Acid dissociation constant and equilibrium constant

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](https://i.stack.imgur.com/HxEqB.png)
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!




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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