Transcript Chapter Fifteen - Salina USD 305

1
Today…
• Turn in:
– Nothing
• Our Plan:
– Notes – Acids, Bases, & pH
– Worksheet #1
• Homework (Write in Planner):
– Worksheet #1 due next class
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
2
Chapter Fifteen
Acids, Bases, and
Acid–Base Equilibria
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Chapter Fifteen
3
Crash Course Introduction
• http://www.youtube.com/watch?v=LS67vS
10O5Y
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Chapter Fifteen
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Review - The Flow Chart
pH
pOH
• What is
pH + pOH = 14
the pH if
- log[H [OH
O ]
- log[OH ]
the
10
10
1]is 3.8 x
10-4 M?
3
+1
-1
-pOH
-pH
[H3O+1]
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[OH-1]
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Chapter Fifteen
5
Sig Figs with pH
• Significant figures with pH are unique!
• pH has as many places AFTER THE
DECIMAL as significant figures in the
concentration.
– Ex: If you have 3.4 x 10-3 M HCl, the pH is
2.47.
– Try it out: The calculator tells you the pH of
1.789 x 10-5 M HNO3 is 4.747389659. What
would you write as the pH?
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Chapter Fifteen
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Review
• See if you remember how to do the math.
Complete each problem.
1. [H3O+1] = 3.28 x 10-4, find pH
2. pH = 2.5, find [OH-1]
3. [OH-1] = 1.23 x 10-8, find [H3O+1]
4. pOH = 2.4, find pH
5. [H3O+1] = 6.980 x 10-3, find pOH
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Chapter Fifteen
7
Definitions of Acid & Base
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The Arrhenius Theory
• Arrhenius theory: an acid forms H+
(H3O+1) in water; and a base forms
OH– in water.
• But not all acid–base reactions involve
water, and many bases (NH3,
carbonates) do not contain any OH–…
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Chapter Fifteen
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The Brønsted–Lowry Theory
• Brønsted–Lowry theory defines acids and
bases in terms of proton (H+) transfer.
• A Brønsted–Lowry acid is a proton donor.
• A Brønsted–Lowry base is a proton
acceptor.
– The conjugate base of an acid is the acid
minus the proton it has donated.
– The conjugate acid of a base is the base plus
the accepted proton.
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Chapter Fifteen
10
Ionization of HCl
HCl acts as an
acid by donating
H+ to H2O
H2O is a base in this
reaction because it
accepts the H+
Conjugate
acid of H2O
Conjugate
base of HCl
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Chapter Fifteen
11
Ionization of Ammonia
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Chapter Fifteen
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Water Is Amphiprotic
H2O acts as an acid when
it donates H+, forming the
conjugate base ___
H2O acts as a base when it
accepts H+, forming the
conjugate acid ___
Amphiprotic: Can act as either an acid or as a base
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Chapter Fifteen
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Example 15.1
Identify the Brønsted–Lowry acids and bases and their
conjugates in:
(a) H2S + NH3
NH4+ + HS–
(b) OH– + H2PO4–
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H2O + HPO42–
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Chapter Fifteen
14
Let’s See if You’ve Got it…
• For each acid, write its
conjugate base:
1. HC2H3O2
2. H3PO4
-1
3. HCO3
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Chapter Fifteen
15
Strength of Acids & Bases
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Chapter Fifteen
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Strength
• It is important to know whether or not an
acid or base is considered strong or weak
because it determines how we do
calculations.
• Today we will only do calculations for
strong acids because strong acids
COMPLETELY DISSOCIATE.
• That means that their concentration is the
SAME AS the [H3O+1].
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Chapter Fifteen
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Strong Acids
• The “strong” acids—HCl, HBr,
HI, HNO3, H2SO4, HClO4—are
considered “strong” because
they ionize completely in water.
• The “strong” acids all appear
above H3O+ in Table 15.1 on p.
620.
• The strong acids are leveled to
the same strength—to that of
H3O+—when they are placed in
water.
• MEMORIZE THE 6 MOST
COMMON STRONG ACIDS!
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Chapter Fifteen
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Periodic Trends in Acid Strength
• The greater the tendency for HX (general acid) to
transfer a proton to H2O, the more the forward
reaction is favored and the stronger the acid.
• A factor that makes it easier for the H+ to leave
will increase the strength of the acid.
• Acid strength is inversely proportional to H—X
bond-dissociation energy. Weaker H—X bond =>
stronger acid.
• Acid strength is directly proportional to anion
radius. Larger X radius => stronger acid.
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Chapter Fifteen
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Periodic Trends in Acid Strength
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Strength of Carboxylic Acids
• Carboxylic acids all have the –COOH group in common.
• Differences in acid strength come from differences in the R
group attached to the carboxyl group.
• In general, the more that electronegative atoms appear in
the R group, the stronger is the acid.
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Chapter Fifteen
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Helpful Note
• Page 1 of the reference sheet
that I gave you does a nice job
of summarizing how acid
strength increases.
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Chapter Fifteen
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Strength of Conjugate Acid–Base Pairs
• A stronger acid can donate H+ more readily than
a weaker acid.
• The stronger an acid, the weaker is its conjugate
base.
• The stronger a base, the weaker is its conjugate
acid.
• An acid–base reaction is favored in the direction
from the stronger member to the weaker member
of each conjugate acid–base pair.
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… the weaker
the conjugate
base.
The stronger
the acid …
And the stronger
the base …
… the weaker
the conjugate
acid.
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Chapter Fifteen
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Example 15.2
Select the stronger acid in each pair:
(a) nitrous acid, HNO2, and nitric acid, HNO3
(b) Cl3CCOOH and BrCH2COOH
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Chapter Fifteen
25
A Little pH Theory
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Self-Ionization of Water
• Even pure water conducts some electricity. This is due to
the fact that water self-ionizes:
• The equilibrium constant for this process is called the ion
product of water (Kw).
Why did we
leave out water?
• At 25 °C, Kw = 1.0 x 10–14 = [H3O+][OH–]
• This equilibrium constant is very important because it
applies to all aqueous solutions—acids, bases, salts, and
nonelectrolytes—not just to pure water.
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Chapter Fifteen
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The pH Scale
Since pH is a logarithmic scale,
cola drinks (pH about 2.5) are
about ____ times as acidic as
tomatoes (pH about 4.5)
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Chapter Fifteen
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Example 15.4
By the method suggested in Figure 15.5, a student
determines the pH of milk of magnesia, a suspension of
solid magnesium hydroxide in its saturated aqueous
solution, and obtains a value of 10.52. What is the molarity
of Mg(OH)2 in its saturated aqueous solution? The
suspended, undissolved Mg(OH)2(s) does not affect the
measurement.
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Chapter Fifteen
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15.5 Example
Is the solution 1.0 x 10–8 M HCl
acidic, basic, or neutral?
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Chapter Fifteen
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That’s enough for one day…
• Worksheet #1 is due next
class!
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
31
Today…
• Turn in:
– Get out WS#1 to check
• Our Plan:
– Questions on WS#1
– Notes – Weak Acids
– Practice from Reference Sheet
– Investigation 14 Pre-Lab
• Homework (Write in Planner):
– Worksheet #2 due next class
– Investigation 14 Pre-Lab
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Chapter Fifteen
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Quick Review
• Given a 5.5 x 10-6 M H2SO4
solution:
1.
2.
3.
4.
5.
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Is H2SO4 an acid or a base?
Is H2SO4 strong or weak?
What is the pH of the solution?
What is the [OH-1]?
What is the conjugate base of
H2SO4?
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
33
Quick Review - Answers
1.Acid
2.Strong
3.5.26
-9
4.1.8 x 10 M
5.HSO4-1
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Chapter Fifteen
34
Equilibrium in Solutions of Weak
Acids and Bases
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Chapter Fifteen
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pH of WEAK ACIDS
• Weak acids do not dissociate completely
• Some of the acid molecules are
converted to hydronium ion, but some
are not
• To solve for hydronium or hydroxide,
we use an equilibrium calculation (RICE
Table)
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Chapter Fifteen
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pH of Weak Acids
• To find pH, you will often have to write the
equation for the dissociation of an acid or
base, so let’s practice.
1. HNO2 is added to water
2. NH3 is added to water
3. Acetic acid is added to water
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Chapter Fifteen
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Ka and Kb
The equilibrium constant for a Brønsted acid is represented
by Ka, and that for a base is represented by Kb.
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO–(aq)
[H3O+][CH3COO–]
Ka = –––––––––––––––––
[CH3COOH]
Notice that H2O is not
included in either
equilibrium expression.
NH3(aq) + H2O(l)
NH4+(aq) + OH–(aq)
[NH4+][OH–]
Kb = –––––––––––––
[NH3]
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Chapter Fifteen
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Acid/Base Strength and Direction
of Equilibrium
• In Table 15.1, HBr lies above CH3COOH in the acid
column.
• Since HBr is a stronger acid than CH3COOH, the
equilibrium for the reaction:
Weaker base
Weaker acid
Stronger base
Stronger acid
lies to the left.
• We reach the same conclusion by comparing the strengths of
the bases (right column of Table 15.1).
• CH3COO– lies below Br– ; CH3COO– is the stronger base:
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Chapter Fifteen
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Equilibrium in Solutions of Weak
Acids and Weak Bases
1.
2.
3.
4.
5.
These calculations are similar to the equilibrium
calculations performed in Chapter 14.
An equation is written for the reversible reaction.
Data are organized, often in RICE format.
Changes that occur in establishing equilibrium are
assessed.
Simplifying assumptions are examined (the “5%
rule”).
Equilibrium concentrations, equilibrium constant, etc.
are calculated.
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Chapter Fifteen
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The 5% Rule
• EXPLAIN IT.
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Chapter Fifteen
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Chapter Fifteen
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pH of WEAK ACIDS
What is the pH of a solution of 0.1 M acetic acid
if the Ka for acetic acid is 1.8 x 10-5?
↔ CH COO
CH3COOH + H2O
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-1
+ H3O+1
Chapter Fifteen
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Example 15.6
Ordinary vinegar is approximately 1 M CH3COOH and as shown in
Figure 15.6, it has a pH of about 2.4. Calculate the expected pH of
1.00 M CH3COOH(aq), and show that the calculated and measured
pH values are in good agreement.
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Chapter Fifteen
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Example 15.7 (application of 5% rule)
• What is the pH of 0.00200 M CH2ClCOOH
(aq)?
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Example 15.8
What is the pH of 0.500 M NH3(aq) ?
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Chapter Fifteen
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Example 15.9
The pH of a 0.164 M aqueous solution of
dimethylamine is 11.98. What are the values
of Kb and pKb? The ionization equation is
(CH3)2NH + H2O
Dimethylamine
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↔ (CH3)2NH2+ + OH–
Kb = ?
Dimethylammonium ion
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Chapter Fifteen
47
Polyprotic Acids
• A monoprotic acid has one ionizable H atom per molecule.
• A polyprotic acid has more than one ionizable H atom per
molecule.
– Sulfuric acid, H2SO4
– Carbonic acid, H2CO3
– Phosphoric acid, H3PO4
Diprotic
Diprotic
Triprotic
• The protons of a polyprotic acid dissociate in steps, each
step having a value of Ka.
• Values of Ka decrease successively for a given polyprotic
acid. Ka1 > Ka2 > Ka3 , etc.
• Simplifying assumptions may be made in determining the
concentration of various species from polyprotic acids.
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Chapter Fifteen
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Example 15.11
Calculate the following concentrations in an aqueous
solution that is 5.0 M H3PO4:
(a) [H3O+] (b) [H2PO4–] (c) [HPO42–] (d) [PO43–]
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Chapter Fifteen
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Ions as Acids and Bases
• HCl is a strong acid, therefore Cl– is so weakly
basic in water that a solution of chloride ions
(such as NaCl) is virtually neutral.
• Acetic acid, CH3COOH, is a weak acid, so acetate
ion, CH3COO–, is significantly basic in water.
• A solution of sodium acetate (which dissociates
completely into sodium and acetate ions in water)
is therefore slightly basic:
CH3COO– + H2O ↔ CH3COOH + OH–
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Chapter Fifteen
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Carbonate Ion
as a Base
A carbonate ion accepts a proton
from water, leaving behind an OH–
and making the solution basic.
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Ions as Acids and Bases
(cont’d)
• Salts of strong acids and strong bases form neutral
solutions: NaCl, KNO3
• Salts of weak acids and strong bases form basic
solutions: KNO2, NaClO
• Salts of strong acids and weak bases form acidic
solutions: NH4NO3
• Salts of weak acids and weak bases form solutions
that may be acidic, neutral, or basic; it depends on
the relative strengths of the cations and the anions:
NH4NO2, CH3COONH4.
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Chapter Fifteen
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Ions as Acids and Bases
(cont’d)
• In order to make quantitative predictions of
pH of a salt solution, we need an
equilibrium constant for the hydrolysis
reaction.
• The relationship between Ka and Kb of a
conjugate acid–base pair is:
Ka x Kb = Kw
• If instead we have values of pKa or pKb:
pKa + pKb = pKw = 14.00 (at 25 °C)
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Chapter Fifteen
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Example 15.14
Calculate the pH of a solution that is 0.25 M CH3COONa(aq).
Example 15.15
What is the molarity of an NH4NO3(aq) solution that has a pH
= 4.80?
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Chapter Fifteen
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The Common Ion Effect
• Consider a solution of acetic acid.
• If we add acetate ion as a second solute (i.e., sodium
acetate), the pH of the solution increases:
LeChâtelier’s principle: What
happens to [H3O+] when the
equilibrium shifts to the left?
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Chapter Fifteen
Consider:
HC2H3O2 + H2O
0.100M – x
55
H3O+ + C2H3O2
+ X M
+X M
AND
HCl + H2O ↔ H3O+ + Cl-
0.100M
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0.100M
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0.100M
Chapter Fifteen
56
The Common Ion Effect (cont’d)
The common ion effect is the suppression of the ionization of
a weak acid or a weak base by the presence of a common ion
from a strong electrolyte.
Acetic acid solution
at equilibrium: a
few H3O+ ions and a
few CH3COO– ions
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When acetate ion is
added, and equilibrium
reestablished: more
acetate ions, fewer
H3O+ ions
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Chapter Fifteen
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Example 15.16
Calculate the pH of an aqueous solution that is both 1.00 M
CH3COOH and 1.00 M CH3COONa.
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Chapter Fifteen
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Example 15.16 A
• Calculate the pH of an aqueous solution that
is 0.15 M NH3 and 0.35 M NH4NO3.
NH3 + H2O ↔ NH4+1 + OH-1
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Kb = 1.8 x 10-5
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Chapter Fifteen
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Stop!
• To review the different
calculations we’ve done so far in
class, let’s try out p. 2 of the
reference sheet.
• Then, do Worksheet #2. It’s due
next class.
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Chapter Fifteen
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Today…
• Turn in:
– Worksheet #2
• Our Plan:
– Questions on WS#2
– Quiz
– Investigation 14
• Homework (Write in Planner):
– Lab Report Due Monday
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Chapter Fifteen
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Today…
• Turn in:
– Lab Report
• Our Plan:
– Quick Review (Weak Acid HO)
– Notes
• Homework (Write in Planner):
– Lab Report Due Friday
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Chapter Fifteen
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Buffer Solutions
• Many industrial and biochemical reactions—especially
enzyme-catalyzed reactions—are sensitive to pH.
• To work with such reactions we often need a solution that
maintains a nearly-constant pH.
• A buffer solution is a solution that changes pH only
slightly when small amounts of a strong acid or a strong
base are added.
• A buffer contains significant concentrations of both
– a weak acid and its conjugate base, or
– a weak base and its conjugate acid.
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Buffer Solutions (cont’d)
• The acid component of the buffer neutralizes
small added amounts of OH–, forming the weaker
conjugate base which does not affect pH much:
HA + OH– ↔ H2O + A–
• The base component neutralizes small added
amounts of H3O+, forming the weaker conjugate
acid which does not affect pH much.
A– + H3O+ ↔ H2O + HA
• Pure water does not buffer at all …
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Chapter Fifteen
Pure water increases in pH by
about 5 pH units when the OH– is
added, and decreases by about 5
pH units when the H3O+ is added.
64
In contrast, the same amounts of
OH– and H3O+ added to a buffer
solution barely change the pH.
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An Equation for Buffer Solutions
• In certain applications, there is a need to repeat the
calculations of the pH of buffer solutions many times. This
can be done with a single, simple equation, but there are
some limitations.
• The Henderson–Hasselbalch equation:
[conjugate base]
pH = pKa + log ––––––––––––––
[weak acid]
• To use this equation, the ratio [conjugate base]/[weak acid]
must have a value between 0.10–10 and both concentrations
must exceed Ka by a factor of 100 or more.
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Buffer Capacity and
Buffer Range
• There is a limit to the ability of a buffer solution to
neutralize added acid or base.
• This buffer capacity is reached before either buffer
component has been consumed.
• In general, the more concentrated the buffer components in
a solution, the more added acid or base the solution can
neutralize.
• As a rule, a buffer is most effective if the concentrations of
the buffer acid and its conjugate base are equal or nearly so.
• Therefore, a buffer is most effective when the desired pH of
the buffer is very near pKa of the weak acid of the buffer.
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Example 15.17
A buffer solution is 0.24 M NH3 and 0.20 M NH4Cl. (a)
What is the pH of this buffer? (b) If 0.0050 mol NaOH is
added to 0.500 L of this solution, what will be the pH?
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15.17 A
• What is the final pH if 0.03 mol HCl is
added to 0.500 L of a buffer solution that is
0.24 M NH3 and 0.20 M NH4Cl?
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Chapter Fifteen
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Example 15.18
What concentration of acetate ion in 0.500
M CH3COOH(aq) produces a buffer
solution with pH = 5.00?
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15.18 A
• What concentration of acetic acid in 0.250
M CH3COONa (aq) is needed to produce a
buffer solution with pH = 4.50?
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Chapter Fifteen
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Acid–Base Indicators
• An acid–base indicator is a weak acid or base.
• The acid form (HA) of the indicator has one color, the
conjugate base (A–) has a different color. One of the
“colors” may be colorless.
• In an acidic solution, [H3O+] is high. Because H3O+ is a
common ion, it suppresses the ionization of the indicator
acid, and we see the color of HA.
• In a basic solution, [OH–] is high, and it reacts with HA,
forming the color of A–.
• Acid–base indicators are often used for applications in
which a precise pH reading isn’t necessary.
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Chapter Fifteen
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Different indicators have different values of Ka, so
they exhibit color changes at different values of pH …
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Example 15.19
A Conceptual Example
Explain the series of color changes of thymol blue indicator
produced by the actions pictured in Figure 15.14:
(a) A few drops of thymol blue are
added to HCl(aq).
(b) Some sodium acetate is added to
solution (a).
(c) A small quantity of sodium
hydroxide is added to solution (b).
(d) An additional, larger quantity of
sodium hydroxide is added to
solution (c).
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Chapter Fifteen
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Stop
• Work on Worksheet #3 up to
problem 8.
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Chapter Fifteen
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Today…
• Turn in:
– Nothing
• Our Plan:
– Titration Curve Activity
– Notes
– Investigation 15 Pre-Lab
• Homework (Write in Planner):
– Finish WS#3 by next Wednesday
–
Investigation
15
Pre-Lab
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
76
Titration Curve Activity
• With a partner, complete p. 6 &
7 of your worksheet packet.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
77
Neutralization Reactions
• At the equivalence point in an acid–base titration, the acid
and base have been brought together in precise
stoichiometric proportions.
• The endpoint is the point in the titration at which the
indicator changes color.
• Ideally, the indicator is selected so that the endpoint and
the equivalence point are very close together.
• The endpoint and the equivalence point for a neutralization
titration can be best matched by plotting a titration curve,
a graph of pH versus volume of titrant.
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
78
Titration Curve, Strong Acid with Strong Base
Bromphenol blue,
bromthymol blue, and
phenolphthalein all
change color at very
nearly 20.0 mL
At about what volume
would we see a color
change if we used methyl
violet as the indicator?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
Example 15.20
79
Calculate the pH at the following points in the titration
of 20.00 mL of 0.500 M HCl with 0.500 M NaOH:
H3O+ + Cl– + Na+ + OH–  Na+ + Cl– + 2 H2O
(a) before the addition of any NaOH
(b) after the addition of 10.00 mL of 0.500 M NaOH
(c) after the addition of 20.00 mL of 0.500 M NaOH
(d) after the addition of 20.20 mL of 0.500 M NaOH
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
Titration Curve, Weak Acid with Strong Base
80
The equivalence-point
pH is NOT 7.00 here.
Why not??
Bromphenol blue was ok
for the strong acid/strong
base titration, but it
changes color far too early
to be useful here.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
81
Example 15.21
Calculate the pH at the following points in the titration of
20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH:
CH3COOH + Na+ + OH–  Na+ + CH3COO– + H2O
(a) before the addition of any NaOH
(b) after the addition of 8.00 mL of 0.500 M NaOH
(c) after the addition of 10.00 mL of 0.500 M NaOH
(d) after the addition of 20.00 mL of 0.100 M NaOH
(e) after the addition of 21.00 mL of 0.100 M NaOH
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
82
15.21 B
• In a titration similar that that of Example
15.21, 20.00 mL of 0.500 M NaOH is
titrated with 0.500 M CH3COOH (aq).
Calculate the pH after the addition of the
following volumes of the titrant
a.
b.
c.
d.
e.
Prentice Hall © 2005
10.00 mL
20.00 mL
30.00 mL
40.00 mL
Sketch the titration curve
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
83
Example 15.22
A Conceptual Example
This titration curve shown in Figure 15.18 involves 1.0 M
solutions of an acid and a base. Identify the type of
titration it represents.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
84
Lewis Acids and Bases
• There are reactions in nonaqueous solvents, in the gaseous
state, and even in the solid state that can be considered
acid–base reactions which Brønsted–Lowry theory is not
adequate to explain.
• A Lewis acid is a species that is an electron-pair acceptor
and a Lewis base is a species that is an electron-pair donor.
CaO(s) + SO2(g) → CaSO3(s)
Sulfur accepts an electron pair
from the oxygen of CaO
• In organic chemistry, Lewis acids are often called
electrophiles (“electron-loving”) and Lewis bases are often
called nucleophiles (“nucleus-loving”).
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
85
Today…
• Turn in:
–Nothing
• Our Plan:
–Investigation 15
• Homework (Write in Planner):
–Lab Report Due Monday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
86
Today…
• Turn in:
–Lab Report
• Our Plan:
–Review/Practice
• Homework (Write in Planner):
–Homework due next class
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
87
Today…
• Turn in:
– Homework
• Our Plan:
– Homework Questions
– Quiz
– Review Activities
• Homework (Write in Planner):
– Test Next Class
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen
88
Today…
• Turn in:
– Nothing
• Our Plan:
– Questions on Review
– Test
• Homework (Write in Planner):
– Enjoy your weekend
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fifteen