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CHEM 102 CLASS NOTES
CHAPTER 15. Acids and Bases
TABLE OF CONTENT
15.1 Heartburn 659
15.2 The Nature of Acids and Bases 660
15.3 Definitions of Acids and Bases 662
15.4 Acid Strength and the Acid Dissociation Constant (Ka) 665
15.5 Autoionization of Water and pH 668
15.6 Finding the and pH of Strong and Weak Acid Solutions 673
15.7 Base Solutions 682
15.8 The Acid–Base Properties of Ions and Salts 685
15.9 Polyprotic Acids 693
15.1 0 Acid Strength and Molecular Structure 698
15.1 1 Lewis Acids and Bases 700
15.1 2 Acid Rain 701
KEY CONCEPTS
Arrhenius definition
Types of acids and bases
definitions of pH, pOH, pKa, and pKb
pH and pOH for weak acid and weak base
pH of salt solutions
Brønsted-Lowry definition
Organic Acids and Amines
calculate pH and pOH
pH of polyprotic acids
acids, bases,
pH scale
pH nad pOH
Acid-Base R
molecular structure and
acid/base strength
Lewis definit
KEY SKILLS
Identifying Brønsted–Lowry Acids and Bases and Their Conjugates (15.3)
• Example 15.1 • For Practice 15.1 • Exercises 35, 36
Using in Calculations (15.5)
• Example 15.2 • For Practice 15.2 • Exercises 47, 48
Calculating pH from or (15.5)
• Examples 15.3, 15.4 • For Practice 15.3, 15.4 • Exercises 49–52
Finding the pH of a Weak Acid Solution (15.6)
• Examples 15.5, 15.6, 15.7 • For Practice 15.5, 15.6, 15.7 • Exercises 61–66
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Finding the Acid Ionization Constant from pH (15.6)
• Example 15.8 • For Practice 15.8 • Exercises 67, 68
Finding the Percent Ionization of a Weak Acid (15.6)
• Example 15.9 • For Practice 15.9 • Exercises 69–72
Mixtures of Weak Acids (15.6)
• Example 15.10 • For Practice 15.10 • Exercises 77, 78
Finding the and pH of a Strong Base Solution (15.7)
• Example 15.11 • For Practice 15.11 • Exercises 79, 80
Finding the and pH of a Weak Base Solution (15.7)
• Example 15.12 • For Practice 15.12 • Exercises 87, 88
Determining Whether an Anion Is Basic or Neutral (15.8)
• Example 15.13 • For Practice 15.13 • Exercises 93, 94
Determining the pH of a Solution Containing an Anion Acting as a Base
(15.8)
• Example 15.15 • For Practice 15.15 • Exercises 97, 98
Determining the Overall Acidity or Basicity of Salt Solutions (15.8)
• Example 15.16 • For Practice 15.16 • Exercises 99, 100
Finding the pH of a Polyprotic Acid Solution (15.9)
• Example 15.17 • For Practice 15.17 • Exercises 109, 110
Finding the in Dilute Solutions (15.9)
• Example 15.18 • For Practice 15.18 • Exercise 113
Finding the Concentration of the Anions for a Weak Diprotic Acid Solution
(15.9)
• Example 15.19 • For Practice 15.19 • Exercises 111, 112
Introduction to acid base reactions
Describe the types of chemical reactions.
There are three main types among chemical reactions.
a) Precipitation Reactions.
b) Acid/base Reactions.
c) Redox Reactions
a) Precipitation Reaction: When solutions of two reactants are mixed, depending on the
solubility of new salt formed, a precipitate is formed. The driving force for the reaction for the
completion is the removal of ionic species from the solution as a solid.
b) Acid/base Reaction: Simplest acid/base reactions occur, when Arrhenius acids and bases are
mixed in a solution, forming water and the salt. H+ and OH- ions are brought together to form
water because in the water H+ + OH- equilibrium water (product) side is favored. The driving
force for the reaction is the formation of water. Beside Arrhenius acids/bases there are other
acids/bases defined by Bronsted and Lewis definitions explaining whole range of acid base
reactions.
c) Redox Reactions: Reduction-Oxidation reactions are completely different type of reactions
where oxidation numbers of elements in reactants and products are changed through direct
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electron transfer.
Describe the following types of acids and bases: a) Arrhenius b)
c)Lewis
Bronsted-Lowry
Arrhenius Definition:
This is the first acid/base concept to be developed to describe typical acid/base reactions.
Arrhenius Acid:
A substance that produces H +, or (protons) H+3O, (hydronium ion) in an aqueous solution.
Arrhenius Base:
A substance that produces OH-, or hydroxide ion in an aqueous solution.
E.g. HCl (acid), NaOH (base).
Bronsted-Lowry Definition:
This is the second acid/base concept to be developed to include proton transfer reactions to base
other than those containing OH-. This concept can be applied to gas phase proton transfer reactions.
Conjugate acid/base concept also developed from this definition.
Bronsted Acid:
A substance that donates protons (H+)
Bronsted Base:
A substance that accepts protons. E.g. HCl (acid), NH3 (base).
Lewis Definition:
Third and the latest acid/base definition by Lewis was successful in including acid and bases without
proton or hydroxyl ions.
Lewis Acid: A substance that accepts an electron pair.
Lewis base: A substance that donates an electron pair.
E.g. Lewis acid - BF3 , Lewis base - NH 3
The Brønsted-Lowry Concept of Acids and Bases
In 1923, within several months of each other, Johannes Nicolaus Brønsted (Denmark) and
Thomas Martin Lowry (England) published essentially the same theory about how acids and bases
behave. Since they came to their conclusions independently of each other, both names have been used
for the theory name, the Brønsted-Lowry theory of acid and bases. The problems with bases,
especially ammonia, in Arrhenius' theory they found the inspiration for the new theory. Arrhenius
theory states: Arrhenius Acid: A substance that produces H +, or (protons) H+3O, (hydronium ion) in
an aqueous solution. Arrhenius Base: A substance that produces OH¯, or hydroxide ion in an aqueous
solution. The definition of base was so limited it could only include hydroxyl compounds as bases
limiting the acid base theory to few reactions.
According to Brønsted: Acids and bases are substances that are capable of splitting off or
taking up hydrogen ions, respectively." Or an acid-base reaction consists of the transfer of a proton
from an acid to a base. Acid and bases are proton transfer agents. Here is a other way to say the same
thing:
 An acid is a substance from which a proton can be removed.
 A base is a substance that can remove a proton from an acid.
Remember: proton, hydrogen ion and H+ all mean the same thing
Very common in the chemistry world is this definition set:
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

An acid is a "proton donor."
A base is a "proton acceptor."
In an acid, the hydrogen ion is bonded to the rest of the molecule normally through a more
electronegative atom than proton such as an oxygen atom. It takes energy (sometimes a little,
sometimes a lot) to break that bond. Sometimes the same group donating the proton may accept a
proton from a stronger donor and become positively charged. The base is a molecule with a built-in
"drive" to collect protons. As soon as the base approaches the acid, it will (if it is strong enough) rip the
proton off the acid molecule and add it to itself. Some bases are stronger than others, meaning some
have a large "desire" for protons, while other bases have a weaker drive. It's the same way with acids,
some have very weak bonds and the proton is easy to pick off, while other acids have stronger bonds,
making it harder to "get the proton."
The important contribution coming from Lowry to the Brønsted-Lowery theory has to do with the state
of the hydrogen ion in solution. In Bronsted's announcement of the theory, he used H+. Lowry
proposed the use of H3O+ that is commonly used today. Lowry noticed the strong acidity is apparently
developed only in mixtures and never in pure compounds. Even hydrogen chloride only becomes an
acid when mixed with water. This can be explained by the extreme reluctance of a hydrogen nucleus to
lead an isolated existence. The effect of mixing hydrogen chloride with water is probably to provide an
acceptor for the hydrogen nucleus so that the ionization of the acid only involves the transfer of a
proton from one octet to another.
ClH + H2O
Cl¯ + O H3+
Sample Acid-base Equations written in the Brønsted-Lowry Style
A reactions that proceeds to right to a large extent:
HCl + H2O
H3O+ + Cl¯
HCl - this is a Brønsted-Lowry acid, because it has a proton available to be transferred.
H2O - this is a Brønsted-Lowry base, since it gets the proton that the Brønsted-Lowry acid lost.
Now, here comes an interesting idea:
H3O+ - this is a Brønsted-Lowry acid, because it can give a proton.
Cl¯ - this is a Brønsted-Lowry base, since it has the capacity to receive a proton.
Notice that each pair (HCl and Cl¯ as well as H2O and H3O+ differ by one proton (symbol = H+). These
pairs are called Brønsted-Lowry conjugate pairs.
Conjugate Acid-Base Pairs
A conjugate pair is an acid-base pair that differs by one proton in their formulas (remember: proton,
hydrogen ion, etc.).
A conjugate pair is always one acid and one base.
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HCl + H2O
H3O+ + Cl¯
Here is the one conjugate pair from the first example reaction:
HCl and Cl¯
Usually, HCl is called an acid and Cl¯ is called its conjugate base, but that can be reversed if the
context calls for it. So, we can correctly speak of Cl¯ as a base and HCl as its conjugate acid.
Notice that the word conjugate is used with one of the pair and the conjugate is not the primary focus of
the context, it is the secondary.
The other conjugate pair is:
H2O and H3O+
Water is the base, since it is minus a proton compared to H3O+, which is the conjugate acid to water.
Remember conjugate pairs differ by only one proton. If you take away the proton (or add it), you get
the other formula.
Here are some more conjugate acid-base pairs to look for:
H2O and OH¯
HCO3¯ and CO32¯
H2PO4¯ and HPO42¯
HSO4¯ and SO42¯
NH4+ and NH3
CH3NH3+ and CH3NH2
HC2H3O2 and C2H3O2¯
This last one in the list above is special. Because it is used so often, it has an abbreviation:
Acetic acid's (HC2H3O2 or CH3COOH) abbreviation is HAc and the acetate ion's (C2H3O2¯ or
CH3COO¯) is Ac¯.
HNO3 + H2O
H3O+ + NO3¯
The acids are HNO3 and H3O+ and the bases are H2O and NO3¯.
Remember that an acid-base reaction is a competition between two bases (think about it!) for a proton.
If the stronger of the two acids and the stronger of the two bases are reactants (appear on the left side of
the equation), the reaction is said to proceed to a large extent.
Reactions that proceed to a small extent:
If the weaker of the two acids and the weaker of the two bases are reactants (appear on the left side of
the equation), the reaction is said to proceed to only a small extent:
HC2H3O2 + H2O
H3O+ + C2H3O2¯
+
NH3 + H2O
NH4 + OH¯
Identify the conjugate acid base pairs in each reaction.
Identify the Bronsted-Lowery acid/conjugate base and base/conjugate acid pairs in the
equilibrium reactions given below
a) HCl(aq) + H 2O(l) ---> H 3+O(aq) + Cl¯(aq)
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b) H2SO4(aq) + H2O(l)
HSO4¯(aq) + H2O(l)
H 3+O(aq) + HSO4¯(aq)
H 3+O(aq) + SO42¯(aq)
This dissociation of HSO4¯(aq) is negligible
c) H2O(l) + H 2O(l)
H 3+O(aq) + OH¯(aq)
This dissociation is called autoionization of water.
d) NH3 (aq) +H 2O(l)
NH 4+ + OH ¯(aq)
The concept of acid\conjugate base pair and base\conjugate acid pair came out of Bronsted
definition describing proton transfer reactions.
a) HCl/Cl¯ is an acid/conjugate base pair
H2O/H3+O is a base/conjugate acid pair in this equilibrium.
b) H2SO4/HSO4¯ is an acid/conjugate base pair and H2O/H3+O is a base/conjugate acid pair in this
equilibrium.
HSO4¯/SO42¯ is an acid/conjugate base pair, if the second dissociation of HSO4¯ (aq) took
place.
c) H2O/OH¯ is an acid/conjugate base pair in this equilibrium.
H2O/ H3+O is a base/conjugate acid pair and H2O/ OH ¯ is an acid/conjugate base pair in this
equilibrium.
H2O/ H3+O is a base/conjugate acid pair and HC2H3O2/C2H3O2¯ an acid/conjugate base pair in
this equilibrium.
d) NH3 / NH4+ is a base/conjugate acid pair and H2O/OH¯ is an acid/conjugate base pair in this
equilibrium.
General rules are:
A weak acid produces a stronger conjugate
base
A weak base produces a stronger conjugate
acid
The Autoionization of Water
Pure water is a rather poor conductor of electricity because it contains very few ions to permit the
conduction of electrical current. However, very small number of water molecules ~ 10-7 molecules per
mole can act both as an acid and a base, and can do so even without the presence of another species.
The reaction between two water molecules is
H2O(l) + H 2O(l)
H 3+O(aq) + OH¯(aq)
is called autoionization of water.
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Equilibrium constant K for this reaction defines a new constant, Kw. This important equilibrium
constant is called the ion-product constant for water. At 25°C, Kw has the value of 1.0 × 10-14. This
mean in water at 25°C the molarity is 1 x 10-7 for H 3+O and 1 x 10-7 for OH-. This is an important
equilibrium constant. You should memorize this expression:
Kw = [H 3+O][OH-]
[H 3+O] = 1 x 10-7 M
[OH¯] = 1 x 10-7 M
Kw = 1 x10-14 at 25 ºC
The pH Scale
Define the pH scale used for aqueous solutions of acids and bases. What is the relation of pH to
Kw and pOH?
pH is a scale used to express H+ ion concentration [H+]. Since [H+] is either large or negative
number, this scale would reduce to a simple number.
The letter p stands for the function log and change sign [p =-log]; H stands for [H +]
pH
= - log [H+]
Note: That K a , K w and K b values could also be converted to pK a , pK w and pK b values
similarly.
[H+] % Ka [H+] is not equal to Ka , but proportional ( %)_to it.
pH
% pKa
same is true for pH and pK a .
The relation of pH, Kw and pOH
Kw = [H+][OH¯] ;
log Kw = log [H+] +log [OH¯]
-log Kw = -log [H+] -log [OH¯] ;
previous equation multiplied by -1
-log Kw = -log [H+] - log [OH¯]
pKw = pH + pOH;
Kw = 1 x 10-14
; pKw =-log 1 x 10-14 =14
14 = pH + pOH
pH = 14 - pOH
pOH = 14 - pH
Strong acids and bases
Strong acid:
HA + H2O

H3O+ + A¯
Ka > 1
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Strong base: BOH
Strong Acids
Hydrioidic
Hydrobromic
Perchloric
Hyrdrochloric
Chloric
Sulfuric
Nitric
HI
HBr
HClO4
HCl
HClO3
H2SO4
HNO3

OH¯ + B+
Kb > 1
Ka ~ 1011 pKa = -11
Ka ~ 109 pKa = -9
Ka ~ 107 pKa = -7
Ka ~ 107 pKa = -7
Ka ~ 103 pKa = -3
Ka ~ 102 pKa = -2
Ka ~ 20 pKa = -1.3
Strong Bases
Lithium hydroxide LiOH
Sodium hydroxide NaOH
Potasium hydroxide KOH
Rubidium hydroxide
Cesium hydroxide CsOH
Boarder-line Bases
Magnesium hydroxide
Calcium hydroxide
Strotium hydroxide
Barium hydroxide
Kb~ 102-103
RbOH
Mg(OH)2
Ca(OH)2
Sr(OH)2
Ba(OH)2
Kb~ 0.01 to0.1
Weak acids and bases
Hydrofluoric acid
Formic acid
Acetic acid
Nitrous acid
Acetyl Salicylic acid
Hydrocyanic acid
Ammonia
Ethyl amine
HF + H2O
H3O+ + F¯
Ka = 6.6x10-4
pKa = 3.18
HCOOH + H2O
H3O+ + HCOO¯
Ka = 1.77x10-4
pKa = 3.75
HAc + H2O
H3O+ + Ac¯
Ka = 1.76x10-5
pKa = 4.75
+
¯
-4
HNO2+ H2O
H3O + NO2
Ka = 4.6x10
pKa = 3.34
C9H8O4 + H2O
H3O+ + C9H7O4¯ Ka = 3x10-4
pKa = 3.52
HCN + H2O
H3O+ + CN¯
Ka = 6.17x10-10
pKa = 9.21
+
¯
NH3 + H2O
NH4 + OH
Kb=1.79x10-5
pKb = 4.74
+
¯
-4
C2H5NH2 + H2O
C2H5NH3 + OH
Kb=5.6x10
pKb = 3.25
Calculating pH of Strong Acids
The key point is that strong means 100% ionized. That means the [H+] of a strong acid is equal to the
concentration of the acid. After all, ALL of the acid dissociates. No acid molecules are left. So, here is
a typical problem:
Calculate the pH of a 0.100 M solution of HCl.
In essence, this becomes calculate the pH when the [H+] equals 0.100 M. So, to solve it, you write:
pH = - log (0.100) = 1.000
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Important thing to remember about strong acids:
In strong acids such as H 2 SO4 there seem to be two H+ ions after dissociation, but only one H+
comes out therefore, [H+] equals to the molarity of the acid.
Calculating pH of Strong Bases
Strong bases is pretty much the same as strong acids EXCEPT you'll be calculating a pOH first, then
going to the pH. This type of problem is where the relation pH + pOH = 14 is important. As in acids,
strong means 100% ionized. That means the [OH¯] of a strong base is equal to the concentration of
the base multiplied by number of OH¯ in the formula of the base. After all, ALL of the OH¯ of the
base dissociates unlike polyprotic acids. This important difference comes from the fact that acids are
normally covalent or molecular compounds that have to dissociate in water and on the other hand bases
are ionic compounds that simply get ions solvated in water.
Important thing to remember about strong bases:
In strong bases the [OH¯] of a strong base is equal to the molarity of the base multiplied by
number of OH¯ in the formula of the base.
Calculate the pH of a 0.100 M solution of NaOH.
In essence, this becomes calculate the pH when the [OH¯] equals 0.100 M. So, to solve it, you write:
pOH = - log (0.100) = 1.000
pH = 14.000 - 1.000 = 13.000
What is the pH of 0.010 M Ba(OH)2?
There are two OH¯ in formula of Ba(OH)2
[OH¯] = 2(0.010 M) = 0.020 M
pOH = -log(0.020) = 1.70;
pH = 14.00 - 1.70 = 12.30
Calculate the pH and pOH of the following strong acids and bases. (K w = 1 x 10 -14)
a) 0.2 M HNO3
b) b) 0.5 M H 2 SO4
c) 1.5 x 10-2 M NaOH.
a) 0.2 M HNO3 : The dissociation is complete for strong acid such as HNO 3.
HNO3(aq) + H 2O(l)
H3+O(aq) + NO3¯(aq)
Therefore, the moles of H+ ions in the solution is equal to moles of HNO3 at the beginning.
[HNO3] = [H+] = 0.2 mole/L
pH
= -log [H+]
= -log(0.2)
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pH
= 0.699
b) 0.5 M H2 SO4 : H2 SO4 has two acidic protons. However, only the first proton dissociates
completely. Second dissociation is insignificant and usually ignored for poly-protic acids. Removing a
H+ from HSO4¯ is much more difficult than the first proton because of the extra negative charge on
HSO4¯. Therefore, the H+ in the solution mainly come from first acidic proton. This is true for other
polyprotic acids as well such as H 3PO 4.
H2 SO4(aq) + H2O(l)
H 3+O(aq) + HSO4¯(aq)
+
Therefore, the moles of H ions in the solution is equal to moles of H2 SO4 at the beginning.
[H2 SO4] = [H+] = 0.5 mole/L
pH = -log [H+]
= -log(0.5)
pH = 0.30
c) 1.5 x 10-2 M NaOH: NaOH also being a strong base dissociates completely in water.
[NaOH] = [HO¯ ] = 1.5 10 -2 mole/L
pOH = -log[HO¯] = -log1.5 10-2
pOH = 1.82
As defined and derived previously:
Kw = pH + pOH; pKw = 14
pH = pKw + pOH
pH = 14 - pOH
pH = 14 - 1.82
pH = 12.18
Ionization or Dissociation Constants of Acids and Bases
The ionization or dissociation constants of acids and bases other than water are dealt with in a
similar way.
[H3+O][ A¯]
+
¯
Acid:
HA + H2O
H3O + A
; Ka = --------------[HA]
base:
+
BOH
OH¯ + B
Kb
[OH¯][ B+]
= --------------[BOH]
Acid Dissociation Constants (Ka):
The ionization of acetic acid in water is described by the equilibrium
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO¯(aq), for which the equiilibrium constant is
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[H3+O ][ CH3COO¯]
CH3COOH(aq); Ka = ----------------------[HC2H3O2]
Base Dissociation Constants (Kb):
We write the equilibrium constant for the ionization of the aqueous base ammonia, NH3, as
NH3(aq) + H2O
NH4+(aq) + OH¯(aq)
+
[NH4 ][ OH¯]
NH3 (aq); Kb =
----------------------[NH3]
Percent Dissociation (% dissociation):
What is the percent dissociation of an acid or base?
The degree of dissociation is the amount of (acid/base) which dissociates (in this case = x) divided by
the initial amount of acid. The percent dissociation is simply this number times 100%
Amount dissociated
---------------------- x 100
Initial amount
Percent dissociation is another way of indicating dissociation equilibrium. It can be calculated
% Dissociation =
from Ka or Kb value.
Write equations for the dissociation equilibrium reactions for the following acids and bases in
water.
a) HCl b) H2SO4 c) NaOH d) H2O(autoionization) e) HC2H3O(acetic acid) f) NH3
Acid/base equilibrium are few among the many equilibrium reactions in chemistry. All the
terms we applied such as dynamic equilibrium, initial concentrations, equilibrium concentration still
apply to them.
a) HCl(aq) + H 2O(l)
H 3+O(aq) + Cl¯(aq)
b) H2SO4(aq) + H2O(l)
H 3+O(aq) + HSO4¯(aq)
HSO4¯(aq) + H2O(l)
H 3+O(aq) + SO42¯(aq)
For this dissociation of
HSO4¯(aq)
is
negligible
c) NaOH (aq) + H2O(l)
d) H2O(l) + H 2O(l)
Na +(aq) + OH ¯(aq)
H 3+O(aq) + OH¯(aq)
This dissociation is called autoionization of water.
e) HC2H3O2(aq) + H2O(l)
f) NH3 (aq) +H 2O(l)
H 3+O(aq) + C2H3O2¯(aq)
NH 4+ + OH ¯(aq)
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Write the dissociation constants, K a and Kb for the acids and bases in previous problem.
Ka:: Acid dissociation constant for the equilibrium reaction.
Kb: Base dissociation constant for the equilibrium reaction.
Acid in water: HA + H 2O
H 3+O + A¯
Base in water:
B + + OH¯
BOH + H 2O
[H3+O ][ A¯]
[B+ ][OH¯ ]
Ka = -----------;
Kb = --------------[HA]
[BOH]
Looking at the equilibrium in previous problems:
[H3+O ][Cl¯]
a)
HCl; Ka = ----------[HCl]
b)
c)
[H3+O ][HSO4¯]
----------------[H2SO4 ]
H2SO4 ;
Ka =
NaOH: Kb =
[Na+][OH¯ ]
----------[NaOH]
d) Autoionization of water:
Kw = [H 3+O][OH¯]
[H3+O ][C2H3O2¯]
e) HC2H3O2; Ka = -----------------[HC2H3O2]
f) NH3 (aq) + H2O(l)
NH 4+ + OH ¯(aq)
[NH4+ ][OH¯ ]
NH3 ; Kb = --------------[NH 3]
Note: that [H2O(l)] = 1 since it's a pure liquid and usually [H 3+O] is written as [H +].
Acid and Base Strength
A strong acid or base is one for which this equilibrium lies far to the right. A larger value of Ka or
Kb indicates an equilibrium favoring product side. A strong acids/base yields a weak conjugate
base/acid. Acidity and basicity increase with increasing K a or K b.
A weak acid or base is one for which the equilibrium lies far to the left. A weak acid/base has a strong
conjugate base/acid. Acidity and basicity decrease with decreasing K a or K b.
Explain the relative terms (WEAKER/STRONGER) used for acidity and basicity of following
acids/bases comparing their Ka and Kb values.
A larger value of Ka or Kb indicates an equilibrium favoring product side. Acidity and
basicity increase with increasing Ka or Kb.
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Explain the following observations:
a) Any compound with a K a value greater than K w of water will be an acid in water.
b) Any compound with a K b value greater than K w of water will be a base in water.
c) pH scale spans values only from 0-14 in water.
a) An acid (compound) with a K a value greater than K w would dissociate in water and produce H
+
3O ions in a concentration higher than in water alone. Therefore, the hydronium ion concentration for
an acid in water would be greater than 1 x 10-7 moles/L (which is found in neutral water) making the
solution acidic.
b) Any base (compound) with a K b value greater than K w of water would produce OH¯ ions in a
concentration higher than that found in water alone. This solution would be basic.
c) pH scale spans values only from 0-14 in water . In aqueous solutions the maximum or minimum
pH values are limited to those values based on the dissociation equilibrium of water.
2 H2O(l)
H 3O+ (aq) + OH¯(aq); Kw = [H 3+O][OH¯]
-14
Kw = 1 x 10
= [H 3+O][OH¯] = 10-7 x 10-7 (for neutral water)
Extreme cases:
Kw = 1 x 10-14
Kw = [H 3O+][OH¯] = 10-14 x 100 (for basic medium)
Kw = 1 x 10-14 =[H 3+O][OH¯]
Kw = 100 x 10-14 (for acidic medium)
pH value is -log[H +] spans only 0-14
a) HCl, H2SO4 and HC2H3O2 are acidic in aqueous solutions because they donate H+ to water.
Acidity of these three acids increases in the order of acidity: HC2H3O2 < H 2SO4 < HCl
Ka: Ka (HCl) < Ka(H2SO4) < Ka(HC2H3O2)
b) NH3 is a stronger base than H 2O since its K b value is greater than that of H 2O. We can use K a
and K b values to obtain relative acidity and basicity of compounds.
Problem Solving Using Ka,Kb and % dissociation
Calculate the % dissociation of 1.00 M
b)
NH3; Kb = 1.8 x 10-5 solutions of following weak acid and bases.
a)
HCN; Ka = 4.9 x 10-10
These problems are somewhat similar to the calculations we did in chapter 14 on chemical
equilibrium. In these cases Ka is the K or Kc for the acid dissociation equilibrium. However, %
percent dissociation is defined as:
Amount dissociated
% Dissociation =
---------------------- x 100
Initial amount
a) 1.00 M solution of HCN; Ka = 4.9 x 10-10
First write the dissociation equilibrium equation:
HCN(aq) + H 2O(l)
H 3+O(aq) + CN¯(aq)
[H3+O ][CN¯]
14
HCN;
Ka =
----------[HCN]
Initial concentration
Change
Equilibrium concentration
;
[ H2O(l)] = 1
HCN
1.00 M
-x
1.0 - x
H+
0.0 M
x
x
CN¯
0.00 M
x
x
[H 3+O ][CN¯]
x2
Ka = -------------------= ----------------;
1.0 - x ~ 1 since x is small
[HCN]
1.0 - x
x2
Ka = ----------; Ka = 4.9 x 10-10
= x2
1.0
x = sqrt 4.9 x 10-10 = 2.21 x 10 -5
Amount dissociated
2.21 x 10 -5
% Dissociation =
---------------------- x 100 = -------------- x 100
Initial amount
1
-5
% Dissociation
=
2.21 x 10 x 100 = 0.00221 %
Note: 1.0 M HCN solution is almost undissociated. However, what will be the % dissociation if you
make solution dilute?
E.g. If you take a 0.01 M HCN solution, what would be the % dissociation.
[HCN]
[H+ ]
[CN¯ ]
Initial concentration
1.00 M
0.0 M
0.00 M
Change
-x
x
x
Equilibrium concentration 0.01 - x
x
x
+
2
[H 3 O ][CN ]
x
Ka = ----------------- = --------------; 0.01 - x ~ 1 since x is small
[HCN]
0.01 - x
x2
x2
Ka = ----------; Ka = 4.9 x 10-10
=
----------0.01
0.01
-10
x = sqrt ( 4.9 x 10
x 0.01)
= sqrt ( 4.9 x 10 -12 ) =
Amount dissociated
2.21 x 10 -6
% Dissociation =
---------------------- x 100 = -------------- x 100
Initial amount
0.01
% Dissociation
=
2.21 x 10 -4 x 100 = 0.0221 %
The dissociation increases with the dilution.
b) 1.00 M solution of NH3; Kb = 1.8 x 10-5:
First write the dissociation equilibrium equation:
NH3 (aq) +H 2O(l)
NH 4+ + OH ¯(aq)
[NH 4+ ][OH¯ ]
2.21 x 10
-6
15
NH3 ; Kb =
--------------[NH3 ]
[NH3 ]
[NH4+ ]
Initial concentration
1.00 M
0.0 M
Change
-x
x
Equilibrium concentration 1.0 - x
x
+
¯
[NH4 ] [OH ]
x2
Kb =
-------------= ---------------[NH3 ]
1.0 - x
x2
Kb = ----------; Kb = 1.8 x 10-5
1.0
x = sqrt 1.8 x 10-5 = 4.24 x 10 -3
Amount dissociated
% Dissociation = -------------------------- x 100 =
Initial amount
% Dissociation
=
[OH¯ ]
0.00 M
x
x
1.0 - x ~. 1.0 since x is small
= x2
4.24 x 10 -3
-------------- x 100
1
4.24 x 10 -3 x 100
= 0.42 %
Calculate the Ka or Kb of following weak acids and bases:
a) 1 M HF, 2.7% dissociated
b) 15 M NH 3, 0.11% dissociated.
These problems are reverse calculations of K a or Kb from % dissociation.
a) 1 M HF, 2.7% dissociated
First write the dissociation equilibrium equation:
HF(aq) + H 2O(l)
H 3+O(aq) + F¯(aq)
+
¯
[H 3 O ][F ]
HCN; Ka = ----------;
[ H2O(l)] = 1 , [H3+O]=[H+ ]
[HF]
[HF]
[H+ ]
[F- ]
Initial concentration
1.00 M
0.0 M
0.00 M
Change
-x
x
x
Equilibrium concentration 1.0 - 0.027
0.027
0.027
(notice the conversion of % dissociation to a fraction (x): 2.7/100=0.027)
[H3+O ][F-]
0.027 x 0.027
K a = ----------=
---------------- = 7.4 x 10 -4
[HF]
1.0 - 0.027
K a = 7.4 x 10 -4
b) 15 M NH3 , 0.11% dissociated.
First write the dissociation equilibrium equation:
NH3 (aq) +H2O(l)
NH4+ + OH¯(aq)
[NH4+ ][OH¯ ]
NH3 ; Kb = ---------------
16
[NH3]
[NH3 ]
[NH4+ ]
[OH¯ ]
Initial concentration
15.00 M
0.0 M
0.00 M
Change
-x
x
x
Equilibrium concentration 15.0 - 0.0165 0.0165
0.0165
(notice the conversion of % dissociation to a fraction (x): 15(0.11/100)= 0.0165)
[NH4+ ][OH¯ ]
0.0165 x 0.0165
Kb =
-------------=
-------------------= 1.8 x 10-5
[NH 3 ]
15.0 - 0.0165
Kb =
1.8 x 10 -5
Calculate the pH of following weak acid and base solutions:
a) 0.20 M HC2H3O2; Ka= 1.8 x 10-5
b) 10.0 M NH 3; Kb = 1.8 x 10-5.
a) 0.20 M HC2H3O2; Ka= 1.8 x 10-5
First write the dissociation equilibrium equation:
HC2H3O2(aq) + H2O(l)
H 3+O(aq) + C2H3O2¯(aq)
HC2H3O2;
Ka =
[H3+O ][C2H3O2¯]
-----------------= 1.80 x 10-5; [ H2O(l)] = 1 , [H 3+O]=[H+ ]
[ HC2H3O2]
Initial concentration
Change
Equilibrium concentration
Ka =
Ka =
[HC2H3O2]
0.20 M
-x
0.20 -x
[H 3+O ][C2H3O2¯]
-------------------[ HC2H3O2]
=
[H+ ]
0.0 M
x
x
[C2H3O2¯ ]
0.00 M
x
x
x2
---------------- approximation : 0.20 - x ~ 0.20
0.20 - x
x2
----------- = 1.8 x 10-5 ; x2 = 1.8 x 10-5 x 0.20
0.20
x = sqrt 3.6 x 10-6 = 0.00189 =1.89 x 10-3
[H+ ] = x = 1.89 x 10-3 M
pH = -log [H+ ] =- log 1.89 x 10-3 = 2.72
pH of 0.20 M HC2H3O2 solution is 2.72
b) 10.0 M NH3; Kb = 1.8 x 10-5.
First write the dissociation equilibrium equation: .
NH3 (aq) +H2O(l)
NH 4+ + OH ¯(aq)
= 3.6 x 10-6
17
NH3 ; Kb =
[NH4+ ][OH¯ ]
--------------[NH3 ]
Initial concentration
Change
Equilibrium concentration
Kb =
Kb =
[NH3 ]
10.00 M
-x
10.0 - 0.0165
[NH4+ ]
0.0 M
x
0.0165
[OH¯ ]
0.00 M
x
0.0165
[NH3][OH¯ ]
x2
-------------------=
---------------- approximation :10.0 - x ~10.0
[NH3 ]
10.0 - x
x2
----------- = 1.8 x 10-5 ; x2 = 1.8 x 10-5 x 10.0 = 1.8 x 10 -4
10.0
x = sqrt 1.8 x 10 -4 = 0.01342 =1.34 x 10 -2
[OH- ] = x = 1.34 x 10 -2 M
pOH = -log [OH- ] =- log 1.34 x 10 -2 =1.87
pH = 14- pOH = 14- 1.87 = 12.12
pH of 10.0 M NH3 solution is 12.12
Molecular Structure and Acid Strength
Using examples describe following types of acids and bases.
a) Binary acids b) Oxyacid c) Organic acids d) Acidic oxides
e) Basic oxides f) Amine g) Polyprotic acids
There are several types of acids and bases depending on their chemical composition and the molecular
connectivity.
a) Binary Acids:
Compounds containing acidic protons bonded to a more electronegative atom.
e.g. HF, HCl, HBr, HI, H2S
The acidity of the haloacid (HX; X = Cl, Br, I, F)
Series increase in the following order:
HF < HCl < HBr < HI
b) Oxyacids:
Compounds containing acidic - OH groups in the molecule.
Acidity of H2SO4 is greater than H2SO3 because of the extra O (oxygens) attached to S pulling
electrons away making H+ to come off easier.
The order of acidity of oxyacids with a halogen (Cl, Br, or I) shows a similar trend.
HClO4 > HClO3 >
HClO2 > HClO
Perchloric chloric
chlorus
hyphochlorus
c) Organic or carboxylic Acids: Compounds containing - COOH or group. This group is found
attached to many organic groups (groups containing carbon and hydrogen) as in acetic acid. One
18
proton is written in front to show the acidity: A monoprotic acid. Generally these acids are weaker
because of their incomplete dissociation of - COOH acidic proton found in the - COOH group.
Organic acids are weak acids with general formula R–COOH)
They dissociate as follows:
R-COOH(aq) + H2O(l)
RCOO¯(aq) + H3O+(aq)
Use the pKa values provided to rank the following diverse acids by their relative acidity. The
acidity of carboxylic acids is described by the acidity constant, Ka, and its negative logarithm, pKa.
Large Ka's and small pKa's denote high acidities. Acid strength is influenced by substituents on the
carboxylic acid molecule. Electron-withdrawing groups disperse the negative character of the
carboxylate ion and increase acidity whereas electronreleasing groups intensify the negative charge
and decrease acidity.
The strength, number, and proximity of electron-withdrawing groups can have dramatic
effects on relative acidities.
19
FCH2CO2H (strongest acid) > ClCH2CO2H > BrCH2CO2H (weakest acid).
Acid
HCOOH (formic acid)
CH3COOH (acetic acid)
CH3CH2COOH (propanoic acid)
Ka (M)
1.78 X 10-4
1.74 X 10-5
1.38 x 10-5
pKa
3.75
4.76
4.86
d) Acidic Oxides:
These are usually oxides of non-metallic elements such as P, S and N. e.g. NO2, SO2, SO3, CO2
They produce oxyacids when dissolved in water,
Sulphur Trioxide (SO3) reacts with water to form Sulphuric Acid:
SO3 + H2O ---> H2SO4
CO2 + H2O ---> H2CO
NO2 + H2O ---> HNO3
e) Basic Oxides:
These are usually oxides of metallic elements such as Na, K, Ca. They produce hydroxyl bases
when dissolved in water.
e.g. CaO + H2O --> Ca(OH)2
Na2O + H2O ---> 2 NaOH
Acidic Oxide + Basic Oxide ---> Salt
Example: SO3 + Na2O ---> Na2SO4
e) Monoprotic Acids:
The term protic refers to acidity or protons. Monoprotic acids have only one acidic proton. e.g.
HCl.
f) Polyprotic Acids: They have more than one acidic proton.
e.g.
H2SO4 - diprotic acid
H3PO4 - triprotic acid.
Phosphoric acid, H3PO4 (Ka1 = 7.11 x 10-3, Ka2 = 6.32 x 10-8, Ka3 = 4.5 x 10-13), is a weak
triprotic acid. A plot of the composition of a solution of this acid as a function of pH is shown below.
(a) On this plot, indicate the three buffer regions for this acid system. (b) The 4 species derived from
this acid are Na3PO4, Na2HPO4, NaH2PO4, and H3PO4.
20
f) Organic Acids and Amines
Amines are a class of organic bases derived from ammonia NH3 by replacing hydrogen by
organic groups. They are defined as bases similar to NH3 by Bronsted or Lewis acid/base definitions.
Amines are organic bases derived fro NH3
R-NH3+(aq) + OH¯(aq)
R-NH2(aq) + H2O(l)
Base
CH3NH2 (methylamine)
(CH3)2NH (dimethylamine)
(CH3)3N (trimethylamine)
CH3CH2NH2 (ethylamine)
Kb (M)
2.34 x 10-11
2.09 x 10-11
1.78 x 10-11
2.00 x 10-11
pKb
10.63
10.68
10.75
10.70
Lewis Acid and Bases
There are numerous possible ways to define and acid and a base. One common way is the Lewis acid
base model:
 An acid is a electron pair acceptor.
 A base is a electron pair donor.
This definition is more broad than the Bronstead definition. Obviously, H+ is an acid and OH- is a base
under either definition, since for an proton to bind to a base, it must accept a pair of electrons
H+(aq) + :OH-(aq) -> H2O(l)
However, the Lewis definition extends beyond just the proton. For example, many metal ions can act as
Lewis acids when they form complex ions
Fe+3(aq) + 6CN-(aq) -> Fe(CN)6-3(aq)
Here, the electrons which form the bond between the iron and the cyanide ion start as lone pairs on the
cyanide. The iron accepts the electrons and is thus an electron pair acceptor and a Lewis acid, the
21
cyanide ions donate a pair of electrons are are thus Lewis bases. Note that there are no protons in this
reaction, but it is still an acid/base reaction.
Example: For the reaction below, identify the acid and the base
Cu+2(aq) + 4NH3(aq) -> Cu(NH3)4+2(aq)
Solution: Cu+2 accepts four pairs of electrons from the ammonia molecules, so it is an acid. The
ammonia molecules are donating their lone pairs, so the ammonia molecules are the bases.
Complex Ions of Coordination Compounds
Many metal ions (usually transition metals but including a few others) form complex ions with a
variety of anions and molecules. These are formed in a similar way as any other molecule is: for
example:
There is a bit of unique nomenclature to complex ions in Coordination Compounds:




The metal is known as the central metal ion
The anions or molecules attached to the metal are called ligands
The coordination number is the number of places on the metal ion where ligands are bound.
The bond between the metal ion and the ligand, where the ligand supplies both electrons, is
known as a coordinate covalent bond.
What acid base concepts (Arrhenius/Bronsted/Lewis) would best describe the following reactions:
a) HCl(aq) + NaOH(aq)
---> NaCl(aq) + H2O(l)
b) HCl(g) + NH3(g)
---> NH4Cl(s)
c) BF3(g) + NH3(g)
---> F3B:NH3(s)
d) Zn(OH)2(s) + OH-(aq)
---> [Zn(OH)4]2- (aq)
Arrhenius Concept:
This is a typical Arrhenius acid/base reaction where an acid [HCl] reacts with a base (NaOH) to
produce H2O and a salt (NaCl).
a) HCl(aq) + NaOH(aq)
--->
NaCl(aq) + H2O(l)
Bronsted Concept:
Arrhenius definition of bases does not include NH3. Therefore, Bronsted definition would best
describe this reaction. HCl is the Bronsted acid, a proton donor. NH3 is the Bronsted base, a proton
acceptor.
b) HCl(g) + NH3(g) ---> NH4Cl(s)
Lewis Concept:
Arrhenius definition would not include either BF3 or NH3. Bronsted definition would not describe BF3.
22
Therefore, Lewis definition has to be used. BF3 is the Lewis acid which accepts an electron pair from
the donor Lewis base: NH3.
c) BF3(g) + NH3(g) ---> F3B:NH3(s)
Lewis Concept:
Neither Arrhenius or Bronsted concept would describe this reaction. According to Lewis
definition Zn(OH)2 is the Lewis acid which accept an electron pair from the donor OH - which is a
Lewis base.
d) Zn(OH)2(s) + OH¯(aq) --->
[Zn(OH)4]2¯ (aq)
Acid-Base Reactions of Salts
Hydrolysis of a Salt
A salt is an ionic compound containing positive ions other than H+ and negative ions other than
OH¯. Most salts will dissociate to some degree when placed in water. In many cases, ions from the salt
will react with water to produce hydronium ions or hydroxide ions. Any chemical reaction in which
water is one of the reactants is called a hydrolysis reaction.
Salts can be thought of as being derived from the neutralization of an acid and a base. A salt
formed from a strong acid and a strong base1 will not hydrolyze (react with water). When placed in
water, these salts dissociate completely, and their ions remain uncombined in solution. An example of
such a salt is NaCl, formed from a strong acid (HCl) and a strong base (NaOH).
Salt of Weak Base
Salts formed from a strong acid and a weak base hydrolyze to form a solution that is slightly
acidic. In this kind of hydrolysis, the water molecules actually react with the cation of the weak base.
For
example, when ammonium chloride, NH4Cl, hydrolyzes, water molecules react with the NH4+ ion:
NH4+ + 2H2O  NH4OH + H3O+
The formation of the H3O+ ion from this reaction makes the solution acidic.
Salt of Weak Acid
Salts formed from a weak acid and a strong base hydrolyze to form a solution that is slightly
basic. In this kind of hydrolysis, it is the anion of the weak acid that actually reacts with the water. For
example, when sodium acetate, HC2H3O2, hydrolyzes, water molecules react with the acetate ion:
C2H3O2- + H2O --> HC2H3O2 + OHThe formation of the OH- ion from this reaction makes the solution basic. Salts formed from a
weak acid and a weak base produce solutions that may be slightly acidic, slightly basic, or neutral,
depending on how strongly the ions of the salt are hydrolyzed.
You could test several different salt solutions using pH paper and phenolphthalein solution to
determine their basicity.
The H+ ion concentration or pH of a solution made after dissolving a salt in water depends on
the strength of the acid and base used to make that salt. General rule to predict pH of a salt solution:
23
1) Salt of strong acid and strong base forms ions which are
neutral.
2) Salt of weak acid and strong base forms ions which are basic.
3) Salt of strong acid and weak base forms ions which are acidic.
4) Salt of weak acid and weak base forms ions which are either
neutral, basic or an acidic depending on the relative strengths of the
acid and the base.
These general rules are based on the conjugate acid/base behavior of weak acids/bases.
General rules are:
A weak acid produces a stronger conjugate
base
A weak base produces a stronger conjugate
acid
Relative strengths of conjugate acids and bases in water
Conjugate Acid
HCl (strong acid, ~100% dissociated in water)
CH3COOH (weak acid)
NH3 (weak base)
Conjugate Base
Cl- (very weak base, ~0% protonated in
water)
CH3COO- (srtong base, grater tendency to
accept a proton from water.)
NH4+ (srtong acid, grater tendency to donate
a proton to water.)
Which of the following salt solutions in water would produce ions which are acidic, basic or
neutral ions?
a) NaCl
b) NaC2H3O2
c) NaHSO4
d) NH4Cl
The H+ ion concentration or pH of a solution made after dissolving a salt in water depends on
the strength of the acid and base used to make that salt. General rule to predict pH of a salt solution:
1) Salt of strong acid and strong base forms ions which are neutral.
2) Salt of weak acid and strong base forms ions which are basic.
3) Salt of strong acid and weak base forms ions which are acidic.
4) Salt of weak acid and weak base forms ions which are
eitherneutral, basic or an acidic depending on the relative strengths
of the acid and the base.
These general rules are based on the conjugate acid/base behavior of weak acids/bases.
a) The salt, NaCl is a salt of strong acid (HCl) and strong base (NaOH), therefore, forms a neutral
solution. The conjugate acid base Cl- and conjugate acid (H2O) are both very weak and does not affect
the pH of the solution.
24
b) The salt, NaC2H3O2, is a salt of weak acid (HC2H3O2) and strong base (NaOH), therefore, forms a
basic solution.
The conjugate base C 2H3O2- is a strong base giving a basic pH to the solution, and the conjugate
acid (H 2O) is weak and does not affect the pH of the solution.
c) The salt, NaHSO4, is a salt of strong acid (H 2SO4) and strong base (NaOH), therefore, should
form a neutral solution. However, HSO4- ion could dissociate further
HSO 4- + H 2O
H 3+O + SO42+
producing H giving the solution an acidic pH.
d) The salt, NH4Cl, is a salt of strong acid (HCl) and strong base (NH 4OH), therefore, forms an
acidic solution.
The conjugate acid NH 4+ is a strong acid giving an acidic pH to the solution, and the conjugate
base (Cl- ) is weak and does not affect the pH of the solution.
Relationship between Ka and Kb of Conjugate Acid-Base Pairs
The relationship between Ka and Kb for any conjugate acid-base pairs is as follows:
(Ka)(Kb) = Kw
Where Ka is the ionization constant of the acid form of the pair, Kb is the ionization constant for the
base form of the pair, and Kw is the ionization constant for water. This equation is used to find either
Ka or Kb when the other is known.
Example: The Ka for acetic acid is 1.7 x 10-5. What is the value of Kb for the acetate ion?
(1.7 x 10-5)(Kb) = 1 x 10-14
Kb = 5.9 x 10-10
Example: The Kb for aniline is 3.8 x 10-10. What is the value of Ka for the anilonium ion?
(Ka)(3.8 x 10-10) = 1 x 10-14
Ka = 2.6 x 10-5
Calculating pH of Salt Solutions
To calculate the pH of a salt solution one needs to know the concentration of the salt solution,
whether the salt is an acidic, basic, or neutral salt, the equation for the interaction of the ion with the
water, the equilibrium expression for this interaction and the Ka or Kb value.
Example: Calculate the pH of a 0.500 M solution of KCN. Ka for HCN is 5.8 x 10-10.
25

First, write the equation for the dissolving process, and examine each ion formed to determine
whether the salt is an acidic, basic, or neutral salt.
KCN(s) --> K+(aq) + CN-(aq)
K+ is a neutral ion and CN- is a basic ion. KCN is a basic salt.

Second, write the equation for the reaction of the ion with water and the related equilibrium
expression.
CN-(aq) + H2O(l) --> HCN(aq) + OH-(aq)
Kb = [HCN][OH-]
[CN-]

Third, use the given Ka for HCN to find the value of Kb for CN-.
(5.8 x 10-10)(Kb) = 1 x 10-14
Kb = 1.7 x 10-5

Make an "ICE" chart to aid in the solution. Let "x" represent the amount of CN- that interacts
with the water.
CN-(aq)
HCN(aq)
OH-(aq)
0.500 M
0M
0M
-x
+x
+x
0.500 M - x
x
x
Initial Concentration
Change in Concentration
Equilibrium
Concentration

Subsititute equilibrium values and the value for Kb to solve for x.
Kb
x2
= 7 x 10 = ------------5
0.500 - x
We will make the assumption that since Kb is so small that the value for x will be very small as well,
thus the term (0.500 - x) is equal to (0.500).
1.7 x 10-5 = x2/(0.500)
x = 2.9 x 10-3
26

Determine the pH of the solution. Since "x" represents the hydroxide ion, OH concentration,
we can convert it into pOH and than find the pH.
pOH = -log(2.9 x 10-3) = 2.54
pH = 14 - 2.54 = 11.46
Calculate the pH of
a) 0.5 M NH4Cl salt solution. (NH 3; Kb = 1.8 x 10-5)
b) 0.5 NaC2H3O2 salt solution. (HC 2H 3O2; Ka = 1.8 x 10-5)
a) What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)
An aqueous solution of NH4Cl is acidic as discussed previously.
NH4Cl dissociates completely in water:
NH4Cl ---> NH4+ (strong conjugate acid) + Cl- (weak conjugate base)
NH4 dissociates further in water producing an acid solution.
NH4+ + H2O
H 3+O + NH3 ;this is a acid dissociation equilibrium or K a
For a weak base (NH3) or acid (HC2H3O2) with Kb or Ka ,respectively, the value of K a or K b
for the corresponding conjugate acid (NH 4+ ) or base (C2H3O2-) can be calculated from the following
expression: Kb x Ka = Kw
We need the Ka of NH4+ (conjugate acid) before we can do the calculation: using K b x K a = K w
(1.0 x 10-14)
Kb(NH3) x Ka (NH4+) = 1.0 x 10-14
1.8 x 10-5 x Ka (NH4+)
= 1.0 x 10-14
1.0 x 10-14
+
Ka (NH4 )
=
-----------= 5.56 x 10-10
1.8 x 10-5
Initial Concentration
Change in Concentration
NH4+ (aq)
0.5 M
H3+O (aq)
0M
NH3(aq)
0M
-x
+x
+x
Equilibrium Concentration
0.5 M - x
x
x
+
2
[H 3 O ] [NH3 ]
x
Ka(NH4+) = -------------------- = ---------------- ; approximation :0.5 - x . 0.5
[NH 4+]
0.5 - x
x2
Ka(NH4+) =
----------- = 5.56 x 10 -10
; x2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10
0.5
x = sqrt 2.78 x 10 -10 = 0.01342 =1.34 x 10 -2
[H+ ] = x = 1.34 x 10 -2 M
pH = -log [H+ ] = - log 1.34 x 10 -2
pH = 4.78
27
pH of 0.5 M NH4Cl solution is 4.78 (acidic)
a) What is the pH of a 0.5 M NaC2H3O2 salt solution? (HC2H3O; Ka = 1.8 x 10-5)
An aqueous solution of NaC 2H 3O is basic as discussed previously in problem 13b.
NaC 2H 3O2 dissociates completely in water:
NaC2H3O2 ---> Na+ (weak conjugate acid) + C 2H 3O 2-(stronger conjugate base).
C2H3O2-(strong conjugate base) reacts further in water producing an basic solution.
C2H3O2- + H2O
HO- + HC2H3O2 ;
this is a acid dissociation equilibrium or Kb
For a weak acid (HC2H3O2) with Ka is given:1.8 x 10-5 Kb for the C2H3O2-(stronger conjugate base) in
the above equilibrium can be calculated from the equation:
( Kb x Ka = Kw)
Kb(C2H3O2-) x Ka (HC2H3O2) = 1.0 x 10-14
Kb (C2H3O2-) x 1.8 x 10-5
= 1.0 x 10-14
1.0 x 10-14
Kb (C2H3O2 ) = ------------ = 5.56 x 10-10
1.8 x 10-5
C2H3O2- + H2O(l)
HO - + HC2H3O2
Initial concentration
Change
Equilibrium concentration
[C2H3O2-]
0.5 M
-x
0.5 - x
[C2H3O2-][ HO-]
-------------------- =
[C2H3O2-]
-
Kb (C2H3O2 ) =
[ HO-]
0.0 M
x
x
[HC2H3O2]
0.00 M
x
x
x2
---------------- ; approximation :0.5 - x ~ 0.5
0.5 - x
x2
Kb (C2H3O2 ) = ----------- = 5.56 x 10-10
; x2 = 5.56 x 10-10 x 0.5 = 2.78 x 10-10
0.5
x = sqrt 2.78 x 10-10 = 0.01342 =1.34 x 10 -2
[HO-] = x = 1.34 x 10 -2 M
-
p[OH] = -log [HO-] = - log 1.34 x 10-2
p[OH] = 4.78
pH = 14- pOH = 14- 4.78 = 9.22
pH of 0.5 M HNa2H3O2 solution is 9.22 (basic)
28
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