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?
right.
To find the volume of a 2.5% (w/v) KOH solution that can be prepared from 125 mL of a 5.0% KOH solution, we need to use the concept of dilution.
Dilution is a process that involves adding a solvent (usually water) to a concentrated solution to reduce its concentration. The formula used for dilution is:
C1V1 = C2V2
where C1 and V1 are the concentration and volume of the initial solution, and C2 and V2 are the concentration and volume of the final solution.
In this case, we can rewrite the formula as:
(5.0%)(125 mL) = (2.5%)(V2)
To solve for V2, we rearrange the equation:
V2 = (5.0%)(125 mL) / (2.5%)
First, convert the percentages to decimal form:
5.0% = 0.05
2.5% = 0.025
Substitute the values:
V2 = (0.05)(125 mL) / (0.025)
Now, calculate:
V2 = 2.5 mL
Therefore, 2.5 mL of 2.5% (w/v) KOH can be prepared from 125 mL of a 5.0% KOH solution.