Discussion Overview
The discussion revolves around the calculation of the molarity of a KOH solution, specifically addressing a discrepancy between the participant's calculated molarity of 0.52 M and a colleague's reported value of 0.78 M. The scope includes technical calculations related to solution preparation and molarity determination.
Discussion Character
- Technical explanation
- Debate/contested
Main Points Raised
- One participant calculates the final molarity of their KOH solution to be 0.52 M based on adding 40.69 lb of 90% anhydrous KOH to 561.86 liters of water.
- Another participant questions the accuracy of the reported molarity of 0.78 M, suggesting that it seems incorrect based on their calculations.
- A participant mentions that "anhydrous KOH" is K2O, prompting further discussion about the implications of this definition on the calculations.
- One participant calculates that the molarity for K2O would be 0.62 M, assuming a solution density of 1.0247 g/mL, and notes that the ratio of the two molarities (0.52/0.78) is exactly 2/3, indicating a potential simple error in conversions.
Areas of Agreement / Disagreement
Participants do not reach a consensus on the correct molarity, with multiple competing views on the calculations and definitions involved. The discussion remains unresolved regarding the accuracy of the reported molarity.
Contextual Notes
There are limitations regarding the assumptions made in the calculations, such as the density of the solution and the definition of "anhydrous KOH." These factors may influence the final molarity results but are not fully resolved in the discussion.