Conversion from dry KOH to liquid 45% KOH

pmason61
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?

Gold Member
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.

hutchphd
Homework Helper
Gold Member
- being extracted from a 55 gallon drum.
What do you mean, "extracted" from the container drum?

Mentor
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.

hutchphd
Homework Helper
Gold Member
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.

Homework Helper
Gold Member
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?

Mentor
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.

Hyperfine
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.

Bystander