1. Chapter 4 Chemical Reactions and Solution Stoichiometry
DE Chemistry
Dr. Walker

2. Solutions Solute Solvent Homogeneous mixture
A solute is the dissolved substance in a solution.
Salt in salt water
Sugar in soda drinks
Carbon dioxide in soda drinks
Solvent
A solvent is the dissolving medium in a solution.
Water in salt water
Water in soda

3. “Like Dissolves Like”
Nonpolar solutes dissolve best in nonpolar solvents
Fats
Benzene
Steroids
Hexane
Waxes
Toluene
Polar and ionic solutes dissolve best in polar solvents
Inorganic Salts
Water
Sugars
Small alcohols
Acetic acid

4. Water Aqueous Solutions Water is a polar molecule
Solutions where a solute is dissolved in water
Water is a polar molecule
Unequal distribution of electrons in molecule (electronegativity difference)
Bent shape due to unpaired electrons on oxygen

5. The Dissolving Process
Hydration
Process of dissolving in water
Ionic compounds dissociate in water
Ions interact with the “ends” of the water molecule
Same process occurs with polar compounds, no dissociation

6. Electrolytes Conductivity – Ability to conduct electric current
Requires ions to be present in solution
Electrolytes - A substance that when dissolved in water produces a solution that can conduct an electric current
Strong electrolytes – substances that ionize completely in water
Some ionic salts, strong acids/bases
Weak electrolytes – substances that ionize partially in water
Some ionic salts, weak acids/bases
Nonelectrolytes – substance that DO NOT ionize in water
Primarily organic compounds

7. Electrolytes vs. Nonelectrolytes
The ammeter measures the flow of electrons (current)
through the circuit.
If the ammeter measures a current, and the bulb
glows, then the solution conducts.
If the ammeter fails to measure a current, and the
bulb does not glow, the solution is non-conducting.

8. Molarity
Molarity is the ratio of moles of solute to liters of solution
-Remember, you may not always get moles and liters
-Grams must be converted to moles (divide by molar mass)
-Milliliters must be converted to liters (divide by 1000)

9. Basic Example
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.

10. Basic Example
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
1.56 g HCl
1 mol HCl
= mol HCl
g HCl
= 1.60 M HCl
26.8 mL
1 L
= L HCl
1000 mL

11. Concentration of Ions in Solution
Same calculations overall
One more step…multiply the molarity by the number of ions present
Give the concentration of each ion present in 0.50 M Co(NO3)2.

12. Concentration of Ions in Solution
Same calculations overall
One more step…multiply the molarity by the number of ions present
Give the concentration of each ion present in 0.50 M Co(NO3)2.
Remember, ionic salts dissociate in water, giving one Co2+ and two NO3-
As a result:
(0.50 M x 1) Co2+ = 0.50 M Co2+
(0.50 M x 2) NO32- = 1.0 M NO3-

13. Solutions of Known Concentration
Standard solution - a solution whose concentration is accurately known
Preparation of Standard solutions
How much (volume) x How strong (concentration) x What does it weigh (mass)?

14. Solutions of Known Concentration
Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of M NaCl solution?

15. Solutions of Known Concentration
Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of M NaCl solution?
Step #1: Ask “How Much?” (What volume to prepare?)
Step #2: Ask “How Strong?” (What molarity?)
Step #3: Ask “What does it weigh?” (Molar mass is?)
1.500 L
0.500 mol
58.44 g
= 43.8 g
1 L
1 mol

16. Stoichiometry Using Molarity
Sometimes, you may have to use a solution in a reaction. To do this successfully, you must be able to determine the number of moles that you’re using!
After determining moles of reactant, determine moles and grams of product
You could be given two solutions, and have to figure out the limiting reagent FIRST!

17. Stoichiometry Using Molarity - Example
What mass of solid aluminum hydroxide can be produced when 50.0 mL of M Al(NO3)3 is added to excess KOH?
Write reaction and balance if necessary
__ Al(NO3)3 + __ KOH __ Al(OH)3 + __ KNO3
Al(NO3) KOH Al(OH)3 + 3 KNO3

18. Stoichiometry Using Molarity - Example
What mass of solid aluminum hydroxide can be produced when 50.0 mL of M Al(NO3)3 is added to excess KOH?
Al(NO3) KOH Al(OH)3 + 3 KNO3

19. Stoichiometry Using Molarity - Example
What mass of solid aluminum hydroxide can be produced when 50.0 mL of M Al(NO3)3 is added to excess KOH?
Al(NO3) KOH Al(OH)3 + 3 KNO3
Find moles of Al(NO3)3
Moles of solute = mol

20. Stoichiometry Using Molarity - Example
What mass of solid aluminum hydroxide can be produced when 50.0 mL of M Al(NO3)3 is added to excess KOH?
Al(NO3) KOH Al(OH)3 + 3 KNO3
Find mass of Al(OH)3
moles Al(NO3)3
1 mole Al(OH)3
78.01 g Al(OH)3
= g Al(OH)3
1 mole Al(NO3)3
1 mole Al(OH)3

21. Dilutions
Dilution of a volume of solution with water does not change the number of moles present
Solving dilution problems
M1V1 = M2V2

22. Dilution Example
Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution?

23. Dilution Example
Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution?
M1V1 = M2V2
(11.6 M)(x Liters) = (3.0 M)(0.250 Liters)
x Liters = (3.0 M)(0.250 Liters)
11.6 M
= L

24. Reaction Types Precipitation reactions Acid-Base reactions
When two solutions are mixed, an insoluble solid forms
Acid-Base reactions
A soluble hydroxide and a soluble acid react to form water and a salt
Oxidation-Reduction reactions (aka “redox”)
Reactions in which one or more electrons are transferred

25. Precipitation Reactions
When two solutions are mixed, an insoluble solid forms
These are usually double replacement reactions where the ions exchange places in aqueous solution to form new compounds
AX + BY  AY + BX
One of the compounds formed is usually a
precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

26. Solubility Rules Soluble Insoluble NO32- NH4+ Alkali metals Halogens
Salts of Ag, Pb, Hg, Ba
CO32-
OH-
PO43-

27. Precipitation Reactions
Ionic compounds dissolve in water and the ions separate and move independently
Notice the four distinct ions in solution after dissociation

28. Precipitation Reactions
After dissociation, we can predict the products of this reaction
Reactants do not reform or there is no reaction
Like charges do not bond
Overall reaction becomes:
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Whether the products are aqueous or insoluble depends
on solubility rules – I’m not making you memorize them

29. Describing Reactions There are three ways to describe reactions:
Lead(II) nitrate + potassium iodide  lead(II) iodide + potassium nitrate
Double replacement (ionic) equation
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
Complete ionic equation shows compounds as aqueous ions
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq)
- notice the precipitate cannot dissociate!!
Net ionic equation eliminates the spectator ions
Pb2+(aq) + 2 I-(aq)  PbI2(s)

30. Another Example
Give the complete ionic equation and net ionic equation for the following reaction:

31. Another Example
Give the complete ionic equation and net ionic equation for the following reaction:
Complete ionic equation
Net ionic equation

32. Neutralization (Acid-Base) Reactions

33. Properties of Acids Acids are proton (hydrogen ion, H+) donors
Acids have a pH < 7
Acids taste sour
Acids react with active metals, producing H2
Acids react with carbonates, neutralize bases

34. Acids are Proton (H+) Donors
Strong acids are assumed to be 100% ionized in solution (good H+ donors).
HCl
H2SO4
HNO3
Weak acids are usually less than 5% ionized in solution (poor H+ donors).
H3PO4
HC2H3O2
Organic acids

35. Types of Acids Bronsted-Lowry: Acids are proton donors,
bases are proton acceptors
Arrhenius: Acids produce H+ ions in water,
bases produce OH- ions in solution

36. Acids React with Active Metals
Acids react with active metals to form salts and hydrogen gas.
Mg + 2HCl  MgCl2 + H2(g)
Zn + 2HCl  ZnCl2 + H2(g)
Mg + H2SO4  MgSO4 + H2(g)

37. Acids Neutralize Bases
Neutralization reactions ALWAYS produce a salt and water.
HCl + NaOH  NaCl + H2O
H2SO4 + 2NaOH  Na2SO4 + 2H2O
2HNO3 + Mg(OH)2  Mg(NO3)2 + 2H2O

38. Properties Of Bases Bases are proton (hydrogen ion, H+) acceptors
Bases have a pH > 7
Bases taste bitter
Solutions of bases feel slippery
Bases neutralize acids

39. Bases are Proton (H+ ion) Acceptors
Sodium hydroxide (lye), NaOH
Potassium hydroxide, KOH
Magnesium hydroxide, Mg(OH)2
Calcium hydroxide (lime), Ca(OH)2
OH- (hydroxide) in base combines with H+ in acids to form water
H+ + OH-  H2O

40. Bases Neutralize Acids
Milk of Magnesia contains magnesium hydroxide, Mg(OH)2, which neutralizes stomach acid, HCl.
2 HCl + Mg(OH)2
MgCl2 + 2 H2O

41. Volumetric Analysis
Technique for determining the amount of a certain substance by doing a titration.
A titration involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).

42. Titrations Vocabulary Titrant - Solution of known concentration
Analyte - Solution of unknown concentration
Equivalence point - Point at which the amount of titrant added to analyte results in perfect neutralization
Indicator - a substance added at the beginning of the titration that changes color at the equivalence point
Endpoint - the point at which the indicator changes color
Ideally, the endpoint and equivalence point occur at the same time!!!

44. Successful Titrations
The following three requirements must be met for a titration to be successful:
The exact reaction between titrant and analyte must be known (and rapid).
The stoichiometric (equivalence) point must be marked accurately.
The volume of titrant required to reach the stoichiometric point must be known accurately.
Titration data is usually determined using dilution calculations

45. More To Come….
… on acid/base chemistry in Chapter 14

46. “Redox” reactions Electrons are transferred (LEO says GER)
Gain electrons = reduction
Lose electrons = oxidation
Spontaneous redox rxns can transfer energy
Electrons (electricity)
Heat
Non-spontaneous redox rxns can be made to happen with electricity

47. Examples of redox rxns Photosynthesis Combustion of fuels
Oxidation of sugars, fats, proteins for energy

48. Oxidation Reduction Reactions (Redox)
Each sodium atom loses one electron:
Each chlorine atom gains one electron:

49. Lose Electrons = Oxidation
LEO says GER :
Lose Electrons = Oxidation
Sodium is oxidized
Gain Electrons = Reduction
Chlorine is reduced

50. Oxidation States Keeps track of electrons in redox reactions
Similar to concept of charge in ionic compounds, but assigns numbers to elements in organic molecules as well

51. Oxidation State Examples
H2SO4
H =
S =
O =

52. Oxidation State Examples
H2SO4
H = + 1
S = + 6
O = -2
(+ 1) + (4 x -2) + (+6) = 0
The sum of all compounds is zero!!

53. More Practice

54. More Practice A) K = +1, Mn = +7, O = -2 F) O = -2, Fe = +8/3
B) Ni = +4, O = G) F = -1, O = -2, Xe = +6
C) Na = +1, Fe = +2, O = -2, H = +1 H) F = -1, S = +4
D) H = +1, O = -2, N = -3, P = +5 I) O = -2, C = +2
E) O = -2, P = +3 J) H = +1, O = -2, C = 0

55. Balancing Redox Reactions
Balanced by “half-reaction” method
Split into oxidation and reduction reactions and balanced separately

56. Steps to Balance Redox Reactions (Acidic)

57. Redox Practice
“Chalk talk”

58. Problem Set Chemical Reactions and Solution Stoichiometry (4)