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Chapter 13Acids and Bases
• What are Acids and Bases?
• Strong and Weak Acids and Bases
• Relative Strengths of Weak Acids
• Acidic, Basic, and Neutral Solutions
• The pH Scale
• Buffered Solutions
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Arrhenius Model• Arrhenius Model Acids and Bases
– Proposed by Svante Arrhenius in late 1800s
– An acid in aqueous solution produces hydrogen ions, H+
HCl(g) H+(aq) + Cl-(aq)
– A base in aqueous solution produces
hydroxide ions, OH-
NaOH(s) Na+(aq) + OH-(aq)
– Also explains neutralization of acids and bases; when H+ reacts with OH-, water is formed
H+(aq) + OH- (aq) H2O(l) 13-
H2O
H2O
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Hydronium Ion
• A fundamental problem with the Arrhenius model is the treatment of the behavior of the hydrogen ion, H+
• Hydrogen ions are better represented as hydronium ions, H3O
+, in solution13-
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Bronsted-Lowry Theory
• An acid is any substance that can
donate an H+ ion to another
substance
• A base is any substance that can
accept an H+ ion from another
substance
• Inclusive of all Arrhenius acids and
bases
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Practice – Bronsted-Lowry Acids
and Bases
• For each reaction, identify the
Bronstead-Lowry acid and base
reactants.
1. OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq)
2. H2SO4(aq) + F-(aq) HSO4-(aq) +
HF(aq)
3. NH4+(aq) + H2O(l) NH3(aq) + H3O
+(aq)
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Practice Solutions – Bronsted-
Lowry Acids and Bases
• For each reaction, identify the
Bronstead-Lowry acid and base
reactants.
1. OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq)
The water (H2O) donates an H+ to the OCl- to form HOCl. The water is an acid. The OCl-
accepts a H+ from the water (H2O), therefore it
is the base.
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Practice Solutions – Bronsted-
Lowry Acids and Bases2. H2SO4(aq) + F-(aq) HSO4
-(aq) + HF(aq)
H2SO4, sulfuric acid, is the acid and it
donates an H+ to the F-. F- accepts H+
from the sulfuric acid and is therefore the
base.
3. NH4+(aq) + H2O(l) NH3(aq) + H3O
+(aq)
NH4+ donates an H+ to the water and
thus is the acid. H2O accepts the H+
and is the base.
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Conjugate Acid-Base Pairs• When an acid donates an H+ to a base, the two
products differ from the reactants by one H+
ion.
• Conjugate acid– The product that forms as a result of gaining an H+
ion
• Conjugate base– The product that forms as a result of losing an H+ ion
HCl(g) + H2O(l) ���� H3O+(aq) + Cl-(aq)
acid base conjugate conjugate
acid base
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Practice – Conjugate Acids and Bases
• Identify the conjugate acid for each
base.
a)F-
b)HCO3-
c)H2O
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Practice Solutions – Conjugate Acids and Bases
• Identify the conjugate acid for each base.
The conjugate acid will have one more H+
ion than the base has.
a) F-
The conjugate acid for F- is HF.
b) HCO3-
The conjugate acid for HCO3- is H2CO3.
c) H2O
The conjugate acid for H2O is H3O+.
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Amphoteric Substances
• A substance that can act as either an acid or a
base
• Water is the most common amphoteric
substance. Another common amphoteric substance is the bicarbonate ion, HCO3
-:
HCO3-(aq) + OH-(aq) ���� CO3
2-(aq) + H2O(l)
acid base conjugate conjugate
base acid
HCO3-(aq) + H3O
+(aq) ���� H2CO3(aq) + H2O(l)
base acid conjugate conjugate
acid base13-
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Acidic Hydrogen Atoms
• If an acid has more than one hydrogen
atom, we need to determine which
hydrogen atoms are acidic.
• Typically, in oxoacids,
any hydrogen atoms bonded to oxygen
atoms are acidic.13-
Figure 13.5
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Strong Acids and Bases
• An acid or a base that is a strong electrolyte and completely ionizes or dissociates in water
– Strong acid examples:• HCl(aq)
• H2SO4(aq)
• HNO3(aq)
– Strong base examples:• KOH(aq)
• Ca(OH)2(aq)
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HCl – Example of a Strong Acid
An example of a strong acid is HCl:
HCl(aq) + H2O(l) ���� H3O+(aq) + Cl-(aq)
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Table 13.1 Common Strong Acids
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sulfuric acidH2SO4
perchloric acidHClO4
chloric acid HClO3
nitric acidHNO3
hydroiodic acidHI
hydrobromic acidHBr
hydrochloric acidHCl
NameFormula
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NaOH - Example of a Strong Base
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Table 13.2 Common Strong Bases
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barium hydroxideBa(OH)2
calcium hydroxideCa(OH)2
magnesium hydroxideMg(OH)2
potassium hydroxideKOH
sodium hydroxideNaOH
lithium hydroxideLiOH
NameFormula
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Weak Acids and Bases
• An acid or base that is a weak
electrolyte and therefore, only
partially ionizes in water
• If an acid or base is not strong, then
it is weak.
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CH3COOH - Example of a Weak Acid
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NH3 - Example of a Weak Base
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Table 13.3 Some Common Weak Acids
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FruitMalic acidHC4H4O5
Candy, wine, grapesTartaric acidH2C4H4O6
Nuts, cocoa, parsleyOxalic acidH2C2O4
Phosphoric acid
Lactic acid
Hypochlorous acid
Hydrofluoric acid
Citric acid
Carbonic acid
Acetic acid
Name
Soda, bloodH3PO4
MilkHC3H5O3
Sanitize pool and drinking water
HOCl
Glass etchingHF
Fruit, sodaH3C6H5O7
Soda, bloodH2CO3
Vinegar, sour wineCH3CO2H
OccurrenceFormula
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Table 13.4 Some Common Weak Bases
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Trimethylamine
Methylamine
Calcium
hypochorite
Calcium carbonate
Ammonia
Name
Rotting fish(CH3)3N
Herring brineCH3NH2
Chlorine source for
swimming poolsCa(ClO)2
Antacids, mineralsCaCO3
Glass cleanersNH3
OccurrenceFormula
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Practice – Strong vs. Weak Acids and Bases
• Identify each of the following as a
strong acid, weak acid, strong base,
or weak base. Write an equation to
describe its reaction in water.
1. HI(aq)
2. NaCH3CO2(aq)
3. NH4+(aq)
4. NH3(aq)
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Practice Solutions – Strong vs. Weak Acids and Bases
1. HI(aq)
Hydroiodic acid is a strong acid (it is on the list of strong acids in Table 13.1). Therefore, it completely dissociates in water.
HI(aq) + H2O(l) H3O+(aq) + I-(aq)
2. NaCH3CO2(aq)Sodium acetate is an ionic compound that partially dissociates in water to form Na+(aq) and CH3CO2
-(aq). The CH3CO2- ion is the
conjugate base of the weak acid, HC2H3O2, so it is a weak base and does not completely dissociate in water:
CH3CO2-(aq) + H2O(l) CH3CO2H(aq) + OH-(aq)
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Practice Solutions – Strong vs.
Weak Acids and Bases 3. NH4
+(aq)
Ammonium ion is the conjugate acid of the weak base, ammonia (NH3), so ammonium is therefore a weak acid. It does not completely dissociate in water.
NH4+(aq) + H2O(l) NH3(aq) + H3O
+(aq)4. NH3(aq)
Ammonia is a weak base, shown in Table 13.4, and does not dissociate completely in water.
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
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Practice – Acids and Bases on the Molecular Level
• One of the diagrams below represents
HClO4, and the other represents an aqueous solution of HSO4
-. Which is
which? Explain your reasoning.
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Practice Solutions – Acids and Bases on the Microscopic Level
• The picture on the left (A) shows no acid molecules and therefore shows an acid that has completely dissociated.
• Diagram A would then be a strong acid and of the two choices, perchloric acid (HClO4) is the strong one.
• Diagram B (the picture on the right) shows acid molecules that have not completely dissociated and are therefore the weak acid, HSO4
-.
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Relative Strengths of Weak Acids
• Acid strength depends on the relative number
of acid molecules that ionize when dissolved in water – the degree of ionization.
• Remember that Keq describes the relative amounts of products over reactants.
• If the value of Keq is larger, then more products exist at equilibrium and the acid has a larger percentage of molecules which have been
ionized.
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Acid Ionization Constants
• The acid ionization constant, Ka, describes the equilibrium that forms when an acid reacts with water.
• The larger the Ka value, the stronger the acid.
• When using Ka values to determine the strengths of conjugate acids and cases, use this rule of thumb:– The stronger the acid, the weaker the
conjugate base.13-
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Table 13.5 Weak Acids & Ka Values
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Weakest Bases
Strongest Bases
CN-
NH3
OCl-
CH3CO2-
HCO2-
NO2-
F-
Conjugate Base
6.2 x 10-10
5.6 x 10-10
4.0 x 10-8
1.8 x 10-5
1.8 x 10-4
5.6 x 10-4
6.3 x 10-4
Ka Value
HCN
NH4+
HOCl
CH3CO2H
HCO2H
HNO2
HF
AcidStrongest Acids
Weakest Acids
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Practice – Significance of Ka
• Which solution has the greater concentration of H3O
+, 0.10 M HOCl [Ka = 4.0 x 10-8] or 0.10 M HCN
[Ka = 6.2 x 10-10]?
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Practice Solutions – Significance of Ka
• Which solution has the greater concentration of H3O
+, 0.10 M HOCl or
0.10 M HCN?
Because we have equal concentrations of each acid, we first need to determine
which acid is stronger.
The stronger acid will dissociate more completely in water, thus producing more
H3O+ ion.
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Practice Solutions – Significance of Ka
• Which solution has the greater concentration of
H3O+, 0.10 M HOCl or 0.10 M HCN?
Looking at the Ka’s of the two acids from Table 13.5:
HOCl Ka = 4.0 x 10-8
HCN Ka = 6.2 x 10-10
Since HOCl has the larger Ka, it is the stronger acid and will therefore have a greater
concentration of H3O+ ion.
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Polyprotic Acids• An acid that contains more
than one acidic hydrogen and can thus donate more than one H+ ion
• The acid donates one H+ ion at a time in steps
• The Ka values for polyprotic acids are often labeled to indicate the particular step in the overall ionization process (Ka1, Ka2, Ka3, etc.)
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Figure 13.10
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Table 13.6 Ka for Polyprotic Acids
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H2C4H4O6
H2SO4
H3PO4
H2C2O4
H2S
H3C6H5O7
H2CO3
Formula
4.8x10-13
4.0x10-7
Ka3
4.3 x 10-5
1.0 x 10-2
6.2 x 10-8
1.5 x 10-4
1.0 x 10-19
1.7 x 10-5
4.7 x 10-11
Ka2
1.0 x 10-3
Strong
6.9 x 10-3
5.6 x 10-2
8.9 x 10-8
7.4 x 10-4
4.5 x 10-7
Ka1
Tartaric acid
Sulfuric acid
Phosphoric acid
Oxalic acid
Hydrosulfuric acid
Citric acid
Carbonic acid
Name
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Practice – Polyprotic Acids in Water
• Oxalic acid, H2C2O4, occurs in plants and foods such as parsley, rhubarb, almonds, and green beans. Its Ka
values are listed in Table 13.6.a) Write equations that show the
ionization of oxalic acid in water.
b) Besides water, which ion or molecule has the highest concentration in solution when oxalic acid is added to water?
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Practice Solutions – Polyprotic Acids in Water
a) Write equations that show the ionization of oxalic acid in water.
H2C2O4(aq) + H2O(l) H3O+(aq) + HC2O4
-(aq)
Ka1 = 5.6 x 10-2
HC2O4- (aq) + H2O(l) H3O
+(aq) + C2O42-(aq)
Ka2 = 1.5 x 10-4
b) Besides water, which ion or molecule has the highest concentration in solution when oxalic acid is added to water?
The Ka1 value for H2C2O4 shows that it is a weak acid and ionizes only to a small extent. That means that most of the H2C2O4 and is present in water in the highest concentration.
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Acidic, Basic, and Neutral Solutions
• Acidic solution– The H3O
+ ion concentration is greater than the OH- ion concentration.
• Basic solution– The OH- ion concentration is greater
than the H3O+ ion concentration.
• Neutral solution– Equal concentrations of OH- and H3O
+
– Neither acidic nor basic
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Ion-Product Constant of Water
• Water reacts with itself in a process called self-ionization, in which an H+ ion is transferred from one water molecule to another:
H2O(l) + H2O(l) OH-(aq) + H3O+(aq)
• The equilibrium constant for this process, called the ion-product constant of water, Kw, is:Kw = [OH-][H3O
+] = 1.0 x 10-14 (at 25°C)
and in pure water, the concentrations would be equal, so:
[OH-] = [H3O+] = 1.0 x 10-7
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Table 13.7 Definitions of Neutral, Acidic, and Basic in Aqueous
Solution
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1.0 x 10-14>1.0 x 10-14<1.0 x 10-14[OH-] > [H3O
+]Basic
1.0 x 10-14<1.0 x 10-14>1.0 x 10-14[OH-] < [H3O
+]Acidic
1.0 x 10-14=1.0 x 10-14=1.0 x 10-14[OH-] = [H3O
+]Neutral
Kw[OH-][H3O+]
Relative
Concen-tration
Type of
Solution
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Practice – Calculating H3O+ and
OH- Ion Concentrations
• Given the concentration of OH- ion in
each solution, calculate the
concentration of H3O+ in that
solution. Identify each solution as
acidic, basic, or neutral.
a) [OH-] = 1.0 x 10-8 M
b) [OH-] = 0.010 M
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Practice Solutions – Calculating H3O
+ and OH- Ion Concentrations• Given the concentration of OH- ion in each solution, calculate the
concentration of H3O+ in that solution. Identify each solution as
acidic, basic, or neutral.
a) [OH-] = 1.0 x 10-8 M
Kw = [H3O+][OH- ]
[H3O+] = Kw = 1.0 x 10-14 = 1.0 x 10-6 M H3O
+
[OH- ] 1.0 x 10-8 Acidic
b) [OH-] = 0.010 M
Kw = [H3O+][OH- ]
[H3O+] = Kw = 1.0 x 10-14 = 1.0 x 10-12 M H3O
+
[OH- ] 1.0 x 10-2 Basic13-
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The pH Scale
• The pH of a solution is the negative logarithm
(base 10) of the H3O+
concentration:
pH = - log [H3O+]
• It is convenient to
express the acidity of aqueous solutions on a
pH scale (shown at right).
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Figure 13.12
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Practice Solutions – Calculating pH
• What is the pH of each of the following solutions? Once you’ve done the calculation, check your answer to make sure it makes sense.
a) 0.00085 M HCl
First, we need to find the [H3O+]:
[H3O+] = 0.00085 M
Next, calculate the pH:
pH = - log [H3O+] = - log (0.00085) = 3.1
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Practice – Calculating pH • What is the pH of each of the following solutions? Once you’ve
done the calculation, check your answer to make sure it makes
sense.
b) 0.010 M NaOH
First, we need to find the [H3O+]:
Kw = [OH- ][H3O+]
[OH-] = 0.010 M
[H3O+] = Kw = 1.0 x 10-14 = 1.0 x 10-12 M
[OH- ] 0.010 M
Next, calculate the pH:
pH = - log [H3O+] = - log (1.0 x 10-12) = 12
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Practice – Calculating pH • What is the pH of each of the following solutions?
Once you’ve done the calculation, check your answer to make sure it makes sense.
c) 1.0 M HNO3
First, we need to find the [H3O+]:
[H3O+] = 1.0 M
Next, calculate the pH:
pH = - log [H3O+] = - log (1.0) = 0
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Calculating pOH
• pOH is defined as the negative
logarithm (base 10) of hydroxide ion
concentration, [OH-]:
pOH = - log [OH-]
The relationship between pH and
pOH is:
pH + pOH = 14
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Calculating Concentrations from pH and pOH
• The equation to find pH is:
pH = - log [H3O+]
• To find the H3O+ ion concentration, we need to
take the inverse log of the negative pH:
[H3O+] = 10-pH
• The equation to find pOH is:
pOH = - log [OH-]
• To find the OH- ion concentration, we need to take the inverse log of the negative pOH:
[OH-] = 10-pOH
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Calculating Concentrations from pH and pOH
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Practice – Calculating OH- and H3O
+ concentration
• When the pH of water in a lake falls
below about 4.5, the lake may be
considered dead because few
organisms can survive in such an
acidic environment. What is the pH
of a lake that has an OH-
concentration of 1.0 x 10-9 M?
Would this lake be considered
dead?13-
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Practice Solutions – Calculating OH- and H3O
+ concentration• When the pH of water in a lake falls below about 4.5, the lake may
be considered dead because few organisms can survive in such an acidic environment. What is the pH of a lake that has an OH-
concentration of 1.0 x 10-9 M? Would this lake be considered dead?
If the pH falls below 4.5, then the lake is dead.[OH-] = 1.0 x 10-9 M
Kw = [H3O+][OH-]
[H3O+] = Kw = 1.0 x 10-14 = 1.0 x 10-5 M
[OH-] 1.0 x 10-9
pH = - log [H3O+] = - log (1.0 x 10-5) = 5
Since the pH of the lake is 5, then the lake is NOT dead.
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Measuring pH• pH meters and pH
indicators are often used to determine the pH of a solution.
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Figure 13.15
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Buffered Solutions• A buffer (also known as a buffer system) is a
combination of a weak acid and its conjugate base (or a weak base and its conjugate acid) in about equal concentrations.
• The main buffer system in the blood is made of H2CO3/HCO3
- :
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Figure 13.18
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Practice – Buffer Systems
• Which of the following systems,
when added to water, can act as a
buffer system? For each buffer
system, write a balanced equation.
a) HCl and NaOH
b) CH3CO2H and NaCH3CO2
c) HBr and KBr
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Practice Solutions – Buffer Systems
• Which of the following systems, when added to water, can act as a buffer system?
For each buffer system, write a balanced equation.
a) HCl and NaOH
This is not a buffer system, because HCl is
a strong acid. Strong acids cannot be components of buffers, because they ionize completely in water and are not in
equilibrium with their conjugate bases.
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Practice Solutions – Buffer Systems
• Which of the following systems, when added to water, can act as a buffer system?
For each buffer system, write a balanced equation.
b) CH3CO2H and NaCH3CO2
This is a buffer system because the acid
CH3CO2H is a weak acid and its conjugate base, CH3CO2
-, forms when NaCH3CO2
dissolves in water. The equilibrium that
forms in water is:
CH3CO2H(aq) + H2O(l) CH3CO2-(aq) + H3O
+(aq)
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Practice Solutions – Buffer Systems
• Which of the following systems, when
added to water, can act as a buffer system? For each buffer system, write a balanced
equation.
c) HBr and KBr
This is not a buffer system, because HBr is a strong acid. Strong acids cannot be
components of buffers, because they ionize completely in water and are not in
equilibrium with their conjugate bases.
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Math Toolbox 13.1: Log and Inverse Log Functions
• Using Log Functions on your calculator:
pH = - log [H3O+]
Step 1: Press the +/- (change of sign) key
Step 2: Press the log key
Step 3: Enter the H3O+ concentration,
and then the ENTER or = key
(On some calculators the steps may be
reversed)13-
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Practice – Using Log Functions
• Use your calculator to find the pH of
the following solutions:
1. [H3O+] = 1.0 x 10-8 M
2. [H3O+] = 6.2 x 10-1 M
3. [H3O+] = 5.0 x 10-4 M
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Practice Solutions – Using Log Functions
• Use your calculator to find the pH of
the following solutions:
1. [H3O+] = 1.0 x 10-8 M
pH = - log [H3O+] = - log (1.0 x 10-8) = 8.0
2. [H3O+] = 6.2 x 10-1 M
pH = - log [H3O+] = - log (6.2 x 10-1) = 0.21
3. [H3O+] = 5.0 x 10-4 M
pH = - log [H3O+] = - log (5.0 x 10-4) = 3.3
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Math Toolbox 13.1: Log and Inverse Log Functions
• Using Inverse Log Functions:
[H3O+] = 10-pH
Step 1: Press the INV, SHIFT, or 2nd
button
Step 2: Press the log button
Step 3: Press the +/- (change of sign) key
Step 4: Enter the pH (or pOH), and then
the ENTER or = key
(On some calculators you may need to perform
Step 4 first, then step 3, then steps 1 and 2)13-
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Practice – Using Inverse Log Functions
• Use your calculator to find the [H3O+]
of the following solutions:
1. pH = 5.00
2. pH = 13.60
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Practice Solutions – Using Inverse Log Functions
• Use your calculator to find the [H3O+]
of the following solutions:
1. pH = 5.00
[H3O+] = 10-pH = 10-5.00 = 1.00 x 10-5 M
2. pH = 13.60
[H3O+] = 10-pH = 10-13.60 = 2.512 x 10-14 M
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