IONIC COMPOUNDS Chapter 5

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IONIC COMPOUNDSChapter 5

• Many reactions involve ionic compounds,
  especially reactions in water aqueous solutions.

KMnO4 in water
2
An Ionic Compound, CuCl2, in Water
3
Aqueous Solutions

• How do we know ions are present in aqueous
  solutions?
• The solutions conduct electricity!
• They are called ELECTROLYTES
• HCl, MgCl2, and NaCl are strong electrolytes.
  They dissociate completely (or nearly so) into
  ions.

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Aqueous Solutions

• HCl, MgCl2, and NaCl are strong electrolytes.
  They dissociate completely (or nearly so) into
  ions.

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Aqueous Solutions

• Acetic acid ionizes only to a small extent, so it
  is a weak electrolyte.
• CH3CO2H(aq) ---gt CH3CO2-(aq) H(aq)

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Aqueous Solutions

• Acetic acid ionizes only to a small extent, so it
  is a weak electrolyte.
• CH3CO2H(aq) ---gt CH3CO2-(aq) H(aq)

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Aqueous Solutions

• Some compounds dissolve in water but do not
  conduct electricity. They are called
  nonelectrolytes.

Examples include sugar ethanol ethylene glycol
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Water Solubility of Ionic Compounds
If one ion from the Soluble Compd. list is
present in a compound, the compound is water
soluble.
Screen 5.4 Figure 5.1
Guidelines to predict the solubility of ionic
compounds
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Water Solubility of Ionic Compounds

• Common minerals are often formed with anions that
  lead to insolubility
• sulfide fluoride
• carbonate oxide

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ACIDS
An acid -------gt H in water

• Some strong acids are
• HCl hydrochloric
• H2SO4 sulfuric
• HClO4 perchloric
• HNO3 nitric

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ACIDS
An acid -------gt H in water

• HCl(aq) ---gt H(aq) Cl-(aq)

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The Nature of Acids
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Weak Acids

• WEAK ACIDS weak electrolytes
• CH3CO2H
• acetic acid
• H2CO3 carbonic acid
• H3PO4 phosphoric acid
• HF
• hydrofluoric acid

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ACIDS

• Nonmetal oxides can be acids
• CO2(aq) H2O(liq) ---gt H2CO3(aq)
• SO3(aq) H2O(liq) ---gt H2SO4(aq)
• and can come from burning coal and oil.

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BASESsee Screen 5.9 and Table 5.2
Base ---gt OH- in water

• NaOH(aq) ---gt Na(aq) OH-(aq)

NaOH is a strong base
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Ammonia, NH3An Important Base
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BASES

• Metal oxides are bases
• CaO(s) H2O(liq)
• --gt Ca(OH)2(aq)

CaO in water. Indicator shows solution is basic.
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Know the strong acids bases!
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Net Ionic Equations

• Mg(s) 2 HCl(aq) --gt H2(g) MgCl2(aq)
• We really should write
• Mg(s) 2 H(aq) 2 Cl-(aq) ---gt
  H2(g) Mg2(aq) 2 Cl-(aq)

The two Cl- ions are SPECTATOR IONS they do not
participate. Could have used NO3-.
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Net Ionic Equations

• Mg(s) 2 HCl(aq) --gt H2(g)
  MgCl2(aq)
• Mg(s) 2 H(aq) 2 Cl-(aq) ---gt H2(g)
  Mg2(aq) 2 Cl-(aq)

We leave the spectator ions out Mg(s) 2
H(aq) ---gt H2(g) Mg2(aq)
to give the NET IONIC EQUATION
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Chemical Reactions in WaterSections 5.2
5.4-5.6CD-ROM Ch. 5

• We will look at EXCHANGE REACTIONS

Pb(NO3) 2(aq) 2 KI(aq) ----gt PbI2(s) 2
KNO3 (aq)
The anions exchange places between cations.
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Precipitation Reactions

• The driving force is the formation of an
  insoluble compound a precipitate.
• Pb(NO3)2(aq) 2 KI(aq) -----gt
• 2 KNO3(aq) PbI2(s)
• Net ionic equation
• Pb2(aq) 2 I-(aq) ---gt PbI2(s)

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Acid-Base Reactions

• The driving force is the formation of water.
• NaOH(aq) HCl(aq) ---gt
• NaCl(aq) H2O(liq)
• Net ionic equation
• OH-(aq) H(aq) ---gt H2O(liq)
• This applies to ALL reactions of STRONG acids and
  bases.

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Acid-Base Reactions CCR, page 162
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Acid-Base Reactions

• A-B reactions are sometimes called
  NEUTRALIZATIONS because the solution is neither
  acidic nor basic at the end.
• The other product of the A-B reaction is a SALT,
  MX.
• HX MOH ---gt MX H2O
• Mn comes from base Xn- comes from acid
• This is one way to make ionic compounds!

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Gas-Forming Reactions

• This is primarily the chemistry of metal
  carbonates.
• CO2 and water ---gt H2CO3
• H2CO3(aq) Ca2 ---gt
• 2 H(aq) CaCO3(s) (limestone)
• Adding acid reverses this reaction.
• MCO3 acid ---gt CO2 salt

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Gas-Forming Reactions

• CaCO3(s) H2SO4(aq) ---gt
• 2 CaSO4(s) H2CO3(aq)
• Carbonic acid is unstable and forms CO2 H2O
• H2CO3(aq) ---gt CO2 (g) water
• (Antacid tablet has citric acid NaHCO3)

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See also Gas Forming Reactions in Biological
Systems Three of the pioneers in working out the
roles of NO forming reactions shared a Nobel
Prize in 1988 for their discoveries.
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Quantitative Aspects of Reactions in
SolutionSections 5.8-5.10
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Terminology

• In solution we need to define the -
• SOLVENT
• the component whose physical state
  is preserved when
  solution forms
• SOLUTE
• the other solution component

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Concentration of Solute

• The amount of solute in a solution is given by
  its concentration.

Concentration (M)
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1.0 L of water was used to make 1.0 L of
solution. Notice the water left over.
CCR, page 177
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PROBLEM Dissolve 5.00 g of NiCl26 H2O in
enough water to make 250 mL of solution.
Calculate molarity.
Step 1 Calculate moles of NiCl26H2O
Step 2 Calculate molarity
NiCl26 H2O 0.0841 M
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The Nature of a CuCl2 SolutionIon Concentrations

• CuCl2(aq) --gt
• Cu2(aq) 2 Cl-(aq)
• If CuCl2 0.30 M, then
• Cu2 0.30 M
• Cl- 2 x 0.30 M

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USING MOLARITY

• What mass of oxalic acid, H2C2O4, is required to
  make 250. mL of a 0.0500 M solution?
• Because
  Conc (M) moles/volume mol/V
• this means that

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USING MOLARITY
What mass of oxalic acid, H2C2O4, is required to
make 250. mL of a 0.0500 M solution?
moles MV

• Step 1 Calculate moles of acid required.
• (0.0500 mol/L)(0.250 L) 0.0125 mol
• Step 2 Calculate mass of acid required.
• (0.0125 mol )(90.00 g/mol) 1.13 g

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Preparing Solutions

• Weigh out a solid solute and dissolve in a given
  quantity of solvent.
• Dilute a concentrated solution to give one that
  is less concentrated.

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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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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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PROBLEM You have 50.0 mL of 3.0 M NaOH and you
want 0.50 M NaOH. What do you do?

• How much water is added?
• The important point is that ---gt

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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) 0.15 mol NaOH
• Amount of NaOH in final solution must also 0.15
  mol NaOH
• 0.15/Volume of final solution 0.5 M/ 1 L
• Volume of final solution
• (0.15 mol NaOH)(1 L/0.50 mol) 0.30 L
• or 300 mL

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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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Preparing Solutions by Dilution

• A shortcut
• Cinitial Vinitial Cfinal Vfinal

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The pH Scale

• pH log (1/ H)
• - log H
• Remember log a b if 10ba
• In a neutral solution, H OH- 1.00
  x 10-7 M at 25 oC
• pH - log H -log (1.00 x 10-7)
  - (-7)
• 7
• See CD Screen 5.17 for a tutorial

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pH, a Concentration Scale

• pH a way to express acidity -- the concentration
  of H in solution.

Low pH high H
High pH low H
Acidic solution pH lt 7 Neutral pH 7 Basic
solution pH gt 7
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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)
• pH 2.80

Whats the origin of the name of the soda 7up ?
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ACID-BASE REACTIONSTitrations

• H2C2O4(aq) 2 NaOH(aq) ---gt
• acid base
• Na2C2O4(aq) 2 H2O(liq)
• Carry out this reaction using a TITRATION.

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Setup for titrating an acid with a base
CCR, page 186
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Titration

• 1. Add solution from the buret.
• 2. Reagent (base) reacts with compound (acid) in
  solution in the flask.
• 3. Indicator shows when exact stoichiometric
  reaction has occurred.
• 4. Net ionic equation
• H OH- --gt H2O
• 5. At equivalence point
• moles H moles OH-

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LAB PROBLEM 1 Standardize a solution of NaOH
i.e., accurately determine its concentration.

• 1.065 g of H2C2O4 (oxalic acid) requires 35.62
  mL of NaOH for titration to an equivalence point.
  What is the concentration of the NaOH?

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1.065 g of H2C2O4 (oxalic acid) requires 35.62 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 reqd
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1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL
of NaOH for titration to an equivalence point.
What is the concentration of the NaOH?

• Step 1 Calculate amount of H2C2O4
• 0.0118 mol acid
• Step 2 Calculate amount of NaOH reqd
• 0.0236 mol NaOH
• Step 3 Calculate concentration of NaOH

NaOH 0.663 M
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LAB PROBLEM 2 Use standardized NaOH to
determine the amount of an acid in an unknown.

• Apples contain malic acid, C4H6O5.
• C4H6O5(aq) 2 NaOH(aq) ---gt
• Na2C4H4O5(aq) 2 H2O(liq)
• 76.80 g of apple requires 34.56 mL of 0.663 M
  NaOH for titration. What is weight of malic
  acid?

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76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration. What is weight of malic
acid?

• Step 1 Calculate amount of NaOH used.
• C V (0.663 M)(0.03456 L)
• 0.0229 mol NaOH
• Step 2 Calculate amount of acid titrated.

0.0115 mol acid
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76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration. What is weight of malic
acid?
Step 1 Calculate amount of NaOH used.
0.0229 mol NaOH Step 2 Calculate amount
of acid titrated 0.0115 mol acid

• Step 3 Calculate mass of acid titrated.

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76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration. What is weight of malic
acid?

• Step 1 Calculate amount of NaOH used.
• 0.0229 mol NaOH
• Step 2 Calculate amount of acid titrated
• 0.0115 mol acid
• Step 3 Calculate mass of acid titrated.
• 1.54 g acid

Step 4 Calculate malic acid.
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pH and H

• If the pH of Coke is 3.12, it is ____________.
• Because pH - log H then
• log H - pH
• Take antilog and get
• H 10-pH
• H 10-3.12 7.6 x 10-4 M

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SOLUTION STOICHIOMETRYGas-forming reactions

• Zinc reacts with acids to produce H2 gas.
• Have 10.0 g of Zn
• What volume of 2.50 M HCl is needed to convert
  the Zn completely?

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GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
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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) 2 HCl(aq) --gt ZnCl2(aq) H2(g)
• Step 2 Calculate amount of Zn

Step 3 Use the stoichiometric factor
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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

Step 4 Calculate volume of HCl reqd
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EXCHANGE Precipitation Reactions
EXCHANGE Acid-Base Reactions
EXCHANGE Gas-Forming Reactions
REACTIONS
REDOX REACTIONS
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REDOX REACTIONS

• Redox reactions are characterized by ELECTRON
  TRANSFER between an electron donor and electron
  acceptor.
• Transfer leads to
• 1. increase in oxidation number of some element
  OXIDATION
• 2. decrease in oxidation number of some element
  REDUCTION

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REDOX REACTIONS

• Cu(s) 2 Ag(aq)
• ---gt Cu2(aq) 2 Ag(s)
• In all reactions if something has been oxidized
  then something has also been reduced

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Why Study Redox Reactions
Batteries
Corrosion
Manufacturing metals
Fuels
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OXIDATION NUMBERS

• The electric charge an element APPEARS to have
  when electrons are counted by some arbitrary
  rules
• 1. Each atom in free element has ox. no. 0.
• Zn O2 I2 S8
• 2. In simple ions, ox. no. charge on ion.
• -1 for Cl-
• 2 for Mg2

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OXIDATION NUMBERS

• 3. F always has an oxidation number of -1 when
  forming compounds with other elements.
• 4. Cl, Br and I have oxidation numbers of -1 when
  forming compounds with other elements, except
  when combined with oxygen and fluorine.
• 5a. O has ox. no. -2
• (except in peroxides in H2O2, O -1)

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OXIDATION NUMBERS

• 5b. Ox. no. of H 1
• (except when H is associated with a metal as in
  NaH where it is -1)
• 6. Algebraic sum of oxidation numbers
• 0 for a compound
• overall charge for an ion

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OXIDATION NUMBERS

• NH3 N
• ClO- Cl
• H3PO4 P
• MnO4- Mn
• Cr2O72- Cr
• C3H8 C

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Recognizing a Redox Reaction

• Corrosion of aluminum
• 2 Al(s) 3 Cu2(aq) --gt 2 Al3(aq) 3 Cu(s)
• Al(s) --gt 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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Recognizing a Redox Reaction

• Corrosion of aluminum
• 2 Al(s) 3 Cu2(aq) --gt 2 Al3(aq) 3 Cu(s)
• Cu2(aq) 2 e- --gt Cu(s)
• Ox. no. of Cu decreases as e- are accepted by the
  ion.
• Therefore, Cu is REDUCED
• Cu is the OXIDIZING AGENT in this balanced
  half-reaction.

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Recognizing a Redox Reaction

• Notice that the 2 half-reactions add up to give
  the overall reaction if we use 2 moles of Al
  and 3 moles of Cu2.
• 2 Al(s) --gt 2 Al3(aq) 6 e-
• 3 Cu2(aq) 6 e- --gt 3 Cu(s)
• --------------------------------------------------
  ---------
• 2 Al(s) 3 Cu2(aq) ---gt 2 Al3(aq) 3 Cu(s)
• Final eqn. is balanced for mass and charge.

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Common Oxidizing and Reducing AgentsSee Table
5.4
Cu HNO3 --gt Cu2 NO2
2 K 2 H2O --gt 2 KOH H2
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Recognizing a Redox ReactionSee Table 5.4
Reaction Type
Oxidation
Reduction

• In terms of oxygen gain loss
• In terms of halogen gain loss
• In terms of electrons loss gain

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Examples of Redox Reactions
Metal halogen 2 Al 3 Br2 ---gt Al2Br6
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Examples of Redox Reactions
Nonmetal (P) Oxygen
Metal (Mg) Oxygen
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Examples of Redox Reactions
Metal acid Mg HCl Mg reducing agent H
oxidizing agent
Metal acid Cu HNO3 Cu reducing agent HNO3
oxidizing agent