Weak base titrated with a strong acid
Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$  (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution. 

As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.   

Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived here.

For the equilibrium     

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

the acid dissociation constant has the form

$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$

As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as

$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$

To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule). 

So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.

I hope this helps, and please correct me if I've made any errors.