Aqueous-solution reactions by rt3463df

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									        Aqueous-solution Reactions
   Classify a reaction by
   Homogeneous chemical reactions:
           gas phase
                   aqueous-solution (common occurrence)
   Heterogeneous (more than one phase) chemical reactions:
           gas / liquid
           gas / solid
           liquid / solid
                            aqueous solution reactions       1
Know the meaning of terms
   Nature of Aqueous Solutions
Nature of compounds
      molecular substances (polar, non-polar, H-bonding)
      ionic substances (acids, bases, salts)
              strong electrolytes (completely ionized in solution)
              week electrolytes (not completely ionized in solution)

                  Know your terms and species (in the solution)
                         aqueous solution reactions                2
        Dissolving a Strong Electrolyte

See them in your imaginationaqueous solution reactions   3
aqueous solution reactions   4
               Strong Electrolytes
Strong acids: HNO3, H2SO4, HCl, HClO4
Strong bases: MOH (M = Na, K, Cs, Rb etc)
Salts: All salts dissolving in water are completely ionized. The ions
may react with water (to be discussed in Chem 123)

                           Stoichiometry & concentration relationship
                           NaCl (s)  Na+ (aq) + Cl– (aq)
                           Ca(OH)2 (s)  Ca+(aq) + 2 OH– (aq)
                           AlCl3 (s)  Al3+ (aq) + 3 Cl– (aq)
                           (NH4)2SO4 (s)  2 NH4 + (aq) + SO42– (aq)
                           aqueous solution reactions                   5
             Concentration of Ions
A bottle labeled as 0.100 M Al2(SO4)3.
        [Al3+] = _____ M (mol / L)
        [SO42–] = _____ M
Assume sea water is 0.438 M NaCl, 0.0512 M MgCl2, and 0.001 M CaCl2
        [Na+] = _____ M
        [Mg2+] = _____ M
        [Ca2+] = _____ M
        [Cl–] = _____ M
                                     Know how to calculate your quantities

                            aqueous solution reactions                  6
When ions form a solid that is not very soluble, a solid is formed. Such
a phenomenon is called precipitation.
The formation of a precipitation is also an equilibrium phenomenon (a
subject to be covered in Chem123)
  Ag+ (aq) + Cl– (aq)  AgCl(s)
  AgCl(s)  Ag+ (aq) + Cl– (aq)                 Ksp = [Ag+][Cl–] is a constant
                                                 the solubility product

                           aqueous solution reactions                        7
               Precipitation Reactions
                         Heterogeneous Reactions
                      Spectator ions or bystander ions

Ag+ (aq) + NO3– (aq) + Cs+ (aq) + I– (aq)  AgI (s) + NO3– (aq) + Cs+ (aq)
                Ag+ (aq) + I– (aq)  AgI (s) (net reaction)
                            Ag+ + I– AgI (s)

Soluble ions                                               Mostly insoluble
                         Mostly soluble ions               Silver halides
Alkali metals, NH4+
nitrates, ClO4-,         Halides, sulfates                 Metal sulfides, hydroxides
acetate                                                    carbonates, phosphates
                              aqueous solution reactions                        8
             Acid-base Reactions
            HCl (g)  H+ (aq) + Cl– (aq)
       NaOH (s)  Na+ (aq) + OH– (aq)
neutralization reaction: H+ (aq) + OH– (aq)  H2O (l)

Explain these reactions
Mg(OH)2 (s) + 2 H+ Mg2+ (aq) + 2 H2O (l)
CaCO3 (s) + 2 H+ Ca2+ (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2 HC2H3O2  Mg2+ (aq) + 2 H2O (l) + 2 C2H3O2 – (aq)
                acetic acid

                          aqueous solution reactions           9
  Oxidation-reduction reactions
Oxidation reaction must be accompanied by reductions
        – redox reactions
Increasing oxidation state is oxidation (loss e–, LEO)
Decreasing oxidation state is reduction (gain e–, GER)

What elements are oxidized and reduced in each reaction?
Work out the oxidation state changes for them as well!
2 KClO3  2 KCl + 3 O2
Fe2O3 + 3 CO  2 Fe + 3 CO2
MnO2 + 4 H+ + 2 Cl–  Mn2+ + 2 H2O + Cl2
                         aqueous solution reactions        10
           Review Oxidation States
   Oxidation states is an assigned number.
   Formal charge concept may be used to assign oxidation states.
   Work out the oxidation states of all elements in these species:
   NH3     N2H4        NH2OH N2    N2O                NO       NO2– NO2 NO3–
   Cl–     Cl2         ClO–  ClO2– ClO2               ClO2–    ClO3– ClO4–
   CO       H2C2O4 C2O42– C2H6 CH4                    CO2      CO32–
   PH3     P4          H3PO4 PO43–
   H2S     HS–         S2–     S6         SO2         SO3–     SO3     S2O32– SO42–
   MnO2 KMnO4 MnO4–            K2CrO4 CrO4–               K2Cr2O7 Cr2O7–

Review stoichiometry         aqueous solution reactions                      11
                        Half Reactions
        These reactions explained during the lecture:
            Zn = Zn2+ + 2 e–
                    Cu2+ + 2 e– = Cu
        net (electron transfer)
             Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s)

         For these half reactions,
             Zn = Zn2+ + 2 e–
                   2 H+ + 2 e– = H2
         get and explain the net reaction yourself

                                aqueous solution reactions   12
Explain how reactions proceed
          Balance Half Equations
1. Identify the key element that undergoes an oxidation state change.
2. Balance the number of atoms of the key element on both sides.
3. Add the appropriate number of electrons to compensate for the
   change of oxidation state.
4. Add H+ (in acid medium), or OH- (in basic medium), to balance the
   charge on both sides of the half-reactions; and H2O, if necessary, to
   balance the equations.                           Page 9 of handout

        MnO4 – + ____ + _____  Mn2+ + __ H2O

                           aqueous solution reactions               13
           Balance these half equations
   Zn (s)  Zn2+(aq)                    I will illustrate the balance of these
                                          half reaction equations and those
   Cu2+ (aq)  Cu (s)
                                          equations in other slides during the
   H2(g)  H+ (aq)                      lecture. If you are not at my lecture,
                                          you should practice their balance to
   I –(aq)  I2
                                          acquire your skills.
   Fe2+ (aq)  Fe3+ (aq)
                                          The textbook gives a slight different
   SO32– (aq) SO42– (aq)                method to balance redox equations
                                          and please find out the difference.
   2 S2O32–(aq)  S4O62–(aq) + 2e–1
                                          Both ways wok. You may use either
   Cr2O72–(aq)  Cr3+ (aq)                method.

                              aqueous solution reactions                     14
Work on these from time to time
      Balance Redox Reaction equations
    Add two half reaction equations so that you can cancel all electrons
    to obtain a balanced redox reaction equation
    H2O2 + I –  I2 + H2O
    MnO4 – + H2O2  Mn2+ + O2
    MnO4 – + SO3 2– Mn2+ + SO42–
    MnO4 – + Fe2+  Mn2+ + Fe3+
    Cr2O7 2– + UO2+  Cr3+ + UO22+

Work on these from
time to time to
refresh your skills.         aqueous solution reactions                15
        Disproportionation Reactions
    Balance disproportionation reaction (the same substance is both
    oxidized and reduced)
    H2O2  2 H2O + O2
    S2O32– SO42– + S (s)
    S2O32– SO2 + S (s)

                               aqueous solution reactions             16
Work on these from time to time
                   Analyze and Learn the Skills
Analyze this example and learn the skills to help you overcome any difficulty.
Task:      Balance the equation: S2O32– SO42– + S (s)
Identify the element oxidized and reduced: oxidation states of S in S2O32– SO42– and S are
+4, +6, and 0 respectively. S is both oxidized and reduced.
The oxidation half reaction:   S2O32– SO42– + 8 e– (2 S from +2  +6, (2×4=8))
The reduction half reaction:   S2O32– + 4 e – (2S from +2  0, (2×2=4)) S
Balance the charge with H+     S2O32–SO42– + 8 e– + 10 H+ (both sides have 2-)
                               S2O32– + 4 e – + 6 H+ S              (both sides
have 0)
        Add water to balance: S2O32– + 5 H2O SO42– + 8 e– + 10 H+
                              S2O32– + 4 e – + 6 H+ S + 3 H2O
        Make # of e the same: S2O32– + 5 H2O SO42– + 8 e– + 10 H+
                              2 S2O32– + 8 e – + 12 H+ 4 S + 6 H2O
    The balanced equation: 3 S2O32– + 2 H+ SO42– + 4 S + 10 H+
                                      much solution reactions
Make sure you fill in the details too aqueousto be included here!!!                 17
         Balance Redox Reaction in
              Basic Solutions
  Redox reactions may have different products depends on the acidity
  (pH) of the solutions.
  In basic solutions, there are more OH– ions than H+ ions. Thus, it is
  sensible to have OH– appearing in the equations than to have H+ ions.
  Balance these reactions in a basic solution:
  MnO4 – + CN –  MnO2 + OCN –
  Any one of several ways to assign oxidation states for CN works.
  MnO4 – + SO32–  MnO2 + SO42–

                              aqueous time to polish
Practice balance these from time tosolution reactions you skills!      18
   Oxidizing and Reducing Agents
Oxidizing agent or oxidant such as O2 or F2 is a substance that is able
to oxidize other substances. It is reduced in the process (gains
electron or decreases oxidation state).
Describe reducing agent or reductant

NH3 N2H4      NH2OH N2         N2O         NO          NO2 –   NO2   NO3 –
                       Reductant                                     Species
Cannot be                                                            Cannot be
reduced                                                              oxidized
further                                                              further
                          aqueous solution reactions                         19
   Titration is a method used in volumetric analysis. The addition of a
   solution is carefully controlled so that stoichiometric amounts can be
   read from a burette.
   Titration can be carried out for instantaneous or rapid reactions such
   as neutralization and oxidation reactions.

  Explain these terms:
      quivalence point,
half equivalence point,
                              aqueous solution reactions                20
             Conductance in Titration
  Conductance measurement of a HCl
  solution titrated by NaOH is shown:              conductance
  measured (total) conductance
  area due to the ions labeled
  Do all ions have the same
  Why or why not?
  Why does the total conductance                            H+                 OH–
  Is conductance of an ion depends on                                    Na+
  the concentration?                                               Cl–
                                                                 V of NaOH added
                               aqueous solution reactions                        21
Explain physical properties of chemicals
  Stoichiometry in Solution Chemistry
For titration calculations, the amount of reactant m is evaluated from
the concentration C and volume V by
For example, when m1 amount of acid is neutralized by m2 amount of
base, m1 = m1. For redox reactions, similar relationship can also be
used, but the stoichiometric relationship should be kept in mind.

         Amount in mmol = C in M * V in mL

                          aqueous solution reactions                22
                  Volumetric Analysis
 A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH
 solution to reach the equivalence point. What is the concentration of
 acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?
 Solution:                                                      If you find this solution difficult
 Net reaction:      OH– + HC2H3O2  H2O + C2H3O2 – (aq)         to understand, use your own
                                                                method to solve it

38.08 mL OH–     0.1000 mol      1 mol HC2H3O2       60.05 g HC2H3O2        1 mL vinegar
5 mL vinegar      1000 mL          1 mol OH–          1 mol HC2H3O2        1.01 g vinegar

= 0.0453 HC2H3O2 in vinegar
                                   The vinegar has 4.53 % of acetic acid by mass.
= 4.53 % HC2H3O2 in vinegar

                            a 7% vinegar?
What’s the concentration of aqueous solution reactions                                 23
Another way of thinking

A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH
solution to reach the equivalence point. What is the concentration of
acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?
Solution: a slight variation and hope you find it easier to follow
Net reaction:     OH– + HC2H3O2  H2O + C2H3O2 – (aq)

                  0.1000 mol     1 mol HC2H3O2               60.05 g HC2H3O2
  38.08 mL OH–
                   1000 mL         1 mol OH–                  1 mol HC2H3O2

  5 mL vinegar    1.01 g vinegar
                   1 mL vinegar
                                                                      Mass of acetic acid
                              4.53 % ofHC2H3O2 in vinegar =
                                                                       Mass of sample
= 0.0453 HC2H3O2 in vinegar
= 4.53 % HC2H3O2 in vinegar
                                aqueous solution reactions                              24
    What the concentration of a 7% vinegar?
          Chemical Analysis Application
How much 0.1000 M KMnO4 solution is required to reach the equivalence
point in a titration of 1.00 g oxidize oxalic acid (oa = H2C2O4.2H2O)?
Solution:         Redox reaction:
2 MnO4–   + 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2                 (balanced? Chk pls)

             1 mol oa   2 mol MnO4         1000 mL
 1.00 g oa                                                   = 31.75 mL MnO4–
             126 g oa    5 mol oa        0.1000 mol

       Molar mass of    Note mole
       H2C2O4.2H2O     ratio in the
       = 126 g mol-1    balanced

Do problem 5.97                 aqueous solution reactions                          25
                                   Back Titration
  A 1.00-g sample containing MnO2 dissolved in solution is treated with 2.00 g
  of oxalic acid (oa = H2C2O4.2H2O). Then 50.00 mL 0.1000 M KMnO4 is
  required for the titration of the excess oa. What is the % MnO2 by mass?
  Solution:          Redox reaction:
  2 MnO4–    + 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2
  H2C2O4 + MnO2 + 2 H+ = Mn2+ + 2 H2O + 2 CO2                       (balanced? Chk pls)

                       0.1000 mol MnO4          5 mol oa            126 g oa
   50.00 mL MnO4                                                                  = 1.575 g oa
                            1000 mL           2 mol MnO4            1 mol oa

                          86.9 g MnO2            100 %
(2.000 - 1.575) g oa                                              = 29.0 % MnO2 by mass
                            126 g oa         1.0 g Sample

Cool head 4 complicated problem, eh! aqueous solution reactions                            26
                              Review 1
    Sufficient amount of AgNO3 is placed in a 10.00 mL of tab water, and
    the AgCl solid is filtered and dried. The solid weighs 0.123 g. What is
    the concentration of chloride ion?
    Solution:        access and limiting reagent, and ppt

   0.123 g AgCl      1 mol AgCl           1 mol Cl-          1000 mL
     10.00 mL       143.4 g AgCl         1 mol AgCl            1L

   = 0.0858 mol / L of Cl–

                                aqueous solution reactions                27
Express [Cl-] in mol and mass %.
                               Review 2
    A 0.1234-g sample (S) NaCl and sugar mixture contains 40% NaCl.
    When dissolved in water, the solution is treated with AgNO3. How
    much dry AgCl is the theoretical yield?
    Solution:         percentage analysis

                      40 g NaCl        1 mol AgCl              143.4 g AgCl
    0.1234 g S         100 g S         58.5 g NaCl              1 mol AgCl

    = 0.1323 g AgCl           theoretical yield

                                  aqueous solution reactions                  28
Use the factors to help you think!
                             Review 3
  A 0.2345-g sample (S) NaCl and CaCl2 mixture contains 40% NaCl.
  When dissolved in water, the solution is treated with AgNO3. How
  much dry AgCl is the theoretical yield?
                         143.4 g AgCl
  0.2345*0.40 g NaCl                              = 0.2299 g AgCl
                         58.5 g NaCl

                        2*143.4 g AgCl
 0.2345*0.60 g CaCl2                              = 0.3668 g AgCl
                         110 g CaCl2

                  (0.2299 + 0.3668) g AgCl = 0.5967 g AgCl

Solve the same for a mixture of NaCl and KCl.
                              aqueous solution reactions             29
Find the percentage of NaCl if 0.5000 g AgCl is collected.
                                   Review 4
    When 0.2345-g NaCl and CaCl2 mixture is dissolved in water, the
    solution is treated with AgNO3. The mass of dry AgCl collected is
    0.5967 g. What is the percentage of NaCl in the mixture?
    Solution: Assume the sample contains x fraction of NaCl
    0.2345 (x) g NaCl 143.4 g AgCl
                       58.5 g NaCl

    + 0.2345 (1 – x) g CaCl2         2*143.4 g AgCl
                                                                 = 0.5967 g AgCl
                                      110 g CaCl2

    Solve for x in the equation: x = 0.40 = 40%

Solve the same problem for 0.6000 instead of 0.5967 g AgCl.
                                    aqueous solution reactions                     30
In this problem, what are the min. and max. masses of AgCl? (0.5748 – 0.6114 g)
                      Review 5 - skills
obtain empirical formula from composition and combustion experiment
find molecular formula from empirical formula and molar mass
evaluate theoretical yield in a reaction, identify the limiting reagent
Identify limiting reagent and evaluation percent yield in a reaction
predict amounts of products or reagents required in reactions
find percent yield of a reaction but pay attention to limiting reagent
give concentration of common ions in a solution of several salts
calculate masses of reagents or products (AgCl) involving common ions
find percentage of a compound in a mixture by titration or precipitation

                               aqueous solution reactions                  31
                  Review 6 – skill 2
Identify oxidation states and recognize reagent oxidized
balance redox reaction equations
evaluate quantities of reagents or products involving redox reactions
stoichiometry calculation in titration experiments paying attention to molar
relationship such as H3PO4, Ca(OH)2, as well as redox reactions
evaluate quantities involving isotope composition of elements
calculate one of density, mass and volume in solid, liquid, and gas
name inorganic compounds
write proper net ionic equation of reaction (know your spectator ions)

                             aqueous solution reactions                  32
     TA Hours & Rooms – fall 03
The Tutors handle the tutorial periods for CHEM 120/121 but they can
also provide assistance to individuals or small groups on a drop-in
basis according to the following schedule. (i.e. One of the Tutors will be
in each room for the specified period. You can drop in at any time
during that period to get some help.)
        Day              Time                Room*
        Monday           12:30-1:20          PHY 313
        Monday           1:30-2:20           MC 4062
        Monday           2:30-3:20           PHY 150
        Wednesday        12:30-3:20          PHY 313
        Wednesday        1:30-2:20           MC 4060
* These rooms are for one-on-one tutorial, not tutors office.
                            aqueous solution reactions                33

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