1 Ch 17 — Acids and Bases Jeffrey Mack California State University, Sacramento Chapter 17 Principles of Chemical Reactivity: The Chemistry of Acids and Bases • In Chapter 3, you were introduced to two definitions of acids and bases: the Arrhenius and the Brønsted–Lowry definition. • Arrhenius acid: Any substance that when dissolved in water increases the concentration of hydrogen ions, H + . • Arrhenius base: Any substance that increases the concentration of hydroxide ions, OH , when dissolved in water. • A Brønsted–Lowry acid is a proton (H + ) donor. • A Brønsted–Lowry base is a proton acceptor. Acids & Bases: A Review • Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O(liq) H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water. Strong & Weak Acids/Bases HNO 3 , HCl, H 2 SO 4 and HClO 4 are classified as strong acids. Strong & Weak Acids/Bases • Strong Base: 100% dissociated in water. NaOH(aq) Na + (aq) + OH - (aq) Other common strong bases include KOH and Ca(OH) 2 . CaO (lime) + H 2 O Ca(OH) 2 (slaked lime) CaO Strong & Weak Acids/Bases • Weak base: less than 100% ionized in water An example of a weak base is ammonia NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq) Strong & Weak Acids/Bases
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1 Ch 17 — Acids and Bases
Jeffrey Mack
California State University,
Sacramento
Chapter 17 Principles of Chemical Reactivity: The Chemistry of Acids and Bases
• In Chapter 3, you were introduced to two
definitions of acids and bases: the Arrhenius and
the Brønsted–Lowry definition.
• Arrhenius acid: Any substance that when
dissolved in water increases the concentration of
hydrogen ions, H+.
• Arrhenius base: Any substance that increases
the concentration of hydroxide ions, OH, when
dissolved in water.
• A Brønsted–Lowry acid is a proton (H+) donor.
• A Brønsted–Lowry base is a proton acceptor.
Acids & Bases: A Review
• Generally divide acids and bases into STRONG or
WEAK ones.
STRONG ACID:
HNO3(aq) + H2O(liq) H3O+(aq) + NO3
-(aq)
HNO3 is about 100% dissociated in water.
Strong & Weak Acids/Bases
HNO3, HCl, H2SO4 and HClO4 are classified as
strong acids.
Strong & Weak Acids/Bases
• Strong Base: 100% dissociated in water.
NaOH(aq) Na+(aq) + OH-(aq)
Other common strong
bases include KOH and
Ca(OH)2.
CaO (lime) + H2O
Ca(OH)2 (slaked lime) CaO
Strong & Weak Acids/Bases
• Weak base: less than 100% ionized in water
An example of a weak base is ammonia
NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)
Strong & Weak Acids/Bases
2 Ch 17 — Acids and Bases
Weak acids are much less than 100% ionized
in water.
Example: acetic acid = CH3CO2H
Strong & Weak Acids/Bases
• Proton donors may be molecular compounds,
cations or anions.
3
3 2 3 3
4 2 3 3
2
2 3 3
HNO (aq) H O(l) NO (aq) H O (aq)
NH (aq) H O(l) NH (aq) H O (aq)
HCO (aq) H O(l) CO (aq) H O (aq)
The Brønsted–Lowry Concept of Acids & Bases Extended
• Proton acceptors may be molecular
compounds, cations or anions.
3
3
3 2
2
2 25
3
2 6
2
3 2
NH (aq) H O(l) NH (aq) OH (aq)
Al H O OH (aq) H O(l)
Al H O (aq) OH (aq)
CO (aq) H O(l) HCO (aq) OH (aq)
The Brønsted–Lowry Concept of Acids & Bases Extended
Using the Brønsted definition, NH3 is a BASE in water and water is itself an ACID
Proton acceptor
Proton donor
33 2NH (aq) H O(l) NH (aq) OH (aq)+ -+ +
The Brønsted–Lowry Concept of Acids & Bases Extended
• Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid)
are all capable of donating one proton and so are called
monoprotic acids.
• Other acids, called polyprotic acids are capable of donating
two or more protons.
The Brønsted–Lowry Concept of Acids & Bases Extended
• A conjugate acid–base pair consists of two species that
differ from each other by the presence of one hydrogen ion.
• Every reaction between a Brønsted acid and a Brønsted
base involves two conjugate acid–base pairs
Conjugate Acid–Base Pairs
3 Ch 17 — Acids and Bases
Conjugate Acid–Base Pairs
Water Autoionization and the Water Ionization Constant, Kw:
The water autoionization equilibrium lies far to the left side. In fact, in pure water at 25 °C, only about two out of a billion (109) water molecules are ionized at any instant.
Even in pure water, there is a small concentration of ions present at all times. [H3O
+] = [OH] = 1.00 107
2 2 3H O(l) H O(l) H O (aq) OH (aq)
14
w 3K H O OH 1.00 10+ - -é ù é ù= = ´ë û ë û
Water & the pH Scale
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION.
Equilibrium constant for autoionization = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 °C
Water & the pH Scale
• In a neutral solution, [H3O+] = [OH]
• Both are equal to 1.00 10 7 M
• In an acidic solution, [H3O+] > [OH]
• [H3O+] > 1.00 10 7 M and [OH] < 1.00 10
7 M
• In a basic solution, [H3O+] < [OH]
• [H3O+] < 1.00 10 7 M and [OH] > 1.00 10
7 M
Water & the pH Scale
The pH Scale
• The pH of a solution is defined as the
negative of the base (10) logarithm (log) of
the hydronium ion concentration.
pH = log[H3O+]
• In a similar way, we can define the pOH of a
solution as the negative of the base - 10
logarithm of the hydroxide ion concentration.
pOH = log[OH]
pH + pOH = pKw = 14
The pH Scale
4 Ch 17 — Acids and Bases
• The concentration of acid, [H3O+] is found by
taking the antilog of the solutions pH.
• In a similar way, [OH] can be found from:
pH
3H O 10 + -é ù =ë û
pOHOH 10 - -é ù =ë û
The pH Scale
Once [H3O+] is known, [OH] can be found
from:
And vice versa.
w
3
K[OH ]
[H O ]
-
+=
w3
K[H O ]
[OH ]
+
-=
The pH Scale
• In Chapter 3, it was stated that acids and bases
can be divided roughly into those that are strong
electrolytes (such as HCl, HNO3, and NaOH)
and those that are weak electrolytes (such as
CH3CO2H and NH3)
• In this chapter we will discuss the quantitative
aspects of dissociation of weak acids and bases.
• The relative strengths of weak acids and bases
can be ranked based on the magnitude of
individual equilibrium constants.
Equilibrium Constants for Acids & Bases
• Strong acids and bases almost completely
ionize in water (~100%):
Kstrong >> 1
(product favored)
• Weak acids and bases almost completely
ionize in water (<<100%):
Kweak << 1
(Reactant favored)
Equilibrium Constants for Acids & Bases
• The relative strength of an acid or base can also be
expressed quantitatively with an equilibrium
constant, often called an ionization constant. For
the general acid HA, we can write:
[ ]
2 3
3
a
HA(aq) H O(l) H O (aq) A (aq)
H O AK
HA
+ -
+ -
+ +
é ù é ùë û ë û=
Conjugate
acid
Conjugate
base
Equilibrium Constants for Acids & Bases
• The relative strength of an acid or base can
also be expressed quantitatively with an
equilibrium constant, often called an
ionization constant. For the general base B,
we can write:
[ ]
2
b
B(aq) H O(l) BH (aq) OH (aq)
BH OHK
B
+ -
+ -
+ +
é ù é ùë û ë û=
Conjugate
base
Conjugate
Acid
Equilibrium Constants for Acids & Bases
5 Ch 17 — Acids and Bases
Acids Conjugate
Bases Increase
strength
Increase
strength
Ionization Constants for Acids/Bases
• The strongest acids are at the upper left.
They have the largest Ka values.
• Ka values become smaller on descending the
chart as the acid strength declines.
• The strongest bases are at the lower right.
They have the largest Kb values.
• Kb values become larger on descending the
chart as base strength increases.
Equilibrium Constants for Acids & Bases
• The weaker the acid, the stronger its
conjugate base: The smaller the value of Ka,
the larger the value of Kb.
• Aqueous acids that are stronger than H3O+
are completely ionized.
• Their conjugate bases (such as NO3) do not
produce meaningful concentrations of OH
ions, their Kb values are “very small.”
• Similar arguments follow for strong bases
and their conjugate acids.
Equilibrium Constants for Acids & Bases
Increasing Acid Strength ¾¾¾¾¾¾¾¾¾¾¾®
Acid HCO3 HClO HF
Ka 4.8 1011 3.5 108 7.2 104
Increasing Base Strength ¬¾¾¾¾¾¾¾¾¾¾¾
Base CO32 ClO F
Kb 2.1 104 2.9 107 1.4 1011
Equilibrium Constants for Acids & Bases
Equilibrium Constants for Acids & Bases
Ka Values for Polyprotic Acids
In general, each successive dissociation
produces a weaker acid.
2 2 3
7
a(1)
2
2 3
19
a(2)
H S(aq) H O(l) H O (aq) HS (aq)
K 1 10
HS (aq) H O(l) H O (aq) S (aq)
K 1 10
Equilibrium Constants for Acids & Bases
6 Ch 17 — Acids and Bases
Logarithmic Scale of Relative Acid
Strength, pKa
• Many chemists use a logarithmic scale to
report and compare relative acid strengths.
pKa = log(Ka)
The lower the pKa, the stronger the acid.
Acid HCO3 HClO HF
pKa 10.32 7.46 3.14
Increasing Acid Strength ¾¾¾¾¾¾¾¾¾¾¾®
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
a
b
K
2 2 3
K
2 2
H S(aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H S(aq) OH (aq)
+ -
- -
¾¾®+ +¬¾¾
¾¾®+ +¬¾¾
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
a
b
K
2 2 3
K
2 2
H S(aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H S(aq) OH (aq)
+ -
- -
¾¾®+ +¬¾¾
¾¾®+ +¬¾¾
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
a
b
K
2 2 3
K
2 2
2 3
H S(aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H S(aq) OH (aq)
2H O(l) H O (aq) OH (aq)
+ -
- -
+ -
¾¾®+ +¬¾¾
¾¾®+ +¬¾¾
+
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
a
b
K
2 2 3
K
2 2
2 3
H S(aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H S(aq) OH (aq)
2H O(l) H O (aq) OH (aq)
+ -
- -
+ -
¾¾®+ +¬¾¾
¾¾®+ +¬¾¾
+
3 2
a b 3 w
2
H O HS H S OHK K H O OH K
H S HS
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
[ ]
[ ]3 2
a b 3 w
2
a b w
H O HS H S OHK K H O OH K
H S HS
K K K
+ - -
+ -
-
é ù é ù é ùë û ë û ë û é ù é ù´ = ´ = =ë û ë ûé ùë û
´ =
a
b
K
2 2 3
K
2 2
2 3
H S(aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H S(aq) OH (aq)
2H O(l) H O (aq) OH (aq)
+ -
- -
+ -
¾¾®+ +¬¾¾
¾¾®+ +¬¾¾
+
When adding equilibria,
multiply the K values.
Equilibrium Constants for Acids & Bases
7 Ch 17 — Acids and Bases
Acid–Base Properties of Salts
Anions that are conjugate bases of strong
acids (for examples, Cl or NO3.
These species are such weak bases that they
have no effect on solution pH.
3 2NO (aq) H O(l) No Reaction
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a
solution.
Hydroxide ions are produced via “Hydrolysis”.
2
3 2 3CO (aq) H O(l) HCO (aq) OH (aq)
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a
solution.
Hydroxide ions are produced via “Hydrolysis”.
A partially deprotonated anion (such as HCO3)
is amphiprotic. Its behavior will depend on the
other species in the reaction.
2
3 2 3CO (aq) H O(l) HCO (aq) OH (aq)
Acid–Base Properties of Salts
Alkali metal and alkaline earth cations have no
measurable effect on solution pH.
Since these cations are conjugate acids of
strong bases, hydrolysis does not occur.
2Na (aq) H O(l) No Reaction
Acid–Base Properties of Salts
Basic cations are conjugate bases of acidic cations
such as [Al(H2O)6]3+.
Acidic cations fall into two categories: (a) metal
cations with 2+ and 3+ charges and (b) ammonium
ions (and their organic derivatives).
All metal cations are hydrated in water, forming ions
such as [M(H2O)6]n+.
3
2 26
2
2 35
4
Al H O (aq) H O(l)
Al H O (OH ) (aq) H O (aq)
Ka 7.9 10
Acid–Base Properties of Salts
8 Ch 17 — Acids and Bases
Salt pH of (aq) solution
CaCl2
NH4Br
NH4F
KNO3
KHCO3
Acid–Base Properties of Salts: Practice
Acid–Base Properties of Salts: Practice
Salt pH of (aq) solution
CaCl2 Neutral
NH4Br Acidic
NH4F Basic
KNO3 Neutral
KHCO3 Basic
• According to the Brønsted–Lowry theory, all acid–
base reactions can be written as equilibria involving
the acid and base and their conjugates.
• All proton transfer reactions proceed from the
stronger acid and base to the weaker acid and
base.
Acid Base Conjugate base of the acid + Conjugate acid of the base
Predicting the Direction of Acid–Base Reactions
• When a weak acid is in solution, the products are a
stronger conjugate acid and base. Therefore
equilibrium lies to the left.
• All proton transfer reactions proceed from the
stronger acid and base to the weaker acid and
base.
Predicting the Direction of Acid–Base Reactions
• Will the following acid/base reaction occur spontaneously?
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
Predicting the Direction of Acid–Base Reactions
• Will the following acid/base reaction occur spontaneously?
Ka = 7.5 105 Ka = 1.8 105
Kb = 5.6 1010 Kb = 1.3 1012
Predicting the Direction of Acid–Base Reactions
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
9 Ch 17 — Acids and Bases
• Will the following acid/base reaction occur
spontaneously?
• Equilibrium lies to the right since all proton
transfer reactions proceed from the stronger
acid and base to the weaker acid and base.
Ka = 7.5 105 Ka = 1.8 105
Kb = 5.6 1010 Kb = 1.3 1012
Stronger Acid + Stronger Base Weaker Base + Weaker Acid
Predicting the Direction of Acid–Base Reactions
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
Strong acid (HCl) + Strong base (NaOH)
Net ionic equation
Mixing equal molar quantities of a strong acid
and strong base produces a neutral solution.
3HCl (aq) NaOH (aq) H O (aq) NaCl (aq)
3 2H O (aq) OH (aq) 2H O(l)
Types Acids–Base Reactions
Weak acid (HCN) + Strong base (NaOH)
Mixing equal amounts (moles) of a strong base
and a weak acid produces a salt whose anion
is the conjugate base of the weak acid. The
solution is basic, with the pH depending on Kb
for the anion.
2HCN (aq) + OH (aq) CN (aq) + H O (l)
2CN (aq) + H O (l) HCN (aq) + OH (aq)
Types Acids–Base Reactions
Strong acid (HCl) + Weak base (NH3)
Mixing equal amounts (moles) of a weak base
and a strong acid produces a conjugate acid of
the weak base. The solution is basic, with the
pH depending on Ka for the acid.
3 3 2 4H O (aq) + NH (aq) H O (l) + NH (aq)
4 2 3 3NH (aq) + H O (l) H O (aq) + NH (aq)
Types Acids–Base Reactions
Weak acid (CH3CO2H) + Weak base (NH3)
Mixing equal amounts (moles) of a weak acid
and a weak base produces a salt whose cation
is the conjugate acid of the weak base and
whose anion is the conjugate base of the weak
acid. The solution pH depends on the relative
Ka and Kb values.
3 2 3 3 2 4CH CO H (aq) + NH (aq) CH CO + NH (aq)
Types Acids–Base Reactions
Weak acid + Weak base
• Product cation = conjugate acid of weak base.
• Product anion = conjugate base of weak acid.
• pH of solution depends on relative strengths of
cation and anion.
Types Acids–Base Reactions
10 Ch 17 — Acids and Bases
Types Acids–Base Reactions Summary
Determining K from Initial Concentrations and
pH
2 2 3 2
2
HNO (aq) H O(l) H O (aq) NO (aq)
0.10 M HNO (aq) pH = 2.17
+ -+ +
[H2S] [H3O+] [HS]
Initial 0.10
Change
Equilibrium
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and
pH
2 2 3 2
2
HNO (aq) H O(l) H O (aq) NO (aq)
0.10 M HNO (aq) pH = 2.17
+ -+ +
[H2S] [H3O+] [HS]
Initial 0.10 0 0
Change 0.10 - x + x + x
Equilibrium 0.10 - x x x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and
pH
2 2 3 2
2
HNO (aq) H O(l) H O (aq) NO (aq)
0.10 M HNO (aq) pH = 2.17
+ -+ +
[H2S] [H3O+] [HS]
Initial 0.10 0 0
Change 0.10 - x + x + x
Equilibrium 0.10 - x x x
23 2
a
2
H O NO x x xK
HNO 0.10 x 0.10 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and
pH 2 2 3 2
2
HNO (aq) H O(l) H O (aq) NO (aq)
0.10 M HNO (aq) pH = 2.17
+ -+ +
[H3O+] = [NO2
] = 6.76 103
23
4
a 3
6.76 10K 4.9 10
0.10 6.76 10
Calculations with Equilibrium Constants
23 2
a
2
H O NO x x xK
HNO 0.10 x 0.10 x
Determining K from Initial Concentrations and
pH
2 2 3
7
2 a
H S(aq) H O(l) H O (aq) HS (aq)
1.00 M H S(aq) K 1.0 10
+ -
-
+ +
= ´
[H2S] [H3O+] [HS]
Initial 1.00
Change
Equilibrium
Calculations with Equilibrium Constants
11 Ch 17 — Acids and Bases
[H2S] [H3O+] [HS]
Initial 1.00 0 0
Change - x + x + x
Equilibrium 1.00 - x x x
2 2 3
7
2 a
H S(aq) H O(l) H O (aq) HS (aq)
1.00 M H S(aq) K 1.0 10
+ -
-
+ +
= ´
23 7
a
2
H O HS x x xK 1.0 10
H S 1.00 x 1.00 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and
pH
Determining K from Initial Concentrations and
pH
27
Since x << 1.00, the equation reduces to:
x1.0 10
1.00
x = 3.2 104 pH = 3.50
23 7
a
2
H O HS x x xK 1.0 10
H S 1.00 x 1.00 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and
pH
In general, the approximation that
[HA]equilibrium = [HA]initial x [HA]initial
is valid whenever [HA]initial is greater than or
equal to 100 Ka.
If this is not the case, the quadratic equation
must by used.
Calculations with Equilibrium Constants
Determining pH after an acid/base reaction:
Calculate the hydronium ion concentration and
pH of the solution that results when 22.0 mL of
0.15 M acetic acid, CH3CO2H, is mixed with 22.0
mL of 0.15 M NaOH.
Calculations with Equilibrium Constants
Calculations with Equilibrium Constants
Determining pH after an acid/base reaction:
Calculate the hydronium ion concentration and
pH of the solution that results when 22.0 mL of
0.15 M acetic acid, CH3CO2H, is mixed with 22.0
mL of 0.15 M NaOH.
Solution: From the volume and concentration of
each solution, the moles of acid and base can be
calculated. Knowing the moles after the reaction
and the equilibrium constants, the concentration
of H3O+ and pH can be calculated.
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
All of the acetic acid is converted to acetate ion.
3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)
3 23
1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH
10 mL 1 L
12 Ch 17 — Acids and Bases
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
All of the acetic acid is converted to acetate ion.
3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)
3 23
1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH
10 mL 1 L
3
3 2
0.0033 mols 10 mLCH CO 0.075 M
44.0 mL 1 L
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
3
3 2
0.0033 mols 10 mLCH CO 0.075 M
44.0 mL 1 L
-é ù = ´ =ë û
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change
Equilibrium
3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
3
3 2
0.0033 mols 10 mLCH CO 0.075 M
44.0 mL 1 L
-é ù = ´ =ë û
3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
[ ] 2 143 2 10w
b 5
a3 2
CH CO H OH Kx 1.00 10K 5.6 10
0.075 x K 1.8 10CH CO
- --
--
é ù ´ë û = = = = = ´- ´é ùë û
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Since Kb100 > [CH3CO2-]initial, the quadratic equation
is not needed.
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
[ ] 2 143 2 10w
b 5
a3 2
CH CO H OH Kx 1.00 10K 5.6 10
0.075 x K 1.8 10CH CO
- --
--
é ù ´ë û = = = = = ´- ´é ùë û
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
210
6
6
pH 9
3
x5.6 10
0.075
x OH 6.4 10
pOH log(6.4 10 ) 5.19
pH 14 pOH 8.81 H O 10 1.5 10 M
-
- -
-
+ - -
= ´
é ù= = ´ë û
= - ´ =
é ù= - = = = ´ë û
13 Ch 17 — Acids and Bases
• Because polyprotic acids are capable of donating
more than one proton they present us with
additional challenges when predicting the pH of
their solutions.
• For many inorganic polyprotic acids, the ionization
constant for each successive loss of a proton is
about 104 to 106 smaller than the previous step.
• This implies that the pH of many inorganic
polyprotic acids depends primarily on the hydronium
ion generated in the first ionization step.
• The hydronium ion produced in the second step can
be neglected.
Polyprotic Acuids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
2–2 3 3
2 3
[HSO ][H O ] x1.2 10
[H SO ] 0.45 – x
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
Since 100 Ka is not << 0.45M, the quadratic equation must
be used
2–2 3 3
2 3
[HSO ][H O ] x1.2 10
[H SO ] 0.45 – x
Polyprotic Acids & Bases
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
(a)
x = [H3O+] = 0.0677 M
pH = 1.17
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
(a) in part a we found that x = [H3O+] = 0.0677 M
(b)
[ ]
[ ]
2
3 2 3 3
2 2
3 3 38
a2
3
2 8
3 a2
HSO (aq) H O(l) SO (aq) H O (aq)
SO H O SO 0.0677K 6.2 10
0.0677HSO
SO K 6.2 10 M
- - +
- + -
-
-
- -
+ +
é ù é ù é ùë û ë û ë û= ´ = =é ùë û
é ù = = ´ë û
Polyprotic Acids & Bases
14 Ch 17 — Acids and Bases
Halide Acid Strengths
• Experiments show that the acid strength increases in the
order: HF << HCl < HBr < HI.
• Stronger acids result when the HX bond is readily broken
(as signaled by a smaller, positive value of H for bond
dissociation) and a more negative value for the electron