Chapter 15: Properties of Acids
• An acid is any substance that releases hydrogen
ions, H+, into water.
• Blue litmus paper turns red in the presence of
hydrogen ions. Blue litmus is used to test for acids.
• Acids have a sour taste;
lemons, limes, and
vinegar are acidic.
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Chapter 15
Properties of Bases
• A base is a substance that releases hydroxide ions,
OH –, into water.
• Red litmus paper turns blue in the presence of
hydroxide ions. Red litmus is used to test for
bases.
• Bases have a slippery,
soapy feel.
• Bases also have a bitter
taste; milk of magnesia
is a base.
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Acid–Base Neutralization
• Recall that an acid and a base react with each other
in a neutralization reaction.
• When an acid and a base react, water and a salt are
produced.
• For example, nitric acid reacts with sodium
hydroxide to produce sodium nitrate and water:
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
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The pH Scale
• A pH value expresses the acidity or basicity of a
solution.
• Most solutions have a pH between 0 and 14.
• Acidic solutions have a pH less than 7.
– As a solution becomes more acidic, the pH decreases.
• Basic solutions have a pH greater than 7.
– As a solution becomes more basic, the pH increases.
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Acid–Base Classifications of Solutions
• A solution can be classified according to its pH.
• Strongly acidic solutions
have a pH less than 2.
• Weakly acidic solutions have
a pH between 2 and 7.
• Weakly basic solutions have
a pH between 7 and 12.
• Strongly basic solutions have
a pH greater than 12.
• Neutral solutions have a pH
of 7.
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Buffers
• A buffer is a solution that resists changes in pH
when an acid or a base is added.
• A buffer is a solution of a weak acid and one of its
salts:
– Citric acid and sodium citrate make a buffer solution.
• When an acid is added to the buffer, the citrate
reacts with the acid to neutralize it.
• When a base is added to the buffer, the citric acid
reacts with the base to neutralize it.
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Arrhenius Acids and Bases
• Arrhenius proposed the following definitions for
acids and bases in 1884:
– An Arrhenius acid is a substance that ionizes in water
to produce hydrogen ions.
– An Arrhenius base is a substance that ionizes in water
to release hydroxide ions.
• For example, HCl is an Arrhenius acid and NaOH
is an Arrhenius base.
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Strengths of Acids
• Acids have varying strengths.
• The strength of an Arrhenius acid is measured by
the degree of ionization in solution.
• Ionization is the process where polar compounds
separate into cations and anions in solution.
• The acid HCl ionizes into H+ and Cl– ions in
solution.
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Strengths of Bases
• Bases also have varying strengths.
• The strength of an Arrhenius base is measured by
the degree of dissociation in solution.
• Dissociation is the process where cations and
anions in an ionic compound separate in solution.
• A formula unit of NaOH dissociates into Na+ and
OH– ions in solution.
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Strong and Weak Arrhenius Acids
• Strong acids ionize extensively to release
hydrogen ions into solution.
– HCl is a strong acid and ionizes nearly 100%.
• Weak acids only ionize slightly in solution.
– HF is a weak acid and ionizes only about 1%.
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Arrhenius Acids in Solution
• All Arrhenius acids have a hydrogen atom bonded
to the rest of the molecule by a polar bond. This
bond is broken when the acid ionizes.
• Polar water molecules help ionize the acid by
pulling the hydrogen atom away:
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq) (~100%)
HC2H3O2(aq) + H2O(l) → H3O+(aq) + C2H3O2–(aq)
(~1%)
• The hydronium ion, H3O+, is formed when the
aqueous hydrogen ion attaches to a water
molecule.
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Strong and Weak Arrhenius Bases
• Strong bases dissociate extensively to release
hydroxide ions into solution.
– NaOH is a strong base and dissociates nearly 100%.
• Weak bases only ionize slightly in solution.
– NH4OH is a weak base and only partially dissociates.
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Arrhenius Bases in Solution
• When we dissolve Arrhenius bases in solution,
they dissociate, giving a cation and a hydroxide
anion.
• Strong bases dissociate almost fully, and weak
bases dissociate very little.
NaOH(aq) → Na+(aq) + OH–(aq)
(~100%)
NH4OH(aq) → NH4+(aq) + OH–(aq)
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(~1%)
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Neutralization Reactions
• Recall that an acid neutralizes a base to produce a
salt and water.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
• The reaction produces the aqueous salt NaCl.
• If we have an acid with two hydrogens (sulfuric
acid, H2SO4), we need two hydroxide ions to
neutralize it.
H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)
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Predicting Neutralization Reactions
• We can identify the Arrhenius acid and base that
react in a neutralization reaction to produce a
given salt such as calcium sulfate, CaSO4.
• The calcium must be from calcium hydroxide,
Ca(OH)2; the sulfate must be from sulfuric acid,
H2SO4.
H2SO4(aq) + Ca(OH)2(aq) → CaSO4(aq) + 2 H2O(l)
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Brønsted–Lowry Acids and Bases
• The Brønsted–Lowry definitions of acids and
bases are broader than that of the Arrhenius
definitions.
• A Brønsted–Lowry acid is a substance that
donates a hydrogen ion to any other substance. It
is a proton donor.
• A Brønsted–Lowry base is a substance that
accepts a hydrogen ion. It is a proton acceptor.
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Brønsted–Lowry Acids and Bases,
Continued
•
Let’s look at two acid–base reactions:
1. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
2. HCl(aq) + NH3(aq) → NH4Cl(aq)
•
HCl donates a proton in both reactions and is a
Brønsted–Lowry acid.
•
In the first reaction, the NaOH accepts a proton
and is the Brønsted–Lowry base.
•
In the second reaction, NH3 accepts a proton and
is the Brønsted–Lowry base.
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Amphiprotic Compounds
• A substance that is capable of both donating and
accepting a proton is an amphiprotic compound.
• NaHCO3 is an example:
HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2CO3(aq)
NaOH(aq) + NaHCO3(aq) → Na2CO3 (aq) + H2O(l)
• NaHCO3 accepts a proton from HCl in the first
reaction and donates a proton to NaOH in the
second reaction.
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Acid–Base Indicators
• A solution that changes color as the pH changes is
an acid–base indicator.
• Shown below are the three indicators at different
pH values.
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Acid–Base Titrations
• A titration is used to analyze an acid solution
using a solution of a base.
• A measured volume of base is added to the acid
solution. When all of the acid has been
neutralized, the pH is 7. One extra drop of base
solution after the endpoint increases the pH
dramatically.
• When the pH increases above 7, phenolphthalein
changes from colorless to pink, indicating the
endpoint of the titration.
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Acid–Base Titrations, Continued
• The beginning, middle, and end of an acid–base,
using phenolphthalein as the indicator.
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Titration Problem
• Consider the titration of acetic acid with sodium
hydroxide. A 10.0 mL sample of acetic acid
requires 37.55 mL of 0.223 M NaOH. What is the
concentration of the acetic acid?
HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l)
• We want concentration acetic acid; we have
concentration sodium hydroxide.
conc NaOH ⇒ mol NaOH ⇒
mol HC2H3O2 ⇒ conc HC2H3O2
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Titration Problem, Continued
• The molarity of NaOH can be written as the unit
factor 0.233 mol NaOH / 1000 mL solution.
37.55 mL solution x
1 mol HC2H3O2
0.233 mol NaOH
x
1 mol NaOH
1000 mL solution
= 0.00837 mol HC2H3O2
1000 mL solution
0.00837 mol HC2H3O2
x
10.0 mL solution
1 L solution
= 0.837 M HC2H3O2
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Another Titration Problem
• A 10.0 mL sample of 0.555 M H2SO4 is titrated
with 0.233 M NaOH. What volume of NaOH is
required for the titration?
• We want mL of NaOH; we have 10.0 mL of
H2SO4.
• Use 0.555 mol H2SO4/1000 mL and 0.233 mol
NaOH/1000 mL.
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Another Titration Problem, Continued
H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2H2O(l)
10.0 mL H2SO4 x
0.555 mol H2SO4
1000 mL H2SO4
x
2 mol NaOH
1 mol H2SO4
x
1000 mL NaOH
0.233 mol NaOH
= 47.6 mL NaOH
• 47.6 mL of 0.233 M NaOH is required to
neutralize 10.0 mL of 0.555 M H2SO4.
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Acid–Base Standardization
• A standard solution is a solution in which the
concentration is precisely known.
• Acid solutions are standardized by neutralizing a
weighed quantity of a solid base.
• What is the molarity of a hydrochloric acid
solution if 25.50 mL are required to neutralize
0.375 g Na2CO3?
2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
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Acid–Base Standardization, Continued
0.375 g Na2CO3 ×
1 mol Na2CO3
2 mol HCl
×
1 mol Na2CO3
105.99 g Na2CO3
= 0.00708 mol HCl
1000 mL solution
0.00708 mol HCl
×
= 0.277 M HCl
25.50 mL solution
1 L solution
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Ionization of Water
• Water undergoes an autoionization reaction,
during which two water molecules react to
produce a hydronium ion and a hydroxide ion.
H2O(l) + H2O(l) → H3O+(aq) + OH-(aq)
or
H2O(l) → H+(aq) + OH-(aq)
• Only about 1 in 5 million water molecules is
present as an ion, so water is a weak conductor.
• The concentration of hydrogen ions, [H+], in pure
water is about 1 × 10-7 mol/L at 25 °C.
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Autoionization of Water
• Since [H+] is 1 × 10-7 mol/L at 25 °C, the
hydroxide ion concentration [OH-] must also be 1
× 10-7 mol/L at 25 °C.
H2O(l) → H+(aq) + OH-(aq)
• At 25 °C:
[H+][OH-] = (1 × 10-7)(1 x 10-7)
= 1.0 x 10-14
• This is the ionization constant of water, Kw.
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[H+] and [OH-] Relationship
• At 25 °C, [H+][OH-] = 1.0 x 10-14. So if we know
the [H+], we can calculate [OH-].
• What is the [OH-] if [H+] = 0.1 M ?
[H+][OH-] = 1.0 x 10-14
(0.1)[OH-] = 1.0 x 10-14
[OH-] = 1.0 x 10-13
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The pH Concept
• Recall that pH is a measure of the acidity of a
solution.
• A neutral solution has a pH of 7, an acidic
solution has a pH less than 7, and a basic solution
has a pH greater than 7.
• The pH scale uses powers of 10 to express the
hydrogen ion concentration.
• Mathematically, pH = –log[H+]
– [H+] is the molar hydrogen ion concentration.
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Calculating pH
• What is the pH if the hydrogen ion concentration
in a vinegar solution is 0.001 M?
• pH = –log[H+]
• pH = –log(0.001)
• pH = – ( –3) = 3
• The pH of the vinegar is 3, so the vinegar is acidic.
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Calculating [H+] from pH
• If we rearrange the pH equation for [H+], we get:
[H+] = 10–pH
• Milk has a pH of 6. What is the concentration of
hydrogen ion in milk?
[H+] = 10–pH = 10–6 = 0.000001 M
[H+] = 1 x 10–6 M
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Advanced pH Calculations
• What is the pH of blood with [H+] = 4.8 x 10–8 M?
pH = –log[H+] = –log(4.8 x 10–8) = – (–3.82)
pH = 3.82
• What is the [H+] in orange juice with a pH of
2.75?
[H+] = 10–pH = 10–2.75 = 0. 0018 M
[H+] = 2.75 x 10–3 M
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Strong and Weak Electrolytes
• An aqueous solution that is a good
conductor of electricity is a strong
electrolyte.
• An aqueous solution that is a poor
conductor of electricity is a weak
electrolyte.
• The greater the degree of ionization
or dissociation is, the greater the
conductivity of the solution.
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Electrolyte Strength
• Weak acids and bases are weak electrolytes.
• Strong acids and bases are strong electrolytes.
• Insoluble ionic compounds are weak electrolytes.
• Soluble ionic compounds are strong electrolytes.
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Strengths of Electrolytes
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Total Ionic Equations
• The concept of ionization allows us to portray
ionic solutions more accurately by showing strong
electrolytes in their ionized form.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
• Write strong acids and bases and soluble ionic
compounds as ions.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
• This is the total ionic equation. Each species is
written as it predominantly exists in solution.
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Net Ionic Equations
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
• Notice that Na+ and Cl- appear on both sides of the
above-noted equation. They are spectator ions.
• Spectator ions are in the solution, but do not
participate in the overall reaction. We can cancel
out the spectator ions to give the net ionic
equation.
• The net ionic equation is as follows:
H+(aq) + OH-(aq) → H2O(l)
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Writing Net Ionic Equations
• Complete and balance the nonionized chemical
equation.
• Convert the nonionized equation into the total
ionic equation.
– Write strong electrolytes in the ionized form.
– Write weak electrolytes, water, and gases in the
nonionized form.
• Cancel all the spectator ions to obtain the net ionic
equation.
– If all species are eliminated, there is no reaction.
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