Sunday, January 9, 2011

Diluting Solutions to Prepare Workable Solutions

A Quick Recap:
Solvent = the liquid that does the dissolving
Solute = the substance which dissolves
Solution = is prepared by dissolving a solute in a solvent.

-         When a solution is diluted more solvent is added to it.
-         The moles of solute is always constant (the only difference is that there is more solvent in the less concentrated solution)

The total number of solutes in the solution remains the same after dilution, but the volume of the solution becomes greater.

Thus we have the equation:
            Moles of solute before = Moles of solute after
M1V1 = M2V2

Lets try an example:

Concentrated HNO3 is 15.4 mole/L.  How would you prepare 2.50 L of 0.375 M HNO3?

M1V1 = M2V2
15.4 mole/L x V1  = 0.375 M x 2.50 L
→ Re organize the equation
 V1  = 0.375 M x 2.50 L
                 15.4 M

V1 = 0.0609 L
V1 = 60.9 mL

*Remember Sig Figs

To calculate how much water we need in order to dilute the solution we take our known volume (V2) and subtract it from V1.
       2.50 L – 0.0609 L  = 2.44L
So, you would need to add 2.44 L of water to make the equation true.

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 By Candace

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