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Calculating equilibrium constant rules

  • Thread starter kasse
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  • #1
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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.
 

Answers and Replies

  • #2
133
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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
 
  • #3
366
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I managed to find the correct solution by multiplying Ka1 and the 1/Ka2. Is 1/Ka2 the same av Kb?
 
  • #5
133
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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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