Surviving C1V1 = C2V2.



C1V1 = C2V2.


C1 = original concentration of the solution, before it gets watered down or diluted.

C2 = final concentration of the solution, after dilution.

V1 = volume about to be diluted

V2 = fianl volume after dilution


By drawing the "X" through the equal sign and filling in the formula with letters of a size permitted by the borders of the "X", it reminds you that :


for all dilution problems C1> C2, and V1< V2.


It makes sense because to dilute, we add water. This increases the volume but lowers concentration.


Examples by Type:


1.         Easiest:             Joe has a 2 g/L solution. He dilutes it and creates 3 L of  a 1 g/L solution.

                                    How much of the original solution did he dilute?

            Solution:           We are looking for V1:

                                    C1V1 = C2V2

                                                2x = 1(3)

                             x = 1.5 L


2.         A little trickier:  Joe has 20 L of a 2 g/L solution. He diluted it, and created 3 L of  a 1 g/L solution.

                                    How did he make such a solution.


We're only going to use part of the 20 L. Remember we have to end up with 3 L after dilution, so not only do we have to start with less than 20 L but also less than 3 L. How much is unknown = V1, and it amounts to the same problem as in example 1, but don't use the 20L.

                                    C1V1 = C2V2

                                                2x = 1(3)

                             x = 1.5 L.


3.         Trickier too:     Joe has 20 L of a 2 g/L solution. To this solution he adds 30 L. What is the final concentration of the solution?


If 30 L is added to the 20 L, then the volume about to be diluted, V1, is 20 L. And adding 30 L of water to a 20 L solution creates a fianl volume, V2,  of 50 L. Our unknown is C2.

                                    C1V1 = C2V2

                                                2(20) = x(20 + 30)

                                    x = 40/50 = 0.80 g/L