What is the relationship between acid dissociation constant and equilibrium constant for an acid's reaction with water? If we have this reaction:

$\ce{CH3COOH + H2O → CH3COO- + H3O+}$

I would calculate the equilibrium constant to 0,005241 M at 25 celcius enter image description here Acid dissociation constant for acetic acid is as known 1,75 × 10$^{−5}$. How can this be true? The equilibrium constant I have calculate is about 300 times larger than the acid dissociation constant!


1 Answer 1


An acid dissociation constant is a particular example of an equilibrium constant. For the reaction: $$\ce{CH3COOH +H2O <=>CH3COO- +H3O+}$$ The equilibrium constant would be: $$\newcommand{\ac}[1]{\text{activity of }\ce{#1}}{\bf K_{eq}}=\frac{\ac{CH3COO-}\cdot\ac{H3O+}}{\ac{CH3COOH}\cdot\ac{H2O}}$$ And the acid dissociation constant would be: $${\bf K_a}=\ce{\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$ Now the relation between both would be: $$\newcommand{\acc}[1]{\gamma_{\ce{#1}}}{\bf K_{eq}}={\bf K_a}\underbrace{\frac{\acc{CH3COO-}\acc{H3O+}}{\acc{CH3COOH}\acc{H2O}}}_{\Gamma_0}\frac1{\ce{[H2O]}}$$ Now: $${\bf K_a}={\bf K_{eq}}[\ce{H2O}]\Gamma_0^{-1}$$ So you're missing some factors.They're not exactly the same.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.