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Author Topic: Making Potassium Acetate Solution  (Read 1084 times)

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deezy

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Making Potassium Acetate Solution
« on: May 18, 2017, 07:26:07 AM »

Not sure if this is the right forum, but needing to know how to formulate Potassium Acetate.

I understand the inputs are potassium hydroxide and acetic acid which makes potassium acetate and water.  Can someone tell me how many lbs of each will produce how many lbs of potassium acetate?  Does any water need to be added to the KOH and acetic acid?

Will the potassium acetate precipitate off and crystallize to become a dry powder or will the end product be in liquid solution? 

Thanks in advance for any input.
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Borek

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Re: Making Potassium Acetate Solution
« Reply #1 on: May 18, 2017, 12:04:50 PM »

Best to react them in solution to avoid overheating, so yes, depending on the concentration of the acetic acid used you may need to add water and you will end with the solution.

Mass ratio can be calculated from the reaction stoichiometry - start with the reaction equation.
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deezy

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Re: Making Potassium Acetate Solution
« Reply #2 on: May 18, 2017, 05:53:00 PM »

Thanks for the response.  Can you correct me if I'm wrong... I am using a 45 percent by weight KOH solution in water.

In the equation below if I use 500 grams of the KOH solution, I am calculating I need 405 grams of Acetic Acid to completely react.  Is that right?

The Reactant masses should equal the Product masses right?  I must be doing something wrong because they are not equal with my calculations.

Will the percentage of Active Ingredient in the Acetic acid affect or change these numbers?  For example using a 99% Acetic acid compared to an 80%?  Just throwing those percentages out there, not sure if those are accurate.

K + OH + H2O + CH3COOH = CH3COOK + 2H20

Mass:
K – 39
OH – 17
H2O – 18
CH3OOH – 60
CH3COOK – 98
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Borek

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Re: Making Potassium Acetate Solution
« Reply #3 on: May 18, 2017, 09:13:06 PM »

In the equation below if I use 500 grams of the KOH solution, I am calculating I need 405 grams of Acetic Acid to completely react.  Is that right?

The number is wrong, hard to comment without seeing what you really did.

Quote
The Reactant masses should equal the Product masses right?

Yes, that's a mass conservation. Please remember there is also water produced, plese remember initial system contains water.

Quote
Will the percentage of Active Ingredient in the Acetic acid affect or change these numbers?

"These numbers" - which numbers? Amount of the acid required to neutralize KOH present in 500 grams of 45% solution is always the same, amount of _solution_ containing given amount of the acetic acid is variable and depends on the acetic acid solution concentration.
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deezy

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Re: Making Potassium Acetate Solution
« Reply #4 on: May 19, 2017, 06:39:29 AM »

Sorry- here are the calculations I did so you can see.  Are these Correct?  Thanks much!

Potassium Hydroxide in Water Solution + Acetic Acid --->  Potassium Acetate + Water

K + OH + H2O + CH3COOH ---> CH3COOK + 2H20

•   Example using 500g of KOH in water solution (KOH in water solution is a 45% Potassium by weight)
•   Mass to Mole, Mole to Mole, Mole to Mass Calculation
o   500g KOH Solution x 1 mol KOH Solution x 1 mol CH3COOH   x   60g CH3COOH =     405g CH3COOH
                                     74g KOH Solution      1 mol KOH Solution   1 mol CH3COOH


•   Same example calculating Products
•   Mass to Mole, Mole to Mole, Mole to Mass Calculation
o   500g KOH Solution x 1 mol KOH Solution x 1 mol CH3COOK   x   98g CH3COOK   =    662g CH3COOK
                                     74g KOH Solution      1 mol KOH Solution   1 mol CH3COOH

o   500g KOH Solution x 1 mol KOH Solution x 2 mol H2O        x         18g H2O        =        243g H2O
                                     74g KOH Solution      1 mol KOH Solution      1 mol H2O

•   Reactants mass equals Products mass
o   500g KOH Solution + 405g CH3COOH = 905g
o   662g CH3COOK + 243g H2O = 905g


Molar Mass:
K – 39
OH – 17
H2O – 18
CH3OOH – 60
CH3COOK – 98
KOH + H2O – 74


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deezy

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Re: Making Potassium Acetate Solution
« Reply #5 on: May 19, 2017, 07:55:14 AM »

Here is second try, I took the water out on the left side of the equation.  Not sure which is correct, because the water in the KOH solution is remaining constant.  Only one water molecule is being produced from the reaction.

Potassium Hydroxide in Water Solution + Acetic Acid --->  Potassium Acetate + Water

K + OH + CH3COOH ---> CH3COOK + H20

•   Example using 500g of KOH in water solution (KOH in water solution is a 45% KOH)
•   Mass to Mole, Mole to Mole, Mole to Mass Calculation
o   500g KOH                   x 1 mol KOH  x 1 mol CH3COOH   x   60g CH3COOH =     535g CH3COOH
                                           56g KOH         1 mol KOH             1 mol CH3COOH


•   Same example calculating Products
•   Mass to Mole, Mole to Mole, Mole to Mass Calculation
o   500g KOH                  x 1 mol KOH x 1 mol CH3COOK   x   98g CH3COOK   =    875g CH3COOK
                                           56g KOH       1 mol KOH             1 mol CH3COOH
 
o   500g KOH                  x 1 mol KOH  x 1 mol H2O        x         18g H2O        =        160g H2O
                                           56g KOH      1 mol KOH                1 mol H2O

•   Reactants mass equals Products mass
o   500g KOH + 535g CH3COOH = 1035g
o   875g CH3COOK + 160g H2O = 1035g

Molar Mass:
K – 39
OH – 17
H2O – 18
CH3OOH – 60
CH3COOK – 98
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deezy

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Re: Making Potassium Acetate Solution
« Reply #6 on: May 19, 2017, 10:59:59 AM »

Adding to the second example with no water on the left side of the equation:

The KOH solution I'm using is a 45% KOH in water, to scale the correct weights to reflect that:

500g KOH x 0.45 = 225g of actual KOH needed to be neutralized

535g of Acetic Acid x 0.45 = 240.75g of Acetic Acid needed to fully neutralize all the KOH.  (this is assuming I use 100% Acetic Acid)

This looks right to me but please correct if I'm wrong.
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Borek

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Re: Making Potassium Acetate Solution
« Reply #7 on: May 19, 2017, 10:20:14 PM »

500g KOH x 0.45 = 225g of actual KOH needed to be neutralized

Yep, that's where you were wrong in your initial calculations.

Note that it means your initial mixture already contains plenty of water.
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