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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!

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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.

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