I am trying to calculate the equilibrium constant of the following equilibrium:
$$\ce{CH3COOH (aq) + OH- (aq) <=> CH3COO- (aq) + H2O (l)}$$
I am aware that this reaction essentially goes to completion (the position of the equilibrium lies far to the right). However, I still would like to calculate the equilibrium constant.
I know that $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ (I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$). I know that $K_\mathrm{a}(\ce{CH3COOH}) = 1.8\times10^{-5}$ and that $K_\mathrm{w} = 10^{-14}$. However, I am struggling to determine what $K_\mathrm{b}(\ce{OH-})$ is.
I learnt that $K_\mathrm{b}$ is called the "base ionization constant." However, $\ce{OH-}$ is a base, but it is already an ion! So, how can something that is already an ion, ionize? According to this webpage, $\ce{OH-}$ has a $K_\mathrm{b}$ of 1.0. However, what is the equilibrium that led to this number? Is it $\ce{OH- + H2O <=> H2O + OH-}$?
Moreover, if the $K_\mathrm{b}$ of $\ce{OH-}$ is 1.0, is $\ce{OH-}$ considered to be a weak base? Generally, strong bases have a $K_\mathrm{b}$ that is much greater than 1.0. On the other hand, if bases that contain hydroxide ions, such as sodium hydroxide and potassium hydroxide, are strong bases, shouldn't the hydroxide ion itself also be a strong base?