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I bought some sodium hydroxide on Amazon and went about titrating solutions I make with it.

As I'm an amateur chemist at home, I have limited tools. A pad of paper, some 5% acetic acid vinegar, and some pH strips.

I assumed the pH of vinegar was 2.5 ± 0.5.

Making a solution of 4g NaOH powder in 100g of (tap) water, I expected a solution of 1 mol/L. (NaOH 40g/mol)

I went about finding the ratio of concentrations, working by factors of 10 and then refining, that pH strips would indicate around 7.

At my great surprise, this turned out to be around 0.9:1 NaOH 1M to vinegar (about 20mL:22mL – the margin of error being around a factor of 1.1)

A vinegar pH of 2.5 suggests an H+ concentration of 3.2 mM; and so that that ratio would be, $1\mathrm M/3.2 \mathrm{mM} = 312:1$. So experimentally, $\mathrm{[OH^-]}$ is about 350 times lower than expected. I'm trying to figure out why that would be.

Am I missing something obvious here? I'm looking for a mistake but can't find one.

Here are my thoughts:

Bought NaOH might not be pure

It would take a "purity" of 0.25% to explain such a difference. Unlikely.

NaOH might not dissociate completely experimentally

NaOH is a strong base. pKb would need to be extra high to explain a factor of 350.

Calculation error

I've checked and rechecked my math, everything seems to hold up. 22mL of vinegar brings 70µmol [H+], 20 mL NaOH solution brings 50µmol [OH-].

Measurement error

I've done the experiment twice. Same kind of values. Adding 80% of vinegar volume leaves the paper unmistakably dark. Adding 120% of equilibrium volume leaves the paper yellow-ish. 90% to 110%: green-ish. To measure quantity added, I use a scale with a resolution of ±0.01g. (Measurements were taken ±0.5g)

Bad pH paper

It would take really bad paper to make a pH 9 liquid appear to be pH 7, wouldn't it? A factor of 350 is about $log_{10}(350) \approx 2.5$ off on the log scale. Unlikely. Besides, aren't the indicator substances pretty standard?

Acidic tap water

I live in a big city, I've read the water analyses. Looks like the water is around pH 6.5. I don't think at this scale it would change much.

Wrong chemistry/theoretical background

It's possible a 4% w/v solution of NaOH doesn't have a pH near 14. Or that mixing it with 1.1 parts of 5% vinegar doesn't make a pH of 7. I have not yet found the basis for that. At equal quantities [H+] and [OH-] in room temperature water, I don't see a reason for the weak acid to be a factor.

So I'm pretty much stumped. Why does my solution have a pH 11.5 instead of 14? Is there an obvious chemical concept I've missed?

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    $\begingroup$ pH of vinegar is irrelevant to your calculations. It is not the amount of H+, it is the amount of CH3COOH that matters. Related: chemistry.stackexchange.com/questions/60407/… $\endgroup$ – Ivan Neretin Apr 9 '20 at 5:01
  • $\begingroup$ I see. So the pH of 2.5 is not equal to all the potential H+ it can ever release because that's what a weak acid is, right? $\endgroup$ – Jonathan Allard Apr 9 '20 at 8:36
  • $\begingroup$ Okay yes, looks like that's pretty much it. So 5% acetic acid (0.83M) and 0.95 NaOH pretty much neutralize each other at a 0.9:1 ratio. - I'd welcome suggestions or edits to make this question more useful for others $\endgroup$ – Jonathan Allard Apr 9 '20 at 8:50
  • $\begingroup$ The ration 0.9:1 has no meaning ! When you neutralize acetic acid with NaOH, you not only destroy the H+ ions, as these ions are pretty rare is the solution. Above all NaOH destroys the non dissociated acidic molecules, which are much more abondant in the solution. As consequence, if you want to make calculations about the neutralization of CH3COOH you should forget about the pH. You should simply consider that the number of moles of NaOH is equal to the total number of moles of CH3COOH (dissociated or not) $\endgroup$ – Maurice Apr 9 '20 at 10:11

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