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Neutralization

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Explain how acid-base titration is used to calculate the concentration of an acid or a base. ... human body kidney stones from insoluble salts. Acid-Base Reactions ... – PowerPoint PPT presentation

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Title: Neutralization


1
Neutralization
  • Pioneer High School
  • Mr. Gonzalez

2
Section 21.1Neutralization Reactions
  • OBJECTIVES
  • Explain how acid-base titration is used to
    calculate the concentration of an acid or a base.

3
Section 21.1Neutralization Reactions
  • OBJECTIVES
  • Explain the concept of equivalence in
    neutralization reactions.

4
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

5
Acid-Base Reactions
  • Neutralization Reaction - 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)

6
Titration
  • Titration is the process of adding a known amount
    of solution of known concentration to determine
    the concentration of another solution
  • Remember? - a balanced equation is a mole ratio

7
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

8
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

9
Neutralization
  • The solution of known concentration is called the
    standard solution
  • added by using a buret
  • Continue adding until the indicator changes color
  • called the end point of the titration
  • Sample Problem 21-2, page 618

10
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-

11
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

12
Equivalents
  • In any neutralization reaction, the equivalents
    of acid must equal the equivalents of base
  • How many equivalents of base are in 2 mol
    Ca(OH)2?
  • The mass of one equivalent is its gram
    equivalent mass (will be less than or equal to
    the formula mass)
  • HCl 36.5 g/mol H2SO4 49.0 g/mol

13
Equivalents
  • Sample Problem 21-3, page 620
  • Sample Problem 21-4, page 620

14
Normality (N)
  • It is 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

15
Normality (N)
  • Normality (N) equiv/L
  • equiv Volume(L) x N and also know NM 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

16
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

17
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

18
Section 21.2Salts in Solution
  • OBJECTIVES
  • Demonstrate with equations how buffers resist
    changes in pH.

19
Section 21.2Salts in Solution
  • OBJECTIVES
  • Calculate the solubility product constant (Ksp)
    of a slightly soluble salt.

20
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

21
Salt Hydrolysis
  • Hydrolyzing salts usually from
  • strong acid weak base, or
  • weak acid strong base
  • Strong refers to the degree of ionization
  • How do you know if its strong?
  • Refer to handout provided

22
Salt Hydrolysis
  • To see if the resulting salt is acidic or basic,
    check the parent acid and base that formed it
  • HCl NaOH ?
  • H2SO4 NH4OH ?
  • CH3COOH KOH ?

23
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 its salts or a weak base and one of its
    salts

24
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!!!

25
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

26
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

27
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

28
Solubility Product Constant
  • Consider AgCl(s) ? Ag(aq) Cl-(aq)
  • The equilibrium expression is
  • Ag x Cl-
  • AgCl

Keq
29
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

30
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

31
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

32
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
    Chateliers principle), and is called the common
    ion effect

33
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
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