Dilutions from Concentrated Stock Solutions  
When a dissolved solvent is added to a solution, the solute is spread out through the large volume and the numebr of moes per unit volume decreases.  The solution is said to be diluted.  Such dilutions are a natural part of chemistry in the lab.  If two solutions of different solutes are moixed, the total volume increases and becomes occupied by both solutes.  As a result the concentration of both solutes is less.  At other times dilution is deliberate and essential.  Stock solutions of common chemicals are often prepared in large volumes in preset concentrations.  The chief supply of HCl in the lab may be 1.00 M HCl.  In your experiment, you may requir for a much less conencrtated solution, so the concentrated solution must be diluted first.
 
  When carrying out a dilution with an acid.  Always add acid to water.  Never add water to the acid.   When acid and water mix a great deal of heat is released.  Acid added to water gives a large heat sink for the heat to dissipate in.  If you add water to acid, the heat generated will cause the water to boil, and may splatter you with the hot acid solution.
 
Dilution Calculation
All the calculations that you need to do are based on a simple fact.  As a solution is diluted, the number of moles of solute doesn't change; the solute simply spreads out through a larger volume.
 
We will use the following symbols: V for volume and M for molarity. Therefore Vc and Mc are the concentrated solutions volume and molarity. Vd and Md are then the diluted solutions volume and molarity.

                                            Vc Mc  =  Vd Md
 

Problem:  How can we prepare 100 of 0.040 K2Cr2O7 from 0.200 M K2Cr2O7?
Solution:
First assemble the data into a table:
                 Vc =  ?               Vd = 100 mL
                Mc = 0.200 M    Md = 0.040 M

Next, use the formula from above.  Vc X 0.20 M = 100 mL X 0.040 M
                                                      Vc = 100 mL X 0.040 M
                                                                      0.20 M
                                                      Vc = 20.0 mL

Place 20.0 mL of 0.200 M K2Cr2O7 into a 100 mL volumetric flask and then add water until the final volume is 100 mL.
 
 

Practise Problem:  Describe how to make 500 mL of 0.20 M NaOH from 0.50 M NaOH.
 
 
 
 
 
 
 
Problem:
How many millilitres of water would have to be added to 100 mL of 0.40 M HCl to give a solution with a concentration of 0.10 M?
 
First assemble the data.         Vc = 100 mL         Vd = ?
                                                Mc = 0.40 M         Md = 0.10 M

Use the equation:   100 mL X 0.40 M = Vd  X 0.10 M
                                                     Vd  = 400 mL

The final volume must be 400 mL.  But since we started with 100 mL we will add 300 mL of water until the volume reaches 400 mL.
 

Practise Problem: How many millilitres of water would have to be added to 300 mL of 0.5 M NaOH to give a solution with a conentration of 0.2 M?
 
 
 
 
 
 
          Go to the Solution Dilution Worksheet