You have confused neutrality point, inflection point and equivalence point. The former is where there are equal numbers of protonated and deprotonated solvent species ($\mathrm{pH}~7$ in water at $\pu{25 ^\circ C}$). The inflection point is where the second derivative of the titration curve (amount of base added plotted against $\mathrm{pH}$) is a maximum (maximum buffering) and the third the point where the amount of analyte and titrant are equivalent. If you take acetic acid and start adding base to it even a small increment of added base will produce a large change in $\mathrm{pH}$. The base isn't reacting with the $\ce{HOAc}$, it is reacting with the solvent. The happens until the inflection point $\mathrm{pH}~4.76$ for acetic acid, is approached in which region the base is reacting with the $\ce{HOAc}$:
$$\ce{HOAc + OH- -> OAc- + H2O}$$
and a small change in $\mathrm{pH}$ requires much more base. Eventually, (about $\mathrm{pH}~4.76,$ that is two $\mathrm{pH}$ units above the inflection point) nearly all the $\ce{HOAc}$ is reacted and additional base just goes to deprotonate water, increasing the $\ce{OH-}$ ion concentration and $\mathrm{pH}$ at a rate much faster than near the inflection point. Thus the fact that you wind up at a $\mathrm{pH}$ near $9$ is simply a consequence of the fact that you are well past the inflection point of your acid which is, because the inflection point is at $4.76$, considered a weak acid. Once all the acid has been deprotonated, additional additions of base go to deprotonate water and the pH rises.
The pH at the equivalence point is the pH at which the solution is neutral. Finding this pH requires that we compute the charge of each species in the solution and totalling them up as a function of pH. We then seek the pH value that zeroes the sum. The species are $\ce{Na+}$, $\ce{OH-}$, $\ce{AcOH}$ and $\ce{AcO-}$. The charge on a liter of water is $Q_w(pH) = 10^{-pH} - 10^{(pH - pK_w)}$ which is the sum of the hydronium and hydroxyl ion charges. For the acetate its a bit trickier. We start with the Henderson Hasselbach equation writing it in the form $r_j = 10^{(pH - pK_j)}$ This is the ratio of the concentration (activity, really) of species which have lost j protons to the that of the species which has lost $j-1$. In the case of acetic acid there is only one proton to lose and so there is only one $r$ but in the case of polyprotic acids there is, of course, 1 $r$ for each proton. Now it shouldn't take you long to figure out that if the ratios are $r_1, r_2, r_3...$ the fraction of undissociated acid molecules is $f_0 = 1/(1 + r_1 + r_1r_2 + r_1r_2r_3)$ As r_1 is the ratio of singly deprotonated to undissociated its clear the the fraction of singly deprotonated is $f1 = r_1f_0$. At this point we have all we need to calculated the charge. Assuming 1 mole $\ce{NaOH}$ and 1 mole $\ce{AcOH}$ in 1 L of water the total charge would be $+1 + 1*Qw(pH) -1*f_1(pH)$ So the question now turns to how we find the value of pH which brings this expression to 0. Clearly the fastest way to a solution is to have and Excel spreadsheet compute it and either grope for the answer (pH 9 gives + $4.75·10^-5$ and pH 9.5 gives $4.75·10^-5$ so clearly the answer lies between those 2 pH values. This suggests a bisecting root finder as a path to a solution and that works very well. Excel has the Solver which does the iterations for you and takes you directly to the solution $pH = 9.38$. Note that the solution for 1 mole each in 10 liters of solution is $pH = 9.38$ which makes it clear that the water is playing a significant role here. Not all the $\ce{OH-}$ are going to deprotonate acid. Many are going to shift the waters charge balance.
