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

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SUMMARY

The discussion focuses on the titration of an unknown acid using NaOH with a concentration of 0.0189M. The participant diluted 0.1g of the acid in 100ml of distilled water and titrated 25ml of this solution, requiring 7.5ml of NaOH to reach the equivalence point. The key equations used are C1V1=C2V2 and M=n/V. The confusion arises regarding the correct volume (V2) to use in calculations, with suggestions that V2 should be 25ml instead of 32.5ml, as the titration only involves the acid solution.

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  • Understanding of titration principles and equivalence points.
  • Familiarity with molarity calculations and the concept of moles.
  • Knowledge of the equations C1V1=C2V2 and M=n/V.
  • Basic laboratory skills in preparing solutions and performing titrations.
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  • Review the concept of equivalence points in titration and how to identify them.
  • Study the implications of dilution on concentration calculations.
  • Learn about the significance of the final volume in titration calculations.
  • Explore common pitfalls in titration calculations and how to avoid them.
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Chemistry students, laboratory technicians, and educators involved in teaching titration techniques and acid-base chemistry.

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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