Why did neutralization of weak acid calculation differ from empirical results?

I recently stumbled upon the problem of needing to neutralize a volume of 0.33M acetic acid with 0.1M NaOH. I was curious how many mL of 0.1M NaOH I would need to bring 1mL of 0.33M Acetic Acid solution to a pH of 7.


For my calculations I used the following data:


Inputs


* Ka = 0.000018
* M of Acetic Acid 0.33
* M of NaOH 0.1
* Volume of Acetic Acid (mL): 1


Outputs:


$$
\sqrt{Ka\*0.33} = 0.002437 M of H+
$$


$$
0.002437 /1000mL \* 1mL = 0.000002 mol H+
$$


Volume needed to Neutralize


$$mol H+/M NaOH = 0.000024 L of NaOH$$


$$-> 0.024372 mL of NaOH$$


However, in practice this did not work. My empirical results ended up:


3.1 mL 0.1M NaOH neutralized 1mL 0.33M Acetic Acid


Why did this calculation differ from my observations? How reliable is a disassociation constant here?




No need to use dissociation constants.
$\pu{1 mL}$ of $\ce{0.33 M}$ acetic acid contains $\ce{3.3 10^{-4} mol}$ acid.
To neutralize this acid, $\ce{3.3 10^{-4} mol NaOH}$ is needed, which is included in a volume $\pu{V = n/c = \frac{3.3·10^{-4} mol}{0.1 mol/L} = 3.3 10^{-3} L = 3.3 mL}$


This amount will not go exactly to $\mathrm{p}H$ $7$, as the final $\mathrm{p}H$ will be around $9$ . But you mentioned that you only want to **approach** neutral $\mathrm{p}H$. The volume of $\ce{NaOH}$ necessary to go from $\mathrm{p}H$ $7$ to $\mathrm{p}H$ $9$ is negligible in this problem.


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