Calculating Molarity of KOH Solution: Discrepancy in Reported Results

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The discussion centers on verifying the final molarity of a solution after adding 40.69 lb of 90% anhydrous KOH to 561.86 liters of water. The initial calculation yields a molarity of 0.52, while a colleague's report states 0.78 M, raising questions about potential errors. The calculations show that 297.01 moles in 561.86 liters results in 0.52 M, leading to skepticism about the higher value. The term "anhydrous KOH" is clarified, with some participants suggesting it could refer to K2O, but this does not significantly affect the results. Notably, the ratio of the two molarity values (0.52/0.78) is exactly 2/3, indicating a possible simple conversion error in the colleague's calculations.
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I'm trying to verify the final molarity of my solution.

I'm adding 40.69 lb of 90% anhydrous KOH to 148.43 (561.86L) gallons of water . I get a final molarity of 0.52. The report I'm reading arrives at a molarity of 0.78. Is this an error?

Thanks to anyone who can provide the calcs.
 
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Is this homework? If so, we can move it to the correct subforum.
 
No it's not homework. I'm checking the calcs in a report by a colleague in nthe alkaline hydrolysis business and his stated 0.78 M KOH seems off to me. I get 297.01 moles in 561.86 liters to be 0.52 molarity
 
pmason61 said:
I'm adding 40.69 lb of 90% anhydrous KOH to 148.43 (561.86L) gallons of water . I get a final molarity of 0.52. The report I'm reading arrives at a molarity of 0.78. Is this an error?
"Anhydrous KOH" is K2O; any thoughts?
 
What's the other 10% of the anhydrous KOH?
 
Bystander said:
"Anhydrous KOH" is K2O; any thoughts?

Doesn't change the result enough. I am getting 0.52 M for KOH and 0.62 M for K2O (assuming the density of the solution to be 1.0247 g/mL, which is more or less OK for 0.52 M, should be a bit higher for 0.62 M).

Note 0.52/0.78 is exactly 2/3, such coincidences often mean some simple error in conversions.
 
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