Why Are A- and HA Concentrations Equal at Half Equivalence Point?

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Discussion Overview

The discussion revolves around the concept of titration, specifically why the concentrations of the conjugate base (A-) and the weak acid (HA) are equal at the half equivalence point. Participants explore the implications of this equality on pH and pKa relationships, as well as the underlying stoichiometry involved in the titration process.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant states that at the half equivalence point, the concentrations of A- and HA are equal, leading to pH being equal to pKa.
  • Another participant suggests considering the neutralization stoichiometry and the dissociation of the weak acid, implying that the weak acid's dissociation is negligible.
  • A participant presents a dissociation model using an ICE table to illustrate the relationship between the concentrations of HA and A- at the half equivalence point.
  • Another participant challenges the use of the ICE table, suggesting that simple stoichiometry should be followed instead.
  • A later reply references the Henderson-Hasselbalch equation to explain that when [A-] equals [HA], the pH equals pKa, indicating a deeper relationship between these quantities.

Areas of Agreement / Disagreement

Participants generally agree on the conclusion that at the half equivalence point, the concentrations of A- and HA are equal and that this leads to pH equaling pKa. However, there is disagreement on the methods used to arrive at this conclusion, particularly regarding the use of stoichiometry versus ICE tables.

Contextual Notes

Some limitations include the assumptions made about the dissociation of the weak acid and the applicability of the Henderson-Hasselbalch equation in different contexts. The discussion does not resolve the debate over the preferred method for analyzing the situation.

Who May Find This Useful

This discussion may be useful for students and educators in chemistry, particularly those studying acid-base titrations and the principles of equilibrium in weak acid systems.

Oerg
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Hi All

I have a question that I require an answer to. For the above titration, concentration of A-(conjugate base of acid) is the same as concentration of HA at half equivalence. It is at this point that pka=ph of the solution. I agree to this conclusion, but why is the concentration of A- equals to concentration of HA at this point? Any help will be deeply appreciated. Thanks.
 
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Write reaction equation, think in terms of the neutralization stoichiometry. This is a weak acid, so it dissociates very slightly on its own - so slightly, that its own dissociation can be ignored.
 
so, let HA be a weak acid

let x be the amount of acid dissociated. so starting with 1 mol of HA

------------HA <-> H+ + A-
conc--------1-x ----x ---x

in order to achieve the required concentration, conc of HA = conc of A-

1-x=x
x=0.5

0.5 mol of H+ is displaced which requires 0.5 mol of Naoh which is also half of total naoh required, am i right?
 
You are mostly right, but you are using ICE instead of following simple stoichiometry.

HA + NaOH -> NaA + H2O
 
You can see this pretty easy from the Henderson Hasslebach equation:

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

When [A-] is equal to [HA], you get log 1 in the above equation which is equal to 0. So pH = pKa. The Henderson Hasslebach equation is derived from the standard Ka equilibrium equation, and you can see that derivation if you want a deeper understanding.
 

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