Very often, solutions are stored in concentrated forms called stock solutions. Now, if a solution with a lower concentration is needed, some of the stock solution is mixed with a solvent and a dilute solution is obtained.

Concentrated solution + Solvent  →   Dilute solution

This process is called dilution and it is very convenient because the concentration of the dilute solution can be easily calculated using the dilution formula:

M₁ and V₁ are the concentration and the volume of the stock solution, and M₂, V₂ are the concentration and the volume of the final solution that we prepare.

For example, how many liters of 0.350 molar sodium hydroxide solution can be prepared from 400.0 mL of 3.60 M solution?

First, write the given data, and for this, we need to correctly assign the M₁, V_1, M_2, and V_2.

0.350 molar solution needs to be prepared, and therefore, it is the M₂ – the concentration of the final solution.

400.0 mL is the volume of the stock solution, so we assign it as V₁

3.60 M is the concentration of the stock solution, so that is M₁, and therefore, V2 is what we need to find.

List, this together and set up the dilution formula:

M₁ = 3.60 M

V₁ = 400.0 mL

M₂ = 0.350 M
_________________

V₂ = ?

 M₁V₁ = M₂V₂

\[{{\rm{V}}_{\rm{2}}}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{M}}_{\rm{2}}}}}\; = \frac{{3.60\cancel{M}\; \times \;{\rm{400}}{\rm{.0}}\;{\rm{mL}}}}{{0.350\;\cancel{M}}}\; = \;4,110\;{\rm{mL}}\;{\rm{ = }}\;{\rm{4}}{\rm{.11}}\;{\rm{L}}\]

We used the volumes in mL and convert them to liters at the end. However, you can also convert the volumes to liters first which will also help avoid potential mistakes because the units of molarity are mol/L.

Let’s do another example. Let’s say we have a concentrated 6.0 M solution of paint and need to prepare 2.0 L of a 0.45 molar solution. How many liters of the stock do we need?

First, assign the notation for the stock and final solutions, and list, the given data and the unknown together:

M₁ = 6.0 M

M₂ = 0.45 M

V₂ = 2.0 L
_________________

V₁ = ?

 M₁V₁ = M₂V₂ 

\[{{\rm{V}}_{\rm{1}}}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{M}}_{\rm{1}}}}}\; = \frac{{0.45\cancel{M}\; \times \;{\rm{2}}{\rm{.0}}\;{\rm{L}}}}{{6.0\;\cancel{M}}}\; = \;0.15\;{\rm{L}}\]

So, we will add 0.15 L of the paint, and add some solvent to make the final volume of the solution to 2.0 L.

The volume of the solvent then is 2.0 – 0.15  = 1.85 L.

Check Also

• Solutions
• Strong and Weak Electrolytes
• Dissociation of Ionic Compounds
• Molecular, Ionic, and Net Ionic Equations
• Molarity 
• Ion Concentration
• Precipitation Reactions
• Definitions of Acids and Bases
• Acid-Base Reactions
• Stoichiometry of Reactions in Aqueous Solutions
• Acid-Base Titrations
• Oxidation State
• Oxidation-Reduction (Redox) Reactions

• Balancing Redox Reactions

• Reactions in Aqueous Solutions Practice Problems