18
Ionic Equilibria I 離子平衡:
維他命C
Acids
and
Bases
Ascorbic acid
檸檬酸
Citric acid
Chapter Goals
1.A Review of Strong Electrolytes (強電解質)
2.The Autoionization of Water (水的自身離子化)
3.The pH and pOH Scales (pH值和pOH值)
4.Ionization Constants for Weak Monoprotic Acids and
Bases (弱單質子酸及鹼的解離常數)
5.Polyprotic Acids (多質子酸)
6.Solvolysis (溶劑分解)
7.Salts of Strong Bases and Strong Acids (強鹼及強酸形成的鹽)
8.Salts of Strong Bases and Weak Acids (強鹼及弱酸形成的鹽)
9.Salts of Weak Bases and Strong Acids (弱鹼及強酸形成的鹽)
10.Salts of Weak Bases and Weak Acids (弱鹼及弱酸形成的鹽)
11.Salts That Contain Small, Highly Charged Cations
2
A Review of Strong
Electrolytes
•This chapter details the equilibria of weak acids
and bases.
–We must distinguish weak acids and bases from
strong electrolytes.
•Weak acids and bases ionize or dissociate
partially, much less than 100%.
–In this chapter we will see that it is often less
than 10%!
•Strong electrolytes ionize or dissociate
completely. (強電解質完全解離)
–Strong electrolytes approach 100% dissociation
in aqueous solutions. (強電解質在水溶液中幾近
100%解離)
3
Aqueous Solutions: An
Introduction
• The reason nonelectrolytes do not conduct
electricity is because they do not form ions in
solution.
•ions conduct electricity in solution
Acid: a substance that produces hydrogen ions, H+,
in aqueous solutions
Base: a substance that produces hydroxide ions,
OH-, in aqueous solutions
Salt: a compound that contains a cation other than
H+, and an anion other than hydroxide ion,
OH- , or oxide ion, O24
Aqueous Solutions: An
Introduction
•Classification of solutes
–strong electrolytes強電解質- conduct electricity
extremely well in dilute aqueous solutions
•Examples of strong electrolytes
1.HCl, HNO3, etc.
– strong soluble acids
2.NaOH, KOH, etc.
– strong soluble bases
3.NaCl, KBr, etc.
– soluble ionic salts
– ionize in water essentially 100%
檸檬酸
5
Aqueous Solutions: An
Introduction
•Classification of solutes
–weak electrolytes - conduct electricity poorly in
dilute aqueous solutions
1.CH3COOH, (COOH)2
•weak acids
2.NH3, Fe(OH)3
•weak bases
3.some soluble covalent salts
•ionize in water much less than 100%
6
Aqueous Solutions: An
Introduction
Strong and Weak Acids
• Acids are substances that generate H+ in aqueous
solutions.
• Strong acids ionize 100% in water.
HCl(g)
100%
H+(aq) + Cl-(aq)
HNO3 + H2O 100%
HNO3
H2O
H3O+(aq) + NO-3(aq)
H+(aq) + NO-3(aq)
7
Aqueous Solutions: An
acids--ionize
Introductionstrong
almost 100%
氫氯酸
氫溴酸
氫碘酸
硝酸
過氯酸
氯酸
硫酸
8
Aqueous Solutions: An
Weak acids--Typically
Introduction
ionize 10% or less!
氫氟酸
醋酸
氰化氫
亞硝酸
碳酸
亞硫酸
磷酸
草酸
9
Aqueous Solutions: An
Introduction
Strong Bases, Insoluble Bases, and Weak
Bases
•
Strong Bases
–Characteristic of common inorganic bases is that
they produce OH- ions in solution.
–Similarly to strong acids, strong bases ionize 100%
in water.
NaOH
H2O
Na+(aq) + OH-(aq)
Ba(OH)2
H2O
Ba+2(aq) + 2OH-(aq)
10
Aqueous Solutions: An
Introduction
氫氧化鋰
氫氧化鈉
氫氧化鉀
氫氧化銣
氫氧化銫
氫氧化鈣
氫氧化鍶
氫氧化鋇
Notice that they are all hydroxides of IA and IIA metals
11
Aqueous Solutions: An
Introduction
Strong Bases, Insoluble Bases, and Weak
Bases
•
Insoluble bases
– Ionic compounds that are insoluble in water,
consequently, not very basic.
– Cu(OH)2, Zn(OH)2, Fe(OH)2氫氧化亞鐵, Fe(OH)3
• Weak bases
– are covalent compounds that ionize slightly in
water.
– Ammonia is most common weak base-- NH3
NH3(g) + H2O(l)
NH4+(aq) + OH-(aq)
12
13
A Review of Strong
Electrolytes
Most Water Soluble Salts
The solubility guidelines from Chapter 4 will help
you remember these salts.
NaCl(s)
Ca(NO3)2
H2O
100%
H2O
100%
Na+(aq) + Cl-(aq)
Ca+2(aq) + 2NO3-(aq)
14
A Review of Strong
Electrolytes
•The calculation of ion concentrations in solutions of
strong electrolytes is easy.
Example 18-1: Calculate the concentrations of ions in
0.050 M nitric acid, HNO3.
HNO3 + H2O
0.050M
100%
H3O+(aq) + NO-3(aq)
0.050M
0.050M
15
A Review of Strong
Electrolytes
Example 18-1 Calculation of Concentrations of Ions
Calculation the molar concentration of Ba2+ and OH- ions in
0.03M barium hydroxide.
Ba(OH)2
H2O
initial
0.03M
change -0.03M
final
0.0M
[Ba2+]=0.03M
Ba2+(aq) + 2OH-(aq)
+0.03M
0.03M
+2x(0.03)M
+0.06M
[OH-]=0.06M
Exercise 4 and 6
16
A Review of Strong
Electrolytes
Example 18-2: Calculate the concentrations of ions in 0.020 M
strontium hydroxide, Sr(OH)2, solution.
Sr(OH)2
H2O
initial
0.02M
change -0.02M
final
0.0M
[Sr2+]=0.02M
Sr2+(aq) + 2OH-(aq)
+0.02M
0.02M
+2x(0.02)M
+0.04M
[OH-]=0.04M
17
The Autoionization of Water
水的自身離子化[反應]
•Pure water ionizes very slightly.
–The concentration of the ionized water is less than
one-millionth molar at room temperature.
•We can write the autoionization of water as a
dissociation reaction similar to those previously done
in this chapter.
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
•Because the activity of pure water is 1, the
equilibrium constant for this reaction is:
Kc=[H3O+][OH-]
18
The Autoionization of Water
•Experimental measurements have determined that the
concentration of each ion is 1.0 x 10-7 M at 25oC.
–Note that this is at 25oC, not every temperature!
•We can determine the value of Kc from this information.
Kc=[H3O+][OH-]
= (1.0x10-7)x(1.0x10-7)
= 1.0x10-14
19
The Autoionization of Water
•This particular equilibrium constant is called the ionproduct for water and given the symbol Kw.
–Kw is one of the recurring expressions for the
remainder of this chapter and Chapters 19 and 20.
Kw=[H3O+][OH-]
= 1.0x10-14
20
Example 18-2 Calculation of Ions Concentrations
Calculation the concentrations of H3O+ and OH- ions in 0.05M
HNO3 solution.
Strong acid HNO3 + H2O
H3O+(aq) + NO3- (aq)
initial
0.05M
change -0.05M
+0.05M
+0.05M
At equil
0.0M
0.05M
+0.05M
[H3O+]=[NO3-]=0.05M
2 H2O(l)
H3O+(aq) + OH-(aq)
0.05M
initial
+xM
change -2xM
+xM
+xM
At equil
(0.05+x)M
Kw=[H3O+][OH-]
1.0x10-14 =(0.05+x)(x) (very small number)
1.0x10-14 =(0.05)(x)
x=2.0x10-13 M=[OH-]
Exercise 14
21
Example 18-3: Calculate the concentrations of H3O+ and OHin 0.050 M HCl.
HCl + H2O
H3O+ + Cl-
initial
0.05M
change -0.05M
+0.05M +0.05M
At equil
0.0M
0.05M +0.05M
[H3O+]=[Cl-]=0.05M
2 H2O(l)
H3O+(aq) + OH-(aq)
0.05M
initial
+xM
change -2xM
+xM
+xM
At equil
(0.05+x)M
Kw=[H3O+][OH-]
1.0x10-14 =(0.05+x)(x) (very small number)
1.0x10-14 =(0.05)(x)
x=2.0x10-13 M=[OH-]
22
The pH and pOH scales
• A convenient way to express the acidity and
basicity of a solution is the pH and pOH scales.
• The pH of an aqueous solution is defined as:
pH=-log[H3O+]
or [H3O+]=10-pH
pOH=-log[OH-]
or [OH-]=10-pOH
23
The pH and pOH scales
•If either the [H3O+] or [OH-] is known, the pH and pOH
can be calculated.
Example 18-4: Calculate the pH of a solution in which
the [H3O+] =0.030 M.
pH=-log[H3O+]
pH=-log(3.0x10-2)
pH=1.52
24
The pH and pOH scales
Example 18-5: The pH of a solution is 4.597. What is the
concentration of H3O+?
pH=-log[H3O+]
4.597=-log[H3O+]
[H3O+]=10-4.597
[H3O+]=2.53x10-5M
25
Example 18-3 Calculation of pH
Calculate the pH of a solution in which the H3O+
concentration is 0.050mol/L.
pH=-log[H3O+]
pH=-log(5.0x10-2)
pH=1.3
Exercise 22
Example 18-4 Calculation of H3O+ concentration from pH
The pH of a solution is 3.301. What is the concentration of
H3O+ in this solution?
pH=-log[H3O+]
3.301=-log[H3O+]
[H3O+]=10-3.301
[H3O+]=5.0x10-4M
Exercise 24
26
The pH and pOH scales
•A convenient relationship between pH and pOH may
be derived for all dilute aqueous solutions at 250C.
[H3O+][OH-]= 1.0x10-14
•Taking the logarithm of both sides of this equation
gives:
log[H3O+]+log[OH-]= -14.0
•Multiplying both sides of this equation by -1 gives:
-log[H3O+]+(-log[OH-])= 14.0
•Which can be rearranged to this form:
pH + pOH = 14.0
27
The pH and pOH scales
•Remember these two expressions!!
– They are key to the next three chapters!
[H3O+][OH-]= 1.0x10-14
pH + pOH = 14.0
28
The pH and pOH scales
•The usual range for the pH scale is 0 to 14.
[H3O+]=1.0 M to
pH=0
to
[H3O+]= 1.0x10-14M
pH=14.0
•And for pOH the scale is also 0 to 14 but inverted from
pH.
–pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.
[OH-]= 1.0x10-14 up to [OH-]=1.0M
pOH=14.0
to pOH=0
29
The pH and pOH scales
Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a
0.020 M HNO3 solution.
– Is HNO3 a weak or strong acid?
– What is the [H3O+] ?
Strong acid HNO3 + H2O
H3O+(aq) + NO3- (aq)
initial
0.02M
change -0.02M
+0.02M
+0.02M
At equil
0.0M
0.02M
+0.02M
[H3O+]= 2x10-2 M
pH=-log(2x10-2 M)
pH=1.70
Kw=[H3O+][OH-]=1.0x10-14
[OH-]=1.0x10-14/[H3O+]
=1.0x10-14/2.0x10-2
=5.0x10-13 M
pOH=-log[OH-]
=-log(5.0x10-13)
=12.30
30
The pH and pOH scales
•To help develop familiarity with the pH and pOH scale
we can look at a series of solutions in which [H3O+]
varies between 1.0 M and 1.0 x 10-14 M.
[H3O+]
[OH-]
pH
pOH
1.0 M
1.0 x 10-14 M
0.00
14.00
1.0 x 10-3 M
1.0 x 10-7 M
1.0 x 10-11 M
1.0 x 10-7 M
3.00
7.00
11.00
7.00
2.0 x 10-12 M
5.0 x 10-3 M
11.70
2.30
1.0 x 10-14 M
1.0 M
14.00
0.00
31
Example 18-6 Calculations Involving pH and pOH
Calculate the [H3O+], pH, [OH-], and pOH for a 0.015
M Ca(OH)2 solution.
Ca(OH)2
initial
0.015M
change -0.015M
At equil
0.0M
[OH-]= 3x10-2 M
pH=14-pOH
=14-1.52
=12.48
Ca2+ + 2OH-(aq)
+0.015M +2x(0.015)M
0.015M +0.03M
pOH=-log(3x10-2 M)
pOH=1.52
Kw=[H3O+][OH-]=1.0x10-14
[H3O+]=1.0x10-14/[OH-]
=1.0x10-14/3.0x10-2
=3.3x10-13 M
Exercise 26 and 37
32
33
胃酸
炭酸飲料
檸檬
醋
蕃茄
啤酒
尿液
牛奶
唾液
血漿
蛋白
鎂乳(瀉藥)
氨水
34
pH meter
Universal indicator
35
Universal pH paper
Ionization Constants for Weak
Monoprotic Acids單質子酸 and Bases
Strong acids ionize completely in dilute aqueous,
whereas weak acids ionize only slightly
•Let’s look at the dissolution of acetic acid, a weak
acid, in water as an example.
•The equation for the ionization of acetic acid is:
CH3COOH + H2O  H3O+ + CH3COO-
•The equilibrium constant for this ionization is
expressed as:
[H3O+] [CH3COO-]
K c=
[CH3COOH] [H2O]
36
Ionization Constants for Weak
Monoprotic Acids and Bases
•The water concentration in dilute aqueous solutions is
very high.
•1 L of water contains 55.5 moles of water.
•Thus in dilute aqueous solutions:
[H2O]  55.5M
37
Ionization Constants for Weak
Monoprotic Acids and Bases
•The water concentration is many orders of magnitude
greater than the ion concentrations.
•Thus the water concentration is essentially that of pure
water.
–Recall that the activity of pure water is 1.
[H3O+] [CH3COO-]
K c=
[CH3COOH] [H2O]
[H3O+] [CH3COO-]
Kc [H2O] =
[CH3COOH]
[H3O+] [CH3COO-]
K
[CH3COOH]
38
Ionization Constants for Weak
Monoprotic Acids and Bases
•We can define a new equilibrium constant for weak
acid equilibria that uses the previous definition.
– This equilibrium constant is called the acid ionization
constant 酸解離常數.
– The symbol for the ionization constant is Ka.
[H3O+] [CH3COO-]
Ka =
[CH3COOH]
=1.8x10-5
for acetic acid
39
Ionization Constants for Weak
Monoprotic Acids and Bases
•In simplified form the dissociation equation and acid
ionization expression are written as:
CH3COOH  H+ + CH3COO[H+] [CH3COO-]
=1.8x10-5
Ka =
[CH3COOH]
40
Ionization Constants for Weak
Monoprotic Acids單質子酸 and Bases
Carbonic acid H2CO3
Citric acid C3H5O(COOH)3
41
Ionization Constants for Weak
Monoprotic Acids and Bases
•From the above table we see that the order of
increasing acid strength for these weak acids is:
HF > HNO2 > CH3COOH > HClO > HCN
•The order of increasing base strength of the anions
(conjugate bases) of these acids is:
F- < NO2- < CH3COO- < ClO- < CN-
42
Ionization Constants for Weak
Monoprotic Acids and Bases
Example 18-8: Write the equation for the ionization of the
weak acid HCN and the expression for its ionization
constant.
+
-
HCN  H + CN
[H+] [CN-]
Ka =
[HCN]
=4.0x10-10
43
Example 18-9: In a 0.12 M solution of a weak monoprotic
acid, HY, the acid is 5.0% ionized. Calculate the
ionization constant for the weak acid.
HY  H+ + Y-
[H+] [Y-]
Ka =
[HY]
• Since the weak acid is 5.0% ionized, it is also 95% unionized.
• Calculate the concentration of all species in solution.
[H+]=[Y-]=0.05 x (0.12M)
=0.006M
=6x10-3M
[HY]=0.95 x (0.12M)
=0.11M
(6.0x10-3)(6.0x10-3)
[H+] [Y-]
Ka =
=
[HY]
(0.11)
Ka =3.3x10-4
44
Example 18-10: The pH of a 0.10 M solution of a weak
monoprotic acid, HA, is found to be 2.97. What is the
value for its ionization constant?
pH = 2.97 so [H+]= 10-pH
[H+]= 10-2.79 [H+]=1.1x 10-3
•Use the [H3O+] and the ionization reaction to determine
concentrations of all species.
HA  H+ + A-
At equil
(0.1-1.1x10-3)M 1.1x10-3M
1.1x10-3M
 0.1M
•Calculate the ionization constant from this information.
(1.1x10-3)(1.1x10-3)
[H+] [A-]
=
Ka =
0.1
[HA]
Ka =1.2x10-5
45
Example 18-7 Calculations of Ka and pKa from equilibrium
Concentrations
Nicotinic acid is a weak monoprotic oganic acid that
we can represent as HA.
HA + H2O  H3O+ + A-
A dilute solution of nicotinic acid was found to contain
the following concentrations at equilibrium at 25oC.
What are the Ka and pKa values? [HA]=0.049M;
[H3O+]=[A-]=8.4x10-4M.
[H3O+] [A-]
(8.4x10-4)(8.4x10-4)
=
Ka =
0.049
[HA]
Ka =1.4x10-5
pKa =-log(1.4x10-5)
=4.85
Exercise 30
46
Example 18-8 Calculations of Ka from Percent Ionization
In 0.01M solution, acetic acid is 4.2% ionized. Calculate
its ionization constant.
CH3COOH  H+ + CH3COO-
• Since the weak acid is 4.2% ionized, it is also 95.8%
unionized.
• Calculate the concentration of all species in solution.
[H+]=[CH3COO-]=0.042 x (0.01M)
=4.2x10-4M
[CH3COOH]=0.958 x (0.01M)
=9.58x10-3M
[H3O+] [CH3COO-]
Ka=
[CH3COOH]
(4.2x10-4)(4.2x10-4)
=
(9.58x10-3)
= 1.8x10-5
Exercise 38
47
Example 18-9 Calculations of Ka from pH
The pH of a 0.115M solution of chloroacetic acid,
ClCH2COOH, is measure to be 1.92. Calculate Ka from
this weak momoprotic acid.
pH=-log[H3O+]
[H3O+]=10-pH=10-1.92=0.012M
HA + H2O  H3O+ + Ainitial
0.115M
change -0.012M
+0.012M +0.012M
At equil 0.103M
0.012M 0.012M
[H3O+] [A-]
Ka=
[HA]
(0.012)(0.012)
=
(0.103)
= 1.4x10-3
Exercise 40
48
Example 18-11: Calculate the concentrations of the various
species in 0.15 M acetic acid, CH3COOH, solution.
CH3COOH  H+ + CH3COOinitial
0.15M
change
+xM
+xM
-xM
At equil (0.15-x)M
xM
xM
[H3O+] [CH3COO-]
Ka=
=1.8x10-5
[CH3COOH]
(x)(x)
x is small enough to
Ka=
=1.8x10-5
(0.15-x)
ignore compare to
0.15M
x2(0.15)x(1.8x10-5)
x22.7x10-6
x=1.6x10-3 M=[H3O+]=[CH3COO-]
[CH3COOH]=0.15-1.6x10-3  0.15M
49
Ionization Constants for Weak
Monoprotic Acids and Bases
•Let us now calculate the percent ionization for the
0.15 M acetic acid. From Example 18-11, we know
the concentration of CH3COOH that ionizes in this
solution. The percent ionization of acetic acid is
[CH3COOH]ionized
% ionization=
x100%
[CH3COOH]original
(1.6x10-3M)
x100%
% ionization=
0.15M
=1.1%
50
Example 18-12: Calculate the concentrations of the
species in 0.15 M hydrocyanic acid, HCN, solution.
Ka= 4.0 x 10-10 for HCN
HCN + H2O  H3O+ + CNinitial
0.15M
change
+xM
+xM
-xM
At equil (0.15-x)M
xM
xM
[H3O+] [CN-]
(x)(x)
-10
Ka =
=4.0x10 = (0.15-x)
[HCN]
x is small enough to
x2(0.15)x(4.0x10-10)
ignore compare to 0.15M
-6
+
x=7.7x10 M=[H3O ]=[CN ]
[HCN]=(0.15-7.7x10-6)  0.15M
[HCN]ionized
% ionization=
x100%
[HCN]original
=(7.7x10-6/0.15)x100%
=0.0051%
51
Ionization Constants for Weak
Monoprotic Acids and Bases
•Let’s look at the percent ionization of two weak acids
as a function of their ionization constants. Examples
18-11 and 18-12 will suffice.
Solution
Ka
[H+]
pH
% ionization
0.15 M acetic acid
1.8 x 10-5
1.6 x 10-3
2.80
1.1
0.15 M HCN
4.0 x 10-10
7.7 x 10-6
5.11
0.0051
• Note that the [H+] in 0.15 M acetic acid is 210 times
greater than for 0.15 M HCN.
52
•All of the calculations and understanding we have at
present can be applied to weak acids and weak bases!
Example 18-13: Calculate the concentrations of the various
species in 0.15 M aqueous ammonia.
NH3 + H2O  NH4+ + OHinitial
0.15M
change
+xM
+xM
-xM
At equil (0.15-x)M
xM
xM
[NH4+] [OH-]
(x)(x)
-5
Kb=
=1.8x10 =
(0.15-x)
[NH3]
x is small enough to
x2(0.15)x(1.8x10-5)
ignore compare to 0.15M
-3
+
x=1.6x10 M=[NH4 ]=[OH ]
[NH3]=(0.15-1.6x10-3)  0.15M
[NH3]ionized
% ionization=
x100%
[NH3]original
=(1.6x10-3/0.15)x100%
=1.1%
53
Example 18-14: The pH of an aqueous ammonia solution is
11.37. Calculate the molarity (original concentration) of
the aqueous ammonia solution.
pH=11.37
pOH+pH=14
pOH=14-11.37=2.63
[OH-]=10-pOH=10-2.63=2.3x10-3M
NH3 + H2O  NH4+ + OHinitial
xM
change 2.3x10-3M
2.3x10-3M 2.3x10-3M
At equil (x-2.3x10-3)M 2.3x10-3M 2.3x10-3M
-3)x(2.3x10-3)
(2.3x10
[NH4+] [OH-]
Kb=
=1.8x10-5 =
(x-2.3x10-3)
[NH3]
x(2.3x10-3)(2.3x10-3)/(1.8x10-5)
x=0.3 M=[NH3]
54
55
Example 18-10 Calculation of Concentrations from Ka
(a) Calculate the concentration of the various species in
0.10M hypochlorous acid, HOCl.For HOCl, Ka=3.5x10-8. (b)
What is the pH of this solution?
HOCl + H2O  H3O+ + OClinitial
0.10M
change
-xM
+xM
+xM
At equil (0.10-x)M
xM
xM
[H3O+] [OCl-]
(x)(x)
Ka=
=
= 3.5x10-8
[HOCl]
(0.10-x)
x2(0.1)(3.5x10-8)
x5.9x10-5=[H3O+]=[OCl-]
[HOCl]=(0.10-5.9x10-5)=0.10M
pH=-log(5.9x10-5)
= 4.23
Exercise 42
56
Example 18-11 Percent Ionization
Calculate the percent ionization of a 0.10 M solution of
acetic acid.
CH3COOH + H2O  H+ + CH3COOinitial
0.10M
change
+xM
+xM
-xM
At equil (0.10-x)M
xM
xM
(x)(x)
[H3O+] [CH3COO-]
-5
Ka=
=1.8x10 = (0.10-x)
[CH3COOH]
x is small enough to
x2(0.1M)x(1.8x10-5)
ignore compare to 0.10M
-3
+
x=1.3x10 M=[H3O ]=[CH3COO ]
[CH3COOH]=0.10-1.3x10-3  0.10M
[CH3COOH]ionized
% ionization=
x100%
[CH3COOH]original
(1.3x10-3M)
=
x100% =1.3%
0.10M
Exercise 48
57
Example 18-12 pKa Value
The Ka values for acetic acid and hydrofluoric acid are
1.8x10-5 and 7.2x10-4, respectively. What are their pKa
value?
For HF
For CH3COOH
pKa=-logKa
pKa=-logKa
=-log(7.2x10-4)
=-log(1.8x10-5)
=3.14
=4.74
Exercise 50
Example 18-13 Acid Strength and Ka Value
Given the following lost of weal acids and their Ka values,
arrange the acids in order of (a) increasing acid strength
and (b) increasing pKa values.
Acid
Ka
(a) Increasing acid strength
HOCl 3.5x10-8
-10
HCN 4.0x10
HCN<HOCl<HNO2
-4
HNO2 4.5x10
(b) Increasing pKa values
HNO2<HOCl<HCN
Exercise 109
58
Polyprotic Acids
•Many weak acids contain two or more acidic
hydrogens.
–Examples include H3PO4 and H3AsO4.
•The calculation of equilibria for polyprotic acids is
done in a stepwise fashion.
–There is an ionization constant for each step.
•Consider arsenic acid, H3AsO4, which has three
ionization constants.
1. Ka1 = 2.5 x 10-4
2. Ka2 = 5.6 x 10-8
3. Ka3 = 3.0 x 10-13
59
Polyprotic Acids
•The first ionization step for arsenic acid is:
H3AsO4  H+ + H2AsO4[H+] [H2AsO4-]
Ka1=
[H3AsO4]
=2.5x10-4
•The second ionization step for arsenic acid is:
H2AsO4-  H+ + HAsO42[H+] [HAsO42-]
Ka2=
[H2AsO4-]
=5.6x10-8
60
Polyprotic Acids
•The third ionization step for arsenic acid is:
HAsO42-  H+ + AsO43Ka3=
[H+] [AsO43-]
[H2AsO42-]
=3.0x10-13
•Notice that the ionization constants vary in the
following fashion:
Ka1>Ka2>Ka3
•This is a general relationship.
–For weak polyprotic acids the Ka1 is always > Ka2,
etc.
61
Polyprotic Acids
Example 18-15: Calculate the concentration of all species
in 0.100 M arsenic acid, H3AsO4, solution.
H3AsO4  H+ + H2AsO4-
At equil
(0.10-x)M
xM
xM
(x)(x)
[H+] [H2AsO4-]
=2.5x10-4
=
Ka1=
[H3AsO4]
(0.10-x)
X2+2.5x10-4X-2.5x10-5=0
-2.5x10-4  (2.5x10-4)2-4(1)(-2.5x10-5)
x=
2x1
x=-5.1x10-3 and x=4.9x10-3
[H+]=[H2AsO4-]=4.9x10-3M
[H3AsO4]=(0.10-4.9x10-3) M =0.095M
62
Polyprotic Acids
 Next,
write the equation for the second step ionization and
represent the concentrations.
H2AsO4-  H+ + HAsO4-2
(4.9x10-3)M (4.9x10-3)M
At equil (4.9x10-3-y)M
yM
yM
(4.9x10-3+y)(y)
[H+] [HAsO4-2]
-8
=5.6x10
=
Ka2=
-3
(4.9x10 -y)
[H AsO -]
[] from 1st step
2
4
y<<4.9x10-3
Thus (4.9x10-3 –y)  4.9x10-3
(4.9x10-3)(y)
=5.6x10-8
Ka2= (4.9x10-3)
y=5.6x10-8=[H+]2nd=[HAsO42-]
Note the [H+]1st >>[H+]2nd
63
 Finally,
Polyprotic Acids
repeat the entire procedure for the third ionization step.
H+
+ AsO43-
(5.6x10-8)M (4.9x10-3+5.6x10-8)M
(5.6x10-8-z)M
zM
zM
[] from 1st and 2nd step
At equil
HAsO42- 
[H+] [AsO4-3]
Ka3=
[HAsO4-]
(4.9x10-3+5.6x10-8+z)(z)
=
(5.6x10-8-z)
Thus (5.6x10-8 –z)  5.6x10-8
z<<5.6x10-8
=3.0x10-13
(4.9x10-3)(z)
=3.0x10-13
Ka3= (5.6x10-8)
z=3.4x10-18=[H+]3rd=[AsO43-]
• Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.
[H+][OH-] =1.0x10-14
(4.9x10-3)x[OH-]=1.0x10-14
[OH-]=2.0x10-12
64
Polyprotic Acids
• A comparison of the various species in 0.100 M H3AsO4
solution follows.
Species
Concentration
H3AsO4
0.095 M
H+
0.0049 M
H2AsO4-
0.0049 M
HAsO42-
5.6 x 10-8 M
AsO43-
3.4 x 10-18 M
OH-
2.0 x 10-12 M
65
Example 18-18 Solutions Strong Polyprotic Acid
Calculate concentration of all species present in 0.10M H2SO4.
Ka2=1.2x10-2.
H2SO4 + H2O  H3O+ + HSO4100% ionization
0.10M 0.10M
0.10M
HSO4- + H2O  H3O+ + SO4-2
0.10M
0.10M
+xM
-xM
+xM
(0.10-x)M
(0.10+x)M xM
[H3O+] [SO4-2]
(0.1+x)(x)
-2
Ka2=
=
=
1.2x10
(0.1-x)
[HSO4-]
x2+0.1x= 1.2x10-3 – (1.2x10-2)x
x=0.01 or x=-0.12
[H3O+]2nd=[SO42-]=0.01M
[H2SO42-]  0M [HSO4-]=0.1-0.01M=0.09 M
[H3O+]=0.10+0.01M=0.11 M
[OH-]=1.0x10-14/0.11=9.1x10-14 M
Exercise 38
66
Solvolysis溶劑分解[作用]
•This reaction process is the most difficult concept in
this chapter.
•Solvolysis is the reaction of a substance with the
solvent in which it is dissolved.
•Hydrolysis refers to the reaction of a substance with
water or its ions.
•Combination of the anion of a weak acid with H3O+
ions from water to form nonionized weak acid
molecules.
67
Solvolysis
•Hydrolysis refers to the reaction of a substance with
water or its ions.
–Hydrolysis is solvolysis in aqueous solutions.
•The combination of a weak acid’s anion with H3O+
ions, from water, to form nonionized weak acid
molecules is a form of hydrolysis.
A- + H3O+  HA + H2O
Recall H2O + H2O  H3O+ + OH-
•The reaction of the anion of a weak monoprotic acid
with water is commonly represented as:
A- + H2O  HA + OH-
The removal of H3O+ upsets the water
equilibrium
68
Solvolysis
•Recall that at 25oC
•in neutral solutions:
[H3O+] = 1.0 x 10-7 M = [OH-]
•in basic solutions:
[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M
•in acidic solutions:
[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M
69
Solvolysis
•Remember from BrØnsted-Lowry acid-base
theory:
–The conjugate base of a strong acid is a
very weak base.
–The conjugate base of a weak acid is a
stronger base.
70
Solvolysis
Hydrochloric acid, a typical strong acid, is
essentially completely ionized in dilute aqueous
solutions.
HCl +H2O
~100%
H3O++Cl-
• The conjugate base of HCl, the Cl- ion, is a very
weak base.
– The chloride ion is such a weak base that it will
not react with the hydronium ion.
Cl- + H3O+ → No rxn. In dilute aqueous solutions
• This fact is true for all strong acids and their anions.
71
Solvolysis
•HF, a weak acid, is only slightly ionized in dilute
aqueous solutions.
•Its conjugate base, the F- ion, is a much stronger base
than the Cl- ion.
•The F- ions combine with H3O+ ions to form nonionized
HF.
–Two competing equilibria are established.
HF +H2O
H3O++F-
H3O++F-
Only slightly
HF +H2O
Nearly completely
72
Solvolysis
•Dilute aqueous solutions of salts that
contain no free acid or base come in four
types:
1. Salts of Strong Bases and Strong Acids
2. Salts of Strong Bases and Weak Acids
3. Salts of Weak Bases and Strong Acids
4. Salts of Weak Bases and Weak Acids
73
Salts of Strong Bases and
Weak Acids
• Salts made from strong acids and strong soluble
bases form neutral aqueous solutions.
• An example is potassium nitrate, KNO3, made from
nitric acid and potassium hydroxide.
KNO3(s) ~100% in H2O K++NO3H2O +H2O
H3O++OH-
HNO3 KOH
The ions that are in solution
The KOH and HNO3 are present in equal
amounts
There is no reaction to upset [H3O+][OH-]
Thus the solution is neutral
74
Salts of Strong Bases and
Weak Acids
•Salts made from strong soluble bases and weak
acids hydrolyze to form basic solutions.
– Anions of weak acids (strong conjugate bases)
react with water to form hydroxide ions.
•An example is sodium hypochlorite, NaClO, made
from sodium hydroxide and hypochlorous acid.
~100% in H2O
NaClO(s)
H2O +H2O
Na++ClOH3O++OH-
HClO NaOH
The ions that are in solution
Which is the stronger acid or base?
75
Salts of Strong Bases and
Weak Acids
NaClO(s) ~100% in H2O Na++ClOH2O +H2O
H3O++OHClO- + H3O+
HClO + H2O
•We can combine these last two equations into one
single equation that represents the total reaction.
ClO- + H2O
HClO + OH-
•The equilibrium constant for this reaction, called the
hydrolysis constant, is written as:
Kb=
[HClO] [OH-]
[ClO-]
76
Salts of Strong Bases and Weak
Acids
• Algebraic manipulation of the previous expression
give us a very useful form of the expression.
• Multiply the expression by one written as [H+]/ [H+].
H+/H+ = 1
[HClO][OH-]
[H+]
Kb=
x
[ClO ]
[H+]
[H+][OH-]
[HClO]
Kb=
x
+
1
[ClO ][H ]
-14
Kw
1
1x10
Kb=
=
x Kw =
Ka for HClO
3.5x10-8
Ka for HClO
[HClO] [OH-]
-7
=
2.9x10
Kb=
[ClO-]
77
Salts of Strong Bases and
Weak Acids
• This same method can be applied to the anion of any weak
monoprotic acid.
A- + H2O
[HA] [OH-]
Kb=
[A-]
HA + OH=
Kw
Ka for HA
Kw=KaKb
78
Salts of Strong Bases and
Weak Acids
Example 18-16: Calculate the hydrolysis constants for the following
anions of weak acids.
The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 104.
F- + H2O
HF + OH-
[HF] [OH-]
Kb=
[F-]
=
1.0x10-14
Kb=
7.2x10-4
=1.4x10-11
Kw
Ka for HF
79
Salts of Strong Bases and
Weak Acids
• The cyanide ion, CN-, the anion of hydrocyanic acid, HCN.
For HCN, Ka = 4.0 x 10-10.
CN- + H2O
HCN + OH-
[HCN] [OH-]
Kw
Kb=
=
Ka for HCN
[CN ]
1.0x10-14
Kb=
4.0x10-10
=2.5x10-5
80
Example 18-17: Calculate [OH-], pH and percent hydrolysis for
the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO,
solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite
solutions.
NaClO(s)
0.10M
~100% in H2O
ClO- + H2O
initial
0.10M
change
-xM
At equil (0.10-x)M
Na++ ClO-
0.10M 0.10M
HClO + OH+xM
xM
[HClO] [OH-]
-7=
Kb=
=2.9x10
[ClO-]
+xM
xM
(x)(x)
(0.1-x)
x=1.7x10-4M=[ClO-]=[OH-]
x<<0.1 so (0.1-x)0.1
pOH=-log(1.7x10-4)=3.77
pH=14-3.77=10.23
% hydrolysis=
% hydrolysis=
[ClO-]hydrolyzed
x100%
[ClO ]original
1.7x10-4
x100% =0.17%
0.10M
81
Salts of Strong Bases and
Weak Acids
• If a similar calculation is performed for 0.10 M NaF solution and
the results from 0.10 M sodium fluoride and 0.10 M sodium
hypochlorite compared, the following table can be
constructed.
Solution
Ka
Kb
[OH-] (M)
pH
% hydrolysis
NaF
7.2 x 10-4
1.4 x 10-11
1.2 x 10-6
8.08
0.0012
NaClO
3.5 x 10-8
2.9 x 10-7
1.7 x 10-4
10.23
0.17
82
Salts of Weak Bases and
Strong Acids
• Salts made from weak bases and strong acids form acidic
aqueous solutions.
• An example is ammonium bromide, NH4Br, made from
ammonia and hydrobromic acid.
~100% in H2O
NH4Br (s)
H2O +H2O
NH4++ BrOH- + H3O+
The ions that are in solution
NH3
Which is the stronger acid or base?
HBr
83
Salts of Weak Bases and
Strong Acids
The relatively strong acid, NH4+, reacts with the OHion removing it from solution leaving excess H3O+
NH4+ + OH-
NH3 + H2O
generates excess H3O+
• The reaction may be more simply represented as:
NH4+ + H2O
•more simply as: NH +
4
NH3 + H3O+
NH3 + H+
• The hydrolysis constant expression for this process is:
[NH3][H3O+]
Ka=
[NH4+]
or Ka=
[NH3][H+]
[NH4+]
84
Salts of Weak Bases and
Strong Acids
• Multiplication of the hydrolysis constant expression by [OH-]/
[OH-] gives:
[NH3][H3O+] [OH-]
Ka=
x
[NH4 ]
[OH-]
[H3O+][OH-]
[NH3]
Ka=
x
+
1
[NH4 ][OH ]
-14
Kw
1
1x10
Ka=
=
x Kw =
Kb for NH3
1.8x10-5
Kb for NH3
[NH3] [H3O+]
-10
=
5.6x10
Ka=
[NH4-]
85
Salts of Weak Bases and
Strong Acids
• In its simplest form for this hydrolysis:
NH4+
NH3 + H+
[NH3] [H+]
-10
=
5.6x10
Ka=
[NH4-]
86
Example 18-18: Calculate [H+], pH, and percent hydrolysis for the
ammonium ion in 0.10 M ammonium bromide, NH4Br,
solution.
+
+
NH4
initial
0.10M
change
-xM
At equil (0.10-x)M
[NH3] [H+]
Ka=
[NH4+]
NH3 + H
+xM
xM
+xM
xM
(x)(x)
=
= 5.6x10-10
(0.1-x)
2=5.6x10-11M
x
x<<0.1 so (0.1-x)0.1
X=7.5x10-6M=[NH3]=[H+]
pH=-log(7.5x10-6)=5.12
% hydrolysis=
% hydrolysis=
[NH4+]hydrolyzed
x100%
[NH4+]original
7.5x10-6
x100% =0.0075%
0.10M
87
Salts of Weak Bases and
Weak Acids
• Salts made from weak acids and weak bases can
form neutral, acidic or basic aqueous solutions.
– The pH of the solution depends on the relative
values of the ionization constant of the weak
acids and bases.
1.Salts of weak bases and weak acids for which
parent Kbase =Kacid make neutral solutions.
– An example is ammonium acetate, NH4CH3COO,
made from aqueous ammonia, NH3,and acetic
acid, CH3COOH.
Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.
88
Salts of Weak Bases and
Weak Acids
• The ammonium ion hydrolyzes to produce H+ ions. Its
hydrolysis constant is:
NH4+
NH3 + H+
[NH3] [H+]
-10
=
5.6x10
Ka=
[NH4-]
• The acetate ion hydrolyzes to produce OH- ions. Its hydrolysis
constant is:
CH3COO- + H2O  CH3COOH + OH[OH-] [CH3COOH]
-10
Ka =
=5.6x10
[CH3COO-]
89
Salts of Weak Bases and
Weak Acids
•Because the hydrolysis constants for both ions are
equal, their aqueous solutions are neutral.
•Equal numbers of H+ and OH- ions are produced.
~100% in H2O
+
NH4 CH3COO
H2O +H2O
NH4++CH3COO-
OH- + H3O+
NH4OH CH3COOH
The ions that are in solution
A weak acid and base are formed in solution
90
Salts of Weak Bases and
Weak Acids
2.Salts of weak bases and weak acids for which
parent Kbase > Kacid make basic solutions.
• An example is ammonium hypochlorite, NH4ClO,
made from aqueous ammonia, NH3,and
hypochlorous acid, HClO.
Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8
91
Salts of Weak Bases and
Weak Acids
3. Salts of weak bases and weak acids for which parent
Kbase < Kacid make acidic solutions.
• An example is trimethylammonium
fluoride,(CH3)3NHF, made from trimethylamine,
(CH3)3N,and hydrofluoric acid acid, HF.
– Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4
92
Salts of Weak Bases and
Weak Acids
• Summary of the major points of hydrolysis up to
now.
1.The reactions of anions of weak monoprotic acids
(from a salt) with water to form free molecular acids
and OH-.
A- + H2O
[HA] [OH-]
Kb=
[A-]
HA + OH=
Kw
Ka for HA
93
Salts of Weak Bases and
Weak Acids
2. The reactions of anions of weak monoprotic acids
(from a salt) with water to form free molecular acids
and OH-.
BH+ + H2O
Ka=
B + H3O+
Kw
Kb (waek base)
94
Salts of Weak Bases and
Weak Acids
• Aqueous solutions of salts of strong acids and strong
bases are neutral.
• Aqueous solutions of salts of strong bases and weak
acids are basic.
• Aqueous solutions of salts of weak bases and strong
acids are acidic.
• Aqueous solutions of salts of weak bases and weak
acids can be neutral, basic or acidic.
The values of Ka and Kb determine the pH.
95