Buffer solution + equilibrium + Organic compound

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Discussion Overview

The discussion revolves around the application of the acid dissociation constant (Ka) in the context of buffer solutions and equilibrium involving organic compounds, specifically acetic acid and its conjugate base acetate. Participants are exploring the correct formulation of Ka in relation to equilibrium concentrations versus initial concentrations modified by dissociation.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the expression for Ka is given as Ka =(ch3coo-)(h+)/(ch3cooh) instead of using initial concentrations modified by dissociation, suggesting a misunderstanding of equilibrium concentrations.
  • Another participant emphasizes the importance of using equilibrium concentrations in the calculation of Ka, indicating that the first equation reflects this approach.
  • A later reply clarifies that the confusion may stem from the representation of sodium acetate (CH3COONa) and its dissociation in aqueous solution, noting that the addition of HCl leads to the protonation of acetate ions.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct approach to calculating Ka, with differing views on the use of equilibrium versus initial concentrations and the implications of dissociation in buffer solutions.

Contextual Notes

There are unresolved assumptions regarding the definitions of concentrations used in the calculations, as well as the extent of dissociation that may affect the interpretation of the equilibrium expressions.

myvow
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For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
20141228_7bca95847216b430816fm7mRBFmSHq75.png
 
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myvow said:
why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x)

First equation uses equilibrium concentrations, second uses initial concentrations modified by the extent of the dissociation, to calculate equilibrium concentrations. Do you understand what these means?
 
myvow said:
For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
20141228_7bca95847216b430816fm7mRBFmSHq75.png

You have now modified your initial post. I do not understand what you have in mind by your x.
You might have been confused by the CH3COONa. That is a conventional formula, it really just represents the composition of sodium acetate but in aqueous solution and for that matter in crystal sodium acetate is all Na+ and CH3COO-. When you add to this "HCl" - I.e. H+ + Cl- - a fraction of the acetate ions CH3COO- become protonated to form CH3COOH and then the calculations you are asked about involve that fraction, which actually determines the pH as you can see from the given formulae.
 
Last edited:

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