# In a weak acid-strong base titration why is [H+] no longer equal to [A-] after addition of the base? (A2 level chemistry)

I understand that before any base is added $$\ce{[H+]}$$ is equal to $$\ce{[A-]}$$ but why is $$\ce{[H+]}$$ no longer equal to $$\ce{[A-]}$$ when the strong base is added?

If all the following reactions below are occurring in solution and both $$\ce{CH3COO-}$$ an $$\ce{H+}$$ are able to react in either one of two of these different reactions then why should the $$\ce{[H+]}$$ no loner equal $$\ce{[A-]}$$ if both have equal opportunities to be consumed? (I am aware that not all the $$\ce{CH3COOH}$$ will have dissociated but surely this would not affect the concentrations of either $$\ce{[H+]}$$ or $$\ce{[A-]}$$?)

$$\ce{Na+ + CH3COO- -> CH3COONa}$$ $$\ce{CH3COO- + H+ -> CH3COOH}$$ $$\ce{H+ + OH- -> H2O}$$

Is it because the salt $$\ce{CH3COONa}$$ is much more likely to dissociate than $$\ce{CH3COOH}$$ and $$\ce{H2O}$$ meaning more $$\ce{CH3COO-}$$ remains in solution than $$\ce{H+}$$?

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$
The base removes the hydrogen ions from solution, and the acid further dissociates to maintain LeChatelier's equilibrium. The $$\ce{H3O+}$$ ion concentration will decrease as it is continually removed whereas the $$\ce{A-}$$ concentration will increase until the acid is fully neutralized into a salty solution.