Calculating equilibrium constant rules

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Discussion Overview

The discussion revolves around calculating the equilibrium constant for the reaction between acetic acid (CH3COOH) and nitrite ion (NO2-), leading to acetate ion (CH3COO-) and nitrous acid (HNO2). Participants explore the relationships between the equilibrium constants of the involved species and the rules governing their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant attempts to find the equilibrium constant by using the acid dissociation constant (Ka) for acetic acid and the base dissociation constant (Kb) for nitrite ion, but expresses uncertainty about the rules for combining these constants.
  • Another participant suggests writing the K expression for the equilibrium and comparing it to the Ka expressions for both CH3COOH and HNO2.
  • A participant claims to have found the correct solution by multiplying Ka1 and the reciprocal of Ka2, questioning whether this reciprocal is equivalent to Kb.
  • Another response clarifies that while the reciprocal of Ka2 is not the same as Kb, they are related through the equation Ka * Kb = Kw, cautioning about changes in indices.
  • A further inquiry is made about whether the correct solution was derived through numerical calculation or by rearranging the Ka expressions to relate them to the original equilibrium constant.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the equilibrium constant, with some finding a solution through specific manipulations of Ka and Kb, while others question the validity of these methods. There is no consensus on the rules for combining these constants.

Contextual Notes

Participants mention the need to be cautious about how indices change when manipulating the constants, indicating potential limitations in their understanding of the relationships between Ka, Kb, and Kw.

kasse
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Homework Statement



Find the eq. constant for the reaction

CH3COOH (aq) + NO2- (aq) = CH3COO- (aq) + HNO2 (aq)

The Attempt at a Solution



I find Ka for the reaction CH3COOH = CH3COO- + H+ and Kb for the reaction NO2- + H+ = HNO2

But what are the rules then? I try multiplying the constants, but this does not give the correct answer.
 
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kasse said:

Homework Statement



Find the eq. constant for the reaction

CH3COOH (aq) + NO2- (aq) = CH3COO- (aq) + HNO2 (aq)

write the K expression for this equilibrium

then compare it to the Ka expressions for both CH3CO2H AND HNO2
 
I managed to find the correct solution by multiplying Ka1 and the 1/Ka2. Is 1/Ka2 the same av Kb?
 
kasse said:
I managed to find the correct solution by multiplying Ka1 and the 1/Ka2. Is 1/Ka2 the same av Kb?

Did you just happen to work out the numbers and find the right answer or did you see that rearranging the Ka expressions for the acids gave Ka(CH3CO2)/Ka(HNO2) which was equal to the K expression for the original equilibrium?
 

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