Quantitative Titration - NaOH and HCl

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Discussion Overview

The discussion revolves around a lab experiment involving the titration of hydrochloric acid (HCl) with sodium hydroxide (NaOH) to determine the molarity of the NaOH solution. Participants explore calculations related to the titration, including the molarity of the acid and base, the stoichiometry of the reaction, and the determination of the molar mass of an unknown salt.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the molarity of HCl is 0.15M and questions the calculation of the NaOH molarity based on the average volume used during titration.
  • Another participant confirms the 1:1 mole ratio between HCl and NaOH in the reaction, providing calculations that support this ratio.
  • Concerns are raised about the accuracy of the diluted acid concentration, with a participant suggesting that the number of moles of acid should remain constant despite dilution.
  • There is a discussion about whether it is necessary to calculate the new concentration after dilution, with one participant suggesting that the approximate values cancel out in the calculations.
  • A later post introduces the idea of determining the molar mass of an unknown salt, questioning how to proceed without knowing the concentration of the solid.
  • Another participant suggests that determining the molar mass typically requires knowing the mass of the solid acid and the number of moles derived from titration.
  • One participant calculates the number of moles of NaOH reacting with a known mass of acid, leading to a discussion about the resulting equivalent mass and its implications for identifying the acid.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy of the calculated concentrations and the necessity of recalculating after dilution. There is no consensus on the best approach to determine the molar mass of the unknown salt, as participants suggest different methods and considerations.

Contextual Notes

Participants note limitations regarding the precision of volume measurements and the implications for calculating concentrations. The discussion also highlights the distinction between molar mass and equivalent mass, particularly in the context of strong versus weak acids.

fishfish
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Homework Statement



Hi, So I have this lab in which we had to titrate HCl with NaOH (of an unknown molarity)

Scan0004.jpg

Scan0005.jpg


So here's the lab.. I decided to scan it because I'm not a good explainer at all, heh.
The molarity of the HCl is 0.15M by the way

What I was wondering is how to calculate #2 under calculations and results... For our titration of the HCl, we had gotten 15.8mL (average) of NaOH used. For calculations, however, I used the concentration of the diluted acid (0.06M for the 25mL of water and HCl combined.. which I don't know whether it's right or not) to calculate that the molarity of the NaOH is 0.095M

Homework Equations



CaVa=CbVb

And would we have to use Kw = [H+][OH-] in any way here?


The Attempt at a Solution



CaVa=CbVb
(0.06M)(0.025L) = (Mb)(0.0158)L)
Mb=0.095M
 
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For question 2 it refers to the chemical equation which is:
HCl + NaOH -> NaCL + H2O

In which case the moles of base used per mole of acid is 1. It is a 1:1 ration, and your titration shows this.
M(base) = n/V(L)
.095(.0158) = n
n = .0015
and
M(acid) = n/V(L)
.06(.025) = n
n= .0015

a 1:1 mole ratio
 
fishfish said:
however, I used the concentration of the diluted acid (0.06M for the 25mL of water and HCl combined.. which I don't know whether it's right or not)

You don't know exact volume (about+about is not exactly 25) so you can't calculate exact concentration of HCl. However, it is all about number of moles of acid - did it changed after dilution?

You may consult www.titrations.info - a lot of information about titrations, acid/base ones included.
 
nickdk said:
For question 2 it refers to the chemical equation which is:
HCl + NaOH -> NaCL + H2O

In which case the moles of base used per mole of acid is 1. It is a 1:1 ration, and your titration shows this.
M(base) = n/V(L)
.095(.0158) = n
n = .0015
and
M(acid) = n/V(L)
.06(.025) = n
n= .0015

a 1:1 mole ratio

Oh really? Haha, I thought it was asking for something more difficult! Thank you :biggrin:
 
Borek said:
You don't know exact volume (about+about is not exactly 25) so you can't calculate exact concentration of HCl. However, it is all about number of moles of acid - did it changed after dilution?

You may consult www.titrations.info - a lot of information about titrations, acid/base ones included.

Wait, since it's not accurate enough, then I assume that I can't follow through with Nickdk's solution then?
 
What he calculated using numbers you gave is acidentally correct, but it is not how it should be done.

You should calculate number of moles of the acid introduced to the solution - that means precisely measured 10 mL of 0.15M solution, or 1.5 mmole.

Now, after solution was diluted to ABOUT 25 mL, its concentration went down to ABOUT 0.06M. However, new concentration is calculated assuming number of moles of acid have not changed - and you need this number of moles to calculate number of moles of base that reacted. And you have already calculated that it was 1.5 mmole.

If so, do you need to calculate new concentration after dilution? You divide number of moles by approximate volume and then you multiply it again by the same approximate value - so approximate volume cancels out and fact that it was an approximate value doesn't matter any longer, question is - was this operation necessary?
 
Last edited by a moderator:
Borek said:
What he calculated using numbers you gave is acidentally correct, but it is not how it should be done.

You should calculate number of moles of the acid introduced to the solution - that means precisely measured 10 mL of 0.15M solution, or 1.5 mmole.

Now, after solution was diluted to ABOUT 25 mL, its concentration went down to ABOUT 0.06M. However, new concentration is calculated assuming number of moles of acid have not changed - and you need this number of moles to calculate number of moles of base that reacted. And you have already calculated that it was 1.5 mmole.

If so, do you need to calculate new concentration after dilution? You divide number of moles by approximate volume and then you multiply it again by the same approximate value - so approximate volume cancels out and fact that it was an approximate value doesn't matter any longer, question is - was this operation necessary?

--
buffer calculator, concentration calculator
pH calculator, stoichiometry calculator

Hi, I'm back again, sorry to be such a bother.. But I was wondering if it were possible to find the molar mass of the unknown salt (as our teacher instructed us to do) just from the given information. Since there's no concentration of the solid, how would it be possible to find the molar mass of it? Would I have to use the concentration of the acid dissolved in the water as the conentration to solve? And if I were to use the concentration of that solution, how would I go about doing that? Any help on this will be appreciated :biggrin:
 
Not sure what you are asking about (no idea what your insctructor told you and what information you can use), but usually these things are done using known mass of the solid acid and determining number of moles of the acid by titration - if you know mass of the sample and number of moles, finding molar mass is a breeze.
 
Last edited by a moderator:
Borek said:
Not sure what you are asking about (no idea what your insctructor told you and what information you can use), but usually these things are done using known mass of the solid acid and determining number of moles of the acid by titration - if you know mass of the sample and number of moles, finding molar mass is a breeze.

--
buffer calculator, concentration calculator
pH calculator, stoichiometry calculator

Well, this is what I have so far :

mass of acid = 0.7g
average volume used for titration= 0.0295L

nNaOH= cv
=(0.095M)(0.0295L) [ the 0.095M is calculated by the way]
=2.8x10-3 mol of base reacting with 0.7g of acid

After this.. I used a ratio to solve for the mass of acid that would react with 1 mol of base ( as #6 called for)

which is about 250g of acid / 1 mol of NaOH

So, after a bit of research, I'm assuming I do no. moles = mass / molar mass..

but the thing is, I don't have the number of moles of acid that I need for the equation..Well that is unless I use the number of moles that reacted with the 0.7 g of acid, then do...

2.8x10-3 mol = 0.7g / Molar mass
Molar mass = 250g/mol

However, this seems a bit too high to belong to any of the strong acids... unless it isn't a strong acid maybe?
 
  • #10
fishfish said:
However, this seems a bit too high to belong to any of the strong acids... unless it isn't a strong acid maybe?

It can be some other acid as well. And 250 is a correct result (that is, assuming numbers you have listed are correct).

Just note that this is not a molar mass, but so called equivalent mass - you don't know if this is mono- or multiprotic acid.
 
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