Chapter 4: Reactions in Aqueous Solutions

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Chapter 4 Reactions in Aqueous Solutions

• Chapter Outline
• 4.1 Solute Concentrations, Molarity
• 4.2 Precipitation Reactions
• 4.3 Acid-base Reactions
• 4.4 Oxidation-reduction Reactions

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Solute Concentrations Molarity

• The concentration of a solute in solution can be
  expressed in terms of its molarity
• Molarity (M) moles of solute/liters of solution
• What is the molarity of a solution containing
  1.20moles of substance A in 2.50L of solution?
• A 1.20moles/2.50l 0.480 mol/l 0.480m

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Molarity (Contd)

• How many grams of K2CrO4 is needed to make 1
  liter of a 0.100M solution.
• 0.100moles K2CrO4 x 194.2g K2CrO4/1 mole K2CrO4
  19.4g K2CrO4
• The molarity of a solution can be used to
  calculate
• The number of moles of solute in a given volume.
• The volume of solution containing a given number
  of moles.

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Example 1

• The bottle labeled concentrated HCl in the lab
  contains 12.0mol HCl per liter of solution. That
  is, HCl 12.0 M.
• How many moles of HCl are there in 25.0ml of this
  solution?
• What volume (V) of concentrated HCl must be taken
  to contain 1.00mol of HCl?

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Example 1 (Contd)

• The required conversion factors are
• 12.0mol HCl/1 L or 1 L/12.0mol HCl
• nHCl 25.0mL x 1L/1000mL x 12.0mol HCl/1 L
  0.300mol HCl
• V 1.00mol HCl x 1 L/12.0mol HCl 0.0833L
  (83.3mL)

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Ionization

• When an ionic solid dissolves in water, cations
  and anions separate
• NaCl(s) ? Na(aq) Cl-(aq)
• In water, this molecule is completely ionized.
  Ionic solids are often referred to as strong
  electrolytes. Their ions are excellent
  conductors of electricity.

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Ionization (Contd)

• Consider ionization of MgCl2
• MgCl2 ? Mg2(aq) 2Cl-(aq)
• 1 mole of MgCl2 yield 1 mole of Mg2 and 2 moles
  of Cl-. It follows then that
• Molarity of Mg2 molarity of MgCl2
• Molarity of Cl- 2 x molarity of MgCl2

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Ionic Solids Containing Polyatomic Ions

• (NH4)3PO4(s) ? 3NH4(aq) PO43-(aq)
• Molarity NH4 3 x molarity of (NH4)3PO4
• Molarity PO43- molarity of (NH4)3PO4

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Example 2

• Give the concentrations in moles per liter of
  each ion in
• (a) 0.080M K2SO4 (b) 0.40M FeCl3
• (a) K2SO4(s) ? 2K(aq) SO42-(aq)
• The conversion factors are 2mol K/1mol K2SO4 and
  1mol SO42-/1mol K2SO4.

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Example 2 (Contd)

• K 0.080mol K2SO4/1L x 2mol K/1mol K2SO4
  0.16M K
• SO42- 0.080mol K2SO4/1L x 1mol SO42-/
• 1mol K2SO4 0.080mol SO42-

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Example 2 (Contd)

• (b) FeCl3(s) ? Fe3(aq) 3Cl-(aq)
• Fe3 0.40mol FeCl3/1L x 1mol Fe3/
• 1mol FeCl3 0.40mol Fe3
• Cl- 0.40mol FeCl3/1L x 3mol Cl-/
• 1mol FeCl3 1.2M Cl-

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Precipitation Reactions

• A precipitation reaction occurs when water
  solutions of two different ionic compounds are
  mixed and an insoluble solid separates out of
  solution.
• The precipitate is itself ionic the cation comes
  from one solution and the anion from another. To
  predict the occurrence of these reactions, we
  must know which ionic substances are insoluble in
  water.

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Predicting Precipitates

• Figure 4.4 Precipitation Diagram.
• If a cation in solution 1 mixes with an anion in
  solution 2 to form an insoluble compound (colored
  squares), that compound will precipitate.
• Cation-anion combinations that lead to the
  formation of a soluble compound (white squares)
  will not give a precipitate.

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Examples

• NiCl2 and NaOH
• A precipitate of Ni(OH)2, an insoluble compound,
  will form.
• NaCl, a soluble compound, will not precipitate.

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Predicting Precipitation Reactions

• Consider the following pairs of solutions
• (a) CuSO4 and NaNO3 (b) Na2CO3 and CaCl2
• (a) Ions present Cu2, SO42- Na, NO3-
• Possible precipitates Cu(NO3)2, Na2SO4
• According to table, both of these are soluble, no
  precipitate forms.

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Predicting (Contd)

• (b) Ions present Na, CO32- Ca2, Cl-
• Possible precipitates NaCl, CaCO3
• Sodium chloride is soluble, but calcium carbonate
  is not. When these two are mixed, calcium
  carbonate precipitates.

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Net Ionic Equations

• The precipitation reaction that occurs when
  solutions of Na2CO3 and CaCl2 are mixed can be
  represented by a simple equation.
• Ca2(aq) CO32-(aq) ? CaCO3(s)
• Note the equation only includes the ions that
  participate in the reaction. Na and Cl are
  spectator ions which are present in solution
  before and after the precipitation of calcium
  carbonate.

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Net Ionic Equations

• A net ionic equation is one in which only the
  ions involved in the reaction are shown. Like
  all equations, net ionic equations must show
• (i) Atom balance there must be the same number
  of atoms of each element on both sides of the
  equation.
• (ii) Charge balance there must be the same total
  charge on both sides of the equation.

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Examples of Net Ionic Equation

• Write a net ionic equation if possible for the
  following pairs of solutions
• (a) NaOH and Cu(NO3)2
• Ions present Na, OH- Cu2, NO3-
• NaNO3 is soluble, but Cu(OH)2 is not.
• Equation Cu2(aq) 2OH-(aq) ? Cu(OH)2(s)

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Examples (Contd)

• (b) BaCl2 and Ag2SO4
• Ions present Ba2, Cl- Ag, SO42-
• Possible precipitates AgCl, BaSO4
• Both compounds are insoluble, so two reactions
  occur.
• Ba2(aq) SO42-(aq) ? BaSO4(s)
• Ag(aq) Cl-(aq) ? AgCl (s)

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Stoichiometry

• We can apply the approach introduced in chapter 3
  to calculate mole-mass relationships in solution
  reactions represented by net ionic equations.
• Limiting reagent problems involving net ionic
  reactions are solved in the same manner.

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Examples

• Consider the net ionic equation derived from the
  reaction that occurs when solutions of NaOH and
  Cu(NO3)2 are mixed.
• Cu2(aq) 2OH-(aq) ? Cu(OH)2 (s)
• What volume of 0.106M Cu(NO3)2 solution is
  required to form 6.52g of solid Cu(OH)2?

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Example (Contd)

• Strategy The first 3 steps in the path are
• mass Cu(OH)2 ? moles Cu(OH)2 ? moles Cu2
• Because 1 mole of Cu(NO3)2 produces one mole of
  Cu2, it follows that
• moles Cu(NO3)2 moles Cu2
• V(CuNO3)2 6.52g Cu(OH)2 x 1mole Cu(OH)2/97.57g
  Cu(OH)2 x 1mole Cu2/1mol Cu(OH)2 x
• 1mole Cu(NO3)2/1mole Cu2 x 1.00L
  Cu(NO3)2/0.106moles Cu(NO3)2 0.630L Cu(NO3)2

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Acid-Base Reactions

• An acid is a species that produces H ions in
  water solution
• A base is a species that produces OH- ions in
  water solution
• Strong Acids (HCl) completely ionize in water
  while weak acids do not.
• HCl (aq) ? H(aq) Cl-(aq)

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Strong Acids and Bases

• There are six common strong acids
• Hydrochloric Acid (HCl)
• Hydrobromic Acid (HBr)
• Hydroiodic Acid (HI)
• Nitric Acid (HNO3)
• Perchloric Acid (HClO4)
• Sulfuric Acid (H2SO4)

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Writing Equations for Strong and Weak Acids

• Strong HB(aq) ? H(aq) B-(aq)
• Weak HB(aq) H(aq) B-(aq)
• Strong Bases
• Lithium Hydroxide (LiOH)
• Sodium Hydroxide (NaOH)
• Potassium Hydroxide (KOH)
• Calcium Hydroxide (Ca(OH)2)

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Strong Bases

• A strong base in water completely ionizes to OH-
  and a cation.
• NaOH(s) ? Na(aq) OH-(aq)
• A weak base produces OH- ions by reacting with
  water.
• NH3(aq) H2O ? NH4(aq) OH-(aq)

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Equations for Acid-Base Reactions

• I. Strong Acid-Strong Base Consider a solution
  of HNO3 (strong acid) mixed with a solution of
  NaOH (strong base).
• HNO3(aq) ? H(aq) NO3-(aq)
• NaOH(aq) ? Na(aq) OH-(aq)
• H(aq) OH-(aq) ? H2O
• This reaction is referred to as neutralization.

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Equations for Acid-Base Reactions (Contd)

• II. Weak Acid-Strong Base Two-step reaction.
• (1) HB(aq) H(aq) B-(aq)
• (2) H(aq) OH-(aq) ? H2O
• Add the equations together
• HB(aq) OH-(aq) ? B-(aq) H2O

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• III. Strong Acid-Weak Base a solution of HCl
  added to a solution of NH3.
• (1) NH3(aq) H2O NH4(aq) OH-(aq)
• (2) H(aq) OH-(aq) ? H2O
• Add the two equations together
• H(aq) NH3 ? NH4

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Acid-Base Titrations

• Acid-base reactions in water are often used to
  determine the concentration of a dissolved
  species (H, OH-) or its percentage in a solid
  mixture.
• Titration is measuring the volume of a standard
  solution (solution of known concentration)
  required to react with a measured amount of
  sample.

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Example of Titration

• CH3COOH(aq) OH-(aq) ? CH3COO- H2O
• The objective of the titration is to determine
  the point at which the reaction is complete (acid
  solution is neutralized). This point is called
  the equivalence point.

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Oxidation-Reduction reactions

• An oxidation-reaction (red-ox) reaction involves
  the transfer of electrons between two species in
  aqueous solution.
• In a redox reaction, one species loses (donates)
  electrons, while the other species gains
  (accepts) electrons.

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Example of Red-Ox Reaction

• Zn(s) 2H(aq) ? Zn2 H2(g)
• A red-ox reaction can be split into two
  half-reactions oxidation and reduction.
• Oxidation Zn(s) ? Zn2 2e-
• Reduction 2H(aq) 2e- ? H2(g)

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Example (Contd)

• Oxidation and reduction always occur together
  you cannot have one without the other.
• There is no net change in the number of electrons
  in a red-ox reaction. Those electrons given off
  in the oxidation are taken on by another species
  in the reduction.

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Terms in Red-Ox Reactions

• The ion or molecule that accepts electrons is
  called the oxidizing agent.
• The ion or molecule that donates electrons is
  called the reducing agent.

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Oxidation Number

• The concept of oxidation number is used to
  simplify the electron bookkeeping in redox
  reactions.
• (1) For a monoatomic ion (Na, S2-), the
  oxidation number is simply the charge of the ion.
• (2) In a molecule or polyatomic ion, the
  oxidation number of an element is a
  pseudo-charge, obtained in a rather arbitrary
  fashion.

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Rules for Assigning Oxidation Numbers

• I. The oxidation number of an element in an
  elementary substance is 0. For example, the
  oxidation number of chlorine in Cl2 or of
  phosphorous in P4 is 0.
• II.The oxidation number of an element in a
  monoatomic ion is equal to the charge of that
  ion. In the ionic compound, NaCl, sodium has an
  oxidation number of 1 and chlorine an oxidation
  number of 1.

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Rules (Contd)

• III. Certain elements have the same oxidation
  number in all or almost all of their compounds.
  Group 1 metals are always 1 ions when they form
  compounds and Group 2 elements are always 2 ions
  when they form compounds. Oxygen is usually 2
  (except peroxide) while hydrogen is ordinarily 1
  (can be 1).

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Rules (Contd)

• IV. The sum of the oxidation numbers in a neutral
  species is 0 in a polyatomic ion, it is equal to
  the charge of that ion.
• What is the oxidation number of S in Na2SO4?
• First look for elements whos oxidation number is
  always or almost always the same.
• In Na2SO4, Na is 1 and O is 2. Sodium sulfate
  must be neutral, so the sum of the oxidation
  numbers must be 0.
• 0 2(1) x 4(-2) x 6

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Working Definitions of Oxidation and Reduction

• Oxidation is defined as an increase in the
  oxidation number.
• Reduction is defined as a decrease in the
  oxidation number.
• Zn(s) 2H(aq) ? Zn2 H2(g)
• Zn is oxidized (oxid. 0 ? 2)
• H is reduced (oxid. 1 ? 0)

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Example

• 2Al(s) 3Cu2(aq) ? 2Al3 3Cu(s)
• Al is oxidized (oxid. 0 ? 3)
• Cu2 is reduced (oxid. 2 ? 0)

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Balancing Half-Reactions

• Before you can balance an overall redox reaction,
  you must balance the half reactions.
• Fe2(aq) ? Fe3(aq)
• (oxid. Fe 2 ? 3)
• Fe2(aq) ? Fe3 e-

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Rules for Balancing Half-Reactions

• 1. Balance the atoms of the element being
  oxidized or reduced.
• 2. Balance the oxidation number by adding
  electrons. Add them to the left for a reduction
  add them to the right for an oxidation.
• 3. Balance charge by adding H ions in acidic
  conditions and OH- in basic conditions.
• 4. Balance hydrogen by adding H2O molecules.
• 5. Check to make sure that oxygen is balanced.

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Example

• Balance the following half-equations
• (a) MnO4-(aq) ? Mn2 (acidic solution)
• (b) Cr(OH)3(s) ? CrO42- (basic solution)
• MnO4-(aq) ? Mn2
• i) Equation is atom balanced
• ii) Because Mn is reduced from 7 to 2, 5
  electrons must be added to the left hand side.
• MnO42-(aq) 5e- ? Mn2(aq)

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Example (Contd)

• iii) There is a charge of 6 on the left and 2
  on the right. To balance, add 8 H to the left
  to give a charge of 2 on both sides.
• MnO42-(aq) 8H(aq) 5e- ? Mn2(aq)
• iv) To balance the 8H ions on the left, add 4
  H2O molecules to the right.
• MnO42-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O

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Example (Contd)

• (b) Cr(OH)3(s) ? CrO42- (basic solution)
• i) There is one Cr atom on each side
• Cr(OH)3(s) ? CrO42- (basic solution)
• ii) Because oxidation of Cr increases from 3
  to 6, add the electrons to the right.
• Cr(OH)3(s) ? CrO42-(aq) 3e-

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Example (Contd)

• Cr(OH)3(s) ? CrO42-(aq) 3e-
• iii) There is a charge of 0 on the left and a
  charge of 5 on the right. To balance charge,
  add 5 OH- ions to the left.
• Cr(OH)3(s) 5OH-(aq) ? CrO42-(aq) 3e-
• iv) There are 8 hydrogens on the left and none on
  the right. Add four waters to balance the
  hydrogens.
• Cr(OH)3(s) 5OH-(aq) ? CrO42-(aq) 4H2O 3e-

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Balancing Redox Equations

• 4 Step Procedure
• 1. Split the equation into two half-reactions,
  one for reduction, the other for oxidation.
• 2. Balance one of the half-equations with respect
  to both atoms and charge as described above.
• 3. Balance the other half-reaction.
• 4. Combine the two half-reactions in such a way
  as to eliminate electrons.

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Example

• Balance the following redox equation
• Fe2(aq) MnO42-(aq) ? Fe3(aq) Mn2(aq)
• in acidic conditions
• (a) (1) oxidation Fe2(aq) ? Fe3(aq)
• reduction MnO42-(aq) ? Mn2(aq)
• (2), (3) The balanced half-reactions, as
  obtained previously, are
• Fe2(aq) ? Fe3(aq) 1e-
• MnO42-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O

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Example (Contd)

• (4) To eliminate electrons, multiply the
  oxidation half-reaction by 5 and add to the
  reduction half-reaction
• 5Fe2(aq) ? Fe3 1e-
• MnO42-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O
• 5Fe2 MnO42-(aq) 8H(aq) ? 5Fe3(aq) Mn2
  4H2O