posted by Madison .
What volume of 2.5% (w/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?
Wouldn't you think you could dilute the 5.0% by a factor of 2 (making 125 mL to 250 mL) and that would reduce the concn to 2.50%?
Yes, but would that give me 63 mL or 250 mL. That's where I get confused.
Think it through. If you DILUTE something by a factor of 2, you take 125 mL and make it 250 mL. That makes it weaker, right? How much weaker? 250/125 = 2 x weaker so if it were 5.00% before, now it is 2.50%. If you ended up with 63 mL, the same grams in less volume is MORE concd.
mL x % = mL x %
mL x 2.50 = 125 x 5.0%
mL = (125 x 5.0)/2.50 = 250 mL
Thanks!I get it now. Getting 250 mL just confused me because it was more than I started with, but that's what I want, just in a weaker concentration, right?