# What is the equilibrium constant of CH3COOH?

Jan 9, 2015

When placed in water, weak acids (generic $H A$) form a homogeneous equilibrium in which acid molecules react with water to form aqueous hydronium ions, ${H}_{3}^{+} O$, and aqueous anions, ${A}^{-}$.

In the case of acetic acid, which is a weak acid, the equilibrium can be described like this:

$C {H}_{3} C O O {H}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C H O {O}_{\left(a q\right)}^{-} + {H}_{3}^{+} {O}_{\left(a q\right)}$

The equilibrium constant is

${K}_{e q} = \frac{\left[{H}_{3}^{+} O\right] \cdot \left[C {H}_{3} C H O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right] \cdot \left[{H}_{2} O\right]}$

Since the concentration of liquid water is left out of the expression, the equilibrium constant for this reaction is called acid dissociation constant, ${K}_{a}$

${K}_{a} = \frac{\left[{H}_{3}^{+} O\right] \cdot \left[C {H}_{3} C H O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}$

The values of the acid dissociation constants for various acids are usually given to you in an exam, acetic acid's equilibrium constant being $1.8 \cdot {10}^{- 5}$; however, if the value is not given to you, you can always use the equilibrium concentrations described in the above equation to solve for ${K}_{a}$.