1. Ch. 9: Introduction to Solutions and Aqueous Reactions
Dr. Namphol Sinkaset
Chem 200: General Chemistry I

2. I. Chapter Outline Introduction Solution Concentrations
Solution Calculations
Aqueous Solutions
Precipitation Reactions
Acid/Base Reactions
Gas-Evolution Reactions
Oxidation-Reduction Reactions

3. I. Aqueous Chemistry
Water-based chemistry is the most well studied – why?
In this chapter, we will focus on reactions that take place in water and look at:
Solution stoichiometry
Common aqueous reactions

4. II. Solution Concentration
There are two parts of a solution.
solute: substance present in smaller amount
solvent: substance present in larger amount
For stoichiometry, the important aspect of a solution is its concentration.
concentration: amount of solute present in a certain volume of solution

5. II. Concentrated vs. Dilute
Concentrated solutions have a lot of solute relative to solvent.
Dilute solutions have a little solute relative to solvent.

6. II. Quantitative Concentrations
The most common concentration unit is molarity, which is moles solute per L of solution.
In a solution, solute is evenly dispersed in the solvent!!

7. II. Solution Preparation
Preparing solutions requires a series of exacting steps.

8. III. Solution Calculations
There are 3 main types of solution-based calculations.
What is the concentration?
Use definition of molarity.
Solution creation/dilution.
Use definition of molarity / M1V1 = M2V2
Stoichiometry.
Use molarity as a conversion factor.
Start w/ problems using definition of molarity.

9. III. Sample Problem
e.g. Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in water to make mL of solution.

10. III. Sample Problem
e.g. How many grams of Na2HPO4 are needed to make 1.50 L of a M Na2HPO4 solution?

11. III. Solution Dilution
A stock solution is a solution of high concentration.
Lower concentration solutions can be made from the stock via dilution.

12. III. Sample Problem
e.g. How many mL of a 2.0 M NaCl solution are needed to make 250 mL of a 0.50 M NaCl solution?

13. III. Solution Stoichiometry
In the stoichiometry we’ve done before, amounts were converted between grams and moles.
In solutions, amounts are converted between volumes and concentrations.
The key is remembering that molarity is a conversion factor between moles and volume!

14. III. Sample Problem
e.g. How many mL of 0.10 M HCl reacts with 0.10 g Al(OH)3 according to the reaction below? Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)

15. III. Sample Problem
e.g. How much PbCl2 forms when 267 mL 1.50 M lead(II) acetate reacts with 125 mL 3.40 M sodium chloride according to the reaction below?
Pb(CH3COO)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaCH3COO(aq)

16. IV. Chemical Reactions
There are countless reactions, but only a few categories of reactions.
With experience, it becomes easier to identify what will happen in a reaction.
First, we take a close look at the solute and solvent in a solution.

17. IV. Forming a Solution Attractive forces hold solute together.
When solute is added to a solvent, new potentially attractive forces arise.
Competition between these forces occurs.

18. IV. Aqueous Solutions Water is a particularly “active” solvent.
As a solvent, water has one important characteristic: it is polar!

19. IV. Interactions in Aqueous Solutions
The polar nature of water allows it to interact with charged species in solution.

20. IV. Water-Ion > Na+Cl-

21. IV. Electrolytes
Compounds that dissociate in water and lead to electrical conductivity are called electrolytes.

22. IV. Strong/Weak Electrolytes
Strong electrolytes dissociate completely in water.
NaCl(s)  Na+(aq) + Cl-(aq)
Logically, weak electrolytes do not dissociate completely in water.
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
H2O
H2O

23. IV. Sample Problem
e.g. How many moles of each ion are in a solution formed by dissolving 354 g of magnesium hydroxide in water?

24. V. Precipitation Reactions
The formation of a precipitate (ppt) is a strong driving force for a reaction.
Precipitate is a fancy word for solid.
The attractions in these solids are too strong for H2O to break up.

25. V. Predicting Precipitates (95%)
Li+, Na+, K+, NH4+ salts are soluble.
NO3-, CH3COO-, ClO4- salts are soluble.
Ag+, Pb2+, Hg22+ salts are insoluble.
Cl-, Br-, I- salts are soluble.
CO32-, S2-, O2-, OH- salts are insoluble.
SO42- salts are soluble except for CaSO4,SrSO4, and BaSO4.
If none of these apply, it’s insoluble.

26. V. Sample Problem
e.g. Predict the precipitates in the following aqueous reactions.
sodium hydroxide + cadmium(II) nitrate
magnesium bromide + potassium acetate
ammonium sulfate + barium chloride
sodium iodide + lead(II) nitrate

27. V. Writing Reactions No
spectator
ions

28. V. Sample Problem
Write balanced molecular, total ionic, and net ionic equations for the reaction between strontium chloride and lithium phosphate.

29. VI. Acids/Bases
There are many definitions of acids and bases, but we will use the Arrhenius definitions for now.
Acids are molecular compounds that produce H+ ions in aqueous solution.
Bases are substances that produce OH- ions in aqueous solution.

30. VI. Aqueous Acids What is H+ comprised of?
Water interacts so strongly w/ H+, that it forms a bond with it.
e.g. HCl(g) + H2O(l)  (H2O)H+(aq) + Cl-(aq)
This is called the hydronium ion, and it’s usually written as H3O+.
Note that H+(aq) = H3O+(aq)
Polyprotic acids have more than one ionizable H+, e.g. H2SO4.

31. VI. Common Acids & Bases

32. VI. Acid Nomenclature
There are two categories of acids that have different naming rules.
Binary acids
Oxoacids
You can easily recognize acids because their formula has H as the first element.

33. VI. Naming Binary Acids Binary acids contain a nonmetal anion.
HCl = hydrochloric acid
HBr = hydrobromic acid
H2Se = hydroselenic acid
HI = hydroiodic acid

34. VI. Naming Oxoacids Oxoacids contain an oxoanion. Set 1 Set 2
HNO3 = nitric acid
H2SO4 = sulfuric acid
HClO3 = chloric acid
HClO4 = perchloric acid
Set 2
HNO2 = nitrous acid
HClO2 = chlorous acid
HClO = hypochlorous acid
H2SO3 = sulfurous acid

35. VI. Acid/Base Reactions
The driving force for this reaction is the formation of water; other product is a salt.
The net ionic equation for acid/base reactions is always the same!!
H+(aq) + OH-(aq)  H2O(l)

36. VI. Acid/Base Titrations
The concentration of an acid or base can be determined experimentally.

37. VI. Titration Terminology
titration: procedure in which one solution of known [ ] is used to determine the [ ] of another solution
indicator: a substance used to visualize the end of a reaction
The equivalance point occurs when moles acid = moles base.
The endpoint occurs when the solution changes color due to the indicator.

38. VI. Titration Problems
The first step in solving a titration problem is writing the titration reaction!!
After identifying the reaction, it becomes a solution stoichiometry problem!
Again, use unit labels on the numbers to guide your calculation.

39. VI. Sample Problem
e.g. To determine the concentration of a solution of H2SO4, you titrate a mL sample of it with M NaOH. If it takes mL of the NaOH solution to reach the endpoint, what’s the concentration of the H2SO4?

40. VII. Gas-Evolution Reactions
Many gas-evolution reactions are also acid-base reactions.
The formation of the gas could be direct or through decomposition of a product.
H2CO3(aq)  H2O(l) + CO2(g)

41. VII. It Makes Bubbles

42. VII. Common Gas Products

43. VII. Sample Problem
Write balanced molecular, total ionic, and net ionic equations for the reaction between aqueous solutions of hydrobromic acid and potassium sulfite.

44. VIII. Reactions Involving e- in Motion
The movement of e- from one atom to another is another driving force for reactions.
Oxidation is the loss of e-.
Reduction is the gain of e-.
They are coupled processes; one cannot occur w/out the other.

45. VIII. Reduction-Oxidation Reactions
These are also known as redox reactions.
Formation of NaCl is an example.
Na  Na+ + e-
½ Cl2 + e-  Cl-
Cl2 oxidizes Na; Cl2 is the oxidizing agent.
Na reduces Cl2; Na is the reducing agent.

46. VIII. NaCl Formation via Redox

47. VIII. Nonpolar to Polar
You don’t need complete e- transfer for a redox reaction.
Can also have just a shift of e- density.
Consider H2(g) + Cl2(g)  2HCl(g)

48. VIII. Oxidation States
Oxidation states (oxidation numbers) allow us to keep track of which atoms are gaining/losing e- in a reaction.
oxidation state (number): the charge an atom would have if e- are transferred completely and not shared
Note that in ionic compounds, we consider e- as totally transferred, so the ionic charge is the oxidation state.

49. VIII. Rules for Assigning O.N.
Atoms in elemental form have O.N. = 0.
Charge on a monatomic ion equals its O.N.
The sum of all O.N. must equal the total charge.
For Group 1, O.N. = +1.
For Group 2, O.N. = +2.
For H, O.N. = +1 w/ nonmetals, -1 w/ metals and B.
For F, O.N. = -1.
For O, O.N. = -1 in peroxides and -2 in all others.
For Group 17, typically O.N. = -1.

50. VIII. Assigning O.N.
e.g. Determine O.N. for all atoms in the following.
KMnO4
NH4+
IF3
ZnCl2

51. VIII. Sample Problem
e.g. Determine the substances that are oxidized and reduced in the reaction below.
5CO(g) + I2O5(s)  I2(s) + 5CO2(g)