Calculating Molarity of KOH Solution: Discrepancy in Reported Results

[pmason61]

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.

 

[jrmichler]

Is this homework? If so, we can move it to the correct subforum.

 

[pmason61]

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

 

[Bystander]

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?

 

[Mayhem]

What's the other 10% of the anhydrous KOH?

 

[Borek]

"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.