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Chemistry
Doubts regarding solubility product problem
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[QUOTE="Borek, post: 5735239, member: 23711"] 1. It doesn't matter what the volume is - as long as you mix two equal volumes when trying to calculate concentration after the dilution you will get something like [itex]\frac V {V+V}[/itex] where the V cancels out leaving you with [itex]\frac 1 2[/itex]. As V cancels out, we can safely ignore it. 2. They never assumed the volume to be 1 L, they just halved the concentration for the reason explained above. 3. But then, actually there is nothing wrong with assuming volume of 1L (or any other). All that is important is that you assume equal volumes and 1L is much easier to use in further calculations than - say - 2.7641 gallons. [/QUOTE]
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Doubts regarding solubility product problem
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