1. CHAPTER 5 Reactions in Aqueous Solution
5.1 Properties of Compounds in Aqueous Solution
5.2 Precipitation reactions
5.3 Acids and Bases
5.4 Reactions of Acids and Bases
5.5 Gas-forming reactions
5.6 Classifying Reactions in Aqueous Solution
5.7 Oxidation-Reduction Reactions
5.8 Measuring Concentrations of Compounds in solution
5.9 pH, a Concentration scale for acids and bases
5.10 Stoichiometry of Reactions in Aqueous Solutions
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Kull Chem 105 Chapter 2

2. 5.1 Properties of Compounds in Aqueous Solution
Solute
Solvent
Solution
Electrodes (conductor of electricity)
Electrolytes (ionic compound soluble in water, conducts electricity)
Ions in Aqueous Solution: electrolytes, complete/nearly complete dissociation
Electrolytes: All ionic compounds soluble in water
Strong – nearly complete dissociation, good conductor
Weak – only partial dissociation, weak conductor
Figure 5.3, page 179 (be able to use)
Non-electrolyte: dissolve in water, but do not conduct
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3. Water Solubility of Ionic Compounds
If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble.
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4. 5.2 Precipitation reactions Writing Equations for Aqueous Ionic Reactions
Produces a water-insoluble product: precipitate
Writing the equation: use state symbols, (s) denotes the precipitate
Three types of equations are used to represent aqueous ionic reactions: molecular, total ionic, and net ionic equations.
molecular equation: shows all reactants and products as if they were intact, un-dissociated
total ionic equation: shows all the soluble ionic substances dissociated into ions. Charges must balance
net ionic equation: it eliminates the spectator ions and shows the actual chemical change taking place.
Spectator ions not involved in chemical change.
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5. Precipitation Reactions
The “driving force” is the formation of an insoluble compound — a precipitate.
Molecular equation
Pb(NO3)2(aq) KI(aq) >
2 KNO3(aq) PbI2(s)
Net ionic equation
Pb2+(aq) I-(aq) ---> PbI2(s)
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6. Precipitation Reactions: Predicting Whether a Precipitate Will Form
Predict whether a reaction occurs, and write balanced total and net ionic equations.
(a) iron(III) chloride(aq) + cesium phosphate(aq) →
(b) sodium hydroxide(aq) + cadmium nitrate(aq) →
(c) magnesium bromide(aq) + potassium acetate(aq) →
(d) silver sulfate(aq) + barium chloride(aq) →
Follow-Up Problem 4.3
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7. Problem 5.2.1 molecular equation
Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq)
Write the Total Ionic and Net Ionic equations
Total
Net
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8. Problem 5.2.2 Write the Total Ionic and Net Ionic equations for
CaCl2(aq) + Na2CO3(aq) →
CaCO3(s) + 2 NaCl(aq)
Total
Net
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9. Problem 5.2.3
This problem ties in concepts from at least 3 chapters: 3, 4, & 5.
Solutions of iron (III) chloride and potassium hydroxide are combined.
Naming, balancing formulas, balancing reactions, solubility
Write the:
Molecular equation
Total ionic equation
Net ionic equation
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10. Review: Combining skills Problem 5.2.4
Combine calcium chloride and potassium phosphate
Write the molecular equation
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11. Review: Combining skills Problem 5.2.4
3 CaCl2(aq) + 2 K3PO4(aq) → Ca3PO4(s) + 6 KCl(aq)
Write the Total Ionic and Net Ionic equations
Combining 5 grams CaCl2 with 3.5 grams of K3PO4 produced only a 67% yield of KCl.
What is the limiting reactant?
What mass of product did you make?
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12. Road Map Where we were Where we are going Total & Net ionic equations
Precipitation reactions
Where we are going
Acids and Bases
Classifying reactions
Oxidation reactions
Measuring concentrations of compounds in solution pH
Stoichiometry of reactions in aqueous solution
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13. 5.3 – Acids and Bases (know table 5.2, pg 187)
Acid: increases the H+ concentration
Base: increases the OH- concentration
Strong acid or base: completely dissociates/ionizes
Weak acid or base: partial ionization
Oxides of nonmetals and metals
Acidic oxides (C,N,S) give H+
Basic oxides (CaO) give OH-
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14. Know the strong acids & bases!
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15. 5.4 Acid-Base Reactions The “driving force” is the formation of water.
NaOH(aq) + HCl(aq) --->
NaCl(aq) + H2O(liq)
Net ionic equation
OH-(aq) + H+(aq) > H2O(liq)
This applies to ALL reactions of STRONG acids and bases.
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16. 5.4 - Reactions of Acids and Bases
An Acid reacting with a Base produces a
salt and water
Acid(aq) + Base(aq)  salt (s) + H2O (l)
HX MOH ---> MX H2O
This is one way to make compounds!
Neutralization reaction: a strong acid with a strong base
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17. 5.4 Acid-Base Reactions: Neutralization
What volume of M Ba(OH)2 would neutralize mL of M HCl solution?
Follow-Up Problem 4.5
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18. 5.5 Gas-forming reactions
This is primarily the chemistry of metal carbonates.
CO2 and water ---> H2CO3
H2CO3(aq) + Ca2+ (aq) --->
2 H+(aq) + CaCO3(s) (limestone)
Adding acid reverses this reaction.
MCO3 + acid ---> CO2 + salt
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19. 5.5 Gas-Forming Reactions
CaCO3(s) + H2SO4(aq) --->
2 CaSO4(s) + H2CO3(aq)
Carbonic acid is unstable and forms CO2 & H2O
H2CO3(aq) > CO2 (g) + water (l)
(Antacid tablet has citric acid + NaHCO3)
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21. CHAPTER 5 Reactions in Aqueous Solution
5.1 Properties of Compounds in Aqueous Solution
5.2 Precipitation reactions
Be familiar with and know how to use the solubility rules
5.3 Acids and Bases
5.4 Reactions of Acids and Bases
5.5 Gas-forming reactions
5.6 Classifying Reactions in Aqueous Solution
Know the strong acids & bases, table 5.2
5.7 Oxidation-Reduction Reactions
5.8 Measuring Concentrations of Compounds in solution
5.9 pH, a Concentration scale for acids and bases
5.10 Stoichiometry of Reactions in Aqueous Solutions
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Kull Chem 105 Chapter 2

22. 5.6 Classifying reactions in aqueous solution
Balance the following reactions and then classify each as precipitation (s), acid-base, or gas-forming reaction (g).
a) Ba(OH)2 (aq) + 2 HCl (aq)  BaCl2 (aq) + 2 H2O (l)
b) 2 HNO3 (aq) + CoCO3 (s)  Co(NO3)2 (aq) + H2O (l) + CO2 (g)
c) 2 Na3PO4 (aq) + 3 Cu(NO3)2 (aq)  Cu3(PO4)2 (s) + 6 NaNO3 (aq)
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23. 5.7 Oxidation-Reduction Reactions
Oxidation: any process in which oxygen is added to another substance; gains oxygen,
OIL: Oxidation Is Loss of electrons
RIG: Reduction Is Gain of electrons
Reducing agent gets oxidized
Oxidizing agent gets reduced
Follow-Up Problem 4.5
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24. 5.7 Oxidation-Reduction Reactions Oxidation Numbers: ReDox Reactions
5-36 (pg 225) Oxidation number
PF6- H2AsO4- UO2- N2O5 POCl5 XeO42-
5-37 (a)
Zn (s)  Zn2+ (aq) 0  2+ (ox)
N5+  N4+ (aq) 5+  4+ (red)
O2-  O2- (aq) no change
H+  H+ (aq) no change
Oxidation (loss of e-)
Reduction (gain of e-)
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25. 5.7 Recognizing a Redox Reaction
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)
Al(s) --> Al3+(aq) + 3 e-
Ox. no. of Al increases as e- are donated by the metal.
Therefore, Al is OXIDIZED
Al is the REDUCING AGENT in this balanced half-reaction.
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26. 5.7 Recognizing a Redox Reaction
Notice that the 2 half-reactions add up to give the overall reaction if we use 2 mol of Al and 3 mol of Cu2+.
2 Al(s) --> 2 Al3+(aq) e-
3 Cu2+(aq) + 6 e- --> 3 Cu(s)
2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)
Final eqn. is balanced for mass and charge.
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27. 5.7 Common Oxidizing and Reducing Agents See Table 5.4
Metals (Cu) are reducing agents
Metals (Na, K, Mg, Fe) are reducing agents
HNO3 is an oxidizing agent
Cu + HNO3 --> Cu NO2
2 K + 2 H2O -->
2 KOH + H2
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28. 5.7 Oxidation-Reduction Reactions
Thermite reaction
Fe2O3(s) Al(s)
----> 2 Fe(s) + Al2O3(s)
X Reactant Product
Fe e-  0
O  -2
Al 0  e-
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29. 5.8 Measuring Concentrations
How many moles of each ion are in each solution?
(a) 2 mol of potassium perchlorate dissolved in water
(b) 354 g of magnesium acetate dissolved in water
(c) x 1024 formula units of ammonium chromate dissolved in water
(d) L of 0.55 M sodium bisulfate
Follow-Up Problem 4.1
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30. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Add water to the 3.0 M solution to lower its concentration to 0.50 M
Dilute the solution!
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31. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
But how much water
do we add?
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32. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
M • V =
(3.0 mol/L)(0.050 L) = mol NaOH
Amount of NaOH in final solution must also = 0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or mL
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33. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Conclusion:
add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.
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34. 5.8 Measuring Concentrations of Compounds in Solution
Concentration (Molarity) = moles/L = [ ]
How many moles of H+(aq) are present in 451 mL of 3.20 M hydrobromic acid?
5-46 What volume 2.06 M KMnO4, in liters, contains 322 g of solute?
0.989 L
Follow-Up Problem 4.2
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35. 5.8 Preparing solutions of Known Concentrations
5-50 What mass of oxalic acid, H2C2O4, is required to prepare 250. mL of a solution that has a concentration or 0.15 M H2C2O4?
3.4 g H2C2O4
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36. 5.9 pH, a Concentration scale for acids and bases
pH: a way to express acidity -- the concentration of H+ in solution.
Low pH: high [H+]
High pH: low [H+]
Acidic solution pH < 7
Neutral pH = 7
Basic solution pH > 7
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37. 5.9 pH, a Concentration scale for acids and bases
pH = -log [H+]
pOH = - log [OH-]
Kw = [H+][OH-] = 1.00 x 10-14
pH + pOH = 14
5-56 A saturated solution of milk of magnesia, Mg(OH)2, has a pH of What is the hydrogen ion concentration of the solution? Is the solution acidic or basic?
3 x 10-11
basic
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38. [H+] and pH
If the [H+] of soda is 1.6 x 10-3 M, the pH is ____?
Because pH = - log [H+] then
pH= - log (1.6 x 10-3)
pH = 2.80
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39. 5.10 Stoichiometry of Reactions in Aqueous Solutions
5-64 Hydrazine, N2H4, a base like ammonia, can react with an acid such as sulfuric acid. What mass of hydrazine reacts with 250. mL of M H2SO4?
2 N2H4 (aq) + H2SO4 (aq) 
2 N2H5+ (aq) + SO42- (aq)
2.34 g N2H4
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40. 5.10 SOLUTION STOICHIOMETRY
Zinc reacts with acids to produce H2 gas.
We have 10.0 g of Zn
What volume of 2.50 M HCl is needed to convert the Zn completely?
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41. 5.10 SOLUTION STOICHIOMETRY
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?
Step 1: Write the balanced equation
Zn(s) HCl(aq) --> ZnCl2(aq) + H2(g)
Step 2: Calculate moles of Zn
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42. Step 3: Use the stoichiometric factor to get Moles of HCl
5.10: Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?
Step 3: Use the stoichiometric factor to get
Moles of HCl
Step 4: Calculate volume of HCl req’d
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43. 5. 10: Standardize a solution of NaOH: i. e
5.10: Standardize a solution of NaOH: i.e., accurately determine its concentration.
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
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44. Step 1: Calculate amount of H2C2O4
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
Step 2: Calculate amount of NaOH req’d
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45. 1. 065 g of H2C2O4 (oxalic acid) requires 35
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
= mol acid
Step 2: Calculate amount of NaOH req’d
= mol NaOH
Step 3: Calculate concentration of NaOH
[NaOH] = M
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46. Apples contain malic acid, C4H6O5. C4H6O5(aq) + 2 NaOH(aq) --->
5.10: Use standardized NaOH to determine the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) NaOH(aq) --->
Na2C4H4O5(aq) H2O(liq)
76.80 g of apple requires mL of M NaOH for titration.
What is weight % of malic acid?
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47. Step 1: Calculate moles of NaOH used.
76.80 g of apple requires mL of M NaOH for titration. What is weight % of malic acid?
Step 1: Calculate moles of NaOH used.
C • V = (0.663 M)( L) = mol NaOH
Step 2: Calculate amount of acid titrated.
Step 3: Calculate mass of acid titrated.
Step 4: Calculate % malic acid.
= mol acid
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48. Challenge
5- 72
5- 75
5-76
First 3 teams (max 4 members) to correctly solve the problem receive 5 bonus points
M NaOH
% Fe
g C6H8O6
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49. Next Lesson
Quiz Chapter 5
Chapter 6
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50. Chapter 6 Principles of Reactivity: Energy and Chemical Reactivity
Energy transfer
Calorie burning, gravitational, chemical, electrostatic
Heat- mostly seen in chemical processes
Thermodynamics – transfer of heat between objects; science of heat and work
Energy: Some basic principles – capacity to do work
Kinetic – energy of motion
Potential – energy of position
Conservation of energy (aka First law of thermodynamics)
Total energy of universe is constant
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51. Energy: Some basic principles – capacity to do work
Systems and surroundings
Thermal equilibrium – system and surroundings reach same temperature
Exothermic (heat out of system)
Endothermic (heat into system)
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