Here is a repeat of the combined equation from above: This is a resonable assumption in this example. Also, note the implicit assumtion that the volumes are additive. Nowhere does it state that the solution was diluted to 20 mL. It simply tells you the starting volume (5.00 mL in 2 and 3) and that 15.0 mL was added. Note how the problem is worded to avoid giving a final volume. X = 8.40 x 10 -4 M (concentration of the stock solution) X = 1.68 x 10 -4 M (concentration of dilution 1) X = 4.20 x 10 -5 M (concentration of dilution 2) X = 1.05 x 10 -5 M (concentration of dilution 3) The beaker is labeled dilution 3.Ĥ) 1.00 mL of the solution in the beaker labeled dilution 3 is transferred over to another beaker called dilution 4 and enough water is added so that the final volume is 5.00 mL and the concentration of the final dilution is 2.10 x 10 -6 M.

The beaker is labeled dilution 2.ģ) 5.00 mL of the solution in the beaker labeled dilution 2 is transferred to another beaker and 15.0 mL of water is added. The beaker is labeled dilution 1.Ģ) 5.00 mL of the solution in the beaker labeled dilution 1 is transferred to another beaker and 15.0 mL of water is added. Consequently, each volume ratio is a factor less than 1.Įxample #2: What is the molarity of dilution 1?ġ) A stock solution of salicylic acid is diluted by transferring 10.0 mL of the stock solution into a beaker and adding 40.0 mL of water. Notice that each volume ratio represents a dilution. Now, substitue into the equation for solution C: Now, substitue into the equation for solution B: To do this, I will eliminate the intemediate concentrations symbolized by x and y. I'm now going to form one equation that relates C i and C f. In place of the specific numbers in step two, I would like to use symbols: I would like to now reconsider the solution. In the above, I carried two guard digits in my values for x and y. Solution C ⇒ (0.51936) (87.0) = (z) (287)Īdvice: if you use the above technique, make sure to carry several guard digits as you do each calculation. Solution: (the solution will be followed by some discussion)ġ) Assign unknowns to the final molarities of each dilution:Ģ) Set up all three dilutions using M 1V 1 = M 2V 2 Solution A ⇒ (5.60) (46) = (x) (116)ģ) Each of these three eqations may now be solved in turn to give the final answer (symbolized by the variable z): What is the concentration of sucrose in solution C? Solution C = 87.0 mL of Solution B is diluted to 287 mL

Solution B = 58.0 mL of Solution A is diluted to 248 mL Solution A = 46.0 mL of the stock solution is diluted to 116 mL

Example #1: The following successive dilutions are applied to a stock solution that is 5.60 M sucrose:

Four Fold Serial Dilutions