# 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?

• Use MathJax formatting for clarity. Aug 11, 2022 at 22:47
• Since when is NaOH a weak base? Aug 11, 2022 at 22:54
• Also no idea what you even wanted to do in this calculation. Perhaps you don't get what "neutralise" means in this context... You don't need any constants, and whatnots, just equal amounts of acid and base. Aug 11, 2022 at 22:58
• I wanted to find out the volume of 0.1M NaOH would cause 1mL of 0.33M Acetic Acid solution to approach neutral ph. Aug 11, 2022 at 22:59
• It looks you errorneously assume you just need to neutralize those H+ that exist in the acid solution as the equilibrium concentration. You forget that this equilibrium provides more H+ when the existing ones are being neutralised. Aug 12, 2022 at 2:55

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.