Surviving C₁V₁ = C₂V₂.

Remember

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> C_2, 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