Masterbatch, dual lye and Vinegar

Shop deals on ebay
MunbynSMF
shop deals on amazon
Soapmaking Forum

Help Support Soapmaking Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
penelopejane

penelopejane

Well-Known Member
Joined
Sep 19, 2015
Messages
5,588
Reaction score
4,526
Location
Sth Coast, NSW, Australia
Hi

I am trying to Masterbatch, dual lye and Vinegar.
Can someone please check if the vinegar neutralising bit is right?

My recipe has 95% NaOH and 5% KOH
NaOH= 128 g
KOH = 7.5g
Total water = 316g

I have masterbatched the lye already so I want to add 50% water.
316 - 128 - 7.5 = 180g vinegar instead of water

28 g of vinegar neutralises 1 g NaOH
95% of 180g = 171 g NaOH/28 = 6 g NaOH extra

28 g of vinegar neutralises 1.4 g KOH
5% of 180 g = 9 g KOH/28 = 0.32 x 1.4g = 0.45 extra.

Is that right?
How in heavens name do I weigh 0.45g KOH?
Anyone know how much an 1/8 of KOH weighs?
 
Do you actually want any potassium acetate? Does it have the same impact as sodium acetate?

Could you add enough NaOH directly to neutralise the vinegar and then add in the other 128g NaOH and 7.5g KOH to make your 50% solution? Would you need a touch more liquid to make sure all the lyes are dissolved, though?
 
penelopejane said:
Hi

I am trying to Masterbatch, dual lye and Vinegar.
Can someone please check if the vinegar neutralising bit is right?

My recipe has 95% NaOH and 5% KOH
NaOH= 128 g
KOH = 7.5g
Total water = 316g

I have masterbatched the lye already so I want to add 50% water.
316 - 128 - 7.5 = 180g vinegar instead of water

28 g of vinegar neutralises 1 g NaOH
95% of 180g = 171 g NaOH/28 = 6 g NaOH extra

28 g of vinegar neutralises 1.4 g KOH
5% of 180 g = 9 g KOH/28 = 0.32 x 1.4g = 0.45 extra.

Is that right?
How in heavens name do I weigh 0.45g KOH?
Anyone know how much an 1/8 of KOH weighs?
Click to expand...

It's looking like your math is correct as long as it is 5% vinegar you are using.

The last bit, I am wondering about. Did you mean 1/8 teaspoon? I am guessing that's what you meant, but can't be sure. So I will respond based on my assumption.

Teaspoon measuring utensils are not always reliably accurate, so that can impact the results.

You actually can find weights of many chemicals online with conversions to include various types of measurement for given weights. With a simple Google search I found that 1 teaspoon of potassium hydroxide pellets weighs 4.5 grams (reference). Therefore, dividing that by 8 means that 1/8th teaspoon of KOH pellets should weigh 0.5625grams. And even that is more than you need. So going one further, do you have a 1/16th measuring spoon? Or even a 1/32nd measuring spoon? I don't, but I have seen them. Still, that seems a bit too complex and the potential for inaccuracy only complicates the complexity.

So you're talking less than half a gram. Does your scale accurately measure to 0.01 accuracy? If it does, that's how you measure 0.45 grams. If it doesn't, you can't.

I have two scales, one that measures to 0.01 grams that I use for these types of measurements. The other scale measures to 0.1g accuracy. That is the one I use for measuring my oils.

My suggestion if you don't have a scale that measures to 0.01g accuracy and you don't want to wait until you can get one (they're actually quite affordable from Amazon and various other vendors), is to use half of your 1/8 teaspoon to measure your extra KOH. It will be less than your desired 0.45g, but at least it won't be more. Then order a scale such as this one or something that will allow you to measure smaller weights more accurately.
 
Last edited by a moderator:
28 g of vinegar neutralises 1.4 g KOH <-- YES!

5% of 180 g = 9 g KOH/28 = 0.32 x 1.4g = 0.45 extra. <--- no, don't do this.

Is that right? How in heavens name do I weigh 0.45g KOH?

***

I think you're confusing things a bit. The relationship is 28 grams (1 ounce by weight) of regular COMMERCIAL vinegar straight from the bottle consumes 1.4 g KOH. You don't need to convert the vinegar weight to a pure acetic acid basis -- you just use the total weight of the commercial vinegar.

KOH consumed by the vinegar = (Wt vinegar, grams) X 1.4 g KOH / 28 g vinegar = 180 X 1.4 / 28 = 9 g KOH

There's your answer.

Here's my article on this: http://classicbells.com/soap/aceticAcid.html

***

If you every really do need to weigh 0.45 grams, you have two options.

Get a scale that can weigh at least to 0.1 gram and weigh out either 0.4 or 0.5 grams. A statistics instructor would say round the "5" up or down to the nearest even number, so I would round the 0.45 to 0.4 g.

Or if you have a scale that only reads to whole grams, you'd just round the 0.45 down to zero, using the statistically correct method for rounding, and add nothing.
 
Last edited:
DeeAnna said:
28 g of vinegar neutralises 1.4 g KOH <-- YES!

5% of 180 g = 9 g KOH/28 = 0.32 x 1.4g = 0.45 extra. <--- no, don't do this.

Is that right? How in heavens name do I weigh 0.45g KOH?

***

I think you're confusing things a bit. The relationship is 28 grams (1 ounce by weight) of regular COMMERCIAL vinegar straight from the bottle consumes 1.4 g KOH. You don't need to convert the vinegar weight to a pure acetic acid basis -- you just use the total weight of the commercial vinegar.

KOH consumed by the vinegar = (Wt vinegar, grams) X 1.4 g KOH / 28 g vinegar = 180 X 1.4 / 28 = 9 g KOH

There's your answer.

Here's my article on this: http://classicbells.com/soap/acetic...extra NaOH KOH 130 / 28 x 1.4 = 9g extra KOH
Click to expand...
 
Last edited:
Earlene
Yes, that's how I got to 1/8 tsp i realise I don't need to weigh it just add < 1/8 tsp. so confusing!
No I haven't got a small gram scale.

TEG
I just assumed you would have to neutralise both from all the posts I've read on the process but nothing dealt with dual lye so I wasn't sure.
 
Last edited:
Thanks for being patient with me -- I see what your point is now. Yes, your original math is correct. I was wrong.

That said, I want to point out something -- None of us know how the acetic acid is going to react with the KOH and NaOH molecules. It seems reasonable that the reaction would be in direct relationship to the proportions of KOH and NaOH in the mixture (what you're assuming with the 95% and 5% proportions), but we really don't know that is the case. The acetic acid could react entirely with the NaOH for all we know.

Bottom line is you have to make an assumption and live with it. When faced with two equally reasonable options, choose the simpler one.

If it were me, given the % of KOH is tiny compared to the % of NaOH, I'd figure the acetic acid is going to react entirely with the alkali that's present in the largest amount (the NaOH) and calculate accordingly.

This makes the math easy ... and is just as defensible of an assumption as the other.

***

"...NaOH 130 / 28 x 1 = 6.5g extra NaOH
KOH 130 / 28 x 1.4 = 9g extra KOH..."

Maybe it's not my day for thinking straight -- I'm not seeing where the 130 grams is coming in. I thought you were using 180 g of commercial vinegar?

If 180 grams of vinegar, then --

NaOH consumed by the vinegar = (Wt vinegar, grams) X 1 g NaOH / 28 g vinegar = 180 X 1 / 28 = 6.4 grams NaOH

I'd add 6.4 g NaOH (or round down to 6 if you will be measuring in whole grams) and leave it at that -- don't bother with adding extra KOH.

And if you think about it -- the rounded answer of 6 grams NaOH and zero grams KOH is the exact answer I'd use even if I did want to portion the acetic acid between the NaOH and KOH as you have been doing.
 
DeeAnna said:
If it were me, given the % of KOH is tiny compared to the % of NaOH.

This makes the math easy ... and is just as defensible of an assumption as the other.

***

"...NaOH 130 / 28 x 1 = 6.5g extra NaOH
KOH 130 / 28 x 1.4 = 9g extra KOH..."

Maybe it's not my day for thinking straight -- I'm not seeing where the 130 grams is coming in. I thought you were using 180 g of commercial vinegar?

If 180 grams of vinegar, then --

NaOH consumed by the vinegar = (Wt vinegar, grams) X 1 g NaOH / 28 g vinegar = 180 X 1 / 28 = 6.4 grams NaOH
Click to expand...

I don't know where the 130 came from either! Thank you for picking it up.

Thank you I will just neutralise the NaOH and see how it goes. I might even do a tester of both and see if I can compare the two.
 
Last edited:
DeeAnna, thank you for clearing that up. I've only done a dual lye vinegar soap once and when I did it wasn't full vinegar, just a small amount left in a bottle so I could toss out the bottle without wasting the vinegar. So I didn't bother to calculate the extra lye since it was such a small amount. But, I remember wondering how I would do this calculation for dual lye vinegar for soap in the future. I had thought I'd have to go the same route as penelopejane in her calculations above. It just seemed logical.

But to see that using the NaOH as the 'simpler' component for the calculations makes just as much sense and is so much less complicated. I suppose if I was making 100 pounds of soap at a time, it might not be the same logic, but for my purposes, it certainly will help to prevent that extra brain stress.

So to get this straight in my mind, when I do this (I plan to soon), the additional lye added will all be NaOH based on an 'as if' calculation of 100% NaOH masterbatch lye, rather than a 95%NaOH masterbatch lye. That's what I see in the calculations above. Just want to clarify for my sometimes slow-to-grasp mind.
 
"...the additional lye added will all be NaOH based on an 'as if' calculation of 100% NaOH masterbatch lye, rather than a 95%NaOH masterbatch lye...."

I'm a bit fuzzy today, so pardon me if I'm slow on the uptake.

If I'm following you correctly, the calculation of NaOH (or KOH) needed to neutralize acetic acid is completely independent of what kind of masterbatched lye solution you're using. The 1 g NaOH / 28 g commercial vinegar is on the basis of solid NaOH, not a masterbatched solution. So you figure out the vinegar you want to use and calculate the grams if solid NaOH that will be neutralized by the vinegar. That's the 6.4 grams of NaOH I'm coming up with.

You'd then need to translate that into how much solution is needed to get this extra NaOH into the soap pot. Again, I'd probably keep it as simple as possible. If it was me, I would be comfortable in this specific instance of simply doubling the number --> 2 X 6.4 = 12.8 grams of masterbatch solution. I'd round the number to whole grams as seems appropriate (I personally would round up to 13 grams, but others say they always round lye weights down -- I encourage everyone to do what seems best.) That's the weight of masterbatched solution I'd add for the vinegar neutralization.

Again, I'm suggesting this because the error involved is honestly pretty small. Small enough it's going to get lost in all the other errors we deal with when making soap -- things like our NaOH and KOH being less than 100% pure, and sap values that are averages, and so on. If you've ever added whole milk to a batch of soap and ignored the contribution of the milkfat to the superfat, you've introduced as much or more error into that batch of soap.
 
earlene said:
So to get this straight in my mind, when I do this (I plan to soon), the additional lye added will all be NaOH based on an 'as if' calculation of 100% NaOH masterbatch lye, rather than a 95%NaOH masterbatch lye. That's what I see in the calculations above. Just want to clarify for my sometimes slow-to-grasp mind.
Click to expand...


Yes but just of the water that remains after you have allowed for the masterbatch lye/water.

My recipe has 95% NaOH and 5% KOH
NaOH= 128 g
KOH = 7.5g
Total water = 316g

I have masterbatched the lye already so I want to add 50% water.
316 (total water) - 128 (NaOH) - 7.5 (KOH) = 180g vinegar instead of water.

Use this 180g of vinegar to work out the extra NaOH to use forgetting the KOH amount.

I think that is what you are asking.
 
Right, that is what I concluded (re: the remaining liquid.)

I haven't masterbatched a dual lye solution yet, but have planned to do that so I could streamline the process when making dual lye soaps. I fact a couple of times I wanted to do dual lye, but decided against it to save myself some extra time. Even though it's not that much work to do the math and so forth, it's still time consuming.

I also haven't masterbatched a vinegar lye solution. I've seen at least one or two other people here talk about it, so was planning on giving both a try at some point. This is what I thought you were talking about, penelope, is that right, or did I misunderstand?

I do masterbatch NaOH with distilled water already.
 
earlene said:
Right, that is what I concluded (re: the remaining liquid.)

I haven't masterbatched a dual lye solution yet, but have planned to do that so I could streamline the process when making dual lye soaps. I fact a couple of times I wanted to do dual lye, but decided against it to save myself some extra time. Even though it's not that much work to do the math and so forth, it's still time consuming.

I also haven't masterbatched a vinegar lye solution. I've seen at least one or two other people here talk about it, so was planning on giving both a try at some point. This is what I thought you were talking about, penelope, is that right, or did I misunderstand?

I do masterbatch NaOH with distilled water already.
Click to expand...

I don't mix the masterlye solutions. I have one masterbatching 50% NaOH and one 50% KOH. That way I get a choice. It's good to masterbatching the KOH because it's such a small amount it's easier to weigh large amounts and dispense double the amount.
 
Ah, I thought that you wanted to mb with vinegar. Very sorry.

I would still in this case (with two mb solutions) pour in more of the NaOH to react with all of the acetate and then use the recipe required amount of KOH. Your mix of unreacted lye would be 95:5 and you would have the sodium acetate without having to think of tiny amounts of KOH.
 
A question for those who have done a dual lye batch of soap:
Does it trace really quickly?

I have made three batches in the last couple of days (30% lye concentration and non accelerating FO) two with NaOH + KOH + Vinegar 50% and one with NaOH + KOH + much smaller amount of vinegar and despite SBing only a tiny, tiny amount until it just reached emulsion once I turned my back to fiddle with one part that I was colouring it was almost plop and pour.
Have others experienced this?
 
I can't say I've seen any great difference. I've made 13 batches with 5% KOH and one at 10%. A few with accelerating FOs, but mostly with non-accel FOs. One 4-color hanger swirl, a couple with pencil lines, some with in-the-pot swirls. If the added KOH was causing unusual acceleration, I think I'd have noticed by now.
 
DeeAnna said:
I can't say I've seen any great difference. I've made 13 batches with 5% KOH and one at 10%. A few with accelerating FOs, but mostly with non-accel FOs. One 4-color hanger swirl, a couple with pencil lines, some with in-the-pot swirls. If the added KOH was causing unusual acceleration, I think I'd have noticed by now.
Click to expand...

So would it be the vinegar?
 
I have to say I don't have a clue about that, PJ. I don't use vinegar or sodium acetate in my soap, so I don't have any experience to give an answer.

It's easy to focus on new additives or techniques as being the cause of soaping troubles, but other issues can sometimes be the cause. I'm reminded of a recent poster who was absolutely certain his new batch of coconut oil was the reason for his soaping problems, but after some discussion and testing it turned out it was his NaOH that was the culprit. So it might be good to broaden your troubleshooting to look at other possibilities, such as older fats that might have higher levels of free fatty acids, etc.
 

Similar threads

Orla
Dual lye, citric acid and sugar
Replies
5
Views
2K
DeeAnna
DeeAnna
cmzaha
Masterbatching Lye/Vinegar
2
Replies
23
Views
3K
DeeAnna
DeeAnna
E
Dual Lye Cold Process Soap with Henna
Replies
2
Views
2K
eulani
E
dibbles
Dual Lye Mistake/Fix
Replies
3
Views
697
dibbles
dibbles
penelopejane
Dual Lye disasters
2
Replies
21
Views
3K
Steve85569
Steve85569

Latest posts

Back
Top