1. CHAPTER 4
AP CHEMISTRY

2. PRECIPITATION PROBLEMS
Water
Highly polar
Ionic and polar compounds are attracted to the positive and/or negative ends of the water molecule. This is called HYDRATION (a hydration reaction is something different)
When a precipitate is formed there is a zero net charge
Solubility rules (handout)
What happens when a ionic compound dissolves in water?
They break up into their ion form
Cation (positive ion) and anion (negative ion)
What is a precipitate?
A solid is formed from two aqueous solutions
Using the solubility rules predict what will happen
CuCl2 + (NH4)2SO4
Ba(NO3)2 + Na2CO3

3. NET IONIC EQUATIONS
Net ionic equations include those species that participate in the reaction
What are spectator ions?
Ions that do not participate in the reaction
You must show that atoms and charges are balanced
Write a net ionic equation for the following:
KOH and FeCl2
MgCl2 and AgNO3
(NH4)2SO4 and LiBr

4. ACID/BASE An acid is a species that
Has hydrogen ions in a water solution
Has a sour taste
Reacts with bases
Changes an indicators color
A base is a species that
Has hydroxide ions in a water solution
Has a bitter taste
Reacts with acids
Strong acids
Ionizes 100% in water
Strong electrolyte
HCl, HClO4, HClO3, HNO3, H2SO4, HBr, HI
Weak acids
Ionizes slightly into ions
HB(aq) <=> H+ (aq) + B-(aq)
HF, any carbon containing acid

5. CONTINUED Weak acid with water CH3COOH + H2O <=> H3O+ + CH3COO-
A strong base in water
Dissociates 100%
KOH --> K+ + OH-
Hydroxides with group one, Ba2+, Sr2+, and Ca2+ (slightly)
Weak bases in water
Dissociates slightly
B + H2O <=> BH+ + OH-
F-, H3PO4, NH3, CH3NH2

6. ELECTROLYTES Solute - substance
Solvent - water, what you put solute in
Electrolytes- conduct electricity
Strong - strong acids and bases, soluble salts
Weak - weak acids and bases, insoluble or slightly soluble salts
Nonelectroytes - molecules, nonpolar covalent compounds
ARRHENIUS
Ions are responsible for conducting electricity
Down played convictions to get PhD then crusaded to get theory accepted
Received Nobel prize 1903

7. EXAMPLES Weak acid - weak base
Can be acidic, basic or neutral depending on which is stronger, acid or base
Strong acid - strong base
Forms a salt and is a strong electrolyte
Neutralization reaction, pH is neutral
H+ (aq) + OH-(aq)  H2O(l)
Weak acid - strong base
Forms a base
CH3COOH + OH-  H2O + CH3COO-
Strong acid - weak base
HCl + NH3  NH4Cl
H+ (aq) + Cl- (aq) + NH3(aq) --> NH4+(aq) +Cl- (aq)
H+ (aq) + NH3(aq) --> NH4+(aq)

8. NET IONIC REACTIONS Carbonic acid with sodium hydroxide H2CO3 + NaOH
H2CO3 + Na+ + OH- -->H2O(l) + Na+ + CO32-
H2CO3(aq) + OH-(aq) --->H2O(l) + CO32-(aq)
Dimethylamine with sulfuric acid
H3C-NH-CH3
Hydroiodic acid with barium hydroxide

9. OXIDATION-REDUCTION REACTIONS
An oxidation-reduction reaction involves transfer of electrons
Addition of O2 or H2 often occur
LEO goes GER
Oxidized or oxidation
Species that lose electrons
oxidation number goes up
Cl- --> Cl2
Reduced or reduction
Species that gain electrons or oxidation number reduced
Cu+ + Al --> Cu + Al3+
Oxidizing agent
What causes oxidation
Reducing agent
What causes reduction

10. OXIDATION NUMBER RULES
Elements found in nature is zero
Cu, O2
Monoatomic ions same as charge
Cl-, Mg2+
Fluorine always 1-
Oxygen 2-
Except with fluorine 2+
Peroxide 1-, H2O2
Superoxide ½ -, KO2
Hydrogen 1+
With group one elements then a hyride 1-
Sum of oxidation numbers of a neutral compound is zero
The most electronegative element number is the same as the ion charge
Sum of oxidation numbers of a polyatomic ion is same as its charge

11. REDOX
What is the oxidation number of carbon in CaC2O4? Of Chromium in Cr2O72-?
The concept of oxidation numbers explain oxidation and reduction
Oxidation: increase in oxidation number (lose electrons)
Reduction: decreases in oxidation number (gain electrons)
Balancing redox equations
The half-equation method
Split the equation into two half-reactions
One for reduction and the other for oxidation
Balance one of the half-reactions; balance atom other than O or H first, next use water to balance oxygen, then add hydrogen ions to balance the hydrogen, add electrons to balance charge
Balance other half-reaction
Combine the two in such away as to eliminate all the electrons

12. PROBLEMS Consider the unbalanced equation
Cr3+(aq) + Cl- (aq) ---> Cr(s) + Cl2(g)
Ions can only be free in an aqueous solution
The reaction between MnO4- and Fe2+ (acidic). The reaction can be represented by the following (unbalanced) equation
MnO4- + Fe2+ ---> Mn2+ + Fe3+
Reaction between MnO4- and I-(basic). The reaction can be represented by the following unbalanced equation
MnO4- + I- --> I2 + MnO2

13. MOLARITY Number of moles of solute per liter of solution
M = number of mol/L
Symbol [ ] means molar concentration
Calculate the molarity of a solution when 16.3 g of ammonium sulfate is added to a mL volumetric flask and distilled water is added until the mL mark is reached.
16.3g/ 1 mol / 1000 mL /________
/ g / L / mL
.247 M
How many moles of 18.0 M sulfuric acid are there in 35.0 mL of the solution?
18.0 mol / 35.0 mL / 1L_______
L / / mL
.630 moles H2SO4

14. EXAMPLES
How many moles of chloride ions are in 1.75 L of 1.0 x 103- M zinc chloride?
Formalin is an aqueous solution of formaldehyde (HCHO). At high concentrations it is used as a preservative for biological specimen's. How many grams of formaldehyde are contained in 2.5 L of 12.3 M formalin?
Moles of solute after dilution = moles of solute before dilution
M1V1 = M2V2
When aqueous solutions of sodium hydroxide and chromium (III) nitrate are mixed a green precipitate is formed mL of M NaOH and mL of M Cr(NO3)3. Calculate the mass of precipitate.
A solution of permanganate reacts with iron (II) in solution to give manganese (II) and iron (III) ions. How many mL of M KMnO4 solution will react completely with mL of M Fe(NO3)2?

15. TITRATION
A volumetric analysis involving titration, where a measured volume of a known concentration of an acid or base is required to find the exact concentration (the volume must be known) of an unknown base or acid
Equivalent point
Where exact stoichometry amounts of each reactant are present. (same number of H+ and OH- ) NO EXCESSES
End point
Where the indicator changes color. Immediately stop adding titrate.

16. TITRATION QUESTION
A g sample of a mixture of CaCl2 and NH4NO3 was analyzed by dissolving the sample in water and then completely precipitating the calcium ion as CaC2O4. The CaC2O4 was dissolved in sulfuric acid (MUST KNOW THIS ACID), and the resulting oxalic acid was titrated with a KMnO4 solution. The titration required mL of M KMnO4 to reach equivalence point. Manganese (II) is the reduction product of MnO4- and CO2 is the oxidation product of H2C2O4.
1) Redox equation
2) How many moles of H2C2O4 were titrated?
3) How many moles of CaCl2 were in the original sample?
4) What was the percent mass of CaCl2 in the original sample? find mass from #3 and divide by .951 and X 100

17. THE END

18. MnO4-  Mn2+
8H+ + MnO4-  Mn2+ + 4H2O
8H+ + 5e- + MnO4-  Mn2+ + 4H2O
H2C2O4  CO2
H2C2O4  2CO2 + 2H+ + 2e-
16H+ + 10e- + 2MnO4-  2Mn H2O
5H2C2O4  10CO2 + 10H+ + 10e-
6H+ + 2MnO4- + 5H2C2O4  2Mn H2O + 10CO2

19. 38.36 mL of M KMnO4
.03836L / moles KMnO4
/ L
1.09 X 10-3 mol KMnO4
1.09 X 10-3 mol KMnO4 / 5 moles H2C2O4
/ 2 moles KMnO4
2.71 X 10-3 moles H2C2O4
END

20. look at the flow of ions from one compound to another
there is one C2O42- in one H2C2O4
that came from the compound CaC2O4, which is a one to one ratio
the Ca came from CaCl2 which only gives you 1 Ca
so the number of moles is the same as part 2