PH of a polyprotic acid using multiple equilibria - is this correct?

Click For Summary

Discussion Overview

The discussion revolves around calculating the pH of a 0.02025 M solution of tartaric acid using the method of multiple equilibria, specifically addressing the contributions of its dissociation constants (Ka1 and Ka2) and the implications of intermediate species in the calculations.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests verification of their method and calculations for determining the pH of tartaric acid.
  • Another participant expresses difficulty in accessing the provided document and suggests using LaTeX or images instead.
  • A participant asserts that the pH of the solution is 2.41, indicating that calculations must align with this result to be considered correct.
  • Another participant notes a discrepancy in their own calculations using a step-by-step equilibrium method, suggesting both methods should yield the same answer.
  • There is a discussion about the importance of considering the presence of intermediate species (HT-) in the calculations, with one participant arguing that it should not be ignored.
  • A later reply acknowledges the oversight regarding the overall reaction and the necessity of accounting for intermediate reactions in the derivation.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the pH, particularly regarding the treatment of intermediate species and the validity of their respective methods. No consensus is reached on the correct pH value or method.

Contextual Notes

Participants highlight limitations in their approaches, including potential missing intermediate reactions and the implications of using different methods for equilibrium calculations.

FredericChopin
Messages
101
Reaction score
0

Homework Statement


Find the pH of 0.02025 M tartaric acid using the method of multiple equilibria given that its Ka1 value is 9.2*10-4 and its Ka2 value is 4.31*10-5.


Homework Equations


Kaoverall = Ka1 * Ka2


The Attempt at a Solution


Please check the Word document.

Is my method and calculation correct?

Thank you.
 

Attachments

Physics news on Phys.org
For some reason I can't open the file. .docx is not the best format. Try to post it here using LaTeX, or - at worst - as an image.
 
Hm... that's strange.

I'll upload images files then (LaTeX is too much trouble...)

They're in order of page number.
 

Attachments

  • Is this correct-0.jpg
    Is this correct-0.jpg
    11.7 KB · Views: 581
  • Is this correct-1.jpg
    Is this correct-1.jpg
    11.7 KB · Views: 556
  • Is this correct-2.jpg
    Is this correct-2.jpg
    19.8 KB · Views: 569
  • Is this correct-3.jpg
    Is this correct-3.jpg
    8.4 KB · Views: 555
Seems like you have completely ignored presence of HT-.

pH of the 0.02025 M tartaric acid is 2.41, as long as your calculations don't reproduce this result they are wrong.
 

Attachments

  • Untitled-1.png
    Untitled-1.png
    34.9 KB · Views: 620
The reason I'm asking is because I answered the same question but instead used the step-by-step equilibrium method and I got a different answer.

I can't see where I went wrong though. Both methods should yield the same answer.

They're in order of page number.
 

Attachments

  • pH Calculations 2-0.jpg
    pH Calculations 2-0.jpg
    13.1 KB · Views: 599
  • pH Calculations 2-1.jpg
    pH Calculations 2-1.jpg
    20.8 KB · Views: 567
  • pH Calculations 2-2.jpg
    pH Calculations 2-2.jpg
    15.2 KB · Views: 536
  • pH Calculations 2-3.jpg
    pH Calculations 2-3.jpg
    14.5 KB · Views: 581
  • pH Calculations 2-4.jpg
    pH Calculations 2-4.jpg
    3.1 KB · Views: 597
Borek said:
Seems like you have completely ignored presence of HT-.

pH of the 0.02025 M tartaric acid is 2.41, as long as your calculations don't reproduce this result they are wrong.

Hm... but wouldn't it cancel out in the equation?
 
No. Just because it cancels out in a partial system of equations doesn't mean it shouldn't be taken into account. Overall equation doesn't say a word about intermediate steps, which doesn't mean they don't exist.

Correct derivation is shown on the page I already linked to.
 
Ah. You're right. I see where I went wrong.

It's not working because I'm writing an overall reaction that doesn't exist. And it wouldn't work unless there's a third reaction. Otherwise, as you said, I would miss the intermediate reactions.

Thank you.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K