Acetic acid has a $K_\mathrm{a}$ of $\pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $\ce{NaOH}$?
Given acetic acid
$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$
$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$, then we get $K_\mathrm{b} = \pu{5.55e-10}$. How do I use this to find $K_\mathrm{eq}$?
I know how to find $K_\mathrm{eq}$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?