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Old 01-29-2009, 09:37 AM   #1
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Default Potassium Bicarb

Hey ya'll, I signed up yesterday and this is my first post, but I've been lurking and I've learned a lot already. Anyhow, I want to try a SNA on a metheglyn I'm starting, but when I went to my brew store they only carried potassium bicarbonate (KHCO3), not potassium carbonate (K2CO3) as called for. Anyhow, I saw on Wikipedia that above 100c, KHCO3 will decompose into potassium carbonate. Looking at this reaction: K2CO3 + CO2 + H2O → 2 KHCO3 , I'm guessing i should use about twice as much KHCO3 by weight, or 10 grams for a 5 gallon batch, and boil it. Does anyone have any better advice for this?


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Old 01-29-2009, 09:47 AM   #2
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Oh, I just found a post by HighTest saying to use 67% more. Does anyone have a preference for heating it or not? Thanks a lot, this place rules!


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Old 01-29-2009, 12:33 PM   #3
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There is no need to use heat - I don't. I just add it to the water I use to prepare the must.

Also, I'm not sure where my brain was when I said use 67% more, but let me correct that number. It should have been 45% more. I'll explain...

The stoichiometry between KHCO3 & K2CO3 is such that for every 1 mole of K2CO3, 2 moles of K is produced. Whereas, in KHCO3, the result is 1 mole of K - that is most like why you were told to use twice as much bicarbonate.

The addition of this compound serves two purposes. First, it acts to buffer the must pH, Secondly, it supplements the must K level (critical to the maintenance of must pH).

However, the weight of 1 mole of KHCO3 is not equivalent to K2CO3, and therein lies the basis for the need to use 45% more not twice as much...

BTW, please point me in the direct of where I noted 67% as I'd like to correct that value - thank you...
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Old 01-30-2009, 08:22 AM   #4
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Thanks for the info, I added some to my must and we'll see how it works! Here's the URL of the post with that info- I believe the weight was correct for a 5gal batch, I guess the percentage was just off. http://www.homebrewtalk.com/f30/ques...um+bicarbonate

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Old 01-30-2009, 12:01 PM   #5
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Thank you for the link. I also had the weight wrong... I'd sure like to know where my brain was during that post...
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Old 01-30-2009, 12:22 PM   #6
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For the purpose of pH control both K2CO3 (potash) and KHCO3 will serve just fine.

1 mole of KHCO3 weighs 100.11 grams.
1 mole of K2CO3 weighs 138.20 grams.

Toa adjust pH in equal measure twice as many moles of KHCO3 have to be added as K2CO3. This means the correct percentage is 69%.
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Old 01-30-2009, 12:39 PM   #7
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Now take that information and calculate the weight of each compound necessary to produce 136ppm K in a volume of 5 gallons. Now what is the % change in weight of the 2 compounds...

Notice that my response in that topic was prefixed, "to achieve the same ionic potassium contribution (136 ppm)..." It did not address pH management.

Reason: add clarification
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Old 01-30-2009, 01:51 PM   #8
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You'd still need twice as much KHCO3 as K2CO3 soI don't quite see how the percentage would be any different...

The exact weight of the required amounts varies with the OG but the ratio between these two substances is still the same. 1:1.44

Note also how Hightest says you need ~45% more, see how close "45% more" is to multiplication by 1.44?

Still not convinced?

-Edit-
Just for the record:
136 ppm K+ ions in 5 gallons at an OG of 1.060:
(((18,9270589 * 1,060 * 136) / 1 000 000) / 39,09) * 100,11 = 6,9877876g KHCO3

136 ppm K+ ions in 5 gallons at an OG of 1.060:
((((18,9270589 * 1,060 * 136) / 1 000 000) / 39,09) / 2) * 138,2 = 4,82325568g K2CO3

(4,82325568 / 6,98778764) * 100 = 69%
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Old 01-30-2009, 03:02 PM   #9
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I'm not sure if we're saying the same thing, but let's use your values for the weights of the two compounds that produce 136ppm K. Now let's also focus on what I actually asked, "Now what is the % change in weight of the 2 compounds" - note the highlighted words.

Calculating the % change between two parameters is not equivalent to asking what % of one is the other... .

To determine a % change you subtract the reference value from the new value, then divide by the reference value, multiply the result by 100, and add a % sign. The result is the percent change between the two.

Using your values (6.98g - 4.82g) / (4.82g) = 0.448 - 44.8% (~45%)

Putting all this into words becomes...
Quote:
If I know I would use 4.82g of K2CO3 to produce 136 ppm K in a 5 gallon must, what weight of KHCO3 will I need to produce the same 136 ppm K? The answer is 44.8% more [(4.82g)(1.448) = 6.979g], which is certainly not twice as much...
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Old 01-30-2009, 07:57 PM   #10
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I corrected 67%. It's all there.


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