Calculating pKa from pH: Understanding Titration

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Discussion Overview

The discussion revolves around calculating the pKa of a weak acid (HA) from a titration with NaOH, specifically focusing on the conditions at the equivalence point and the implications for pH and pKa calculations. The scope includes mathematical reasoning and conceptual clarification related to acid-base titration principles.

Discussion Character

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

Main Points Raised

  • One participant presents a calculation for pOH based on the pH at the equivalence point and attempts to derive pKa using the equation pH = pKa + log([A]/[HA]).
  • Another participant challenges the logic of subtracting moles of OH- that reacted with HA, suggesting that OH- cannot be both present in the solution and react with the acid simultaneously.
  • A different participant questions the volume used in calculations, emphasizing the need to account for the total volume after adding NaOH to the HA solution.
  • One participant comments on the geographical context of the problem, suggesting that the pH at the equivalence point can vary significantly and that understanding the relationship between pH and pKa is crucial.
  • There is a suggestion to explicitly write out the formula for Ka to clarify the relationship between the concentrations of the species involved.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions made regarding the moles of OH- and their role in the titration process. There is no consensus on the correct approach or resolution of the calculations presented.

Contextual Notes

Limitations include potential misunderstandings about the roles of species at the equivalence point, the need for clarity in volume calculations, and the variability of pH at the equivalence point depending on the specific acid-base system involved.

Frankenstein19
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Homework Statement


When 100.0 mL of weak acid HA was titrated with 0.09381M NaOH, 27.63mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What is the pKa of HA

Homework Equations


pH=pKa+log([A]/[HA]

[OH]-10^(-pOH)

The Attempt at a Solution



14-10.99=3.01 this gives me pOH

From this I can calculate the concentration of OH: 0.000977237 M

Since we are at the equivalence point, to get the moles I multiplied the above concentration by 0.02763L and got 2.7*10^(-5)

I know that initially I had 0.0025919703 moles of HA

I subtract the moles of OH that react with HA and get 0.0025919703-2.7*10^(-5)=2.56399*10^(-3)

The moles of OH are the moles of A: 2.7*10^(-5)

Using pH=pKa+log([A]/[HA]

I get 10.99=pKa-1.97

Giving me a pH of 12.96 instead of 9.69 which is the correct answer

What am I doing wrong
 
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Frankenstein19 said:
I subtract the moles of OH that react with HA and get 0.0025919703-2.7*10^(-5)=2.56399*10^(-3)

Doesn't make much sense. OH- can't be at the same time present in the solution (pOH) and be the one that reacted with the acid. Either, either.

The moles of OH are the moles of A: 2.7*10^(-5)

No.

Hint: pH at the equivalence point is that of a salt of a weak acid. Can you write reaction that is responsible for the OH- production in such a solution?
 
Frankenstein19 said:
to get the moles I multiplied the above concentration by 0.02763L and got 2.7*10^(-5)
Are you using the correct volume. Remember you have added the NaOH solution to the HA solution.
But Borek's point still applies.
 
Can I ask what country you are studying in? This question has certain geographical hallmarks.I would probably fail the exam there because I would write as answer " good luck with finding the pH at the equivalence point!".I would know where the equivalence point is alright but the pH there is varying wildly. Anyway you have worked out the molarity of the NaA solution, now suppose we dissolved NaA In water to make that molarity, to ask what the pH of that solution would be, knowing that pKa, would make sense. So would the inverse question, knowing the pH what is the pKa?

It is too late at night for me to concentrate on your calculation, but there might be a mistake where you say the moles of OH-, are the moles of A?? - rather they are the moles of moles of HA aren't they? I suggest you write out the formula for Ka explicitly and then play with that – you should get a formula with a square in it that you may remember.
 

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