Chapter 19 - Neutralization

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Chapter 19 - Neutralization
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Section 19.1Neutralization Reactions

• OBJECTIVES
• Explain how acid-base titration is used to
  calculate the concentration of an acid or a base

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Section 19.1Neutralization Reactions

• OBJECTIVES
• Explain the concept of equivalence in
  neutralization reactions.

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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

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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 19.1, page 458 lists some salts

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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
• Sample Problem 19-1, page 460

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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

Simulation
Simulation Worksheet
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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 19.4, page 461

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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 19-2, page 461

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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-

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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

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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

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Equivalents

• Sample Problem 19-3, page 462
• Sample Problem 19-4, page 462

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Normality (N)

• Useful 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

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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

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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

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Normality (N)

• Doing titrations with normality use NA x VA
  NB x VB
• Sample Problem 19-6, page 464
• Sample Problem 19-7, page 464
• Sample Problem 19-8, page 464

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Section 19.2 Salts in Solution

• OBJECTIVES
• Demonstrate with equations how buffers resist
  changes in pH

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Section 19.2 Salts in Solution

• OBJECTIVES
• Calculate the solubility product constant (Ksp)
  of a slightly soluble salt

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Salt Hydrolysis

• A salt
• comes from the anion of an acid (Cl-)
• comes from the cation of a base (Na)
• formed from a neutralization reaction
• some neutral others acidic or basic
• Salt hydrolysis- salt reacts with water to
  produce acid or base solution

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Salt Hydrolysis

• Hydrolyzing salts usually made from
• strong acid weak base, or
• weak acid strong base
• Strong refers to the degree of ionization (100)
• What pH will result from the above
• combinations?

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Salt Hydrolysis

• To see if the resulting salt is acidic or basic,
  check the parent acid and base that formed it
• NaCl HCl NaOH
• NH4OH H2SO4 NH4OH
• CH3COOK CH3COOH KOH

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Strong Acids HCl HClO4 H2SO4 HI HNO3 HBr
Strong Bases Mg(OH)2 NaOH Ca(OH)2 KOH
To determine if a salt is made From a combination
which is acid/base weak/weak
weak/strong strong/strong strong/weak
Na Cl
OH
H
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Lab 42 Salt Hydrolysis Universal Indicator
Colors
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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
• HA / A-
• or a weak base and one of its salts
• NH3 / NH4

Simulation
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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
• they make each other in the process!

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Unbuffered reaction between and acid an base
Buffered solution and reaction of an acid with a
base
HA / A-
HCl NaOH ? NaCl HOH
HCl

A-
HA

Cl-
pH
pH
Add strong acid
Add strong acid
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Buffers

• Example Ethanoic (acetic) acid and sodium
  ethanoate (also called sodium acetate)
• HC2H302 / NaC2H302 becomes
• HC2H302 / C2H302 1-
• Weak acid weak base
• The buffer capacity is the amount of acid or base
  that can be added before a significant change in
  pH

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Buffers

• Buffers that are crucial to maintain the pH of
  human blood
• carbonic acid - hydrogen carbonate
• H2CO3 / HCO3 -
• 2. dihydrogen phosphate - monohydrogen phoshate
• H2PO4- / HPO4 2-
• Table 19.2, page 469 has some important buffer
  systems
• Sample Problem 19-9, page 468

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Calculating ksp or Solubility Product
Constant Copy Example 10/11 pg 470/471 into
your notes What does a high Ksp mean? What does
a low Ksp mean?
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Solubility Product Constant

• Salts differ in their solubilities
• Table 19.3, page 470
• Most insoluble salts will actually dissolve to
  some extent in water
• said to be slightly, or sparingly, soluble in
  water

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Solubility Product Constant

• Consider AgCl(s)
• The equilibrium expression is

Ag(aq) Cl-(aq)
Ag x Cl-
Keq
AgCl
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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

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Solubility Product Constant

• Values of solubility product constants are given
  for some sparingly soluble salts in Table 19.4,
  page 471
• Although most compounds of Ba are toxic, BaSO4 is
  so insoluble that it is used in gastrointestinal
  examinations by doctors! - p.632

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Solubility Product Constant

• To solve problems
• a) write equation,
• b) write expression, and
• c) fill in values using x for unknowns
• Sample Problem 21-10, page 634
• Sample Problem 21-11, page 634

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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 (II) ion is the common ion

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Pb2 CrO42-
PbCrO4 ?
Add Pb(NO3)2

• This causes a shift in equilibrium (due to Le
  Chateliers principle), and is called the common
  ion effect

?shift
PbCrO4 ? Pb2 CrO42-
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Common Ion Effect

• Sample ProblemThe Ksp of silver iodide is
  8.3x10-17. What is the iodide concentration of a
  1.00L saturated solution of AgI to which 0.020
  mol of AgNO3 is added?
• 1. Write the equilibrium equation.
• AgI (s) Ag1 I1-
• 2. Write the Ksp expression Ksp
  Ag11 I1-1 8.3x10-17
• Ksp (x)1 (x)1
  8.3x10-17
• Ksp (x 0.020)1 (x)1
  8.3x10-17
• Ksp (0.020) (x)
  8.3x10-17
• x 4.2x10-15

Note X is So small that it can be ignored
The of iodide ion is 4.2x10-15 M
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• The solubility product constant (Ksp) can be used
  to predict whether a precipitate will form or not
  in a reaction
• if the calculated ion-product concentration is
  greater than the known Ksp, a precipitate will
  form

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ksp Sample
Problem A student prepares a solution by
combining 0.025 mol CaCl2 with 0.015 mol
Pb(NO3)2 in a 1 L container. Will a precipitate
form? WHAT WE KNOW!!!
X ppt
0.025 mol/L
0.015 mol/L
ksp Pb21 Cl1-2
The values we need to solve the problem!
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1. Calculate the concentration of the ions used to
   make the precipitate.

CaCl2
Ca 2
2Cl-

0.025 mol/L
2 mol Cl-
0.025 mol CaCl2
1L
1 mol CaCl2
0.015 mol Pb(NO3)2
1 mol Pb2
1L
1 mol Pb(NO3)2
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2) Calculate the Ksp for the precipitate
PbCl2 ? Pb2 2Cl-
Ksp Pb 21 Cl-2
Ksp ( 0.015 M)1 ( 0.050 M) 2
Ksp for PbCl2 3.75 x 10 -5
Calculated Ksp 3.75 x 10 -5 is gt Known Ksp 1.7
x 10 -5 for PbCl2
See pg 471 Table 19-4 for known values
Calc gt Known PPT Knowngt Calc No PPT Known
Calc No PPT
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