What are the final concentrations after mixing these two solutions?
September 9, 2017 2:25 PM   Subscribe

If I mix 15 mL of solution A consisting of 5 mg/mL of compound A, with 15 mL of solution B consisting of 5 mg/mL of compound B, what will be my final concentrations of compounds A and B in the resulting 30 mL volume? Similarly, what if I mix 15 mL of solution A with 15 mL of solution C consisting of 2.1 mg/mL of compound A and 5 mg/mL of compound B.

Any links to equations, online calculators, google docs with equations, or excel files would be appreciated.
posted by pwb503 to Science & Nature (4 answers total)
 
15 mL of 5 mg/mL means there's a total of 5*15=75mg of A. Same with B.
Mix them together, and (assuming no reactions that produce or break apart the solvent) you've got 30mL of mixture with 75 mg of each compound in it. 75mg/30mL=2.5 mg/mL. Put another way, compound A is now spread out over 30 mL instead of just 15mL, so the concentration is halved.

To solve the other problem, same sort of thing. Figure out the total mg of each compound, and divide by the total volume.
posted by notsnot at 2:44 PM on September 9 [2 favorites]


What notsnot said. The formula is c = m/V, rearranged, m = cV.

This does assume that the density doesn't change. At such low concentrations it shouldn't, unless the compounds are in difference solvents e.g. Ethanol and water.
posted by kjs4 at 2:50 PM on September 9 [2 favorites]


For this, I use (C1)(V1)=(C2)(V2), where C is concentration and V is volume.
In your case, (5 mg/ml)(15 ml)=(x)(30 ml).
Solving for x yields 2.5 mg/ml, as noted above.
posted by Knowyournuts at 4:32 PM on September 9 [2 favorites]


Are both solutions in the same solvent? I know that mixing x ml of water with y ml of alcohol does not result in (x+y) ml of alcohol and water, so you might need to account for that?
posted by rustcrumb at 11:22 AM on September 10


« Older How did they measure and test typing speed before...   |   Do I have a rat or mice infestation? Newer »

You are not logged in, either login or create an account to post comments