1. Dilution Chapter 15 Ch 15 ppt 3 - Dilution.ppt

2. Today you will learn: What does it mean to dilute a solution?
How is the amount of solute affected by dilution?
How is the molarity affected by dilution?
How to calculate the new concentration when a solution is diluted.
Stay the same.
Ch 15 ppt 3 - Dilution.ppt

3. Dilution of Solutions
Dilution- the procedure for preparing a less concentrated solution from a more concentrated one.
In the lab, the solutions you use, usually have to be diluted from a stock solution.
Ch 15 ppt 3 - Dilution.ppt

4. Dilution of Solutions
When you want to dilute a solution, what happens to the number of moles of solute present in the solution?
Do they increase?
Decrease?
Stay the same?
Stay the same.
Ch 15 ppt 3 - Dilution.ppt

5. Dilution of Solutions
Figure 4.19
Ch 15 ppt 3 - Dilution.ppt

6. Dilution of Solutions

7. Dilution of solutions M1  V1 = M2 V2
Since moles are constant before and after dilution, we can use the following formula for calculations.
M1  V1 = M2 V2

8. Dilution of Solutions
Describe how you would prepare 800mL of a 2.0M H2SO4 solution, starting with a 6.0M stock solution of sulfuric acid.
800 mL x L =
,000 mL = 0.800L
M1  V1 = M2  V2 V1 = (M2  V2) / M1
V1 = (2.0 M x L) / 6.0 M
V1 = (1.6 M x L) / 6.0 M
V1 = 0.27 L
L (= 270 mL) of the 6.0M H2SO4 solution
2. Add water to make a final volume of 800mL

9. Neutralization HOH = H2O
When a strong acid and a strong base are mixed, the resulting solution contains water and a salt. This is a double replacement reaction.
Ex: HCl + NaOH  HOH + NaCl
Ex: 2 H3PO4 + 3 Ca(OH)2  6 HOH + 1 Ca3(PO4)2
HOH = H2O