Chem Lab Titration: Finding Unknown Molarities

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Discussion Overview

The discussion revolves around a chemistry lab experiment involving titrations to determine the molarities of unknown solutions. Participants are exploring the calculations and formulas used in the titration process, particularly focusing on the relationship between the molarity, volume, and number of moles of the acid and base involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes their approach to calculating the molarity of NaOH using the equation M1V1=M2V2, but expresses confusion over the results from a checking program.
  • Another participant suggests that the issue may lie with the checking program rather than the calculations performed.
  • A different participant introduces a more complete formula that includes the number of moles of acid and base, indicating that N1 and N2 are necessary for the calculations and can be determined from the balanced equation of the titration.
  • One participant provides a calculation for the moles of acid and NaOH, asserting that the moles of acid equal the moles of NaOH and that this can be used to find the molarity of NaOH through algebra.
  • Another reiterates the importance of including N1 and N2 in the calculations and explains that (M1)(V1) equals N1, emphasizing the relationship between these variables.

Areas of Agreement / Disagreement

Participants generally agree on the need to include the number of moles in the calculations, but there is no consensus on the correctness of the checking program or the specific calculations being performed. Multiple viewpoints on the approach to the problem remain present.

Contextual Notes

Some participants note the necessity of writing the balanced equation for the titration to determine the number of moles, which remains an unresolved step in the discussion.

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We recently did an experiment in chem lab with titrations and we are supposed to find out the molarities of some unknown solutions. I thought that I was doing my calculations right, but the checking program that I am using keeps telling me that I am wrong.
We made a standard solution that was 0.09769 M and was 250 mL. We used 25mL aliquots of this solution to titrate against an unknown concentration of NaOH. The mean volume of the base that was used was 23.10 mL. I thought, since our acid was KHP with one titratable proton, that I could just use the equation M1V1=M2V2 like so: (0.09769M)(25.00mL)=M2(23.10mL)...can you please explain where my thinking where wrong?
 
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You did nothing wrong. Perhaps the problem is with the checking program.
 
I think the correct/complete formula is :

(M1)(V1)/N1 = (M2)(V2)/N2

where:
M1, M2 = Molarity of the acid and the base
V1, V2 = Volume of the acid used and the mean volume of the base (Na OH)
N1, N2 = No. of moles of acid and base

Your equation is perfectly correct, but only N1 and N2 are missing.

N1 and N2 can be found out by writing the balanced equation of the Titration.

I hope i have helped. If i am wrong please correct me.
 
How many moles of acid = (0.09769 moles/liter)*(0.025 liter)

How many moles of NaOH = (M)*(0.02310 liter)

The moles of acid equals the moles of NaOH. You have the information to find M in moles/liter. Very simple algebra.
 
HALO3 said:
I think the correct/complete formula is :

(M1)(V1)/N1 = (M2)(V2)/N2

where:
M1, M2 = Molarity of the acid and the base
V1, V2 = Volume of the acid used and the mean volume of the base (Na OH)
N1, N2 = No. of moles of acid and base

Your equation is perfectly correct, but only N1 and N2 are missing.

N1 and N2 can be found out by writing the balanced equation of the Titration.

I hope i have helped. If i am wrong please correct me.

Actually, (M1)(V1) = N1 if M1 is in moles/liter and V1 is in liters. Thus the expression (M1)(V1)/N1 is equal to 1. The same goes for M2, V2 and N2. The resulting expression (my favorite, 1=1) is always true.
 

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