1. Review of Basic Concepts, Molarity, Solutions and Dilutions
Chapter 4
4.5

2. Aqueous Solutions
In Chemistry, many reactions take place in water. This is also true for Biological processes.
Reactions that take place in water are said to occur in an aqueous solution.
Three types of reactions take place in aqueous solutions: Precipitation, Acid-Base and Redox.

3. Properties of Aqueous Solutions
Solution- a homogeneous mixture of two or more substances.
Solute- a substance in a solution that is present in the smallest amount.
Solvent- a substance in a solution that is present in the largest amount.
In an aqueous solution, the solute is a liquid or solid and the solvent is always water.

4. Properties of Aqueous Solutions
All solutes that dissolve in water fit into one of two categories: electrolyte or non-electrolyte.
Electrolyte- a substance that when dissolved in water conducts electricity
Non-electrolyte- a substance that when dissolved in water does not conduct electricity.
To have an electrolyte, ions must be present in water.

5. Electrolytic Properties of Aqueous Solutions
NaCl in water.
What happens?
NaCl(s) → Na+(aq) + Cl–(aq)
Completely dissociates

6. Strong vs. Weak Electrolytes
How do you know when an electrolyte is strong or weak?
Take a look at how HCl dissociates in water.
HCl(s) → H+(aq) + Cl–(aq)
There is a single arrow showing the conversion of HCl to H+ and Cl-. This shows that in water, HCl can dissociate completely. Complete dissociation also means that the ions are hydrated and that they will never reform again. Indicates a strong electrolyte.

7. Electrolytic Properties of Aqueous Solutions
Photo of hydrated ions.

8. Electrolytic Properties of Aqueous Solutions

9. Hydrated Ions

10. Electrolytic Properties of Aqueous Solutions
What about weak electrolytes?
What makes them weak?
Ionization of acetic acid
CH3COOH(aq) ↔ CH3COO–(aq) + H+(aq)
Double arrow means that the reaction is reversible. If the reaction is reversible then acetic acid is being broken down and also being reformed while in water. Since it can reform, there is not complete ionization. Complete ionization defines if an electrolyte is strong or weak.
Sometimes there can be a state of chemical equilibrium……breakdown=formation.

11. Electrolytic Solutions

12. electrolytes
Table 4.3
Strong acids—completely ionize, therefore it is also a strong electrolyte.
Weak Acids---Do not completely ionize, therefore it is also a weak electrolyte.

13. Precipitation Reactions
Precipitation Reaction- a reaction that results in the formation of an insoluble product.
These reactions usually involve ionic compounds.
Formation of PbI2:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Double displacement reaction. Can you see where the ions switched partners?

14. Preciptate

15. Precipitate

16. Precipitation Reactions
How do you know whether or not a precipitate will form when a compound is added to a solution?
By knowing the solubility of the solute!
Solubility- The maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.
Three levels of solubility: Soluble, slightly soluble or insoluble.

17. Precipitation Reactions

18. Determining Solubility
Determine the of solubility for the following:
(1) Ag2SO4
(2) CaCO3
(3) Na3PO4
Only Na3PO4 is soluble.

19. Concentration of Solutions
Concentration of a Solution- amount of solute present in a given quantity of solvent or solution.
We will use Molarity, M for measurement. Molarity can also be called Molar Concentration.
Molarity– the number of moles of solute per liter of solution.
Molarity- moles of solutes/ liters of solution
Or n/v
Moles- grams of sample/ molecular weight of sample
Or g/ mw
To study stoichiometry, we must know how much of the reactants are present in a solution and how to control the amounts of reactants used to bring about a reaction in aqueous solution.

20. Concentration of Solutions
How many moles are there in 24.0g of C?
moles = g/mw
moles = 24.0g C/ 12.0g C
moles = 2.0
There are 2.0 moles of C in 24.0g of C.

21. Concentration of Solutions
How many grams are in 2.0 moles of Boron?
moles= g/MW
2.0 moles = grams/ 10.81g Boron
2.0 moles x 10.81g Boron = grams
Grams = 21.62
There are g of Boron in 2.0 moles of Boron.

22. Concentration of Solutions
What is the Molarity of a 1L solution containing 9.0g HCl?
9.00g HCl x 1 mol HCl/ 18.00g HCl
= 0.5 mol HCl
M = n/v
M = 0.5 mol HCl/ 1L
M = 0.5
The concentration of the solution is 0.5M.

23. Preparation of Solutions
Now that you know how to calculate M, n and v, what does that mean?
You can make your own solutions!
What are the steps in making a proper solution?
Weigh out sample.
Place in correct size flak.
Add dH20 to dissolve sample.
Fill to known line.
Use 250mL flask.

24. Preparation of Solutions

25. Concentration of Solutions
How many grams of Potassium Dichromate, K2Cr2O7, are required to prepare a 250mL solution with a concentration of 2.16M?
250mL x 1L/ 1000mL = .250L
M= n/v
n= M x v
n= 2.16M x .250L
n= 0.54 mol
moles = g/MW
Grams = moles x MW
Grams = 0.54 mol K2Cr2O7 x g K2Cr2O7
Grams = 159
159 grams of K2Cr2O7 are needed to prepare the requested solution.

26. Concentration of Solutions
In a biochemical assay, a chemist needs to add 0.381g of glucose to a reaction mixture. Calculate the volume in millimeters of a 2.53M glucose solution that she should use for this addition.
moles = g/MW
moles = 0.381g C6H12O6/ 180.2g C6H12O6
moles = x 10 –2 mol C6H12O6
M = n/v
v = n/M
v = x 10 –2 mol C6H12O6 / 2.53M C6H12O6
v = 8.36mL
She should use 8.36mL of the 2.53M glucose solution.

27. Preparation of Solutions
Explain the process of creating 1L of 3.0M KCl.
M = n/v
n = M x v
n = 3.0M x 1L
n = 4.0 mol of KCl needed
moles= g/MW
Grams = moles x MW
Grams = 4.0 mol KCl x 36.0g KCl
Grams = 144g KCl
Weigh out 144g of KCl. Put in a 1L flask. Add enough dH20 to dissolve KCl. Fill flask to 1L meniscus.

28. Dilution of Solutions
Dilution- the procedure for preparing a less concentrated solution from a more concentrated one.
Dilutions can be made in increments of 10, 20, 50 or any other value.
Serial Dilution- the process of diluting a solution by removing part of it, placing this in a new flask and adding water to a known volume in the new flask.
In the lab, the solutions you use, usually have to be diluted from a stock solution.

29. Dilution of Solutions
When you want to dilute a solution, what happens to the number of moles present in the solution?
Do they increase?
Decrease?
Stay the same?
Stay the same.

30. Dilution of Solutions
Figure 4.19

31. Dilution of Solutions

32. Dilution of solutions
Since moles are constant before and after dilution, we can use the following formula for calculations.
MiVi = MfVf

33. Dilution of Solutions
Describe how you would prepare 800mL of a 2.0M H2SO4 solution, starting with a 6.0M stock solution of .
800mL x 1L/ 1000mL
= 0.800L
MiVi = MfVf
6.0M x Vi = 2.0M x 0.800L
6.0M x Vi = 1.6M x L
Vi = 1.6M x L/ 6.0M
Vi = 0.26L
0.26L of the 6.0M H2SO4 solution should be diluted to give a final volume of 800mL.