Chapter 16: Aqueous Ionic Equilibria - PowerPoint PPT Presentation

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Chapter 16: Aqueous Ionic Equilibria

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Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria – PowerPoint PPT presentation

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Title: Chapter 16: Aqueous Ionic Equilibria


1
Chapter 16 Aqueous Ionic Equilibria
  • Common Ion Effect
  • Buffer Solutions
  • Titrations
  • Solubility
  • Precipitation
  • Complex Ion Equilibria

2
Solubility Equilibria
  • All ionic compounds dissolve in water to some
    degree
  • We can apply the concepts of equilibrium to salts
    that dissolve to a small extent (insoluble)
  • We can use the equilibrium constant for the
    process to measure relative solubilities in water

3
Solubility Product
  • Equilibrium constant for the dissociation of a
    solid salt into its aqueous ions solubility
    product, Ksp
  • For an ionic solid MnXm, the dissociation
    reaction is
  • MnXm(s) ? nMm(aq) mXn-(aq)
  • The solubility product would be
  • Ksp MmnXn-m

4
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5
Molar Solubility
  • Solubility is the amount of solute that will
    dissolve in a given amount of solution at a
    particular temperature (in grams or moles)
  • The molar solubility is the number of moles of
    solute that will dissolve in a liter of solution.
  • The molarity of the dissolved solute in a
    saturated solution.

6
Calculating Ksp from Solubility 1. Copper (I)
bromide has a measured solubility of 2.0 x 10-4 M
at 25C. Calculate Ksp. 2. Calculate Ksp for
bismuth sulfide (Bi2S3) which has a solubility of
1.0 x 10-15
7
Calculating Solubility from Ksp The Ksp for
copper (II) hydroxide is 1.6 x 10-19 at 25C.
Calculate the solubility, Cu2, OH-.
8
Ksp and Relative Solubility
  • Molar solubility is related to Ksp
  • However, you cannot always compare solubilities
    of compounds by comparing their Ksps
  • In order to compare Ksps, the compounds must
    have the same dissociation stoichiometry

9
The Effect of Common Ion on Solubility
  • Addition of a soluble salt that contains one of
    the ions of the insoluble salt, decreases the
    solubility of the insoluble salt
  • For example, addition of NaCl to the solubility
    equilibrium of solid PbCl2 decreases the
    solubility of PbCl2
  • PbCl2(s) ? Pb2(aq) 2 Cl-(aq)

addition of Cl- shifts the equilibrium to the left
10
  1. Calculate the solubility of Ag2CrO4 in pure water
    (Ksp 9.0 x 10-12)
  2. Calculate the solubility of Ag2CrO4 in a 0.1M
    solution of AgNO3

11
The Effect of pH on Solubility
  • For insoluble ionic hydroxides, the higher the pH
    (and the greater OH-, the lower the solubility
    of the ionic hydroxide
  • OH- acts as a common ion,
  • M(OH)n(s) ? Mn(aq) nOH-(aq)
  • For insoluble ionic compounds that contain anions
    of weak acids, the lower the pH, the higher the
    solubility
  • M2(CO3)n(s) ? 2 Mn(aq) nCO32-(aq)
  • H3O(aq) CO32- (aq) ? HCO3- (aq) H2O(l)

12
Precipitation
  • Precipitation will occur when the concentrations
    of the ions exceed the solubility of the ionic
    compound
  • If we compare the reaction quotient, Q, for the
    current solution concentrations to the value of
    Ksp, we can determine if precipitation will occur
  • Q Ksp, the solution is saturated, no
    precipitation
  • Q lt Ksp, the solution is unsaturated, no
    precipitation
  • Q gt Ksp, the solution would be above saturation,
    the salt above saturation will precipitate
  • Some solutions with Q gt Ksp will not precipitate
    unless disturbed these are called
    supersaturated solutions

13
a supersaturated solution will precipitate if a
seed crystal is added
precipitation occurs if Q gt Ksp
14
Selective Precipitation
  • A solution containing several different cations
    can often be separated by addition of a solution
    that will form an insoluble salt with one of the
    ions, but not the others

15
If 10.0 mL of 0.50 M Pb(NO3)2 and 20.0 mL 1.0 M
NaI are mixed, will a precipitate form?
16
Chapter 18Electrochemistry
17
Redox Reaction
  • Elements change oxidation number
  • e.g., single displacement, and combustion, some
    synthesis and decomposition
  • Oxidation--oxidation number increases
  • Reduction--oxidation number decreases
  • Both must occur in a reaction--two half reactions
  • oxidizing agent is reactant molecule that causes
    oxidation
  • contains element reduced
  • reducing agent is reactant molecule that causes
    reduction
  • contains the element oxidized

18
Rules for Assigning Oxidation States
  • Free elements have an oxidation state 0
  • Monatomic ions have an oxidation state equal to
    their charge.
  • The sum of the oxidation states of all the atoms
    in a compound is 0.
  • The sum of the oxidation states of all the atoms
    in a polyatomic ion equals the charge on the ion.
  • The oxidation number of fluorine is always -1 in
    compounds with other elements.

19
Rules for Assigning Oxidation States
  • 6. Chlorine, bromine and iodine always have
    oxidation numbers of -1 except when bonded to O
    or F.
  • 7. The oxidation number of oxygen is almost
    always -2 the oxidation number of hydrogen is
    almost always 1.
  • Exceptions
  • --When oxygen is in the form of a peroxide
    (O22-), the oxidation number is -1.
  • --When hydrogen forms a binary compound with a
    metal, the oxidation number is -1 and the
    compound is called a hydride.

20
Oxidation and Reduction
  • Oxidation occurs when an atoms oxidation state
    increases during a reaction
  • Reduction occurs when an atoms oxidation state
    decreases during a reaction

Reducing agent
Oxidizing agent
CH4 2 O2 ? CO2 2 H2O -4 1
0 4 2 1 -2
21
Identify the element that is oxidized and the
element that is reduced in each of the following
reactions. What is the oxidizing and the
reducing agent in each reaction?
  • 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
  • MnO2 4 HBr MnBr2 Br2 2 H2O

22
Common Oxidizing Agents
23
Common Reducing Agents
24
Balancing Redox Reactions
  • 1. Assign oxidation numbers
  • --determine element oxidized and element reduced
  • 2. Separate the reaction into oxidation and
    reduction half- reactions.
  • 3. Balance half-reactions by mass
  • a. First balance elements other than H and O
  • b. Balance O using H2O
  • Balance H using H
  • 4. Balance each half-reaction by charge by
    adding electrons to the reactants side of the
    reduction and the product side of the oxidation.
  • 5. Multiply half-reactions by integers to make
    electrons the same in both half-reactions
  • Add half-reactions and cancel the electrons to
    produce a balanced equation.
  • For reactions that occur in acidic solutions,
    skip to step 9.
  • For reactions that occur in basic solutions, add
    the same of OH- as H to both sides of the
    equation.
  • 9. Check that reaction is balanced for mass and
    charge.

25
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