Titration of unknown acid in lab, how to determine molar mass?

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Discussion Overview

The discussion revolves around a chemistry lab problem involving the titration of an unknown acid to determine its molar mass. Participants explore the calculations related to the titration process, including the volumes and concentrations involved, and seek clarification on the correct approach to determine the molar mass from the titration data.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant describes their titration setup, noting the mass of the unknown solid and the volumes used in the titration with NaOH.
  • Another participant questions the initial volume used for calculations, suggesting that the volume for molarity calculations should be 25 ml instead of the total volume after adding NaOH.
  • A participant expresses confusion about the dilution effect when adding NaOH to the acid solution, questioning the logic behind using only the initial 25 ml volume for calculations.
  • There is a challenge regarding the interpretation of the final volume in relation to the titration process, with participants discussing whether the final volume should include both the acid and the added NaOH.
  • One participant provides a link to an external resource, implying that there may be misunderstandings about the concepts involved in titration calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct volume to use in their calculations, with some advocating for using only the initial 25 ml of acid while others argue that the final volume should account for the added NaOH. The discussion remains unresolved with multiple competing views on the approach to the problem.

Contextual Notes

Participants express uncertainty about the definitions of volumes used in the calculations and how they relate to the titration process. There are unresolved questions about the dilution effects and the correct interpretation of the equivalence point in the context of the titration.

redioactif
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Hi,

Homework Statement


It is in my chemistry lab course. I had 0.1g of an unknown solid that I diluted in 100ml of distilled water. Then, I took 25ml and titrated it with NaOH (Concentration = 0.0189M). To reach equivalent point (the point where concentration of NaOH is equal to the concentration of my unknown IN THE SOLUTION), it took 7.5ml of NaOH.

Homework Equations


C1V1=C2V2
M=n/V , where n is number of moles, and V is volume of solution in liter.


The Attempt at a Solution


Basically, I did C1V1=C2V2.
C1= 0.0189M V1=7.5ml C2=? V2=32.5ml (25ml + 7.5ml)
I then get C2=0.00436.

When I solve of n (in M=n/V) and then do grams/moles to get molar mass, I get a molar mass that is impossible, or waay off... So I'm not sure anymore if I am approching this correctly...Is my V2 correct? How about the V I take for M=n/V, is it 0.100L (initial volume at which I diluted the solid).

Thanks in advance for any help.

PS: This is my first post on this forum, and it seems really a great forum!
 
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redioactif said:
Hi,

Homework Statement


It is in my chemistry lab course. I had 0.1g of an unknown solid that I diluted in 100ml of distilled water. Then, I took 25ml and titrated it with NaOH (Concentration = 0.0189M). To reach equivalent point (the point where concentration of NaOH is equal to the concentration of my unknown IN THE SOLUTION), it took 7.5ml of NaOH.

Homework Equations


C1V1=C2V2
M=n/V , where n is number of moles, and V is volume of solution in liter.


The Attempt at a Solution


Basically, I did C1V1=C2V2.
C1= 0.0189M V1=7.5ml C2=? V2=32.5ml (25ml + 7.5ml)
I then get C2=0.00436.

When I solve of n (in M=n/V) and then do grams/moles to get molar mass, I get a molar mass that is impossible, or waay off... So I'm not sure anymore if I am approching this correctly...Is my V2 correct? How about the V I take for M=n/V, is it 0.100L (initial volume at which I diluted the solid).

Thanks in advance for any help.

PS: This is my first post on this forum, and it seems really a great forum!

I don't think your V2 value is right, as you are only titrating 25 ml at the start. Also don't forget that the 25 ml is not the complete sample.
 
OK. If I put my V2 to 25ml, it gives an answer that makes a lot of sens.
However, I don't understand the logic behind it. I mean, the final volume of titration is NOT 25ml, but rather 25ml + 7.5 ml. The initial concnetration of NaOH that I have is for 7.5ml, but as we add that 7.5 to 25, it gets diluted, and the final volume would not be 25...
Can someone please help me understand this, or is my reasoning completley wrong?
 
redioactif said:
OK. If I put my V2 to 25ml, it gives an answer that makes a lot of sens.
However, I don't understand the logic behind it. I mean, the final volume of titration is NOT 25ml, but rather 25ml + 7.5 ml. The initial concnetration of NaOH that I have is for 7.5ml, but as we add that 7.5 to 25, it gets diluted, and the final volume would not be 25...
Can someone please help me understand this, or is my reasoning completley wrong?

The base (NaOH) is only in the 7.5 ml sample, isn't it. There is no additional base in the sample of 25 ml that you made up to titrate against this.
 
No there isn't. I had 25ml of acid only.

I added 7.5ml to reach equivalence point in the titration.

Apparently, my V2 has to be 25ml, but I don't understand why: shouldn't the final volume should include the added NaOH volume + the acid volume (25 + 7.5) ? I'm really confused!
 

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