Weak Acid/Base Equilibria: C6H5OH & C6H5NH2 K_a & K_b

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Discussion Overview

The discussion revolves around the equilibrium expressions for the weak acid C6H5OH and the weak base C6H5NH2 in water, focusing on their respective Ka and Kb values. Participants explore the theoretical framework for deriving these expressions and the implications of the equilibrium constants.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests writing the equilibrium reactions for the ionization of C6H5OH and C6H5NH2 in water to derive Ka and Kb expressions.
  • Another participant proposes that the equilibrium constant expression will include the concentration of water, which is constant and cancels out, leading to a simplified form of Ka and Kb.
  • A third participant directs others to a resource for definitions of Ka and Kb and suggests ignoring second dissociation steps.
  • One participant emphasizes the importance of showing work and implies that prior knowledge from the text is necessary to engage effectively in the discussion.

Areas of Agreement / Disagreement

Participants express differing approaches to deriving the equilibrium expressions, with no consensus on the best method or the completeness of the explanations provided. The discussion remains unresolved regarding the specific steps to take.

Contextual Notes

Some participants reference the need for prior knowledge and understanding of the concepts involved, indicating that assumptions about the audience's familiarity with the material may vary.

Who May Find This Useful

Students studying weak acid/base equilibria, educators looking for examples of equilibrium expression derivation, and individuals interested in chemical equilibrium concepts.

APool555
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C_6H_5OH acts as a weak acid and C_6H_5NH_2 acts as a weak base in water. Write equalibrium expressions for K_a and K_b.
 
Last edited:
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Hi my friend. What you should do here is write first the equilibrium reaction for these two substances ionisation in water, then deduce an expression for Kc-eqm constant. This expression will include the concentration of water, which is constant and thus cancels out of the equation. Then replace Kc by simply Ka or Kb, a new sort of equilibrium constant for these reactions. What you shoulb find is the Ka/Kb is simply the product of the concentrations of the conjugate base/acid and the new acid/base divided by the initial concentration of base/acid. Yours, Joe
 
First, show your work, if you had read the text you should at least know how to start.
 

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