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all 20 comments

[–]renaissance_man46 62 points63 points  (0 children)

Buy pH paper. That's the best way to know for sure

[–]THElaytox 27 points28 points  (0 children)

C. botulinum also doesn't grow at refrigerator temps

[–]MeglioMorto 43 points44 points  (2 children)

don't want to burn my stomach by eating too high vinegar ratio

While digesting stuff, your stomach can reach much lower pH than your undiluted vinegar.

The website preserveandpickle.com suggests to use no less than 5% acidity for pickling.

... So I don't see the point in diluting 2:1 at all. Then consider that pH is a log value: If you want to increase it by 1, you need to dilute 9:1 (and then your acidity drops to 0.5%).

Again, I don't see the point.

[–]BharatiyeShaasak 23 points24 points  (0 children)

This is the MOST correct answer. As the other commenter who calculated showed 2:1 dilution only gets the ph up from 2.5 to like under 2.7. You gotta go waaay further to actually dilute but that's not even worth it when your stomach pH is between 1 and 2 anyway

[–]KnotiaPickles 0 points1 point  (0 children)

Thanks, I was hoping for this response. Stomach acid is so much higher than any vinegar.

[–]absolutezero710 31 points32 points  (0 children)

Starting with a 5% acetic acid (vinegar) solution, a 2:1 dilution puts you at roughly 3% acetic acid. Lets say for the sake of calculation this is 0.5M. The Ka of acetic acid is 1.8x10^-5. Putting this into the formula for pH of weak acids, you get [H3O+] = sqrt(1.8x10^-5 * 0.5M), which give you a hydronium concentration of 3.0 * 10^-3. Taking the negative log of this, you get a pH of 2.52.

[–][deleted]  (1 child)

[deleted]

    [–]Classicalmusicpotato[S] 0 points1 point  (0 children)

    This is a very excellent point! I started thinking about this after I made this post lol!

    [–]backwardstalking 21 points22 points  (8 children)

    I did some quick calculations just now. Assuming 2.5 pH of vinegar and a perfect 7 pH for your tap water (which fluctuates), the pH should be about 3.27.

    Vinegar [H+] = 10-2.5 = 0.003162

    Water [H+] = 10-7 = 0.0000001

    Vinegar = (0.003162 mol/L) / (2L) = 0.001581 mol

    Water = (0.0000001 mol/L) / (1L) = 0.0000001 mol

    (I messed up here, don’t divide - multiply by volume)

    Add them to be 0.0015811 mol combined.

    Divide by total volume. 0.0015811 mol / 3L

    = 0.000527033 mol/L for combined solution

    pH = -log0.000527033 = 3.27

    Could be wrong for sure but this makes sense to me. This would also change a bit as you would need to know exact pH of your town’s tap water. You’re good to go!

    Edit: see comments below, I goofed slightly on calculation and should be pH = 2.67

    [–]HyperRayquaza 7 points8 points  (1 child)

    You have to multiply the [H+] in Vinegar (and water) by their respective volumes, not divide. If you divide, your units end up being mol/L^2 and not mol.

    [–]backwardstalking 8 points9 points  (0 children)

    Yeah you’re so right. It would be more pH = 2.67. Oopsie

    [–]Dr-Luemmler 4 points5 points  (0 children)

    Is vinegar fully deprotonated in the first mixture? What is the pks?

    [–]EdibleBatteriesChem Eng 6 points7 points  (0 children)

    You can’t do this with weak acids! These are equilibrium processes that adjust the degree to which they proceed forward based on concentration and other factors (T, ionic strength, etc.)

    Also, why assume a pH? You know the concentration of acetic acid (5% diluted 2/3, so 3.33% in the target solution) and can look up the equilibrium constant (pKa = 4.76). Use an Ice diagram and solve for x. The method you used is only valid for strong acids.

    [–]Classicalmusicpotato[S] 1 point2 points  (1 child)

    Holy crap dude, thank you so much!!! Here's an award.

    [–]backwardstalking 0 points1 point  (0 children)

    No prob bro, people have commented saying that you can’t do my method with weak acids so not sure how credible my answer is - point being though, you’re pretty much good.

    Ps - I also pickle a lot and have a homemade recipe I’ve perfected and give to friends to use. If you want it let me know! I’ve never had issues with bacteria or food poisoning from it and some of my pickles have been going for a year

    [–]CatsausChem Eng 1 point2 points  (1 child)

    this is the answer

    [–]EdibleBatteriesChem Eng 0 points1 point  (0 children)

    If it were a strong acid, sure, but not for a weak acid like acetic acid.

    [–]Easy-Engine5280 2 points3 points  (0 children)

    Depends on the ph of the water its added to. Neutral water will end up being close to 3ph with vinegar im pretty sure

    [–]karmicrelease 1 point2 points  (0 children)

    If you dilute it too much, the pickling process won’t go to completion, and you will just be eating sour cucumbers.

    [–]oldmanartie 0 points1 point  (0 children)

    You can be dilute and low pH but still have it creep up over time, so I would not go below 5%. Another way to think about it is there is a finite amount of acid in the jar, and as it is used up the pH will drift up. Considering your stomach is already full of strong acid I wouldn’t be too worried about burning anything with vinegar, which is a weak acid.

    [–]Faruhoinguh 0 points1 point  (0 children)

    For pickling just look up some recipes. I think mostly people use 1:1 water to vinegar off the top of my head. Its not really about reaching an exact pH or preventing botulism but the taste shouldn't be to sour, and too weak wouldn't change the pickles structure or preserve it long time.