Reactions in Aqueous Solution class notes

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					Reactions in
Aqueous Solution
Chapter 4
 Why have a unit on (aq)?

 Almost two thirds of our planet is covered by
  water, and water is the most abundant
  substance in our bodies.
   Because water is so common, we tend to take its
    unique chemical and physical properties for
   Water possesses many unusual properties
    essential to support life on Earth.
 One of the most important properties of water is
  its ability to dissolve a wide variety of
   The water in nature, therefore, whether it is the
    purest drinking water from the tap or water from a
    clear mountain stream, invariably contains a variety
    of dissolved substances.
 Solutions in which water is the dissolving
  medium are called aqueous solutions.
 Many of the chemical reactions that take place
  within us and around us involve substances
  dissolved in water.
     Nutrients dissolved in blood are carried to our cells,
      where they enter into reactions that help keep us
     Automobile parts rust when they come into frequent
      contact with aqueous solutions that contain various
      dissolved substances.
      Spectacular limestone caves are formed by
       the dissolving action of underground water
       containing carbon dioxide, CO2(aq):
Reactions in Aqueous Solution

  Most of the reactions considered in Chapter 3 involved
   pure substances reacting with each other.
     Simple types of chemical reactions and how they are
  Most of the reactions that occur in the world involve
   ions or molecules dissolved in water (aqueous
  In addition, we will extend the concepts of
   stoichiometry that we learned in Chapter 3 by
   considering how solution concentrations can be
   expressed and used.
  Solution – homogeneous mixture
    Homogeneous – the same
    Heterogeneous - different

  Solute – what is being dissolved
    Smaller amount

  Solvent – what is doing the dissolving
    Larger amount
  The concentration of a solute in solution can be
   expressed in terms of molarity.

      molarity (M) =    moles of solute
                        Liter of solution

  The symbol [ ] is commonly used to represent
   the molarity of a substance in solution.
Molarity (cont)

  When an ionic solid dissolves in water,
   the cations and anions separate from one
    This process can be represented by a
     chemical equation in which the reactant is
     the solid and the products are the positive
     and negative ions in water solution.

  Ionic solids are commonly described as strong
     The ions they contain are good conductors of electricity in
      aqueous solutions.

  Strong electrolytes – soluble ionic compounds (those
   containing metals, nonmetals, and the ammonium ion)
  Weak electrolytes – partially soluble ionic compounds,
   ionizable molecular compounds (acetic acid)
  Nonelectrolytes – solids (insoluble ionic compounds),
   molecular compounds
 What is the molarity of a solution containing
  0.32 moles of NaCl in 3.4 liters?

 M = mols/Liters

 molarity =0.32 moles NaCl
               3.4 L

 = 0.094 M NaCl
 What is the molarity of a solution made by
  dissolving 2.5 g of NaCl in enough water to
  make 125 ml of solution?

 2.5 g NaCl x 1 mole NaCl
  -------------- -----------------
      1           58.5 g NaCl = 0.0427mole

 molarity =0.0427 mole NaCl
                0.125 L      = 0.34 M NaCl
  What do we use it for?
 Conversions:
   What mass of K3PO4 is required to prepare 4.00
    Liters of 1.50 M solution?

  4.00 L     1.50 mol        272 g
  -------- x ------------- x -------- = 1632 g K3PO4
    1        1 L             1 mol

  Dissolve 1632 g K3PO4 in enough water to make 4.00
   L of solution.
 How would you prepare 400. ml of 1.20 M
  solution of sodium chloride?
 Remember:
 1.20 M NaCl = 1.20 moles NaCl
                     1.00 L solution
 400. ml x   1 L solution x 1.20 moles NaCl
     1        1000 ml        1.00 L solution
   = 0.480 moles NaCl
 0.480 moles NaCl x 58.5 g NaCl
         1           1 mole NaCl
   =28.1 g NaCl

   Dissolve 28.1 g NaCl in enough water to make 400
    mL of solution.

  How would you prepare 0.125 M NaOH

    molarity = moles

    0.125 moles NaOH    40 g NaOH
        1 L solution   1 mole NaOH

  5 g of NaOH dissolved in 1 L of Water
  Similar to molarity – except mols per kg

  Abbreviated as m

  The concentration of a solute in solution can be
   expressed in terms of molality.

       molality (m) =       moles of solute
                            kg of solution

 Tell me if the following problems are
  molarity or molality problems, then solve
  for the unknown.
   What mass of water is required to dissolve
    100 g NaCl to prepare 1.50 m solution?
   What volume of 0.750 M solution can be
    prepared using 90.0 g of NH4Cl?
  Moles of ions
 What is the formula for:
   Magnesium chloride
   Sodium phosphate

 When these dissolve how many ions are
  present for each molecule above?

 How many moles of ions do you have?
Precipitation Reactions
  A precipitation reaction is a reaction which
   results in the formation of an insoluble product,
   or precipitate.

  A precipitation is an insoluble solid that
   separates form the solution.

  Precipitations reactions usually involve ionic
   compounds. They result from double
   displacement reactions.

  Know your solubility Rules!!!
Solubility Rules
 Group 1 ions, Hydrogen, and Ammonium
  form soluble salts
     Na NO3, NaF , K3PO4
     NH4F, (NH4)CO3

 NO3-    All nitrates are soluble.
     Na NO3, NH4 NO3 , Ca(NO3)2

 CH3COO- All acetates are soluble.
     Na CH3COO, NH4 CH3COO , Ca(CH3COO)2
 Solubility Rules

 Cl- All chlorides are soluble
      except AgCl, Hg2Cl2,
      CaCl2, NH4 Cl , Al Cl3
 Other Halides generally follow like Chloride
  except Fluoride
   CaI2 yes           CaF2 not
More Solubility Rules
 SO42- Most sulfates
        are soluble;
        exceptions include
        SrSO4, BaSO4,

  CuSO4, CaSO4, Na2SO4
 More Solubility Rules
 S2-   All sulfides are insoluble
        except those with Group
        1 & 2 elements, H+ and NH4+.

     Na2S, (NH4)2S, and CaS   soluble
     Fe2S3, Al2S3, and CuS    insoluble
More Solubility Rules
 OH-   All hydroxides are insoluble
        except those of the Group 1
        elements, Sr(OH)2 , Ba(OH)2 ,
        and Ca(OH)2 somewhat

     KOH, Sr(OH) 2         soluble
     Al(OH)3 and Fe(OH)3   insoluble
More Solubility Rules
 PO4-3, CO32-, and CO3-2
            All phosphates,
             carbonates, and sulfites
             are insoluble except those
             with Group 1 elements, H+
             or NH4+.

     Na2CO3 , (NH4)2CO3 , K2CO3   soluble
     CaCO3 , Hg2CO3 , Al2(CO3)3   insoluble
  Precipitation Reactions

 Example
   When lead (II) nitrate reacts with sodium iodide

   Molecular equation
   Pb(NO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaNO3 (aq)

   PbI2 is insoluble therefore forms a solid precipitate.

  Decide if the products produced are
   soluble or insoluble. If soluble, then no
   precipitate forms:

    CuSO4 and NaNO3
    Na2CO3 and CaCl2
 Decide if the products produced are
  soluble or insoluble. If soluble, then no
  precipitate forms:

   PbSO4 and RbNO3
   MgCO3 and KCl
Reactions in Aqueous Solution

  Since the chemical equations for these
   reactions involve ions in solution they are
   referred to as an ionic equation.
    An equation written showing reactions
     between ionic compounds/atoms
Net Ionic Equation

  A chemical equation for a reaction
   involving ions in which only those species
   that actually react are included.

    Atom balance- There must be the same
     number of atoms on both sides.
    Charge balance- There must be the same
     total charge on both sides.
  Net Ionic Equation
 Example: Molecular Equation
  Pb(NO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaNO3 (aq)

   Steps: 1 – break each aqueous item into its ions –
     leave solids and liquids together
  Pb+2 (aq) + NO3- (aq) + Na+ (aq) + I- (aq) → PbI2 (s) + Na+
  (aq) + NO3- (aq)

   Steps: 2 – use coefficients to balance the number of
     ions and the charges on each side of the arrow
    Ionic Equation
   Pb+2 (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 I- (aq) → PbI2   (s)   + 2 Na+
    (aq) + 2 NO3- (aq)
   Net Ionic Equation
   Steps: 3 – cancel out ions that appear in identical forms
     among both the reactants and products of an ionic equation

 Ionic Equation
 Pb+2 (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 I- (aq) → PbI2   (s)   + 2 Na+ (aq)
  + 2 NO3- (aq)
 Net ionic equation
Pb+2 (aq) + 2 I- (aq) → PbI2   (s)

What was cancelled out are called spectator ions
 - they are present during a reaction, but do not play a
 role in the reaction
 HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)

 H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  H2O (l) + Na+ (aq)
  + Cl- (aq)

 H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  H2O (l) + Na+ (aq)
  + Cl- (aq)

 H+ (aq) + OH- (aq)  H2O (l)
 1. Molecular: AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
 2. Molecular: Mg(NO3)2 (aq) + Na2CO3 (aq)  MgCO3 (s) + 2
  NaNO3 (aq)
 3. Molecular: strontium bromide (aq) + potassium sulfate (aq) 
  strontium sulfate (s) + potassium bromide (aq)
 4. Molecular: manganese(II) chloride (aq) + ammonium carbonate
  (aq)  manganese(II) carbonate (s) + ammonium chloride (aq)
 5. Molecular: chromium(III) nitrate (aq) + iron(II) sulfate (aq) 
  chromium(III) sulfate (aq) + iron(II) nitrate (aq)
 6. Molecular: K3PO4 (aq) + Al(NO3)3 (aq) 
  AlPO4 (s) + 3 KNO3 (aq)
 7. Molecular: BeI2 (aq) + Cu2SO4 (aq) 
  BeSO4 (aq) + 2 CuI (s)
 8. Molecular: Ni(NO3)3 (aq) + 3 KBr (aq) 
  NiBr3 (aq) + 3 KNO3 (aq)
 9. Molecular: 2 CoBr3 (aq) + 3 K2S(aq) 
  Co2S3 (s) + 6 KBr (aq)
 10. Molecular: barium nitrate + ammonium
  phosphate 
   Combination note taking hand-out (review)

 You need informal notes about the       •On this side, you need to show a
  following on this side:                 visual representation for your informal
    What does Aqueous mean?              notes.
    What are aqueous reactions?
    Why do we have a unit on
     reactions in aq solutions?
    What is a solution?
    What is the difference between
     homogeneous and heterogenous?
    What is the difference between a
     solvent and a solute?
    What is molarity?
    What is molality?
    What is an electrolyte?
    How can you tell is something is a
     strong electrolyte? Weak
     electrolyte? Or non-electrolyte?
    What is a precipitation reaction?
    What is an ionic equation?
    What is a net ionic equation?
 1. Write a balanced molecular equation for each
    of the following pairs of substances.
             BaCl2 with Na2SO4
             K3PO4 with Ca(NO3)2
             Cr(NO3)3 with NaOH
             NaCl with Mg(NO3)2
 2. Using your solubility chart predict if a
    precipitate will form or not.
 3. Write the balanced ionic equation.
 4. Cancel out spectator ions.
 5. Write the net ionic equation.
  Predict products (if needed)
  and write the net ionic
 Pb(NO3)2 (aq)+ 2 HCl (aq) PbCl2 (s) + 2 HNO3

 BaBr2 (aq) + Na2SO4 (aq) 
  Acid-base reactions
 We commonly encounter acids and bases in
  our every day life as can be seen by these
  household items.

 Most Acid-Base reactions are carried out in
Acid-Base Reactions
  Acids
    Acidic solutions have a sour taste.
    Vinegar, lemon juice, and soda are acidic.
    An acid is a species that produces H+ ions in water
  Bases
    Basic solutions have a slippery feeling
    Ammonia, most detergents and cleaning agents are
    A base is a species that produces OH- ions in water
   Acid-Base Reactions (cont)
 Two types – depends on how ionize in water

 Strong acids ionize completely, forming H+ ions and
    HCl (aq)                H+ (aq) + Cl - (aq)

 A weak acid is only partially ionized to H+ ions in
    HF (aq)                 H+ (aq) + F - (aq)
       Molecules containing an ionizable hydrogen atom

       General formula is HB (aq)            H+ (aq) + B - (aq)
Acid-Base Reactions (cont)

  Two types – depends on how ionize in

  Strong bases ionize completely, forming OH -
   ions and cations.
    NaOH (s)          Na+ (aq) + OH- (aq)

  A weak base produces OH - ions in a different
   way. They react with H2O molecules, acquiring
   H+ ions and leaving OH – ions behind.
    NH3 (aq) + H2O          NH4+ (aq) + OH - (aq)
  Common Strong Acids And Bases
All aids and bases other than those listed below can be considered as

         Acid    Name of Acid           Base       Name of Base
    HCl          Hydrochloric acid     LiOH        Lithium hydroxide

    HBr          Hydrobromic acid      NaOH        Sodium hydroxide

    HI           Hydroiodic acid       KOH         Potassium hydroxide

    HNO3         Nitric acid           Ca(OH)2     Calcium hydroxide

    HClO4        Perchloric acid       Sr(OH)2     Strontium hydroxide

    H2SO4        Sulfuric acid         Ba(OH)2     Barium hydroxide
 Weak bases

 A common class of weak bases includes
  organic molecules (containing carbon) known
  as the amines
   Amines – considered a derivative of ammonia in
    which one or more hydrogen atoms have been
    replaced by hydrocarbon groups

   CH3NH2 (aq) + H2O     CH3NH3+ (aq) + OH - (aq)
     methylamine
Equations for Acid-Base Reactions

  When an acidic water solution is mixed
   with a basic water solution, an acid-base
   reaction occurs.
  The nature of the reaction and the
   equation depend on whether the acid and
   base involved are strong or weak.
 Equations for Acid-Base Reactions
 Strong acid – strong base (neutralization
     Net ionic equation:

         H+ (aq) + OH – (aq)           H2O

         Example: HNO3 + NaOH  H2O (l) + NaNO3
             H+ (aq) + OH – (aq)  H2O
               Because strong acid, it is completely converted to H+ +
                  NO3 – ions
               Strong base , it is completely converted to Na+ and OH –
               When the solutions are mixed, the H+ and OH- ions react
                  with each other to form H2O molecules
Equations for Acid-Base Reactions
       Weak acid – strong base (2-step process)
         Net ionic equation:

            HB (aq)               H+ (aq) + B – (aq)
                Ionization of the HB molecule

            H+ (aq) OH – (aq)               H 2O
                Then, neutralization of the H+ ions by the OH- ions
         Overall equation: adding the two equations together
          and cancel what is the same on both sides of the
          arrow (spectator ions)

            HB (aq) + OH – (aq)             B – (aq) + H2O
Equations for Acid-Base Reactions

    Strong acid – weak base (2-step
        Net ionic equation:

            NH3 (aq) + H2O               NH4+ (aq) + OH – (aq)
                Reaction of NH3 with H2O to form products

            H+ (aq) OH – (aq)                    H2O
                Then, neutralization of the H+ ions by the OH- ions
        Overall equation

            H+ (aq) + NH3 (aq)                   NH4+ (aq)
 Table 4.2 summary of
Strong acid-    H+
strong base     OH- H+(aq) + OH–(aq)  H2O(l)

Weak acid –     HB
                      HB (aq) + OH – (aq)  B – (aq) + H2O (l)
strong base     OH-

Strong acid –   H+
weak base       B     H+(aq) + B(aq)  BH+(aq)
Relate this to electrolytes

  As said before – electrolytes are good
   conductors of electricity
    Therefore, strong acids and bases are
     strong electrolytes
    In contrast, weak acids and bases are poor
     conductors because their water solutions
     contain relatively few ions
       They are weak electrolytes

  Dilution – process f adding more solvent
   to solution to prepare a more dilute
  Concentrated solution – large amount of
  Dilute solution – small amount of solute

  M1V1 = M2V2

  How many mL of 16 M HNO3 will you use
   to prepare 750. mL of 0.69 M HNO3?
  How many mL of water will you use?
 What is the molarity of a solution
  prepared by diluting 125 ml of 0.400 M
  KCl to 875 mL of KCl solution?

  What is the difference between a strong
   and weak acid and base?
Acid-Base titrations

  Titration – measuring the volume of a
   standard solution (solution of known
   concentration) required to react with a
   measured amount of sample
    Objective – determine the point at which the
     reaction is complete (called the equivalence
       Reached when the number of moles of OH-
        added is exactly equal to the number of moles of
        acid originally present

  Your group will conduct an acid-base titration
    You will have one of the following:
         Strong acid - strong base
         Weak acid - strong base
         Weak base - weak acid
         Strong acid – weak base
    Your goal is to create a titration curve showing
     amount of base used versus the pH of the solution
   Record what acid and bases your group is using
   Take the initial pH of the acid in your beaker
   Slowly!!! Add base to the acid.
   Every 2 mL stop and record the pH.
   When the solution starts to show pink – add the base
    slower (drop by drop).
   Record the pH after every 10 drops this time.
   Make sure you note the exact amount of base used
    when the solution turned completely pink and stayed
   Then continue adding the base until gone.
   Stopping every 2 mL to stir and record pH.
For help with titration
Strong acid and strong base
Strong acid and weak base
weak acid and strong base
Weak acid and weak base
Calculating the molarity of
your acid
  The molarity of the bases were 1M
  Use this number to convert from the mL of
   base used to reach to equivalence point to
   moles of your base
  Since at your equivalence point the mols of
   base = mols of acid, the number of moles of
   base is the number of moles of acid present in
   your acid
  Now, divide by the initial volume of the acid
   (50mL) to determine what your molarity was
Titration simulation

  Qualitative analysis of an
  unknown solution of ions

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