# 8.2 Calculating dilutions

Remember that during dilution when additional solvent is added the amount of solute remains constant. It is only the concentration that changes. This means that the ratio between the concentration and volume remains constant before and after dilution. This can be exploited to work out a final concentration following dilution or the volume of additional solvent to add to achieve a target concentration. This ratio can be expressed as:

*C*_{1}*V*_{1}* = C*_{2}*V*_{2}

where

*C*_{1} = concentration of the original solution

*V*_{1} = volume of the original solution

*C*_{2} = concentration of the final solution

*V*_{2} = volume of the final solution.

As this is a ratio, it does not matter what units you use for concentration or volume as long as you **use identical units on each side of the equation**.

#### Example 1

What volume of 2 M NaCl would you need to prepare 1 l of a 0.01 M solution?

When first doing this sort of calculation it helps to define the terms by writing out the equation in full. Here we are trying to determine what volume of the original solution (*V*_{1}) we need.

*C*_{1} = 2 M

*V*_{1} = ?

*C*_{2} = 0.01 M

*V*_{2} = 1 l

Since the units for concentration are the same on both sides, no conversion is required. Our *V*_{2 }is in litres so that is the unit our answer will be in.

*C _{1}V_{1} = C_{2}V_{2}*

2 M *V _{1}* = 0.01 M × 1 l

*V _{1} *=

*V _{1}* = 0.005 l = 5 ml

#### Example 2

What is the final concentration of a solution if 50 ml of a 1 M solution is added to 0.24 l?

Remember you need to total final volume here and you have to convert the volume to the same units on each side of the equation.

*C _{1}* = 1 M

*V _{1}* = 50 ml

*C _{2}* = ?

*V _{2}* = 50 ml + 240 ml = 290 ml

*C _{2 }*=

*C _{2}* = 0.17 M = 170 mM

#### Example 3

Tube 1 contains a 2 M solution; 1 ml is removed to tube 2 containing 9 ml of water. Tube 2 is mixed, and 1 ml is removed from this tube and added to tube 3 containing 19 ml of water. Tube 3 is mixed, and 1 ml is removed from this tube and added to tube 4 containing 4 ml of water. What is the concentration of the solution in tube 4?

Remember for serial dilutions we multiply the dilution factors.

Dilution 1 = 1 / (1 + 9) = 1/10

Dilution 2 = 1 / (1 + 19) = 1/20

Dilution 3 = 1 /(1 + 4) = 1/5

Total dilution = 1/10 × 1/20 × 1/5 = 1/1,000

Concentration of tube 4 = 2 M × 1/1,000 = 0.002 M = 2 mM