1. Types of Chemical Reactions & Solutions
Chapter 4

2. What is a Solution?
Soluble – a substance that can dissolve in a given solvent
Miscible: two liquids that can dissolve in each other
Example: water and antifreeze
Insoluble – substance cannot dissolve
Immiscible: two liquids that cannot dissolve in each other
Example: oil & water

3. Why Do Some Substances Dissolve and not Others?
To dissolve, solute particles must dissociate from each other and mix with solvent particles
Attractive forces between solute and solvent must be greater than attractive forces within the solute
Process of surrounding solute particles with solvent particles is called SOLVATION
In water, it is also called HYDRATION

4. Solvation/Hydration

5. Aqueous Solutions of Molecular Compounds
Water is also a good solvent for many molecular compounds (Example: sugar)
Sugar has many O-H bonds (polar)
When water is added, the O-H bond becomes a site for hydrogen bonding with water
Water’s hydrogen bonds pulls the sugar molecules apart
Oil is not a good solute because it has many C-H bonds (not polar) and few or no O-H (polar) bonds

6. Factors that Affect Solvation Rate
Increase Solvation Rate (Dissolve Faster) by:
Agitation (stirring)
Increase surface area (make particles smaller)
Temperature (make it hotter)
All these increase the number of collision between water and the solute

7. The Nature of Aqueous Solutions
Composition of a solution can vary by changing amount dissolved:
Electrical Conductivity
Pure Water is a poor conductor
Good conductors have strong electrolytes
Weak conductors have weak electrolytes
Non-conductors contain non-electrolytes

8. The Nature of Solutions
Originally identified by Arrhenius
Conductivity arises from the presence of IONS in the solution
Ions are charged particles
Ionic theory starts to make sense
Arrhenius postulates:
The extent to which a solution can conduct electricity depends DIRECTLY on the number of ions present.

9. Strong Electrolytes Strong electrolytes COMPLETELY ionize in solution
Example: NaCl  Na+ & Cl-
Arrhenius first associated acidity to the presence of H+ ions (acidus = sour)
Acids ionize to form H+ ions
HCl, HNO3, H2SO4 are strong acids
Strong Acids Completely Ionize
Strong Bases also completely ionize
OH- compounds completely ionize

10. Weak Electrolytes
Weak Electrolytes exhibit a small degree of ionization in water
Weak Acids, Weak Bases
Example: Acetic Acid is a weak acid (~1/100 molecules dissociates)
Example Ammonia is a weak base (~1/100 molecules dissociates)
Non-Electrolytes dissolve in water, but do not dissociate into ions
Examples: sugar, alcohol
Do not conduct electricity in solution

11. Composition of Solutions
The nature of the chemical reaction frequently depends on the amounts of chemicals present:
Molarity = moles solute/1 liter solution
Concentration is determined BEFORE it dissolves
1.0M NaCl is made by measuring 1.0 Moles of NaCl and adding enough water to make 1 L of solutions
1.0M does not mean it contains 1.0 mole of NaCl units
It contains 1.0 mole of Na+ ions and 1.0 mole of Cl- ions
Finding moles from molarity:
Liters solution x molarity = moles of solute

12. Example
What is the concentration of each type of ion in the solution 0.50 M Co(NO3)2
Solid compound dissolves into Co2+ ions and NO3- ions
Co(NO3)2 (s) - Co2+ (aq) + 2 NO3- (aq)
Solution contains 0.50 moles Co2+ ions
Solution contains 1.0 moles NO3- (2*0.50)
This is a CRITICAL conceptual idea to remember for future problems

13. How to Make a Solution of Known Concentration
Example: Make 1.00 L of M K2Cr2O7. How do you do this?
Determine moles of K2Cr2O7 needed:
1.00L solution x mol K2Cr2O7/L solution = mol K2Cr2O7
Convert moles K2Cr2O7 grams
0.200 mol K2Cr2O7 x g K2Cr2O7/mol K2Cr2O7 = 58.8g K2Cr2O7
Measure out 58.8g K2Cr2O7. Transfer it to a 1.00L volumetric flask. Add distilled water to the mark on the flask.

14. Dilution
Dilution: Adding water to a prepared (or stock) solution in order to achieve a desired molarity.
Key: Moles of solute after dilution = moles of solute before solution
M1V1 = M2V2
Proper Procedure:
Use measuring or volumetric pipettes to accurately measure solute
Measuring pipette – has graduated lines
Volumetric pipette – has ONE measurement to fill to

15. Dilution Procedure
How to make 500mL or 1.00M acetic acid from a 17.4M stock solution
Calculate volume of stock solution needed:
Figure out moles of acetic acid:
500mL solution x 1L solution/1000mL Solution x 1.00 mole HC2H3O2 = mole HC2H3O2
V = mol HC2H3O2 /17.4mol HC2H3O2 /1L solution = L or 28.7mL of solution
500 mL of 1M solution – 28.7mL HC2H3O2 = 471.3mL H2O measured into flask
Add 28.7mL HC2H3O2 to H2O

16. Types of Chemical Reactions
Last Year, you learned:
Single replacement
Double replacement
combustion,
Acid/Base
No Longer a Sufficient Concept
We must expand upon what you know to better understand what is happening in a reaction

17. New Types of Reactions New Categories of Reactions
Precipitation
Acid/Base
Oxidation-Reduction
Virtually all reactions can fit into these classifications

18. Precipitation Reactions
A precipitation reaction forms a solid that falls (precipitates) from the solution.
Example:
K2CrO4 (aq) + Ba(NO3)2 (aq)  products including a yellow solid
Actually looks more like:
2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) -- products
How can they form a yellow solid?

19. Precipitation Reactions
Predicting products is very hard
Actual reaction products must be confirmed experimentally before you can really conclude the reaction
Predicting from what we know:
Compound must be electrically neutral
Must contain both cations and anions
What possible combinations exist?
K2CrO4, KNO3, BaCrO4, Ba(NO3)2
Can’t be K2CrO4 or Ba(NO3)2 – these are reactants
KNO3 will always be soluble so precipitate must be BaCrO4

20. Precipitation Reactions
How Do We Know That?
Based on Simple Solubility Rules
Terms:
Soluble – the salt will dissolve in water to a great extent
Slightly Soluble = Insoluble – only a tiny, insignificant portion of the salt dissolves in water

21. Simple Solubility Rules
Text page 150 – Memorize them!
Most nitrates (NO3) are soluble
Most salts of alkali metals and ammonium ions are soluble
Most Chloride, Bromide, and Iodide salts are soluble, EXCEPT Silver, Lead, Mercury
Most Sulfates are soluble, EXCEPT Barium, Lead, Mercury, and Calcium
Most Hydroxide salts are slightly soluble, EXCEPT Sodium and Potassium which are highly soluble. Barium, Tin, and Calcium are marginally soluble
Most Sulfide, carbonates, chromates, and phosphates are only slightly soluble

22. Describing Solution Reactions
Convert the Formula Equation to Complete Ionic Equation
List all the ions on both sides
Solids (precipitates) are not ions
All strong electrolytes are shown as ions in (aq)
This will reveal that some ions do not participate in the reaction and are spectator ions
Be Able to Identify spectator Ions
Create a Net Ionic Equation
Re-write the complete ionic equation, but remove the spectator ions from both sides

23. Example Step 1: Formula Equation
K2CrO4 (aq) + Ba(NO3)2 (aq)  BaCrO4 (s) + 2KNO3 (aq)
Step 2: Complete Ionic Equation
2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) --
BaCrO4 (s) + 2K+ (aq) + 2NO3- (aq)
Step 3: Eliminate Spectator Ions
2K+ (aq) + CrO42- (aq) + Ba2+ (aq) + 2 NO3 (aq) --
BaCrO4 (s) + 2K+ (aq) + 2NO3- (aq)
Step 4: Net Ionic Equation
CrO42- (aq) + Ba2+ (aq) -- BaCrO4 (s)

24. Predict Products of Reactions:
CaCl2(aq) + 2Ag2SO4(aq)  ??
H2SO4 + Na2CO3  ??
Na2CrO4 + AgNO3  ??
Write the total ionic equation for the reaction of hydrofluoric acid with potassium hydroxide.
When aqueous solutions of iron(III) sulfate (Fe2(SO4)3) and sodium hydroxide were mixed, a precipitate formed. What is the precipitate?

25. Stoichiometry for Solution Reactions
Identify all the species (ions or compounds) present in the reaction and determine what reaction occurs
Write the balance NET IONIC Equation
Calculate Moles of Reactants
Determine Limiting Reactant
Calculate Moles of Product or products
Convert to grams or other units

26. Acid/Base Reactions Arrhenius Acids:
H+ ions = Acid
OH- ions = base
Refinement of Concept by Bronsted and Lowry:
Acid is a proton donor
Base is a proton acceptor

27. Acid/Base Reactions What’s the Difference? Example:
What does a H+ ion look like?
A bare proton
But you can have bases that are not OH-
Example:
KOH (aq) + HC2H3O2 (aq)  ??
K+ (aq) + OH- (aq) + HC2H3O2 (aq) are the species present before any reaction occurs
A precipitation reaction could occur between K+ and OH- but KOH is soluble
Or is there another possible proton donor to OH-?
Weak Acid – does not ionize

28. Acid/Base Reactions YES: HC2H3O2 molecules
Hydroxide ion is such a strong base that for purposes of stoichiometric equations, it can be assumed to react completely with any weak acid encountered
Actual net ionic equation is:
OH- + HC2H3O2  H2O + C2H3O2-
Acid/Base reactions are called neutralization reactions

29. Acid/Base Reaction Calculations
List species present in combined solution BEFORE any reaction occurs
Write a balance NET ionic equation
Calculate moles of reactants using volumes and molarities
Determine limiting reactant where appropriate
Calculate moles of required reactant or product
Convert to grams or volume as required

30. Acid/Base Titrations Titration is a volumetric analysis:
Uses a buret
Uses volume of a KNOWN solution (titrant)
Delivered into an unknown solution (analyte)
When titrant added is exactly reacted with analyte you have the equivalence point or stoichiometric point
Equivalence point is marked with an indicator
When the indicator changes color, you have reached the endpoint of the titration

31. Acid/Base Titrations
Once you have reached the ENDPOINT, it is a stoichiometry problem.
How much standard solution (titrant) was used?
How many moles
What was the reaction?
How many moles of analyte was neutralized?
What volume of analyte was neutralized?
Calculate molarity of the analyte.
Review Sample Exercise 4.15

32. ReDox in Acidic Solutions
Write separate equations for the half-reactions
For each half-reaction:
Balance all elements except H and O
Balance O with water
Balance H using H+
Balance charge using electrons
Multiply half-reaction by integer to equalize electron totals
Add half-reactions and cancel identical species
Check that elements & Charges are balanced

33. ReDox in Basic Solutions
Write separate equations for the half-reactions
For each half-reaction:
Balance all elements except H and O
Balance O with water
Balance H using H+
Balance charge using electrons
To both sides of the equation, add OH- ions to equal H+ ions
Form H2O and eliminate from both sides
Multiply half-reaction by integer to equalize electron totals
Add half-reactions and cancel identical species
Check that elements & Charges are balanced

34. ReDox in Basic Solutions
Book: Page 177, Example 4.20
As(s) + CN-(aq) + O2(g) Ag(CN)-2(aq)
Balance oxidation ½ reaction first
Balance as if H+ ions were present. Balance C and N first
2CN-(aq) + Ag(s)  Ag(CN)-2(aq)
Balance the charge
2CN-(aq) + Ag(s)  Ag(CN)-2(aq) + e-
Balance the reduction ½ reaction (O2)
O2(g)  2H2O(l)

35. ReDox in Basic Solutions
O2(g)  2H2O(l)
Balance the Hydrogen with H+
O2(g) + 4H+(aq)  2H2O(l)
Balance the Charge
4e- + O2(g) + 4H+(aq)  2H2O(l)
Multiply Balanced oxidation ½ reaction by 4
4(2CN-(aq) + Ag(s)  Ag(CN)-2(aq) + e-)
8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 4e-

36. ReDox in Basic Solutions
Add the ½ reactions and cancel identical species:
8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 4e-
4e- + O2(g) + 4H+(aq)  2H2O(l)
4e- + O2(g) + 4H+(aq) 8CN-(aq) + 4Ag(s) 
4Ag(CN)-2(aq) + 4e- + 2H2O(l)

37. ReDox in Basic Solutions
Add OH- to both sides to cancel the H+ ions:
O2(g) + 4H+(aq) 8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 2H2O(l)
+ 4OH OH-
Eliminate water molecules formed:
O2(g) + 4H2O(l) 8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) + 2H2O(l) +4OH-
-2H2O H2O
Final Balanced ReDox Reaction:
O2(g) + 2H2O(l) 8CN-(aq) + 4Ag(s)  4Ag(CN)-2(aq) +4OH-
Double-Check to see that elements balance, charges balance.