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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 $\ce{H+}$?

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$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

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.

I can't make much sense of your chemical equations, but it looks to me like they are incorrect. There is no reason for formation of the ionic compounds in solution. They should actually be written in reverse. And you have no strong base to react with the hydronium ions.

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