Chapter 21 - Neutralization

							  Chapter 21
“Neutralization”
       Section 21.1
  Neutralization Reactions
 OBJECTIVES:

 –Explain how acid-base
  titration is used to calculate
  the concentration of an acid or
  a base.
       Section 21.1
  Neutralization Reactions
 OBJECTIVES:

 –Explain the concept of
  equivalence in neutralization
  reactions.
     Acid-Base Reactions
 Acid + Base  Water + Salt
 Properties related to every day:
  –antacids depend on neutralization
  –farmers use it to control soil pH
  –formation of cave stalactites
  –human body kidney stones from
    insoluble salts
        Acid-Base Reactions
 NeutralizationReaction - a reaction
 in which an acid and a base react in
 an aqueous solution to produce a
 salt and water:
  HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
  H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2 H2O(l)
  – Table 21.1, page 614 lists some salts
              Titration
          is the process of adding
 Titration
  a known amount of solution of
  known concentration to
  determine the concentration of
  another solution
 Remember? - a balanced
  equation is a mole ratio
 Sample Problem 21-1, page 616
             Titration
 The concentration of acid (or base)
 in solution can be determined by
 performing a neutralization reaction
  –An indicator is used to show
   when neutralization has occurred
  –Often use phenolphthalein-
   colorless in neutral and acid;
   turns pink in base
Steps - Neutralization reaction
#1. A measured volume of acid of
  unknown concentration is added to
  a flask
#2. Several drops of indicator added
#3. A base of known concentration is
  slowly added, until the indicator
  changes color; measure the volume
 Figure 21.4, page 617
           Neutralization
 Thesolution of known
 concentration is called the
 standard solution
  – added by using a buret
         adding until the indicator
 Continue
 changes color
  – called the “end point” of the titration
 Sample   Problem 21-2, page 618
          Equivalents
 One mole of hydrogen ions reacts
 with one mole of hydroxide ions
 – does not mean that 1 mol of any acid
   will neutralize 1 mol of any base
 – because some acids and bases can
   produce more than 1 mole of
   hydrogen or hydroxide ions
 – example: H2SO4(aq)  2H+ + SO42-
           Equivalents
 Made   simpler by the existence of a
  unit called an equivalent
 One equivalent (equiv) is the
  amount of acid (or base) that will
  give 1 mol of hydrogen (or
  hydroxide) ions
  – 1 mol HCl = 1 equiv HCl
  – 1 mol H2SO4 = 2 equiv H2SO4
           Equivalents
   any neutralization reaction, the
 In
 equivalents of acid must equal the
 equivalents of base
  – How many equivalents of base are in
    2 mol Ca(OH)2?
 Themass of one equivalent is it’s
 gram equivalent mass (will be less
 than or equal to the formula mass):
  HCl = 36.5 g/mol; H2SO4 = 49.0 g/mol
           Equivalents
 Sample Problem 21-3, page 620
 Sample Problem 21-4, page 620
          Normality (N)
 Itis useful for us to know the
  Molarity of acids and bases
 Often more useful to know how
  many equivalents of acid or base a
  solution contains
 Normality (N) of a solution is the
  concentration expressed as
  number of equivalents per Liter
              Normality (N)
 Normality  (N) = equiv/L
 equiv = Volume(L) x N; and also
  know: N=M x eq; M = N / eq
 Sample Problem 21-5, page 621
 Diluting solutions of known Normality:
          N1 x V1 = N2 x V2
 N1 and V1 are initial solutions
 N2 and V2 are final solutions
              Normality (N)
 Titration  calculations often done
  more easily using normality instead
  of molarity
 In a titration, the point of
  neutralization is called the
  equivalence point
  – the number of equivalents of acid and
    base are equal
          Normality (N)
 Doing titrations with normality use:
  NA x VA = NB x VB
 Sample Problem 21-6, page 623
 Sample Problem 21-7, page 623
 Sample Problem 21-8, page 624
       Section 21.2
      Salts in Solution
 OBJECTIVES:
 –Demonstrate with equations
  how buffers resist changes in
  pH.
       Section 21.2
      Salts in Solution
 OBJECTIVES:

 –Calculate the solubility
  product constant (Ksp) of a
  slightly soluble salt.
          Salt Hydrolysis
A  salt is an ionic compound that:
   –comes from the anion of an acid
   –comes from the cation of a base
   –is formed from a neutralization
    reaction
   –some neutral; others acidic or basic
 “Salt hydrolysis” - a salt that reacts
  with water to produce acid or base
          Salt Hydrolysis
    Hydrolyzing salts usually from:
    1. strong acid + weak base, or
    2. weak acid + strong base
    Strong refers to the degree of
     ionization
    How do you know if it’s strong?
    – Refer to handout provided
        Salt Hydrolysis
 Tosee if the resulting salt is
 acidic or basic, check the
 “parent” acid and base that
 formed it:
  HCl + NaOH 
  H2SO4 + NH4OH 
  CH3COOH + KOH 
             Buffers
 Buffers are solutions in which the
 pH remains relatively constant
 when small amounts of acid or
 base are added
 –made from a pair of chemicals:
  a weak acid and one of it’s
  salts; or a weak base and one
  of it’s salts
              Buffers
A  buffer system is better able to
  resist changes in pH than pure water
 Since it is a pair of chemicals:
   –one chemical neutralizes any acid
    added, while the other chemical
    would neutralize any additional
    base
   –AND, they produce each other in
    the process!!!
             Buffers
 Example:   Ethanoic (acetic) acid
  and sodium ethanoate (also called
  sodium acetate)
 Examples on page 628 of these
 The buffer capacity is the amount
  of acid or base that can be added
  before a significant change in pH
                Buffers
 Buffers that are crucial to maintain the
  pH of human blood:
   1. carbonic acid (H2CO3) & hydrogen
    carbonate (HCO31-)
   2. dihydrogen phosphate (H2PO41-) &
    monohydrogen phoshate (HPO42-)
 Table 21.2, page 629 has some
  important buffer systems
 Sample Problem 21-9, page 630
   Solubility Product Constant
 Salts differ in their solubilities
  –Table 21.3, page 631
 Most “insoluble” salts will
  actually dissolve to some
  extent in water
  –said to be slightly, or
   sparingly, soluble in water
  Solubility Product Constant
 Consider: AgCl(s)  Ag+(aq) + Cl-(aq)
 The “equilibrium expression” is:



         [ Ag+ ] x [ Cl- ]
   Keq =
            [ AgCl ]
  Solubility Product Constant
 But, the [ AgCl ] is constant as long
  as some undissolved solid is
  present
 Thus, a new constant is developed,
  and is called the “solubility product
  constant” (Ksp):
Keq x [ AgCl ] = [ Ag+ ] x [ Cl- ] = Ksp
  Solubility Product Constant
 Values  of solubility product
  constants are given for some
  common slightly soluble salts in
  Table 21.4, page 632
 Although most compounds of Ba are
  toxic, BaSO4 is so insoluble that it is
  used in gastrointestinal
  examinations by doctors! - p.632
   Solubility Product Constant

 To  solve problems: a) write
  equation, b) write expression, and
  c) fill in values
 Sample Problem 21-10, page 634
 Sample Problem 21-11, page 634
           Common Ion Effect
   A “common ion” is an ion that is
    common to both salts in solution
     – example: You have a solution of lead
       (II) chromate. You now add some
       lead (II) nitrate to the solution.
        • The lead is a common ion
     – This causes a shift in equilibrium (due
       to Le Chatelier’s principle), and is
       called the common ion effect
       Common Ion Effect
 Sample   Problem 21-12, page 636
 The solubility product constant (Ksp)
  can be used to predict whether a
  precipitate will form or not:
  – if the ion-product concentration is
    greater than the allowed Ksp, then a
    precipitate will form
 Sample   Problem 21-13, page 637