Dilution of Solutions
** Concentrated solutions are often stored in the laboratory stockroom for use as needed. Frequently we dilute these “stock” solutions before working with them.
** Dilution is the procedure for preparing a less concentrated solution from a more concentrated one.
** Suppose that we want to prepare 1 L of a 0.400 M KMnO4 solution from a solution of 1.00 M KMnO4 . For this purpose we need 0.400 mole of KMnO4. Because there is 1.00 mole of KMnO4 in 1 L of a 1.00 M KMnO4solution, there is 0.400 mole of KMnO4 in 0.400 L of the same solution:
Therefore, we must withdraw 400 mL from the 1.00 M KMnO4 solution and dilute it to 1000 mL by adding water (in a 1-L volumetric flask). This method gives us 1 L of the desired solution of 0.400 M KMnO4 .
** In carrying out a dilution process, it is useful to remember that adding more solvent to a given amount of the stock solution changes (decreases) the concentration of the solution without changing the number of moles of solute present in the solution.
** In other words,
moles of solute before dilution = moles of solute after dilution
** Molarity is defined as moles of solute in one liter of solution, so the number of moles
of solute is given by:
** Because all the solute comes from the original stock solution, we can conclude that n remains the same; that is,
Mi and Mf are the initial and final concentrations of the solution in molarity, respectively
Viand Vf are the initial and final volumes of the solution, respectively.
** Of course, the units of Vi and Vf must be the same (mL or L) for the calculation to work.
** To check the reasonableness of your results, be sure that Mi > Mfand Vf > Vi.
Describe how you would prepare 5.00 × 102 mL of a 1.75 M H2SO4solution, starting with an 8.61 M stock solution of H2SO4.
Because the concentration of the final solution is less than that of the original one, this is a dilution process. Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same.
We prepare for the calculation by tabulating our data:
Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with sufficient water to give a final volume of 5.00 × 102 mL in a 500-mL volumetric flask to obtain the desired concentration.
The initial volume is less than the fi nal volume, so the answer is reasonable.
Reference: Chemistry / Raymond Chang ,Williams College /(10th edition).