Why equilibrium favours weak acid or weak base

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Discussion Overview

The discussion revolves around the concept of chemical equilibrium, specifically focusing on why equilibrium favors weak acids or weak bases. The context is primarily homework-related, with participants exploring the reasoning behind equilibrium positions in acid-base reactions.

Discussion Character

  • Homework-related, Exploratory

Main Points Raised

  • One participant expresses uncertainty about why equilibrium favors weak acids or weak bases, indicating a need for clarification.
  • Another participant suggests expressing the equilibrium constant (K) of the reaction using the acid dissociation constants (Ka) for both acids involved.
  • A mathematical expression for K is provided, specifically k=10^{-0.6}, though its relevance to the original question is not fully explored.
  • A later reply confirms that the provided expression is indeed the answer, but does not elaborate on the reasoning behind it.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the underlying reasons for why equilibrium favors weak acids or weak bases, as the initial question remains largely unanswered despite some mathematical input.

Contextual Notes

Participants do not clarify the assumptions behind the use of pKa values or the implications of the provided K expression, leaving some aspects of the discussion unresolved.

Titan97
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Homework Statement


IMG_20151011_101552_639.JPG


Which side does the equilibrium favour?

Homework Equations


Equilibrium favours weak acid or weak base (I don't know why :frown:)

The Attempt at a Solution


Using the given pka values, since the reactant is the weaker acid, reactants are favoured. (Backward reaction is favoured).
But why does equilibrium favour weak acid or weak base?
 
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Try to express the K of the reaction using Ka values for both acids.
 
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##k=10^{-0.6}##
 
That's your answer, isn't it?
 
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Yes.
 

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