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Chapter 21 “Neutralization”. Milbank High School. 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:

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section 21 1 neutralization reactions
Section 21.1Neutralization Reactions
    • Explain how acid-base titration is used to calculate the concentration of an acid or a base.
section 21 1 neutralization reactions3
Section 21.1Neutralization Reactions
    • Explain the concept of equivalence in neutralization reactions.
acid base 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 reactions5
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)

    • Table 21.1, page 614 lists some salts
  • 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
  • Sample Problem 21-1, page 616
  • 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
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
  • 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
  • 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-
  • 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
  • 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 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

  • Sample Problem 21-3, page 620
  • Sample Problem 21-4, page 620
normality n
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
normality n15
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 n16
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 n17
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
Section 21.2Salts in Solution
    • Demonstrate with equations how buffers resist changes in pH.
section 21 2 salts in solution19
Section 21.2Salts in Solution
    • Calculate the solubility product constant (Ksp) of a slightly soluble salt.
salt hydrolysis
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 hydrolysis21
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 it’s strong?
    • Refer to handout provided
salt hydrolysis22
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 


  • 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
  • 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!!!
  • 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 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
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 constant28
Solubility Product Constant
  • Consider: AgCl(s) Ag+(aq) + Cl-(aq)
  • The “equilibrium expression” is:

[ Ag+ ] x [ Cl- ]

[ AgCl ]

Keq =

solubility product constant29
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 constant30
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 constant31
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
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 effect33
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