Honors Chemistry
Chapter 4: Reactions in Aqueous
Solutions
4.1 Aqueous Solutions
• Solution – homogeneous mixture
• Solute – gets dissolved, often smaller quantity
• Solvent – dissolves solute, often greater quantity
• Aqueous solution – solvent is water!
• Electrolyte – conducts electricity when
dissolved in water
• Strong electrolyte – good conductor
• Weak electrolyte – poor conductor
• Nonelectrolyte – does not conduct when dissolved
4.1 Aqueous Solutions
• Dissolution process
• Solvation – solute surrounded by solvent
molecules
• Hydration – solvation with water molecules
• Strong
• NaCl (s)  Na+ (aq) + Cl- (aq)
• Weak
• HF (g)  H+ (aq) + F- (aq)
• Equilibrium
4.2 Precipitation Reactions
• Formation of an insoluble product
• Solubility Rules
• Always soluble
– Alkalai metal salts and ammonium salts
– Nitrates, bicarbonates, and chlorates
• Usually soluble
– Halides (except Ag+, Hg2+, Pb2+)
– Sulfates (except Ag+, Ca2+, Sr2+, Ba2+, Hg2+, Pb2+)
• Usually insoluble
– Carbonates, phosphates, chromates, sulfides,
hydroxides
4.2 Ionic Equations
• Molecular Equation
• Formulas written out as normal
• Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) + 2 KNO3 (aq)
• Ionic Equation
• Dissolved substances shown as free ions
• Pb2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + 2 I- (aq) 
PbI2 (s) + 2 K+ (aq) + 2 NO3- (aq)
• Net Ionic Equation
• Remove “spectator ions”
• Pb2+ (aq) + 2 I- (aq)  PbI2 (s)
4.2 Ionic Equations
• Try this…
• Write a molecular equation, ionic equation, and
net ionic equation for the reaction between silver
nitrate and iron (III) chloride
• 3 AgNO3 (aq) + FeCl3 (aq)  3 AgCl (s) +
Fe(NO3)3 (aq)
• 3 Ag+ (aq) + 3 NO3- (aq) + Fe3+ (aq) + 3 Cl- (aq)
 3 AgCl (s) + Fe3+ (aq) + 3 NO3- (aq)
• Ag+ (aq) + Cl- (aq)  AgCl (s)
4.3 Acid-Base Reactions
• Arrhenius definition
• Acid
• produces H+ ion in aqueous solution
• HCl (g)  H+ (aq) + Cl- (aq)
• CH3COOH (l)  H+ (aq) + CH3COO- (aq)
• Base
• produces OH- ion in aqueous solution
• NaOH (s)  Na+ (aq) + OH- (aq)
• NH3 (g) + H2O (l)  NH4+ (aq) + OH- (aq)
4.3 Acid-Base Reactions
• Properties of Acids
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Sour taste (vinegar, citrus fruits)
React with metals to liberate hydrogen gas
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
React with carbonates and bicarbonates
2 HCl (aq) + CaCO3 (s)  CaCl2 (aq) + H2O (l) + CO2 (g)
• Properties of Bases
• Bitter taste
• Feel slippery (saponification)
• Indicators – change color in acid / base solutions
4.3 Brønsted Theory
• Define based on movement of H+ ion
• Acid = proton donor
• Base = proton acceptor
• HCl (g) + H2O (l)  H3O+ (aq) + Cl- (aq)
• acid
base
conjugate acid conjugate base
• H3O+ is the hydronium ion
• Same as H+ (aq)
• Shows a water of hydration
• NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
• base
acid
CA
CB
4.3 Brønsted Theory
• Strong acid
• HNO3 + H2O  H3O+ + NO3-
• Weak acid
• HBr + H2O  H3O+ + Br-
• Diprotic acid
• H2SO4 + H2O  H3O+ + HSO4• HSO4- + H2O  H3O+ + SO42-
• Triprotic acid
• 3 ionizations, e. g., H3PO4
4.3 Autoprotolysis of Water
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HOH (l)  H+ (aq) + OH- (aq)
Water is both a weak acid and a weak base
Using Brønsted theory,
H2O + H2O  H3O+ (aq) + OH- (aq)
B
A
CA
CB
[H3O+] = 1.0 x 10-7 M in pure water
Define pH = -log [H3O+]
pH = 7 in pure water
Acid = lower pH; Base = higher pH
4.3 Acid-Base Neutralization
• Acid + base  salt + water
• HCl (aq) + NaOH (aq)  NaCl (aq) + H2O
• Ionic equation
• H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq)
+ Cl- (aq) + H2O (l)
• Net ionic equation
• H+ (aq) + OH- (aq)  H2O (l)
• All neutralizations reduce to this
• Try this: H2SO4 (aq) + Mg(OH)2 (aq) 
4.4 Redox Reactions
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Oxidation-reduction reaction
Electron transfer reactions
Oxidation – loss of e- (higher charge)
Reduction – gain of e- (lower charge)
2 Ca (s) + O2 (g)  2 CaO (s)
Ca goes from neutral to 2+ ..... Oxidation
O goes from neutral to 2- ..... Reduction
4.4 Half-Reactions
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Separate oxidation from reduction
2 Ca (s) + O2 (g)  2 CaO (s)
2 Ca  2 Ca2+ + 4 eoxidation
O2 + 4 e-  2 O2reduction
Ca is reducing agent
• Donates e-, gets oxidized
• O is oxidizing agent
• Accepts e-, gets reduced
4.4 Half-Reactions
• Try this...
• Write half reactions for
Zn + 2 HCl  ZnCl2 + H2
• Zn + 2 H+ + 2 Cl-  Zn2+ + 2 Cl- + H2
• Cl- ion is a spectator
• Ox: Zn  Zn2+ + 2 e• Red: 2 H+ + 2 e-  H2
• Zn is reducing agent, H+ is oxidizing agent
4.4 Oxidation Numbers
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Charge an atom has within a molecule
Free element: ox # = 0
Monatomic ions: ox # = charge
Oxygen: always 2- (except peroxides)
Hydrogen: +1 (except in alkalai hydrides)
Fluorine: always -1
Neutral molecule: all ox #’s add up to 0
Ox #’s can occasionally be fractions
4.4 Oxidation Numbers
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Try this…
Find oxidation numbers for all elements:
Al2O3
H2SO4
PO43Fe3O4
4.4 Types of Redox Reactions
• Combination Reactions (synthesis)
• Two or more substances combine, single product
• S + O2  SO2
• Decomposition Reactions
• Compound breaks down into components
• 2 KClO3  2 KCl + 3 O2
• Displacement Reactions
• One ion displaces another in a compound
• Zn + 2 HCl  ZnCl2 + H2
4.4 Displacement Reactions
• Hydrogen Displacement
• IA and IIA metals displace H+ from water
• 2 Na + 2 HOH  2 NaOH + H2
• Most metals displace H+ from acid
• Metal Displacement
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Cu + AgNO3  Cu(NO3)2 + Ag
Activity series shows who can displace whom
p. 134
Cu is higher in series and so can displace Ag
4.4 Redox Reactions
• Halogen Displacement
• F > Cl > Br > I
• 2 NaCl + F2  2 NaF + Cl2
• NaCl + Br2  NR
• Disproportionation Reactions
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Same element is oxidized and reduced
2 H2O2  2 H2O + O2
O starts with 1- charge
Ends with 2- charge in H2O and 0 charge in O2
4.5 Concentration of Solutions
• Molarity (M)
• Number of moles solute in 1 L of solution
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n
M = ---V
• Find the Molarity of 15 g NaCl in 250 mL.
• 15 g
1 mol
------ x --------- = 0.256 mol
1
58.5 g
• M = 0.256 mol / 0.250 L = 1.03 M
4.5 Concentration of Solutions
• How would you make 175 mL of a 0.500 M
solution of CaCl2?
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n
0.500 M = ----------0.175 L
• n = 0.0875 mol
• 0.0875 mol
111 g
--------------- x -------- = 9.71 g
1
1 mol
4.5 Dilution of Solutions
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Reduce the concentration by adding water
Moles of solute stay constant
From M = n/V, we get n = MV
Since n for soln 1 equals n for soln 2,
M1V1 = M2V2
Try this...
• How would you prepare 500 mL of a 0.75 M HCl
solution from a 11.7 M stock HCl solution?
4.6 Gravimetric Analysis
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Lab technique based on measuring mass
Used to find percent composition
E.g., find % Cl in NaCl
AgNO3 + NaCl  NaNO3 + AgCl
Collect AgCl precipitate and mass
Determine mass of Cl in precipitate
Use it to find percent Cl in original sample
We’ll do this in lab
4.7 Acid-Base Titration
• Titration
• solution of known concentration is reacted with
solution of unknown concentration
• Reaction must go to completion
• Calculate the unknown concentration
• Standard Solution
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Solution whose concentration is accurately known
KHP (Potassium Hydrogen Phthalate)
Used to standardize base solution
Base solution then used to titrate an unknown acid
4.7 Acid-Base Titration
• Equivalence Point
• Point where acid and base have neutralized
• Added in exact stoichiometric ratio
• Need an indicator to show end point
• Titration Equation
• mol acid = mol base at equivalence point
• MaVa = MbVb
• If we know the Molarity and volume of base, we
can measure the volume of acid used and
calculate its Molarity
4.8 Redox Titration
• Titration can be done with redox reactions
• Often need to use a reactant whose color
changes during the reaction
• Takes the place of an indicator
• Works just like acid-base titration
• End of chapter 4 – Finally!