Why do strong acid-weak base reactions "essentially go to completion"?
This is a matter of equilibrium. Let's consider the  acid-base reaction of $\ce{HCl}$ and $\ce{NH3}$:

$$\ce{NH3 + HCl &lt;=&gt; NH4+ + Cl-}$$

Which can be broken apart into:

$$\ce{NH3 + H2O &lt;=&gt; NH4+ + OH-}$$

$$\ce{HCl + H2O &lt;=&gt; H3O+ + Cl-}$$

For these reactions at equilibrium:

$$K_\mathrm{eq}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \hspace{1cm} K_\mathrm{b}(\ce{NH3})=\ce{\frac{[NH4+][OH-]}{[NH3]}} \hspace{1cm} K_\mathrm{a}(\ce{HCl})=\ce{\frac{[H3O+][Cl-]}{[HCl]}} $$

$$K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) = \ce{\frac{[H3O+][Cl-]}{[HCl]}} \cdot \ce{\frac{[NH4+][OH-]}{[NH3]}}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \cdot \ce{[H3O+][OH-]} = K_\mathrm{w}K_\mathrm{eq}$$

$$K_\mathrm{w}K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) \Rightarrow$$

$$K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot \frac{K_\mathrm{b}(\ce{NH3})}{K_\mathrm{w}}=\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \Rightarrow$$

$$\mathrm{p}K_\mathrm{eq} = -\log \left(\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \right) = \log \left(\frac{K_\mathrm{a}(\ce{NH4+})}{K_\mathrm{a}(\ce{HCl})} \right) = \mathrm{p}K_\mathrm{a}(\ce{NH4+}) - \mathrm{p}K_\mathrm{a}(\ce{HCl})=9.24-(-8)=17.24$$

Just from the $\mathrm{p}K_\mathrm{a}$ difference of the acid and conjugate acid of our reaction, we figured out that at equilibrium the products are present in a $10^{17.24}:1$ ratio to the reactants, meaning the reaction has essentially gone to completion.