Equilibrium – Acids and Bases
Review of Acids and Bases
• Arrhenius Theory of Acids and Bases
▫ An acid is a substance that dissociates in water to
produce one or more hydrogen ions (H+)
▫ A base is a substance that dissociates in water to
form one or more hydroxide ions. (OH-)
▫ Examples:
 Acid:
 Base:
HCl(aq) H+(aq) + Cl-(aq)
LiOH  Li+(aq) + OH-(aq)
Limitations:
• Classified based on chemical formula
• Some substances do not have OH- in their
chemical formulas but still yield OH- when they
react with water. E.g. NH3 (ammonia)
• Solution?
• Bronsted-Lowry Theory of Acids and
Bases
▫ An acid is a proton (H+) donor and must have
H in its formula.
▫ A base is a proton acceptor and must have a
lone pair of electrons to form a bond with H+
• Two molecules or ions that are related by the
transfer of a proton are called a conjugate
acid-base pair.
▫ Conjugate acid of a base is the particle that
results when the base receives the proton from the
acid.
▫ Conjugate base of the acid is the particle that
results when the acid donates a proton.
Practice
• Identify the conjugate acid/base pairs in the
following:
NH3(aq) + H2O(l)  NH4+(ag) + OH-(aq)
• Amphiprotic: Can act as either an acid or a
base i.e has both a lone pair and an H-atom
▫ Ex: H2O
HCO3-(aq) ) + H2O(l)  H2CO3(aq) + OH-(aq)
HCO3-(aq) + H2O(l)  CO32-(aq) + H3O+(aq)
Strong Acids and Bases
• Completely dissociate in water into their ions
(quantitative reactions)
100%
HCl(aq) + H2O(aq)  H3O+(aq) + Cl-(aq)
100%
LiOH + H2O(aq)  LiOH(aq) + OH-(aq)
• As a result the [H3O+] in a solution of a strong
acid is equal to the concentration of the acid.
• Strong acids include HClO4 (perchloric), HI,
HBr, HCl, H2SO4 (sulfuric), and HNO3 (nitric)
• Strong bases include all oxides and hydroxides
of alkali metals as well as alkaline earth metal
oxides and hydroxides below beryllium.
• The stronger the acid, the weaker it’s conjugate
base and vice versa
Weak Acids and Bases
• Do NOT completely dissociate in water into their ions
1%
CH3COOH(aq) + H2O(aq) ↔ H3O+(aq) + CH3COO-(aq)
1%
NH3(aq) + H2O(aq) ↔ NH4+(aq) + OH-(aq)
• As a result, the concentration [H3O+] in a solution of a
weak acid is always less than the concentration of the
dissolved acid.
Percent Ionization
• % Ionization for strong acids is 100%
• % Ionization for weak acids is < 100%
Polyprotic Acids
• Monoprotic acids contain only a single
hydrogen ion that can dissociate.
▫ Example: HCl
• Polyprotic acids contain more than one
hydrogen ions that can dissociate.
▫ Example H2SO4, H3PO4
Autoionization of Water
• Water dissociates:
H2O(l) + H2O(l) <--> H3O+(aq) + OH-(aq)
What is the equilibrium constant (K) of this
reaction?
Kw =
[H3O+][OH-]
Kw is the ion product constant of water
Kw = 1.0 x 10-14 @ SATP
• [H3O+] > [OH-]  acidic
• [H3O+] < [OH-]  basic
• [H3O+] = [OH-]  neutral
Practice
• There is a 0.25 mol/L solution of HBr(aq)
a) Calculate the hydrogen ion concentration
b) Calculate the hydroxide ion concentration
• Strong acid – ionizes completely
• Kw =
[H3O+][OH-] = Kw = 1.0 x 10-14
Practice
• In a 0.13 mol/L solution of NaOH, what is the
[H+] and [OH-]?
• NaOH is hydroxide of an alkali metal so it is a
STRONG base meaning [OH-]= [base]
• Kw =
[H3O+][OH-] = Kw = 1.0 x 10-14
The pH Scale
•
•
•
•
•
•
•
Measures the acidity of a solution.
Measure [H+] (or [H3O+]) in a solution.
Ranges from 0 to 14
Distilled water is 7 (neutral)
Acids < 7
Bases > 7
A logarithmic scale
▫ A pH of 1 is ten times more acidic then a pH of 2
pH equations
• pH = -log[H3O+]
• [H3O+] = 10-pH
• pOH = -log[OH-]
• [OH-] = 10-pOH
• pH + pOH = 14
Practice
• Calculate the pH of a solution of 1.24 x 10-4 M HCl
• pH = -log[H3O+]
• pH = -log[1.24 x 10-4 ]
• pH = 3.91
Practice
• If the normal pH of blood is 7.3, then find the pOH,
[H3O+] and [OH-]
• pH + pOH = 14
• 7.3 + pOH = 14
• pOH = 6.7
• [H3O+] = 10-pH
• [H3O+] = 10-7.3
• [H3O+] = 5 x 10-85
• [OH-] = 10-pOH
• [OH-] = 10-6.7
• [OH-] = 2 x 10-7
Acid- Base Strength & Dissociation
• Recall: Strong acids and bases dissociate
quantitatively (>99.9%) in water
• Weak acids and bases dissociate partially in
water
• When a weak acid or base is added to water
dynamic equilibrium is established
The Acid-Dissociation Constant, Ka
For Weak Acids:

(aq)
HA(aq)  H2O(l )  H3O

A

(aq)

[H3O ][A ]
Ka 
[HA]
All concentrations are those at equilibrium
Note: the smaller the value of Ka, the weaker the acid
Determine the Ka of propanoic acid (C2H5COOH(aq))
given that a 0.10 mol/L solution has a pH of 2.96.
(Hint: use an ICE table)
[H3O ][A  ]
Ka 
[H3O+] = 10-pH
[H3O+] = 10-2.96
[H3O+] =0.00110
I
C
E
C2H5COOH(aq)
0.10 mol/L
-x
[HA]

 C2H5COOo mol/L
+x
+ H3O+
0 mol / L
+x
The Base-Ionization Constant, Kb
For Weak Bases:
B(aq)  H2O(l )  BH


(aq)
 OH

(aq)

[BH ][OH ]
Kb 
[B]
All concentrations are those at equilibrium
Note: the smaller the value of Kb, the weaker the base
Calculate the pH of a 3.6 X 10-3 mol/L solution of
quinine (C20H24N2O2(aq)). Kb = 3.3 X 10-6
C20H24N2O2(aq) + H2O  HC20H24N2O2 + + OHI
C
E
[BH  ][OH  ]
Kb 
[B]
Relationship between Ka, Kb, & Kw
Example: Consider acid HCN and conjugate base CNHCN(aq) + H2O(l)  H3O+(aq) + CN- (aq)
Ka = [H3O+] [CN -]
[HCN]
Kb = [HCN] [OH -]
[CN-]
Ka Kb = [H3O+] [CN -] [HCN] [OH -]
[HCN]
[CN-]
Ka Kb = [H3O+] [OH -]
Ka Kb = Kw
Practice
• The Kb for hydrazine, N2H4(g), a rocket fuel, is
1.7 x 10-6. What is the Ka of its conjugate acid,
N2H5 (aq)?
• K a Kb = Kw
• Ka (1.7 x 10-6)= 1.0 x 10-14
• Ka = 6.0 x 10-9
Practice
• Chloracetic acid, HC2H2O2Cl(aq) is a weak acid.
Determine the pH of a 0.0100 mol/L solution of
chloracetic acid if the Kb of the conjugate base is Kb=
7.35 x 10-12 .
I
C
E
HC2H2O2Cl (aq)
 C2H2O2Cl-
+ H3O+

• K a Kb = Kw
• Ka (7.35 x 10-12 )= 1.0 x 10-14
• Ka =0.00136


[H3O ][A ]
Ka 
[HA]
Neutralization Reactions
• A salt is an ionic compound that results from a
neutralization reaction
• Acid + base  salt + water
• Salts are strong electrolytes that completely
ionize in water
• Salts can affect the pH of a solution
Neutral Salt Solutions
•
•
•
•
Strong acid + strong base
Both will dissociate completely
Therefore…
Salts containing an anion from a strong acid and
cation from a strong base will be neutral
• Ex: NaOH + HCl  NaCl + H2O
Acidic Salt Solutions
• Strong acid + weak base
• The acid dissociates completely, but the base
only dissociates partially
• Therefore…
• Salts containing an anion from a strong acid and
a cation from a weak base will be acidic
• Ex: HCl + NH3  NH4Cl  NH4+ + Cl• NH4 will act as a weak acid
Basic Salt Solutions
• Weak acid + strong base
• The base will dissociate completely but the acid
will only dissociate partially
• Therefore…
• Salts containing an anion from a weak acid and a
cation from a strong base will be basic
• Ex: HC2H3O2 + NaOH  NaC2H3O2 + H2O
 Na+ + C2H3O2• C2H3O2- will act as a weak base
Buffers
• Resist changes in pH when a moderate amount of
acid or base is added
• Must contain enough acid to react with any base
that is added, and enough base to react with any
acid that is added
• The acid and base components must not react in a
neutralization reaction
• Solutions of a weak acid and the salt of its conjugate
base OR a weak base and the salt of its conjugate
acid
Acetic Acid/Sodium Acetate Buffer
• Consider a buffered solution made by adding similar molar
concentrations of acetic acid (CH3COOH) and its salt, sodium
acetate (CH3COONa)
• Sodium acetate ionizes completely in water:
CH3COONa( s )  CH3COO  ( aq)  Na ( aq)
• When an acid is added to the buffer, the acetate ion reacts with
the hydronium ion to neutralize the solution
 H3O

CH3COOH ( aq)  OH

CH3COO

( aq)
( aq)
 CH3COOH ( aq)  H 2O(l )
• When a base is added to the buffer, the acetic acid reacts with
the hydroxide ions to neutralize the solution
( aq)
 CH3COO

( aq)
 H 2O(l )
Buffer Examples
• It is extremely important for blood to remain
near it’s optimal pH of 7.4
• Any change greater than 0.2 is life-threatening
• If the blood were not buffered, the acid
absorbed by consuming a glass of orange juice
would probably kill you