Chapter 16
Aqueous Ionic Equilibrium
FY HSU
The Danger of Antifreeze
- ethylene glycol.
ethylene glycol.

Every year, thousands of dogs and cats die
from consuming a common household
product: antifreeze that was improperly
stored or that leaked out of a car radiator.

Most brands of antifreeze contain ethylene glycol.
–

sweet taste that can attract curious dogs and cats—and
sometimes even young children,
In the liver, ethylene glycol is oxidized to glycolic
acid (HOCH2COOH),
Why Is Glycolic Acid Toxic?


In high enough concentration in the bloodstream,
glycolic acid overwhelms the buffering ability of the
HCO3− in the blood, causing the blood pH to drop.
An important buffer in blood is a mixture of carbonic
acid (H2CO3) and the bicarbonate ion (HCO3-).
Acidosis


Low blood pH results in acidosis (酸中毒)
, a condition in which the acid affects the equilibrium
between hemoglobin (Hb) and oxygen:
The excess acid causes the equilibrium to shift to the left,
reducing the blood’s ability to carry oxygen.
Buffer Solutions
A buffer resists pH change by neutralizing
added acid or added base
A buffer contains either:
1. significant amounts of a weak acid and its
conjugate base or
2. significant amounts of a weak base and its
conjugate acid.

EX: Blood has a mixture of H2CO3 and HCO3−
Making an acid buffer solution
• It must contain significant amounts of
both a weak acid and its conjugate
base.
 If a strong base is added, it is neutralized
by the weak acid (HC2H3O2) in the buffer.
 If the amount of NaOH added is less
than the amount of acetic acid present,
the pH change is small.
Acidic Buffer Solution
Con’t
 If a strong acid is added, it is
neutralized by the weak acid
(NaC2H3O2) in the buffer.
 If the amount of HCl added is less
than the amount of NaC2H3O2
present, the pH change is small.
How a buffer works
Summarizing Buffer Characteristics:
▶ Buffers resist pH change.
▶ A buffer contains significant amounts of either
1) a weak acid and its conjugate base,
or 2) a weak base and its conjugate acid.
▶ The weak acid neutralizes added base.
▶ The base neutralizes added acid.
Calculating the pH of a Buffer Solution
EXAMPLE 16.1: Calculate the pH of a buffer
solution that is 0.100 M in HC2H3O2 and 0.100
M in NaC2H3O2
SOLUTION:
Solution:
The Henderson–Hasselbalch Equation

a mixture of a hypothetical weak acid, HA (such as
CH3COOH), and its salt, NaA (such as NaCH3COO ).
The Henderson–Hasselbalch Equation
EXAMPLE 16.2 Calculating the pH of a Buffer
Solution as an Equilibrium Problem and with
the Henderson–Hasselbalch Equation
Sol: HC7H5O2 is the acid and NaC7H5O2 is the base.
Henderson–Hasselbalch equation
EXAMPLE 16.3 Calculating the pH Change in a
Buffer Solution after the Addition of a Small
Amount of Strong Acid or Base
Sol:
(OH-會被buffer 的酸給作用完)
Con’t
The Henderson–Hasselbalch Equation
EXAMPLE 16.4 Using the Henderson–Hasselbalch
Equation to Calculate the pH of a Buffer Solution
Composed of a Weak Base and Its Conjugate Acid
Sol:
The Henderson–Hasselbalch Equation
Buffering Effectiveness

A good buffer should be able to neutralize moderate amounts
of added acid or base.

However, there is a limit to how much can be added before the
pH changes significantly.

The buffering capacity is the amount of acid or base a buffer
can neutralize.

The buffering range is the pH range the buffer can be effective.

The effectiveness of a buffer depends on two factors
(1) the relative amounts of acid and base
(2) the absolute concentrations of acid and base.
Effectiveness of Buffers

A buffer will be most effective when the [base]/[acid] = 1.
Effectiveness of Buffers

A buffer will be effective when 0.1 < [base]/[acid] < 10.
Buffer Range:
•A buffer will be most effective when the [acid] and the
[base] are large.
EXAMPLE 16.5 Preparing a Buffer
SOLUTION

The best choice is formic acid because its pKa lies closest to the
desired pH. You can calculate the ratio of conjugate base (CHO2-)
to acid (HCHO2) required by using the Henderson–Hasselbalch
equation as follows:
16.4 Titrations and pH Curves




In an acid–base titration , a basic (or acidic) solution of
unknown concentration reacts with an acidic (or basic)
solution of known concentration.
The known solution is slowly added to the unknown
one while the pH is monitored with either a pH meter or
an indicator (指示劑)
The equivalence point —the point in the titration when
the number of moles of base is stoichiometrically equal
to the number of moles of acid
A plot of the pH of the solution during a titration is
known as a titration curve or pH curve.
Titrations
Consider the titration of 25.0 mL of 0.100
M HCl with 0.100 M NaOH.
Initial pH (before Adding Any Base)
pH after Adding 5.00 mL NaOH
Titration of a Strong Acid with a Strong Base:
Summarizing the Titration of a Strong
Acid with a Strong Base:
▶ The initial pH is simply the pH of the strong acid solution to
be titrated.
▶ Before the equivalence point, H3O+ is in excess. Calculate the
[H3O+] by subtracting the number of moles of added OHfrom the initial number of moles of H3O+ and dividing by the
total volume.
▶ At the equivalence point, neither reactant is in excess and the
pH = 7.00.
▶ Beyond the equivalence point, OH- is in excess. Calculate the
[OH-] by subtracting the initial number of moles of H3O+
from the number of moles of added OH- and dividing by the
total volume.
EXAMPLE 16.6 Strong Acid–Strong
Base Titration pH Curve
A 50.0 mL sample of 0.200 M sodium
hydroxide is titrated with 0.200 M nitric acid.
Calculate pH:
(a) after adding 30.00 mL of HNO3
(b) at the equivalence point
SOLUTION
(a)
(b)
At the equivalence point, the
strong base has completely
neutralized the strong acid. pH=7
The Titration of a Weak Acid with a
Strong Base

Consider the titration of 25.0 mL of 0.100 M
HCHO2 with 0.100 M NaOH.
Initial pH (before Adding Any Base)
pH after Adding 5.00 mL NaOH
pH after Adding 25.0 mL NaOH
(Equivalence Point)
Con’t
pH after Adding 30.00 mL NaOH
Titration of a Weak Acid with a Strong
Base:
EXAMPLE 16.7 Weak Acid–Strong
Base Titration pH Curve
A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200
M KOH. Calculate:
(a) the volume required to reach the equivalence point
(b) the pH after adding 5.00 mL of KOH
Solution
(a)
(b)
Titration Curve: Weak Base with Strong
Acid
半當量點（half-equivalence
point）:
溶液中之[HA]＝[A－]
[H3O+]＝Ka
The Titration of a Polyprotic Acid



When a diprotic acid is titrated with a
strong base, if Ka1 and Ka2 are
sufficiently different, the pH curve will
have two equivalence points
The first equivalence point in the titration
curve represents the titration of the first
proton
The second equivalence point
represents the titration of the second
proton.
Indicators: pH-Dependent Colors

The endpoint is the
point in the titration
at which the indicator
changes color
Indicators: pH-Dependent Colors
From the Henderson–Hasselbalch equation
TABLE 16.1 Ranges of Color Changes
for Several Acid–Base Indicators
16.5 Solubility Equilibria and the
Solubility Product Constant
An equation can represent the equilibrium
between the compound and the ions present in a
saturated aqueous solution:
EX:
 The equilibrium constant for a chemical equation
representing the dissolution of an ionic compound
is the solubility product constant (Ksp).

TABLE 16.2 Selected Solubility
Product Constants ( Ksp ) at 25°C
Notice that Ksp is not the molar solubility, but the solubility product constant
Ksp and Molar Solubility

The molar solubility is the solubility in units
of moles per liter (mol/L).
EX:
∴The molar solubility of AgCl is
1.33x10-5 mol/L
EXAMPLE 16.8 Calculating Molar
Solubility from Ksp

Calculate the molar solubility of PbCl2 in pure water.
Sol:
EXAMPLE 16.9 Calculating Ksp from
Molar Solubility

The molar solubility of Ag2SO4 in pure water is 1.2x10-5
M. Calculate Ksp.
Sol:
Ksp and Relative Solubility



Molar solubility is related to the value of Ksp, but
molar solubility and Ksp are not the same thing.
In fact, “smaller Ksp” doesn’t always mean
“lower molar solubility.”
Solubility depends on both Ksp and the form of
the equilibrium constant expression.
The Effect of a Common Ion on Solubility


The common ion effect affects solubility equilibria as it
does other aqueous equilibria.
The solubility of a slightly soluble ionic compound is
lowered when a second solute that furnishes a
common ion is added to the solution
EXAMPLE 16.10 Calculating Molar
Solubility in the Presence of a Common Ion

What is the molar solubility of CaF2 in a solution
containing 0.100 M NaF?
Solution
The Effect of pH on Solubility
EXAMPLE 16.11 The Effect of pH on
Solubility
Determine whether each compound is more soluble in an
acidic solution than in a neutral solution.
(a) BaF2 (b) AgI (c) Ca(OH)2
16.6 Precipitation


Precipitation will occur when the concentrations of the ions
exceed the solubility of the ionic compound.
If we compare the reaction quotient, Q, for the current
solution concentrations to the value of Ksp, we can
determine if precipitation will occur.
–
–
–

Q = Ksp, the solution is saturated, no precipitation.
Q < Ksp, the solution is unsaturated, no precipitation.
Q > Ksp, the solution would be above saturation, the salt above
saturation will precipitate.
Some solutions with Q > Ksp will not precipitate unless
disturbed; these are called supersaturated solutions.
Supersaturated solution
EXAMPLE 16.12 Predicting Precipitation
Reactions by Comparing Q and Ksp
Sol:
Selective Precipitation



A solution may contain several different
dissolved metal cations that can often be
separated by selective precipitation.
A successful reagent can precipitate with more
than one of the cations, as long as their Ksp
values are significantly different.
Selective precipitation is also called fractional
precipitation.
Selective Precipitation
EXAMPLE 16.13 Finding the Minimum
Required Reagent Concentration for
Selective Precipitation
The magnesium and calcium ions present in seawater
( [Mg2+] = 0.059 M and [Ca2+] = 0.011 M ) can be
separated by selective precipitation with KOH. What
minimum [OH-] triggers the precipitation of the Mg2+ ion?
Sol:

EXAMPLE 16.14 Finding the
Concentrations of Ions Left in
Solution after Selective Precipitation
Sol:
Mg(OH)2:
16.7 Qualitative Chemical Analysis



An analysis that aims at identifying the
cations present in a mixture is called
qualitative cation analysis.
Qualitative (定性) signifies that the interest is
in determining what is present, not how
much is present
Quantitative (定量) are those that determine
how much of a particular substance or
species is present.
FIGURE 16.15 Qualitative Analysis In
qualitative analysis
Specific ions are precipitated successively by the
addition of appropriate reagents.
Group 1 : Insoluble Chlorides


If aqueous HCl is added to an unknown solution of
cations, and a precipitate forms, then the unknown
contains one or more of these cations: Pb2+, Hg22+, or
Ag+
Analyzing for Pb2+
–
–
–
Precipitated PbCl2 is slightly soluble in hot water.
The precipitate is washed with hot water, then aqueous
K2CrO4 is added to the washings.
If Pb2+ is present, a precipitate of yellow lead chromate
forms, which is less soluble than PbCl2.
Analyzing for Ag+ and Hg22+



Any undissolved precipitate is treated with
aqueous ammonia.
If AgCl is present, it will dissolve, forming
Ag(NH3)2+ (the dissolution may not be
visually apparent).
If Hg22+ is present, the precipitate will turn
dark gray/ black,(forms Hg metal and
HgNH2Cl.)
Group 2: Acid-Insoluble Sulfides
H2S is added to the supernatant liquid –
insoluble sulfide formation is the positive test
(Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+)
 The concentration of HS– is so low in a strongly
acidic solution (HCl) that only the most insoluble
sulfides precipitate

Group 1 & Group 2
Group 3 & Group 4 & Group 5
Group 5: Alkali Metals and NH4+



Group five cations are Na+, K+, NH4+.
All these cations form compounds that are soluble in
water—they do not precipitate.
They are identified by the color of their flame.
16.8 Complex Ion Equilibria

Transition metal ions tend to be good
electron acceptors (good Lewis acids).
–
They often bond to one or more H2O molecules to form a
hydrated ion.
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
H2O is the Lewis base, donating electron pairs to form
coordinate covalent bonds
Complex ion

A complex ion is a polyatomic cation or anion consisting of a
central metal atom or ion that has other groups called ligands
bonded to it
 The metal ion acts as a Lewis acid (accepts electron pairs).
 Ligands act as Lewis bases (donate electron pairs).
 The equilibrium involving a complex ion, the metal ion, and the
ligands may be described through a formation constant, Kf:
Ag+(aq)
+ 2
Cl–(aq)
[AgCl2]–(aq)
[AgCl2]–
Kf = –––––––––– = 1.2 x 108
[Ag+][Cl–]2
ligands
TABLE 16.3 Formation Constants of
Selected Complex Ions in Water at 25 ℃
EXAMPLE 16.15 Complex Ion Equilibria

You mix a 200.0 mL sample of a solution that is 1.5x10-3
M in Cu(NO3)2 with a 250.0 mL sample of a solution that
is 0.20 M in NH3. After the solution reaches equilibrium,
what concentration of Cu2+(aq) remains?
Sol:
Con’t
The Effect of Complex Ion Equilibria on
Solubility
Adding NH3 to a solution in equilibrium with AgCl(s)
increases the solubility of Ag+.
The Effect of Complex Ion Equilibria on
Solubility
Solubility of Amphoteric Metal
Hydroxides


Many metal hydroxides are insoluble.
All metal hydroxides become more soluble in acidic
solution.
–

Some metal hydroxides also become more soluble in
basic solution.
–


Shifting the equilibrium to the right by removing OH−
Acting as a Lewis base forming a complex ion
Substances that behave as both an acid and base are
said to be amphoteric.
Some cations that form amphoteric hydroxides include
Al3+, Cr3+, Zn2+, Pb2+, and Sb2+.
aluminum hydroxide
Al(OH)3 is soluble at high pH and soluble at low pH
but insoluble in a pH-neutral solution.
aluminum hydroxide

Al3+ is hydrated in water to form an acidic solution.
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)

酸性
中性
鹼性
Addition of OH− drives the equilibrium to the right
and continues to remove H from the molecules.