# Dilution of Solutions

####
**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 KMnO

_{4}solution from a solution of 1.00 M KMnO_{4}. For this purpose we need 0.400 mole of KMnO_{4}. Because there is 1.00 mole of KMnO_{4}in 1 L of a 1.00 M KMnO_{4}solution, there is 0.400 mole of KMnO_{4}in 0.400 L of the same solution:
Therefore, we
must withdraw 400 mL from the 1.00 M KMnO

_{4}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 KMnO_{4}.####
**Dilution
law **

** 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,

M

_{i}and M_{f}are the initial and final concentrations of the solution in molarity, respectively
V

_{i}and V_{f}are the initial and final volumes of the solution, respectively.
** Of course, the
units of V

_{i}and V_{f}must be the same (mL or L) for the calculation to work.
** To check the
reasonableness of your results, be sure that M

_{i }> M_{f}and V_{f}> V_{i}.####
**Solved
problems**

**Describe how you would prepare 5.00 × 10**

^{2}mL of a 1.75 M H_{2}SO_{4}solution, starting with an 8.61 M stock solution of H_{2}SO_{4}.**Strategy:**

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.

**Solution:**

We prepare for the calculation by tabulating
our data:

Thus, we must
dilute 102 mL of the 8.61 M H

_{2}SO_{4}solution with sufficient water to give a final volume of 5.00 × 10^{2}mL in a 500-mL volumetric flask to obtain the desired concentration.**Check:**

The initial volume is less than the fi nal
volume, so the answer is reasonable.

*Reference: Chemistry*

*/ Raymond Chang ,*

*Williams College*

*/*

*(10th edition).*

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