Calculating equilibrium constant rules

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To calculate the equilibrium constant for the reaction CH3COOH + NO2- = CH3COO- + HNO2, the correct approach involves using the acid dissociation constant (Ka) for acetic acid and the base dissociation constant (Kb) for nitrite. The relationship between Ka and Kb is defined by the equation Ka * Kb = Kw, where Kw is the ion product of water. The correct equilibrium constant can be found by multiplying Ka for acetic acid by the reciprocal of Ka for HNO2, which is equivalent to Kb. Understanding how to manipulate these constants is crucial for accurately determining the equilibrium constant for the reaction.
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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