 M means 6.0 moles per litre of HCl. How many litres would 0.27 moles of HCl be equal to?
 
  • #13
.27m*1L=6.0Mx(V2) .27=6x x=.045 can you confirm this snazzy? The original problem stated what volume of 6 M HCl would be needed to neutralize 15 grams of KOH.
 
  • #14
Looks right to me.
 
  • #15
thanks a lot mate I really appreciate it now just 6 more problems to do :D
 

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