How Do You Calculate the pH at Equivalence Points in Malonic Acid Titration?

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Discussion Overview

The discussion revolves around calculating the pH at the equivalence points during the titration of malonic acid, a dibasic acid, with sodium hydroxide. Participants explore the theoretical and practical aspects of determining pH values at these points, including the buffering capacity and the nature of the resulting solutions.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant states the pKa values of malonic acid and attempts to calculate the pH at the first equivalence point, suggesting a pH of 7.77 based on their calculations.
  • Another participant challenges the correctness of the first endpoint pH value provided in the discussion, stating it is wrong and indicating that the solution at this point is an amphiprotic salt.
  • Subsequent posts express uncertainty about the correctness of the initial calculations and seek clarification on how to approach the problem.
  • There is a suggestion that hints were provided to guide the discussion, but the effectiveness of communication is questioned.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct pH values at the equivalence points, with multiple competing views and uncertainties present throughout the discussion.

Contextual Notes

Participants express confusion regarding the calculations and the expected pH values, indicating potential missing assumptions or misunderstandings about the nature of the solutions at the equivalence points.

nmr
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Malonic acid can ionise in two stages as it is a dibasic acid.
The values of pKa1 and pKa2 of malonic acid are 2.85 and 5.70 respectively. Calculate the pH of the first and second equivalence points of the titration of 10cm^3 of malonic acid of concentration 0.1 M with sodium hydroxide of concentration 0.1 M.

my attempt at a solution:
maximum buffering capacity before first endpoint: pH=pKa1 = 2.85

[OH-] = [(Kw/Ka1)*(0.05)]^(0.5)
therefore pOH = 6.23
pH = 14-6.23 = 7.77

im stuck here as the answer given for the pH of the first endpoint is 3.50 and I have no idea how to work it out
 
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First endpoint pH given as an answer is wrong, so don't worry if you can't reproduce it.

At first endpoint you have a solution of an amphiprotic salt.
 
so is my pH value correct ?
 
Nope.
 
how would i go about solving for the pH then?
 
I have pointed you in the right direction in my first answer.

If you plan to ignore hints please tell, so that I don't waste my time answering.
 
ah ok sorry
 

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