2. One of the most important substances on Earth.
Can dissolve many different substances.
A polar molecule because of its unequal charge distribution.
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3. Nature of Aqueous Solutions
Solution = Homogeneous mixture of 2 or more pure substances
Components of solution: Solvent + Solute(s)
Solute – substance being dissolved = substance we have less of (smaller moles).
Solvent = substance that we have more of (greater moles); In aqueous solutions, the solvent is liquid water.
Electrolyte – substance that when dissolved in water produces a solution that can conduct electricity.
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4. Electrolytes
Strong Electrolytes – conduct current very efficiently Completely ionized in water. All water soluble ionic compounds (solubility rules), 7 strong acids (HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4) are strong electrolytes
Weak Electrolytes – conduct only a small current. A small degree of ionization in water. All weak acids, NH3 are weak electrolytes
Nonelectrolytes – no current flows . Dissolves but does not produce any ions. All molecular solutes such as glucose, sugar are nonelectrolytes
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5. Chemical Reactions of Solutions
We must know:
The nature of the reaction.
The amounts of chemicals present in the solutions.
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6. Molarity
Molarity (M) = moles of solute per volume of solution in liters:
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7. EXERCISE!
A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?
1.57 M
500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).
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8. Concentration of Ions For a 0.25 M CaCl2 solution: CaCl2 → Ca2+ + 2Cl–
Ca2+: 1 × 0.25 M = 0.25 M Ca2+
Cl–: 2 × 0.25 M = 0.50 M Cl–.
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9. Which of the following solutions contains the greatest number of ions?
CONCEPT CHECK!
Which of the following solutions contains the greatest number of ions?
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
a) contains mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct.
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10. Notice
The solution with the greatest number of ions is not necessarily the one in which:
the volume of the solution is the largest.
the formula unit has the greatest number of ions.
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11. Dilution
The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution.
Dilution with water does not alter the numbers of moles of solute present.
Moles of solute before dilution = moles of solute after dilution
M1V1 = M2V2
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12. Add water to the solution.
CONCEPT CHECK!
A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?
Add water to the solution.
Pour some of the solution down the sink drain.
c) Add more sodium chloride to the solution.
d) Let the solution sit out in the open air for a couple of days.
e) At least two of the above would decrease the concentration of the salt solution.
For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct.
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13. EXERCISE!
What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution? 60.0 mL
The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL)
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14. Precipitation Reactions Acid–Base Reactions
Oxidation–Reduction Reactions
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15. Precipitation Reaction
A double displacement reaction in which a solid (precipitate) forms and separates from the solution is called a precipitation reaction. A double displacement reaction is also called a metathesis reaction. As the name implies, 2 species displace each other. The reactants are typically either 2 ionic compounds or an ionic compound & an acid. The 2 compounds ‘swap’ cations to form 2 new compounds. When at least 1 of the products formed is not water soluble, the reaction is recognized as a precipitation reaction. When ionic compounds dissolve in water, the resulting solution contains the separated ions.
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16. The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
Ba2+(aq) + CrO42–(aq) → BaCrO4(s)
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17. Physical states in equations for reactions in aqueous solution
For Acids, Water Soluble compounds & ions in solution: physical state = (aq)
For Water insoluble compounds: physical state = (s)
For gases: physical state = (g)
For H2O ALONE: physical state = (l) for liquid
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18. Simple Rules for Solubility
Most nitrate (NO3-) salts are soluble.
Compounds containing Alkali metal (group 1A) ions and NH4+ are soluble.
Most compounds containing Cl-, Br-, and I- salts are soluble (except those in which the cations are Ag+, Pb2+, Hg22+).
Most compounds containing SO4-2 as anion are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4, AgSO4).
Most OH- are insoluble (LiOH, NaOH, KOH, RbOH, CsOH are soluble, Ba(OH)2, Ca(OH)2 , Sr(OH)2 are also considered soluble).
Most compounds containing S2-, CO32-, CrO42-, PO43- are insoluble, except for those containing the cations in Rule 2.
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19. CONCEPT CHECK!
Which of the following ions form compounds with Pb2+ that are generally soluble in water? a) S2– b) Cl– c) NO3– d) SO42– e) Na+
a), b), and d) all form precipitates with Pb2+. A compound cannot form between only Pb2+ and Na+.
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20. Formula Equation (Molecular Equation)
Balanced equation with physical states included
Use solubility rules to determine which compounds are aqueous and which compounds are solids.
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
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21. Complete Ionic Equation
When all strong electrolytes in the molecular equation are written in dissociated form (separate cation & anion; do not forget to write the charges of the separated ions), the resulting equation is a complete ionic equation.
AgNO3 (aq) is replaced by Ag+(aq) + NO3-(aq);
NaCl(aq) is replaced by Na+(aq) + Cl-(aq); NaNO3 (aq) is replaced by Na+ (aq) + NO3-(aq)
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-(aq)
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22. Ag+(aq) + Cl-(aq) AgCl(s)
Net Ionic Equation
When the ‘common terms’ in the complete ionic equation are cancelled, we obtain the net ionic equation. This equation only includes only those solution components that undergo a change.
For example, the ‘common terms’ that cancel in the example in the previous slide are NO3-, Na+. Therefore the net ionic equation is:
Ag+(aq) + Cl-(aq) AgCl(s)
The ions (the ‘common terms’) that cancel are called the Spectator ions (ions that do not participate directly in the reaction).
Therefore in the worked example, Na+ and NO3- are spectator ions.
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23. CONCEPT CHECK!
Write the correct formula equation, complete ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide.
Formula Equation:
CoCl2(aq) + 2NaOH(aq) Co(OH)2(s) + 2NaCl(aq)
Complete Ionic Equation:
Co2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq)
Co(OH)2(s) + 2Na+(aq) + 2Cl-(aq)
Net Ionic Equation:
Co2+(aq) + 2Cl-(aq) Co(OH)2(s)
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24. Solving Stoichiometry Problems for Reactions in Solution
Identify the species present in the combined solution, and determine what reaction occurs.
Write the balanced net ionic equation for the reaction.
Calculate the moles of reactants.
Determine which reactant is limiting.
Calculate the moles of product(s), as required.
Convert to grams or other units, as required.
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25. What precipitate will form? lead(II) phosphate, Pb3(PO4)2
CONCEPT CHECK!
(Part I)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What precipitate will form?
lead(II) phosphate, Pb3(PO4)2
What mass of precipitate will form?
1.1 g Pb3(PO4)2
The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s).
mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form.
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26. Steps: Recall from chapter 3
Write balanced equation
If it is a limiting reactant problem, identify your limiting reactant as you did in chapter 3
Calculation of moles: 2 options:
a. If mass is given, moles = Mass in g/Molar mass
b. If Molarity & volume is given, moles of solute present in solution = Molarity x Volume in L
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27. Acids & Bases (Arrhenius Definition)
Acids: substances that increases the concentration of H+ when dissolved in water; Recall chapter 2: compounds formed by combining cation = H+ with any anion.
Bases : substances that increases the concentration of OH- when dissolved in water; Recall chapter 2: compounds formed by combining anion = OH- with any cation.
H+(aq) + OH–(aq) H2O(l)
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28. Acid-Base Reactions = Neutralization Reactions (Arrhenius Definition)
In the Arrhenius definition, an acid- base reaction is a double displacement reaction.
The products are salt & H2O. A salt is an ionic compound whose cation is not H+ & whose anion is not OH-.
Acid + Base Salt + H2O(l)
When a strong acid & a water soluble metal hydroxide react, the net ionic equation simplifies to
H+(aq) + OH–(aq) H2O(l)
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29. Equations for Neutralization Reactions
Write your balanced molecular equation with physical states.
Do not dissociate solids, H2O, weak acids when you write your complete ionic equation.
Example 1: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l): Mol. Eq.
C.I.Eq:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)
Net ionic equation: H+(aq) + OH-(aq) H2O(l)
Example 2: HF(aq) + NaOH(aq) NaF(aq) + H2O(l): Mol. Eq.
C.I.Eq: HF(aq) + Na+(aq) + OH-(aq) Na+(aq) + F-(aq) + H2O(l)
Net ionic equation: HF(aq) + OH-(aq) F-(aq) + H2O(l)
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30. Performing Calculations for Acid–Base Reactions
Write the balanced equation for this reaction.
Calculate moles of reactants.
Determine the limiting reactant, where appropriate.
Calculate the moles of the required reactant or product.
Convert to grams or volume (of solution) or molarity, as required.
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31. Acid–Base Titrations
Titration – delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).
Equivalence point – enough titrant added to react exactly with the analyte.
Endpoint – the indicator changes color so you can tell the equivalence point has been reached.
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32. CONCEPT CHECK!
For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint? 1.00 mol NaOH
The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid.
1.00 mol of sodium hydroxide would be required.
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33. Additional Exercises
What is the volume of M Ba(OH)2 needed to neutralize 60.0 mL of M H3PO4?
What is the mass of Ca(OH)2 needed to neutralize 25.0 mL of M H2SO4
What is the molarity of NaOH is 25.0 mL of NaOH neutralized 50.0 mL of M HNO3.

34. Reactions in which one or more electrons are transferred.
Examples: Reactions between metals & nonmetals forming ionic compound; Combustion reactions.
Redox Reactions = Reduction-Oxidation Reactions = Electron Transfer Reactions
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35. Oxidation is the loss of electrons: OIL
Reduction is the gain of electrons: RIG
The two are dependant and you cannot have one without other
The reactant that loses electrons is said to undergo oxidation.
The reactant that gains electrons is said to undergo reduction
The reactant that undergoes oxidation causes the other reactant to undergo reduction and is hence called the REDUCING AGENT.
The reactant that undergoes reduction takes away electrons from the reactant causing its oxidation. It is therefore the OXIDIZING AGENT.
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36. A redox reaction is identified by looking at changes in OXIDATION NUMBER.
Oxidation numbers are imaginary numbers obtained from a set of rules (MEMORIZE).
Every atom in a molecule or polyatomic ion is assigned an oxidation number
OXIDATION NUMBERS SHOULD NOT BE CONFUSED WITH CHARGES FOR POLYATOMIC SPECIES
When a reactant gives up electrons, the oxidation number of at least one of its constituent elements becomes less negative or more positive (larger).
When a reactant gains electrons, the oxidation number of at least one of its constituent elements becomes more negative or less positive (smaller).
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37. Rules for Assigning Oxidation States (MEMORIZE)
Oxidation state of an atom in an element = 0
Oxidation state of monatomic ion = charge of the ion
Oxygen = -2 in most compounds & polyatomic ions (except in peroxide where it = -1)
Hydrogen = +1 in molecular compounds, acids
Fluorine = -1 in all compounds
Sum of oxidation states of all atoms in formula = 0 in compounds
Sum of oxidation states of all atoms in formula = charge of the ion in polyatomic ions
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38. EXERCISE!
Find the oxidation states for each of the elements in each of the following compounds:
K2Cr2O7
CO32-
MnO2
PCl5
SF4
K = +1; Cr = +6; O = –2
C = +4; O = –2
Mn = +4; O = –2
P = +5; Cl = –1
S = +4; F = –1
K2Cr2O7; K = +1; Cr = +6; O = -2
CO32-; C = +4; O = -2
MnO2; Mn = +4; O = -2
PCl5; P = +5; Cl = -1
SF4; S = +4; F = -1
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39. Redox Characteristics
Transfer of electrons
Transfer may occur to form ions
Oxidation – increase in oxidation state (loss of electrons); reducing agent
Reduction – decrease in oxidation state (gain of electrons); oxidizing agent
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40. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
CONCEPT CHECK!
Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent.
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Cr2O72-(aq) + 2OH-(aq) CrO42-(aq) + H2O(l)
2CuCl(aq) CuCl2(aq) + Cu(s)
Zn – reducing agent; HCl – oxidizing agent
c) CuCl acts as the reducing and oxidizing agent
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41. Activity Series: Arrangement of metals in increasing order of ease of oxidation. Li: best reducing agent = most easily oxidized; Gold: worst reducing agent = least easily oxidized.

42. Activity Series & Hydrogen Displacement Reactions
Metal(s) + H2O(l or g)  Metal hydroxide + H2(g)
Metals that undergo the above reaction: All metals above Co in the activity series
Metal(s) + HCl(aq)  Metal chloride + H2(g)
Metals that undergo the above reaction: All metals above H2 in the activity series

43. Activity Series & Metal Displacement Reactions
Any metal will react with salts of less active metals.
The cation of the less active metal will be reduced to the element. The more active active metal will be oxidized to its corresponding cation.
Example:
3Ca(s) + 2Al(NO3)3(aq)  3Ca(NO3)2(aq) + 2Al(s)
3Ca(s) + 2Al+3(aq)  3Ca+2(aq) + 2Al(s)
Metals will not react with salts of more active metals.
3Ca(NO3)2(aq) + 2Al(s) no reaction
3Ca+2(aq) + 2Al(s)  no reaction

44. Halogen Displacement Reactions
Halogens are reduced to the corresponding halide anions. Therefore they are oxidizing agents. The halogen that is most readily reduced (best oxidizing agent) is F2 & the one that is least readily reduced is I2.
Therefore the activity series for halogens is F2 > Cl2 > Br2 > I2.
Any halogen will react with a halide of a halogen that is less reactive.
F2 + 2Cl-  2F- + Cl2
Cl2 + 2F-  no reaction