Conversion from dry KOH to liquid 45% KOH

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Discussion Overview

The discussion revolves around the conversion of dry potassium hydroxide (KOH) to an equivalent volume of liquid KOH solution, specifically a 45% concentration. Participants are verifying calculations related to the purity and molarity of KOH in the context of an industrial process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates that 42 lbs of 90% purity dry flake KOH corresponds to 34 liters of 45% liquid KOH based on their assumptions about molarity and purity.
  • Another participant challenges the calculation, stating that the adjustment for purity is incorrect and questions the interpretation of the quoted molarity.
  • A participant seeks clarification on the term "extracted" in relation to the 55-gallon drum, indicating potential confusion about the process.
  • It is noted by one participant that 45% KOH is more accurately described as a solution rather than liquid KOH, and they suggest that the molarity may be slightly different than stated.
  • Additional information is provided regarding the density of 45% KOH and the formula weight of KOH, which may be relevant for calculations.
  • Two participants inquire about the measurement basis for the "45%" concentration, asking whether it is weight per weight or weight per volume, highlighting the importance of this distinction in calculations.
  • One participant expresses concern about the calculations being performed in an industrial context, suggesting that the individuals involved should ensure they understand the necessary calculations for safety.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and definitions related to the KOH solution, indicating that there is no consensus on the accuracy of the initial calculations or the interpretation of the concentration measurements.

Contextual Notes

Participants highlight potential limitations in the assumptions made regarding purity, molarity, and the form of measurement for the KOH solution, which could affect the calculations presented.

pmason61
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Hoping to get my assumptions and math verified here.

We have a process that requires 42 lbs of dry flake KOH (90% purity). Looking to find the equivalent volume (liters) of 45% liquid KOH (11.1 molar strength) - being extracted from a 55 gallon drum.

1 liter of liquid KOH has 11.1 x 56 grams/liter = 621 grams of KOH (1.37 lbs). Adjusting to 90% purity - 1.23 lbs.

So the equivalent volume of liquid KOH required would be 42/1.23 = 34 liters of 45% liquid KOH. Is this correct?
 
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Do you know that the solution used 90% purity flake KOH? The quoted molarity refers to pure KOH. So this statement is not correct in my opinion: Adjusting to 90% purity - 1.23 lbs.
 
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pmason61 said:
- being extracted from a 55 gallon drum.
What do you mean, "extracted" from the container drum?
 
Not "liquid KOH", but "solution of KOH".

As far as I know 45% w/w KOH is 11.7 M, so the numbers seem to be slightly off.

Broadly speaking 34 L of 11.1 M KOH solution should contain the same amount of KOH that is present in 42 lbs of 90% purity flakes. Assuming I guessed correctly what you mean - your answer seems OK.
 
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Some useful information to use would be or include this:
Density for 45% KOH, 1.510 grams per cubic centimeter (Or 1.456 at 25C; depends on temperature)
KOH formula weight is 56.105 grams per mole.
 
What is the form of measure for the "45%" Potassium Hydroxide? Is this weight per weight, or is this weight per volume? I have seen both in use. You need to know which is yours to be?
 
symbolipoint said:
What is the form of measure for the "45%" Potassium Hydroxide? Is this weight per weight, or is this weight per volume? I have seen both in use. You need to know which is yours to be?
For the record: 45% w/v is around 8M.
 
pmason61 said:
Hoping to get my assumptions and math verified here.

We have a process that requires 42 lbs of dry flake KOH (90% purity). Looking to find the equivalent volume (liters) of 45% liquid KOH (11.1 molar strength) - being extracted from a 55 gallon drum.

1 liter of liquid KOH has 11.1 x 56 grams/liter = 621 grams of KOH (1.37 lbs). Adjusting to 90% purity - 1.23 lbs.

So the equivalent volume of liquid KOH required would be 42/1.23 = 34 liters of 45% liquid KOH. Is this correct?
This gives every indication of being an industrial scale process. Yet neither you nor your colleagues know how to do the calculations?

If that is indeed the case, I suggest you abandon whatever you are attempting until you learn how to work in the necessary environment in a safe manner.
 
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