Hey, good question!

It is true that when you titrate a strong acid and strong base, you get an equivalence point at pH 7 becasue [OH-] = [H+], and the acids cancel each other out.

However, in the cases of weak acids and bases, you get some additional dissociation of the conjugate acid/conjugate base which will react further, shifting the pH of the equivalence point.

The simple way I like to remember is the strong acid or base will overpower the weak one.

I.e. a strong acid and weak base has an acidic equivalence point
A strong base and weak acid has a basic equivalence point
A strong base and a strong acid completely eliminate eachother
For Weak acids and weak bases it depends on which is the strongest acid or base in the mixture

For a more concrete proof of this, when considering the titration of 1 M acetic acid (CH3COOH) with 1 M NaOH solution, we get an equivalence point where [OH-] = [CH3OOH], therefore all of your CH3OOH gets deprotonated to CH3OO-, and all of the OH- gets consumed.

However, recall that acetic acid is a weak acid, therefore CH3OO- is a weak base with Kb ~ 1x10^-9, and will redissociate in solution as follows:

pKb = [OH-][CH3OOH]/[CH3OO-]

Because we have 1 M CH3OO-, and it dissociates to form a small (x) amount of 1:1 ratio of OH- to CH3OOH, we can simplify to:

pKb = 1x10^-9 = (x)(x)/(1-x)

Because the acid dissociation constant is so small, we assume x << 1 and eliminate it in the denominator, leaving:

1x10^-9 = x^2 where x = [OH-]

We can then solve for x, which is equal to the concentration of OH-.

This gives us:

[OH-] = 1x10^-4.5
pOH = 4.5
pH = 9.5 at the equivalence point.

I hope this helps!