Lecture Presentation James F Kirby Quinnipiac University Hamden CT 20 15 Pearson Education Inc 161 amp 162 Some Definitions Arrhenius An acid is a substance that when dissolved in water increases the concentration of hydrogen ions ID: 746754
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Slide1
Chapter 16Acid–Base Equilibria
Lecture Presentation
James F. KirbyQuinnipiac UniversityHamden, CT
© 2015 Pearson Education, Inc.Slide2
16.1 & 16.2 Some DefinitionsArrhenius
An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions.A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.Brønsted–Lowry
An acid is a proton donor.A base is a proton acceptor.Slide3
Brønsted–Lowry Acid and BaseA Brønsted–Lowry acid must have at least one removable (acidic) proton (H
+) to donate.A Brønsted–Lowry base must have at least one nonbonding pair of electrons to accept a proton (H+).Slide4
What Is Different about Water?Water can act as a Brønsted–Lowry base and accept a proton (H+) from an acid, as on the previous slide.
It can also donate a proton and act as an acid, as is seen below.This makes water amphiprotic.Slide5
Conjugate Acids and BasesThe term conjugate means “joined together as a pair.”
Reactions between acids and bases always yield their conjugate bases and acids.Slide6
Solution
Plan
The conjugate base of a substance is simply the parent
substance minus
one proton, and the conjugate acid of a substance is the
parent substance
plus one proton
.
Solve
(a)
If
we remove a proton from HClO
4
, we obtain
ClO
4
−, which is its conjugate base.
(a
)
What is the conjugate base of HClO4, H2S, PH4+, HCO3−?(b) What is the conjugate acid of CN−, SO42− , H2O, HCO3−?
Sample Exercise 16.1 Identifying Conjugate Acids and Bases
The other conjugate bases are HS−
PH3
CO
3
2−Slide7
Solution
(b)
If
we add a proton to
CN
−
,
we get HCN, its conjugate acid
.
(
b
)
What
is the conjugate acid of
CN
−
, SO
4
2−
, H2O, HCO3−?Sample Exercise 16.1 Identifying Conjugate Acids and Bases
The other
conjugate acids are
HSO
4
−
H
3
O
+
H
2
CO
3
,
Notice
that the
hydrogen carbonate ion (
HCO
3
−
)
is amphiprotic. It can
act as
either an acid or a base
.Slide8
The hydrogen sulfite ion (
HSO
3
−
) is
amphiprotic. Write an equation for the reaction of
HSO
3
–
with
water
(
a)
in which
the ion acts as an acid and
(b)
in which the ion acts as a base. In
both cases identify the conjugate acid–base pairs.
Sample Exercise
16.2 Writing Equations for Proton-Transfer ReactionsSlide9
Solution
Analyze and Plan
We are asked to write two equations representing
reactions between HSO
3
−
and water, one in which
HSO
3
−
should
donate a
proton to water, thereby acting as a
Brønsted
–Lowry acid, and one in which HSO3− should accept a proton from water, thereby acting as
a base. We are also asked to identify the conjugate pairs in each equation.
The hydrogen sulfite ion (
HSO3−) is amphiprotic. Write an equation for the reaction of HSO3– with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs.
Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions
Solve
(a)
HSO
3
−
(
aq
)
+
H2O(l) SO32 − (
aq
)
+ H
3
O
+
(
aq
)
The
conjugate pairs in this equation are HSO3− (acid) and SO32 − (conjugate base), and H2O (base) and H3O+ (conjugate acid).Slide10
Solution
(
b)
HSO
3
−
(
aq
)
+
H
2
O
(l) H2SO
3(aq) +
OH− (aq)
The conjugate pairs in this equation are H2O (acid) and OH− (conjugate base), and HSO3− (base) and H2SO3 (conjugate acid).
(b) in which the ion acts as a base. In both cases identify the conjugate
acid–base pairs.
Sample Exercise
16.2
Writing
Equations for Proton-Transfer ReactionsSlide11
Relative Strengths of Acids and Bases
Acids above the line with H2O as
a base are strong acids; their conjugate bases do not act as acids in water.Bases below the line with H2
O as an acid are strong bases; their conjugate acids do not act as acids in water.
The substances between the lines with H
2
O are conjugate acid–base pairs in water.Slide12
Acid and Base StrengthIn every acid–base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.
HCl(aq) + H2O(l) → H
3O+(aq) + Cl(
aq)H2O is a much stronger base than Cl, so the equilibrium lies far to the right (K >> 1).CH
3
COOH(
aq
) + H
2
O(
l
) ⇌
H
3
O
+
(aq) + CH3COO–(aq)Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1).Slide13
For the following proton-transfer reaction use Figure 16.4 to predict whether the
equilibrium lies
to the left
(
K
c
< 1)
or to the right
(
K
c
> 1):
HSO
4
−
(
aq) + CO32 −
(aq) SO42 − (aq) + HCO3− (aq)Sample Exercise 16.3 Predicting the Position of a Proton-Transfer EquilibriumSlide14
Solution
Analyze
We are asked to predict whether an equilibrium lies to the right, favoring products,
or to the
left, favoring reactants
.
Plan
This is a proton-transfer reaction, and the position of the equilibrium will favor the
proton going
to the stronger of two bases. The two bases in the equation are
CO
3
2
−
,
the base in
the forward reaction, and SO42
−, the conjugate base of HSO4
−. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base.
HSO4− (aq) + CO32 − (aq
) SO42 − (aq)
+ HCO3− (
aq
)Slide15
Solve
The
CO
3
2
−
ion appears lower in the right-hand column in Figure 16.4 and is
therefore a
stronger base than
SO
4
2 −
.
Therefore,
CO
32− will get the proton preferentially to become HCO3−
, while SO42 −
will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is,
Kc > 1):Comment Of the two acids HSO4– and HCO3–, the stronger one (HSO4–) gives up a proton more readily, and the weaker one (HCO3–) tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base.Slide16
16.3 Autoionization of WaterWater is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.This is referred to as autoionization.Slide17
Ion Product ConstantThe equilibrium expression for this process isK
c = [H3O+][OH]
(The term H2O is excluded from the equilibrium constant expression because we exclude the concentrations of pure solids and liquids) This special equilibrium constant is referred to as the ion product constant
for water, Kw.Slide18
Ion Product ConstantThis special equilibrium constant is referred to as the
ion product constant for water, Kw.At 25
°C, Kw = 1.0 10
14Kw = [1.0 10
7
][
1.0 10
7
]
K
w
= [H
3
O
+][OH]Slide19
Aqueous Solutions Can Be Acidic, Basic, or NeutralIf a solution is neutral, [H+] = [OH–
].If a solution is acidic, [H+] > [OH–].If a solution is basic, [H+] < [OH
–].Slide20
Suppose that equal volumes of the middle and right samples in the figure were mixed. Would the resultant solution be acidic, neutral or basic?
basicSlide21
Solution
Analyze
We are asked to determine the concentrations of H
+
and OH
−
ions in a neutral solution at 25
°C.
Plan
We will use Equation 16.16 and the fact that, by definition
, [H
+
]
=
[OH−] in a neutral solution
.Solve We will represent the concentration of H+ and OH
− in neutral solution with x
. This gives [H+][OH−] = (x)(x) = 1.0 × 10−14 x2 = 1.0 × 10−14 x = 1.0 × 10−7M = [H+] = [OH−]In an acid solution [H+] is
greater than 1.0 × 10−7 M; in a basic solution
[H+] is less than 1.0
× 10−7 M.
Calculate the values of
[H
+
]
and
[OH
−
]
in a neutral aqueous solution at 25
°C
.
Sample Exercise
16.4
Calculating
[H
+
]
for Pure WaterSlide22
Calculate the concentration of H
+
(
aq
)
in
(
a)
a solution in which
[OH
−
]
is 0.010
M
,
(
b) a solution in
which [OH−] is 1.8
× 10−9M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 °C.Sample Exercise 16.5 Calculating [H+] from [OH−]Slide23
Solution
Analyze
We are asked to calculate the
[H
+
]
concentration in
an aqueous
solution where the
hydroxide concentration
is known
.
Plan
We can use the equilibrium-constant expression for the
autoionization
of
water and the value of
Kw to solve for each unknown concentration.
Calculate the concentration of H
+(aq) in (a) a solution in which [OH−] is 0.010 MSample Exercise 16.5 Calculating
[H+] from [OH−]Slide24
Solve
(a)
Using
Equation 16.16, we
have
This
solution is basic
because
Calculate the concentration of H
+
(
aq
)
in
(
a)
a solution in which
[OH
−] is 0.010 MSample Exercise 16.5 Calculating [H+] from [OH−]Slide25
(
b
)
In
this instance
This solution is
acidic
because
Calculate the concentration of H
+
(
aq
)
in
(
b)
a
solution in
which [OH−] is 1.8 × 10−9M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 °C.Sample Exercise 16.5 Calculating
[H+] from [OH−]Slide26
16.4 The pH Scale
pH is a method of reporting hydrogen ion
concentration.pH = –log[H+]Neutral pH
is 7.00.Acidic pH is below 7.00.Basic pH is above 7.00.Slide27
Solution
Analyze
We are asked to determine the pH of aqueous solutions
for which
we have already calculated
[H
+
].
Plan
We can calculate pH using its defining equation, Equation 16.17
.
pH
=
–
log[H
+
]
(a) In the first instance we found
[H+] to be 1.0 × 10−12
MCalculate the pH values for the two solutions of Sample Exercise 16.5.Sample Exercise 16.6 Calculating pH from
[H+]Slide28
Sample Exercise
16.6
Calculating
pH from
[H
+
]
Solve
pH =
–
log[H
+
]
(a)
In
the first instance we found
[H+]
to be 1.0
× 10−12 M, so thatpH = −log(1.0 × 10−12) = −(−12.00) = 12.00Because 1.0 × 10−12 has two significant figures, the pH has two decimal places, 12.00.Slide29
(b)
For
the second solution,
[H
+
]
= 5.6
× 10
−6
M
.
Before
performing the
calculation, it is helpful to estimate the
pH.
To do so, we note that [H+
] lies between 1 × 10−6 and 1
× 10−5. Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH
: pH = –log[H+]Calculate the pH values for the two solutions of Sample Exercise 16.5.
Sample Exercise 16.6 Calculating pH from
[H+]Slide30
(b)
For
the second solution,
[H
+
]
= 5.6
× 10
−6
M
.
pH
=
−log(5.6 × 10
−6
)
=
5.25Check After calculating a pH, it is useful to compare it to your estimate
. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have reconsidered our
calculation or estimate or both.Calculate the pH values for the two solutions of Sample Exercise 16.5.Sample Exercise 16.6
Calculating pH from [H+]Slide31
16. 4 (cont) Other “p” ScalesThe “p” in pH tells us to take the –log of a quantity (in this case, hydrogen ions).
Some other “p” systems arepOH: –log[OH]
pKw: –log KwSlide32
Relating pH and pOHBecause[H
3O+][OH] = Kw = 1.0
1014we can take the –log of the equation–log[H3O
+] + –log[OH] = –log Kw = 14.00
which results in
pH + pOH =
p
K
w
= 14.00Slide33
Solution
Analyze
We need to calculate
[H
+
]
from
pOH
.
Plan
We will first use Equation 16.20,
pH
+ pOH = 14.00,
to calculate pH
from pOH.
Then
we will use
pH =
–log[H+]to
determine the concentration of H+.A sample of freshly pressed apple juice has a pOH of 10.24. Calculate [H+].
Sample Exercise 16.7 Calculating [H+] from pOHSlide34
pH
= 14.00
− pOH
pH = 14.00
−
10.24 =
3.76
A sample of freshly pressed apple juice has a
pOH
of 10.24. Calculate [
H
+
].
Sample Exercise
16.7
Calculating
[H
+
] from pOHSlide35
Next
we use Equation 16.17
:
pH
=
−log[H
+
]
=
3.76
Thus,
log[H
+
]
=
–3.76
To find [H
+], we need to determine the antilogarithm of −3.76. Your calculator will show this command as 10x or INV log (these functions are usually above the log key). We use this function to perform the calculation:
[H+] = antilog (−3.76)
= 10−3.76 = 1.7 × 10−4 MComment The number of significant figures in [H+] is two because the number of decimal places in the pH is two.Check Because the pH is between 3.0 and 4.0, we know that [H+] will be between 1.0 × 10−3 M and 1.0 × 10−4 M. Our calculated [H+] falls within this estimated range.
A sample of freshly pressed apple juice has a
pOH of 10.24. Calculate [H+].
Sample Exercise
16.7
Calculating
[H
+
]
from
pOHSlide36
How Do We Measure pH?Indicators, including litmus paper, are used for less accurate measurements; an indicator is one color in its acid form and another color in its basic form.
pH meters are used for accurate measurement of pH; electrodes indicate small
changes in voltage to detect pH.Slide37
Which of these is best suited to distinguish between a solution that is slightly acidic and one that is slightly basic?Slide38
16.5 Strong AcidsYou will recall that the seven strong acids
are HCl, HBr, HI, HNO3
, H2SO4, HClO3,
and HClO4.These are, by definition, strong electrolytes and exist totally as ions in aqueous solution; e.g., HA + H2O → H
3
O
+
+ A
–
So, for the
monoprotic
strong acids,
[H
3
O
+
] = [acid]Slide39
Solution
Analyze
and Plan
Because HClO
4
is a strong acid, it is
completely ionized
, giving
[H
+
]
=
[ClO
4
−
]
= 0.040 M.
SolvepH = −log(0.040) =
1.40Check Because [H
+] lies between 1 × 10−2 and 1 × 10−1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places.
What is the pH of a 0.040
M
solution
of HClO
4
?
Sample Exercise
16.8
Calculating
the pH of a Strong AcidSlide40
Strong BasesStrong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr
2+, and Ba2+).Again, these substances dissociate completely in aqueous solution; e.g., MOH(aq)
→ M+(aq) + OH–(
aq) or M(OH)2(
aq
) → M
2+
(
aq
) + 2 OH
–
(
aq
)Slide41
Strong BasesWhich solution has a higher pH, a 0.001 M solution of
NaOH or a 0.001 M solution of Ba(OH)2?
Both
NaOH
and Ba(OH)
2
are soluble hydroxides.
The hydroxide concentration of
NaOH
are 0.001M and
Ba(OH)
2
is 0.002 M
Ba(OH)
2
has a higher hydroxide concentration, it is more basic and has a higher
pH.
Slide42
Solution
Analyze
We are asked to calculate the pH of two solutions of strong bases
.
Plan
We can calculate each pH by either of two equivalent methods. First, we could use
Equation 16.16
to calculate
[H
+
]
and then use Equation 16.17 to calculate the
pH.
Alternatively, we
could use [OH
−
]
to calculate pOH and then use Equation 16.20 to calculate the pH.
What is the pH of
(a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2?Sample Exercise 16.9 Calculating the pH of a Strong BaseSlide43
Solution
NaOH
dissociates in water to give one
OH
−
ion
per formula unit. Therefore, the
OH
−
concentration for
the solution in (a) equals the stated concentration of
NaOH
, namely 0.028
M
.
Method 1:
What is the pH of
(a)
a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2?Sample Exercise 16.9 Calculating the pH of a Strong BaseSlide44
Solution
Method 2
:
pOH
=
−log(0.028)
= 1.55
pH
= 14.00
−
pOH =
12.45
What is the pH of
(a)
a 0.028
M
solution of
NaOH
,
(b) a 0.0011 M solution of Ca(OH)2?Sample Exercise 16.9 Calculating the pH of a Strong BaseSlide45
(
b)
Ca(OH)
2
is a strong base that dissociates in water to give
two
OH
–
ions per formula unit. Thus, the concentration of
OH
–
(
aq
) for the solution in part (b) is 2 × (0.0011 M) = 0.0022 M
.
Sample Exercise 16.9
Calculating the pH of a Strong BaseSlide46
(
b)
Method
1:
Method
2:
pOH
= −
log(0.0022)
= 2.66
pH
= 14.00 − pOH = 11.34
Sample Exercise 16.9
Calculating the pH of a Strong BaseSlide47
16. 6 Weak AcidsFor a weak acid, the equation for its dissociation is HA(
aq) + H2O(l) ⇌ H3O+(
aq) + A–(aq)Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant, Ka
: Ka = [H3O+][A–]
[
HA]Slide48
16. 6 Weak Acids
The greater the value of Ka, the stronger is the acid.
Based on the table which element is most commonly bonded to the acidic hydrogen?Slide49
Comparing Strong and Weak AcidsWhat is present in solution for a strong acid versus a weak acid?Strong acids completely dissociate to ions.
Weak acids only partially dissociate to ions.Slide50
Calculating Ka from the pHThe pH of a 0.10
M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate Ka for formic acid at this temperature.
We know that
[H3O+][HCOO–][HCOOH]
K
a
=
To calculate
K
a
, we need the equilibrium concentrations of all three things.
We can find [H
3
O
+
], which is the same as [HCOO
–], from the pH.[H3O+] = [HCOO–] = 10–2.38 = 4.2 × 10–3Slide51
Calculating Ka from pH
Now we can set up a table for equilibrium concentrations. We know initial HCOOH (0.10 M) and ion concentrations (0
M); we found equilibrium ion concentrations (4.2 × 10–3 M); so we calculate the change, then the equilibrium HCOOH concentration.
[HCOOH],
M
[H
3
O
+
],
M
[HCOO
],
M
Initially
0.10
0
0
Change
4.2
10
3
+4.2
10
3
+4.2
10
3
At equilibrium
0.10
4.2
10
3
= 0.0958 = 0.10
4.2
10
3
4.2
10
3Slide52
Calculating Ka from pH
[4.2 10
3][4.2 103][0.10]
Ka =
= 1.8
10
4
This allows us to calculate
K
a
by putting in the equilibrium concentrations
.Slide53
Solution
Analyze
We are given the molar concentration of an aqueous solution of weak acid and the pH
of the
solution, and we are asked to determine the value of
K
a
for the acid.
Plan
Although we are dealing specifically with the ionization of a weak acid, this problem is
very similar
to the equilibrium problems we encountered in Chapter 15. We can solve this
problem using
the method first outlined in Sample Exercise 15.8, starting with the chemical reaction
and a
tabulation of initial and equilibrium concentrations
.
A student prepared a 0.10
M
solution of formic acid (HCOOH) and found its pH at 25 °C to be 2.38. Calculate Ka for formic acid at this temperature.Sample Exercise 16.10 Calculating Ka from Measured pHSlide54
Solution
Solve
The first step in solving any equilibrium problem is to write the equation for the
equilibrium reaction
. The ionization of formic acid can be written as
HCOOH(
aq
) H
+
(
aq
)
+
HCOO
–
(aq)The equilibrium-constant expression is
A student prepared a 0.10
M
solution of formic acid (HCOOH) and found its pH at 25 °C to be 2.38. Calculate Ka for formic acid at this temperature.Sample Exercise 16.10 Calculating Ka from Measured pH
From the measured pH, we can calculate [H+] pH = –log [H+] = 2.38log[H+] = –2.38 [H+
] = 10–2.38 = 4.2 × 10–3 MSlide55
To determine the concentrations of the species involved in the equilibrium, we imagine that
the solution
is initially 0.10 M in HCOOH molecules. We then consider the ionization of the
acid into
H
+
and
HCOO
–
.
For each HCOOH molecule that ionizes, one H
+
ion and one
HCOO
–
ion are
produced in solution. Because the pH measurement indicates that [H+
] = 4.2 × 10–3
M at equilibrium, we can construct the following table:
Sample Exercise 16.10 Calculating Ka from Measured pHSlide56
Notice
that we have neglected the very small concentration of H
+
(
aq
)
due to H
2
O
autoionization
. Notice
also that the amount of HCOOH that ionizes is very small compared with the
initial concentration
of the acid. To the number of significant figures we are using, the
subtraction yields
0.10
M
:
(0.10 – 4.2
× 10–3) M ≃ 0.10 MWe can now insert the equilibrium concentrations into the expression for Ka:Check The magnitude of our answer is reasonable because Ka for a weak acid is usually between 10–2 and 10–10.Sample Exercise 16.10
Calculating Ka from Measured pHSlide57
Calculating Percent IonizationPercent ionization = 100
In this example, [H3O+]eq = 4.2 10
3 M [HCOOH]initial = 0.10 M
[H3O+]eq[HA]initial
Percent ionization =
100
4.2
10
3
0.10
= 4.2%Slide58
Solution
Analyze
We are given the molar concentration of an aqueous solution of weak acid and the
equilibrium concentration
of
H
+
(
aq
)
and asked to determine the percent ionization of the acid.
Plan
The percent ionization is given by Equation 16.27
.
As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH)
contains 4.2
×
10
–3
M H+(aq). Calculate the percentage of the acid that is ionized.Sample Exercise 16.11 Calculating Percent IonizationSlide59
Solve
As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH)
contains 4.2
×
10
–3
M
H
+
(
aq
).
Calculate the percentage of the acid that is ionized.
Sample
Exercise
16.11
Calculating Percent IonizationSlide60
Method to Follow to Calculate pH Using KaWrite the chemical equation for the ionization equilibrium.
Write the equilibrium constant expression.Set up a table for Initial/Change in/Equilibrium Concentration to determine equilibrium concentrations as a function of change (x).
Substitute equilibrium concentrations into the equilibrium constant expression and solve for x. (Make assumptions if you can!)Slide61
ExampleCalculate the pH of a 0.30 M solution of acetic acid, HC2H3
O2, at 25 C. HC2H3O2
+ H2O ⇌ H3O+ + C2H3O2– Ka
= [H3O+][C2H3O2–] / [HC2H3O2]
CH
3
COOH (
M
)
H
3
O
+
(
M
)
CH
3COO– (M)
Initial
Concentration (
M
)
0.30
0
0
Change in Concentration (
M
)
–
x
+
x
+
x
Equilibrium Concentration (
M
)
0.30 –
x
x
xSlide62
Example (concluded) Ka = [H
3O+][C2H3O2–] / [HC
2H3O2] = (x)(x) / (0.30 – x)
If we assume that x << 0.30, then 0.30 – x becomes 0.30. The problem becomes easier, since we don’t have to use the quadratic formula to solve it.Ka = 1.8 × 10–5
=
x
2
/ 0.30, so
x
=
2.3 × 10
–3
x
= [H
3
O+], so pH = –log(2.3 × 10–3) = 2.64Slide63
Calculate the pH of a 0.20
M
solution of
HCN. (Refer to
Table
16.2 or Appendix D for
the value of
K
a
.)
Sample
Exercise
16.12
Using
K
a
to Calculate pHSlide64
Solution
Analyze
We are given the molarity of a weak acid and are asked
for the
pH.
From Table 16.2,
K
a
for HCN is
4.9
×
10
–10
.
Plan
We proceed as in the example just worked in the text, writing
the chemical equation and constructing a table of initial and
equilibrium concentrations in which the equilibrium concentration of H+ is
our unknown.
Sample Exercise 16.12 Using Ka to Calculate pHSlide65
Solve
Writing both the chemical equation for
the ionization reaction
that forms
H
+
(
aq
) and
the equilibrium-constant
(
K
a
) expression for
the reaction:
Next
, we tabulate the concentrations of
the species involved in the equilibrium reaction, letting
x = [H+] at equilibrium
:Sample Exercise 16.12 Using Ka to Calculate pHSlide66
Sample Exercise 16.12
Using
K
a
to Calculate pH
Substituting the equilibrium concentrations
into
the equilibrium-constant expression
yieldsSlide67
Sample Exercise 16.12
Using
K
a
to Calculate pH
We next make the simplifying approximation
that
x
, the amount of acid that dissociates, is
small
compared with the initial
concentration of
acid, 0.20 – x ≃ 0.20. Thus,Slide68
Solving for
x
, we
have
Sample Exercise 16.12
Using
K
a
to Calculate pH
A
concentration of 9.9
×
10
–6
M
is much
smaller
than 5% of 0.20, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the pH of the solution:Slide69
Strong vs. Weak Acids—Another ComparisonStrong Acid: [H+]eq
= [HA]initWeak Acid: [H+]eq < [HA]initThis creates a difference in conductivity and in rates of chemical reactions.Slide70
Polyprotic AcidsPolyprotic acids have more than one acidic proton.It is always easier to remove the first proton than any successive proton.
If the factor in the Ka values for the first and second dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation.Slide71
16.7 Weak BasesAmmonia, NH3, is a weak base.Like weak acids, weak bases have an equilibrium constant called the base dissociation constant
.Equilibrium calculations work the same as for acids, using the base dissociation constant instead.Slide72
Base Dissociation ConstantsSlide73
ExampleWhat is the pH of 0.15 M NH3?
NH3 + H2O ⇌ NH4+ + OH–
Kb = [NH4+][OH–] / [NH3] = 1.8 × 10–5
NH
3
(
M
)
NH
4
+
(
M
)
OH
–
(
M)Initial
Concentration (M)
0.15
0
0
Change in Concentration (
M
)
–
x
+
x
+
x
Equilibrium Concentration (
M
)
0.15 –
x
x
xSlide74
Example (completed)1.8 × 10 – 5 = x
2 / (0.15 – x)If we assume that x << 0.15, 0.15 – x = 0.15.
Then: 1.8 × 10–5 = x2 / 0.15and: x = 1.6 × 10–3
Note: x is the molarity of OH–, so –log(x) will be the pOH (pOH = 2.80) and [14.00 – pOH] is pH (pH = 11.20).Slide75
The solubility of CO
2
in water at 25
°C
and 0.1
atm
is 0.0037 M. The common practice is to assume that
all the
dissolved CO
2
is in the form of carbonic acid
(H
2
CO
3
)
,
which is produced in the
reactionCO2(aq) + H2O(
l) H2CO3(aq)What is the pH of a 0.0037 M solution of H2CO3?Sample Exercise 16.14 Calculating the pH of a Solution of a Polyprotic AcidSlide76
Solution
Analyze
We are asked to determine the pH of a 0.0037
M
solution of
a
polyprotic
acid
.
CO
2
(
aq
) + H
2
O(
l
) H
2CO3(aq)What is the pH of a 0.0037 M solution of H2CO3?
Sample Exercise 16.14
Plan
H
2
CO
3
is a diprotic acid; the two acid-dissociation constants
,
K
a
1
and Ka2 (Table 16.3), differ by more than a factor of 10
3
. Consequently
, the
pH can be determined by considering only
K
a
1
,
thereby treating
the acid as if it were a
monoprotic
acid.Slide77
Solve
Proceeding as in Sample
Exercises 16.12
and 16.13, we can write the
equilibrium reaction
and equilibrium
concentrations as
The
equilibrium-
constant expression is
Solving
this quadratic equation, we
get
Alternatively, because
K
a
1 is small, we can make the simplifying approximation that
x is small, so thatSolving for x, we have
Sample
Exercise 16.14x = 4.0 × 10–5 M
0.037 – x ≃ 0.0037Slide78
Because we get the same value (to
2
significant figures
) our simplifying assumption
was justified
. The pH is
therefore
Sample
Exercise 16.14
Calculating the pH of a Solution of a
Polyprotic
AcidSlide79
Comment
If we were asked
for [CO
3
2−
] we
would need
to
use
K
a
2
. Let’s illustrate
that
calculation. Using our
calculated values
of [HCO3−] and [H
+] and setting [CO3
2−] = y, we have
Assuming that y is small relative to4.0 × 10−5, we have
Sample Exercise 16.14Slide80
Types of Weak BasesTwo main categories Neutral substances with an Atom that has a nonbonding pair of electrons that can accept H+ (like ammonia and the
amines) Anions of weak acidsSlide81
16.8 Relationship between Ka and Kb
For a conjugate acid–base pair, Ka and Kb
are related in this way:Ka × Kb =
KwTherefore, if you know one of them, you can calculate the other.
© 2015 Pearson Education, Inc.Slide82
Acid–Base Properties of SaltsMany ions react with water to create H+ or OH–. The reaction with water is often called hydrolysis
.To determine whether a salt is an acid or a base, you need to look at the cation and anion separately.The cation can be acidic or neutral.The anion can be acidic, basic, or neutral.Slide83
AnionsAnions of strong acids are neutral. For example, Cl– will not react with water, so OH
– can’t be formed.Anions of weak acids are conjugate bases, so they create OH– in water; e.g., C2H3O2– + H
2O ⇌ HC2H3O2 + OH–Protonated anions from polyprotic acids can be acids or bases: If Ka > Kb,
the anion will be acidic; if Kb > Ka, the anion will be basic.Slide84
CationsGroup I or Group II (Ca2+, Sr2+, or Ba2+) metal cations are neutral.
Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH4+.Transition and post-transition metal cations are acidic. Why? (There are no H atoms in these cations!)Slide85
Hydrated CationsTransition and post-transition metals form hydrated cations.The water attached to the metal is more acidic than free water
molecules, making the hydrated ions acidic.Slide86
Salt Solutions—Acidic, Basic, or Neutral?Group I/II metal cation with anion of a strong acid: neutral
Group I/II metal cation with anion of a weak acid: basic (like the anion)Transition/Post-transition metal cation or polyatomic cation with anion of a strong acid: acidic (like the cation)Transition/Post-transition metal cation or polyatomic cation with anion of a weak acid: compare K
a and Kb; whichever is greater dictates what the salt is.Slide87
Factors that Affect Acid StrengthH—A bond must be polarized with δ+ on the H atom and
δ– on the A atom
Bond strength: Weaker bonds can be broken more easily, making the acid stronger.Stability of A–: More stable anion means stronger acid.Slide88
Binary AcidsBinary acids consist of H and one other element.
Within a group, H—A bond strength is generally the most important factor.Within a period, bond polarity is the most important factor to determine acid strength.Slide89
OxyacidsOxyacids consist of H, O, and one other element, which is a nonmetal.Generally, as the electronegativity
of the nonmetal increases, the acidity increases for acids with the same structure.Slide90
Oxyacids with Same “Other” ElementIf an element can form more than one oxyacid, the oxyacid with more O atoms is more acidic; e.g., sulfuric acid versus sulfurous acid.Another way of saying it: If the oxidation number increases, the acidity increases.Slide91
Carboxylic AcidsCarboxylic acids are organic acids containing the —COOH group.Factors contributing to their acidic behavior:
Other O attached to C draws electron density from O—H bond, increasing polarity.Its conjugate base (carboxylate anion) has resonance forms to stabilize the anion.Slide92
Lewis Acid/Base ChemistryLewis acids are electron pair acceptors.Lewis bases are electron pair donors.All Brønsted–Lowry acids and bases are also called Lewis acids and bases.There are compounds which do not meet the Brønsted–Lowry definition which meet the Lewis definition.Slide93
Comparing Ammonia’s Reaction with H+ and BF3