Why does equilibrium favour weak acid or weak base?
Here is a simple intuitive explanation. In your above example, there is actually 2 reactions occurring, the forward and the reverse reaction. Let us consider the forward reaction. Here, both $\ce{NaOH}$ and $\ce{HCl}$ are strong bases and acids. What this means is that HCl has a strong ability to donate a proton and NaOH has a strong ability to accept protons. This means that when NaOH and HCl react together, the reaction will basically go to completion. So you will end up with up with very little NaOH  and HCl and the equilibrium will favour the products.

Now lets look at the reverse reaction which is: $$\ce{Na+ + Cl- + H2O -&gt; NaOH + HCl}$$ Here $\ce{Cl-}$ is the conjugate base of HCl which is  a strong acid. This means that $\ce{Cl-}$ is a really weak base, meaning that it is pretty bad at pulling protons from molecules. Also water is the conjugate acid of NaOH, which is a strong base. This means that water is a really weak acid, meaning that it is also pretty bad at kicking out one of its protons. This means that in this reaction, barely any of the products will react and you will mainly end up with NaCl and water.

As you can see in both reaction the equilibrium will favour the weak acid and base as they aren't able to react that well together. This effect can also be explained by using equilibrium constants but I won't go into that.