Neutralizing HCl with NaOH: Calculating Volume Needed

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In summary, You need to neutralize 100 ml of 2.5 M HCl with .25 M NaOH. You need to calculate how many moles of NaOH you need to do this, how many litres of NaOH you need, and how many grams of NaOH you need.
  • #1
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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  • #2
Edit: Bad science, corrected in my next post.
 
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  • #3
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?"
 
  • #4
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...
 
  • #5
[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.
 
  • #6
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.
 
  • #7
umm what if I have grams of something?
 
  • #8
Not hard to convert grams to moles using the molecular weight.
 
  • #9
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
 

1. What is a titration and how does it work?

A titration is a laboratory technique used to determine the concentration of a substance in a solution by reacting it with a known concentration of another substance. This is typically done by slowly adding the known substance, called the titrant, to the unknown substance until a reaction occurs and the endpoint is reached. The endpoint is when the reaction is complete and the concentration of the unknown substance can be calculated using the known concentration of the titrant and the volume of titrant added.

2. What equipment is needed for a titration?

The equipment needed for a titration includes a burette, a pipette, a volumetric flask, a conical flask, a magnetic stirrer, and a pH meter. The burette is used to accurately measure and dispense the titrant, while the pipette is used to accurately measure the volume of the unknown substance. The volumetric flask is used to prepare a known concentration of the titrant, and the conical flask is where the reaction takes place. The magnetic stirrer is used to mix the solutions, and a pH meter may be needed to monitor the pH of the reaction.

3. How do you choose the appropriate indicator for a titration?

The appropriate indicator for a titration depends on the nature of the reaction and the substances being titrated. Some common indicators include phenolphthalein, methyl orange, and bromothymol blue. These indicators change color at specific pH values, so it is important to choose the indicator that corresponds to the expected pH at the endpoint of the titration. It is also important to choose an indicator that has a clear color change and is easy to detect.

4. What are the sources of error in a titration?

There are several sources of error in a titration, including human error, equipment error, and environmental factors. Human error can occur when reading the meniscus on a burette or pipette, or when adding the titrant too quickly or too slowly. Equipment error can occur when the equipment is not calibrated correctly or is malfunctioning. Environmental factors such as temperature and air bubbles can also affect the accuracy of a titration.

5. How do you calculate the concentration of the unknown substance from a titration?

The concentration of the unknown substance can be calculated using the following formula: CxVx=CtVt, where Cx is the concentration of the unknown substance, Vx is the volume of the unknown substance, Ct is the concentration of the titrant, and Vt is the volume of titrant added at the endpoint. This formula is based on the principle of stoichiometry, which states that the number of moles of a substance in a reaction is directly proportional to its volume in a solution. By rearranging this formula, the concentration of the unknown substance can be calculated.

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