# 9170官方金沙|官方主頁

1. <output id="cgay6"><nav id="cgay6"><div id="cgay6"></div></nav></output>

<acronym id="cgay6"><label id="cgay6"><xmp id="cgay6"></xmp></label></acronym>
2. <track id="cgay6"><ruby id="cgay6"></ruby></track>
<td id="cgay6"><ruby id="cgay6"></ruby></td>
<acronym id="cgay6"><label id="cgay6"><xmp id="cgay6"></xmp></label></acronym>

# Chemistry - Amount of Acid to Reach pH

#### mindless1

##### Diamond Member
How many grams of L-ascorbic acid powder do I need to make 1L water drop to a pH of 3?

Even better, would you show me how to calculate that?

#### Paperdoc

##### Golden Member
To do this you need to know the value for pKa of Ascorbic Acid. (By the way, L-Ascorbic acid is a particular 3-D arrangement of the molecule parts, and the L- version is really the only one that commonly occurs, so NOT specifying the "L-" part changes nothing for this discussion.) The value for pKa here is 4.17 (looked it up on Google).

When dissolved in water, Ascorbic acid partly dissociates - that is SOME of the molecules break up into H+ ions and a negative ion composed of the rest of the molecule - we call that item the ascorbate ion. For here I'll call it Asc-. If we write a chemical "equation" to represent the "reaction" of that dissociation, it shows the whole molecule on the left side, and the two parts of the dissociated molecule on the right. Then we also can write for that the Equilibrium Constant equation definition, which is simply the PRODUCT of the concentrations of every item on the right-hand side (the H+ and Asc- ions) divided by the product of the concentrations of all the items on the left-hand side (only one thing - the original undissociated Ascorbic Acid.) In chemistry the convention for writing "the concentration of..." is to place square brackets around the symbol for the item. So the Equilibrium Constant definition for this dissociation reaction is

Ka = ([H+] x [Asc-]) / [HAsc]

Now, the definition of pKa is that it is the NEGATIVE of the logarithm (base 10) of the Ka value. That is, for this material, the pKa = -log(Ka), and the pKa value is 4.17 so the value of Ka is 10 to the power of (-4.17) which works out to 6.76 x 10^(-5). Because we are limited here in scientific notation, that item 10^(-5) means "ten to the power of negative 5". Now we have to solve the math problem.

You question was, what is the amount of Ascorbic acid to dissolve in 1 litre of water to make a solution with a pH of 3.0. In chemistry we express the concentration of the solution in units of Moles per litre (Molarity). A "mole" is one molecular weight in grams. For Ascorbic acid, that value is 176.12 grams. Those "concentration of..." things in square brackets above are in this Molarity unit. So we'll figure out how many moles per litre is required, and convert that to grams at the end.

Next item is: what does pH mean? Similarly to the pKa thing above, pH is the NEGATIVE of the logarithm (base 10) of the concentration of H+ ions in the solution. That is, pH = - log [H+]. For your target of pH = 3.0, that means that the concentration of H+ ions needs to be

[H+] = 10^(-3) = 1.00 x 10^(-3)

So, assume we make our solution using y moles of Ascorbic Acid in the litre - we want to find y. Some of that is lost because it breaks down into ions - let's say the number of moles that breaks down is a. So at equilibrium we have a moles of H+ ions on the right-hand side, along with also a moles of Asc- in the litre. And on the left-hand side we have (y - a) moles of undissociated Ascorbic Acid (HAsc). Now we've already established that the concentration of H+ needs to be 1.00 x 10^(-3) from the pH target, and that is also the value for the concentration of the Asc- ions. So we go to the definition of the Dissociation Constant Ka

Ka = ([H+] x [Asc-]) / [HAsc] = 6.76 x 10^(-5)

and plug in the things we know,
[H+] = [Asc-] = 1.00 x 10^(-3)
[HAsc] = (y - 1.00 x 10^(-3))

6.76 x 10^(-5) = {1.00 x 10^(-3) x 1.00 x 10^(-3)} / {y - 1.00 x 10^-3)}

Forgoing algebraic precision, I made a little spreadsheet and tried values of y around 0.1 to get the correct value for Ka on the left. It came out at about y = 0.149 Moles

At a Molecular Weight of 176.12 grams per Mole, that means your solution should start out with 26.24 grams of dry Ascorbic Acid powder dissolved up to a total of 1000 ml of solution.

Last edited:

#### BarkingGhostar

##### Diamond Member
My wife found this thread funny as hell. She teaches chemistry at GSU and just gave a test on this very subject matter. I said, "Sounds like a test question' and she fully agreed. I'm betting we'll see a lot more of this with at-home schooling from Pre-K to senior years in undergraduate schools.

uclaLabrat

#### mindless1

##### Diamond Member
It's for making hot sauce. Recently had first freeze here, am making a large batch that gets away from my normal vinegar base. pH has to be low enough to promote safe storage, keep bacteria in check.

I have pH test strips but they are decidedly less useful in testing a red sauce, since they turn red based on acidity.

Last edited:
uclaLabrat

#### BarkingGhostar

##### Diamond Member
Hey, I believe you, but the wife ... she thinks you're lying. Hahaha So, Pics or lying ...

#### mindless1

##### Diamond Member
Here's some, but the rest are still hang drying in the garage.

#### Paperdoc

##### Golden Member
Now that you tell us what you are doing, let me alert you to a possible problem. Obviously you are aware that safe storage of foods made from vegetable ingredients requires, among other things, a pH low enough to prevent bacterial growth. The critical thing is that this means the pH of the FINAL mixture with all its components mixed and stabilized. Every one of those components can change the concentration of H+ ions (i.e, the pH) by either contributing some or capturing some and removing them from the solution. So, just STARTING from a solution containing only clear water and enough Ascorbic Acid to make it pH 3.0 does not guarantee that it will stay at that pH after all the other ingredients are added in. If you really want to be sure, AFTER you have it all mixed up and bottled, take a small sample out. Try to find a friend who works in a lab, or maybe a high school science teacher, who has access to a pH meter and calibration buffer solutions. Get that person to measure the actual pH of the sample. Should only require an ounce or two of sauce, and it's a simple test at virtually no cost if you don't have to hire a lab.

I understand your problem with pH test tapes which use dyes that change colour. In university chemistry lab one day our prof came to a group of us and challenged us to measure how much acid was in the bottle of Coca-Cola he gave us. The presumption was we would use a common technique of titrating (measuring into a sample of Coke) some base (a solution of sodium hydroxide) until it reached neutral pH. Detection of the "end point" normally is done witg a small bit of a dye added to the mix that changes colour sharply as pH changes. After some thought we asked him how he expected us to do this with a dark brown liquid. His reply: "Hmm. We did this before and it worked. Oh! That was with 7-up!"

mindless1

#### mindless1

##### Diamond Member
^ The base sauce I am making has nothing but slightly acidic peppers and salt. Surely this will keep the pH low?

I found that to be the case when I was making vinegar based sauce with white habaneros, which don't contribute any significant color.

Ultimately I may decide after making a sample at the same ascorbic acid ratio, that it is also too sour, in which case I may just reduce the acid ratio and freeze the bulk of it and thaw some as needed, to be kept refrigerated at the expected consumption rate.

What do you think of the inexpensive pH meters on Amazon like this one?

Last edited:

#### mindless1

##### Diamond Member
Wait a minute... just remembered that heat supposedly destroys vitamin C. Does this mean that heating it also changes its pH lowering ability too?

#### Paperdoc

##### Golden Member
Well, loss of the Ascorbic Acid in the solution by whatever means would certainly raise the pH. Most reports of this focus on how much vitamin C that is naturally present in a food is removed by cooking. I did not research the mechanism, but I am sure it is due to reaction with oxygen. Vitamin C (ascorbic acid) is a well-known anti-oxidant. That is, it will scavenge oxygen in the mixture and remove it by reacting with it, and that also removes some of the ascorbic acid. Not surprisingly, if you have a large volume of water for the total solution, more can be consumed that way. Also, a long cooking time may allow more oxygen to be added to the mixture, calling for more "scavenging". To minimize this loss of ascorbic acid in your work, I suggest this. If you are preparing and cooking a liquid mixture and adding ascorbic acid to that, do not add the ascorbic acid until the very end. Ideally, wait until the mixture cools and stir it in after that. Do not use a lot of stirring, but enough to ensure the powder is distributed uniformly and dissolves. The ascorbic acid is not essential for the cooking process, but is important for the storage period. Also, realize that you will be using a LOT more ascorbic acid than is present normally in foods, so minor loss will not make a significant difference.

I have no experience with those small pH meters, so I cannot tell you how reliable they are. I can suggest, however, that they are at least as good as (and probably better than) those pH test tapes that develop a colour you must match to a reference chart.

#### mindless1

##### Diamond Member
From what I found so far, this "destroyed" aspect from heat in water means it "oxidizes" (?) into dehydroascorbic acid and then to diketogulonic acid. This could be wrong, I saw it mentioned in a forum, not an authoritative source.

There is that other aspect mentioned for cooking, of it being removed by simply dissolving into water that is discarded, but I discard no water, just pepper skins and seeds, so in my case the main issue then is as you mentioned, not adding it ahead of the simmer time the sauce needs to cook down, as my sauce has no thickeners in it besides the pulp of the peppers themselves.

I won't need to stir it in at all, can just add to the almost full jugs of sauce after cooled, and shake them with practically no additional air exposure.

#### Paperdoc

##### Golden Member
I don't know the sequence of reactions for ascorbic acid with oxygen, but that sounds like a sequence to me. Those are just the steps in the process. It is consumed by reacting with (consuming) oxygen, so you may regard that as "destroying" the ascorbic acid. Whether those products generated in this process can still be effective as acids that contribute H+ ions to the solution, I do not know. Dissociation of the acids to release H+ ions is separate from the oxygen-scavenging actions.

Gentle stirring at the end will not add a lot of atmospheric oxygen to the mixture before sealing.

#### uclaLabrat

##### Diamond Member
If it breaks down into other acids it won't impact the pH much. Ascorbic acid can probably break down a number of ways i haven't looked up but if dehydroascorbic acid is one, that's just a dehydration that shouldn't impact much.

I think you're fine.

mindless1

#### jayabansal

##### Junior Member
Following is the easiest and a quick solution to the above question.

HC6H7O6 ?? C6H7O6- + H+
pH = 3
[H+] = 10-3 M
Ka of ascorbic acid = 8.0 × 10–5 M

Substitute the values to calculate the concentration of acid:

Volume = 1 L
Molar mass of ascorbic acid = 176.12 g/mol
Calculate mass of ascorbic acid:

mindless1

#### Paperdoc

##### Golden Member
First, to mindless1, the OP, MY APOLOGIES! I made a serious error in my first post, and I only found it because jayabansal got such a different answer I had to find out why.

My major error was I read the exponent of my calculations wrong, and found the answer IF the Ka for Ascorbic acid had been 6.76 x 10^(-6). BUT it is supposed be 10^(-5)! So, doing it my way properly, my answer should have been that the initial concentration of Ascorbic acid should be 0.0158 Moles per litre, and that comes to 2.78 g of Ascorbic acid powder per litre of solution. I was off by a whopping factor of 10 almost!

Now, that still does not agree exactly with jayabansal, and there are two reasons. To start, I searched for values for Ka and found various values from sources I expected to be reliable. The one I used, from the Merck Index, says pK1 = 4.17, so 10^(-4.17) = 6.76 x 10^(-5). Another says at 10C the pKa = 4.7, so Ka = 10^(-4.7) = 2.00 x 10^(-5). A third says pKa = 4.10, and hence Ka = 7.94 x 10^(-5). Quite a range! If we use my calculation process but jayabansal's value of 8.0 x 10^(-5), my process says the answer is 0.0135 Molar, or 2.38 grams dry Ascorbic acid powder.

The other difference is in the process details. As i said in my fist post, the formula I used insisted on including the fact that a small amount of the original Ascorbic acid placed in the solution is NOT in its original state. It is dissociated, and hence the denominator in my formula uses the unknown "y" for the original amount and subtracts off the amount "lost" by dissociation. This makes for a messy bunch of algebra that I evaded by having a spreadsheet do the work and trying successive approximations until I got the value for y that gave the target result for Ka. jayabansal, on the other hand, made use of the approximation commonly used that the amount dissociated is quite small and should be ignored in the calculation. If you do that, and use the 8.0,,, etc. value for Ka, the simplified calculation process yields an initial Ascorbic acid concentration of 0.0125M, or 2.20 grams of dry powder as jayabansal got. Note that using that approximation yielded results that differ only a little: 2.38 g versus 2.20 g. That's why people feel justified in simplifying with that approximation technique in the calculation.

Bottom line: I made a significant error, and jayabansal's post has prompted me to find and correct it. So depending on which value of Ka one believes, and on whether or not one uses the simplifying approximation in the calculation, the answer is between 2.78 g and 2.20 g of dry powder. Most of that range is due to the variability of reference data for the value of pKa.

#### mindless1

##### Diamond Member
Thanks for the correction. It did seem like 26g was a lot, would've been about 5 heaping teaspoons full at the density of the powder I have, with about 1/2 tsp = 2.2g.

I ended up doing it the other way around. I'd stated I couldn't use pH test strips in a red sauce, but then I can use them in ascorbic acid added to straight water to do a color change test strip test, then use that amount for same total volume of water with the pepper paste in it, but ultimately I ended up overshooting a bit and went with 1-1/2 tsp (6.6g) per half gallon container.

That worked out fine, didn't want to add much more ascorbic acid or else it would have been too tart. I'd have just frozen it for storage before it came to that.

Last edited:

#### Paperdoc

##### Golden Member
Glad to hear you got it worked out.