Daniel L. Reger
Scott R. Goode
David W. Ball
www.cengage.com/chemistry/reger
Chapter 15
Solutions of
Acids and Bases
Arrhenius Acids and Bases
• Acid: a substance that produces H3O+
when dissolved in water.
• Base: a substance that produces OHwhen dissolved in water.
• Arrhenius acids and bases are limited to
water solutions.
Bronsted-Lowry Acids and Bases style
• Acid: proton donor.
• Base: proton acceptor.
• Bronsted-Lowry acid base
reaction: proton transfer from acid
to base.
Conjugate Acid-Base Pairs
• Conjugate Acid-Base Pairs: two
species that differ by a proton.
• Acid : the species that contains the
proton that is transferred.
• Conjugate base: the species formed by
loss of the proton.
HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
• HF and F make up an acid-base
conjugate pair. F- is the conjugate base
of HF.
Acid-Base Conjugate Pairs
HF(aq) + H2O(l) ⇌ F (aq) + H3O (aq)
HF is the acid, F- is its conjugate base
(or F- is a base and HF is its conjugate
acid).
• The reverse reaction is
+
F (aq) + H3O (aq) ⇌ HF(aq) + H2O(l)
+
H3O is the acid, H2O is its conjugate
base.
•
-
+
Acid-Base Conjugate Pairs
Acid
Base
HCl
Cl
H2SO4
HSO4HSO4SO42NH4+
NH3
H3O+
H2O
H2O
OHH2O is amphoteric - it acts as both
an acid and a base.
Acid-Base Reactions
• An Arrhenius system
acid + base  water + salt
is limited to reactions in water.
• A Bronsted-Lowry system
+
HA + B ⇌ A + BH
is more general - a proton is
transferred from the acid (HA) to the
base (B).
Bronsted-Lowry Acid-Base Reactions
acid + base ⇌ conj. base + conj. acid
⇌
F
HCl + H2O ⇌
Cl
H2O + NH2- ⇌ OHHF + NH3
-
+ NH4+
+ H3O+
+ NH3
Identifying Conjugate Pairs
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)
HCN is the acid, CN- is its conjugate base.
H2O is the base, H3O+ is its conjugate acid.
CH3COO-(aq)+H2O(l) ⇌ CH3COOH(aq)+OH-(aq)
CH3COO- is the base, CH3COOH is its
conjugate acid. H2O is the acid, OH- is its
conjugate base.
Test Your Skill
Identify the acid-base conjugate pairs
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)
CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)
Autoionization of Water
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
• H3O+ and H2O are a conjugate acidbase pair.
• H2O and OH are a conjugate acidbase pair.
Autoionization of Water
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
•
The equilibrium constant for this
reaction is called Kw.
+
• Kw = [H3O ][OH ]
• Kw changes with temperature and is
-14
equal to 1.0 × 10 at 25o C.
Calculating Hydrogen and Hydroxide Ion
Concentrations
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
• In pure water,
[H3O+]
+
-
= [OH ].
-
Kw = [H3O ][OH ] = 1.0 × 10
Kw =
+
2
+
[H3O ]
-
= 1.0 × 10
-14
-7
• [H3O ] = [OH ] = 1.0 × 10 M
-14
Acidic or Basic Solutions
• Adding an acid or a base to water
causes the H3O+ and OHconcentrations to change.
• Calculate the hydroxide ion
concentration of a solution in which
the hydrogen ion concentration is
-3
3.6 × 10 M.
Test Your Skill
• Calculate the hydrogen ion
concentration in a solution in which
the hydroxide ion concentration is
0.025 M.
pH Scale
• pH = -log10[H3O+]
Calculate the pH of a 0.0034 M
solution of [H3O+].
• pH = -log(0.0034) = 2.47
• pH values are generally given to two
decimal places. A pH of 2.47 has two
significant figures; the 2 serves to
locate the decimal point of the original
number.
Calculating Hydrogen Ion Concentration from
pH
• Calculate the hydrogen ion concentration
in a solution that has a pH of 3.52.
p-notation
• The p-notation can be used for other
quantities.
• pOH = -log[OH-]
• pKw = -log Kw = -log(1.0 × 10-14) = 14.00
Relating pH and pOH
[H3O+][OH-] = Kw = 1.0 × 10-14
log([H3O+][OH-]) = log Kw
log[H3O+] + log[OH-] = log Kw
-log[H3O+] - log[OH-] = -log Kw
pH + pOH
= pKw
pH + pOH
= 14.00
Relating pH and pOH
Acid
pH
Neutral
Base
<7
=7
>7
pOH >7
=7
<7
pH + pOH = 14.00
Strong Acids and Bases
• HA(aq) + H2O(l)
100%
H3O+(aq) +A-(aq)
• Strong acids ionize completely in
solution.
• Memorize the six common strong acids:
HCl, HBr, HI, HNO3, HClO4, and H2SO4.
pH of a Strong Acid Solution
• Calculate the pH of a 0.050 M HCl
solution.
Strong Bases
• Strong bases quantitatively produce
hydroxide ions in water.
• The most common strong bases are the
group IA and soluble IIA oxides and
hydroxides.
Strong Bases
NaOH(s) Na (aq) + OH (aq)
+
-
Li2O(s) + H2O(l) 2Li (aq) + 2OH (aq)
+
-
pH of a Strong Base Solution
•
Calculate the pH of a 0.035 M Ba(OH)2
solution.
Test Your Skill
What is the concentration of a solution of
HCl if the pH is 3.75?
What is the concentration of a solution of
KOH if the pH is 11.60?
Weak Acids and Bases
• Weak acids and weak bases are those
that do not ionize completely in water.
• A weak acid exists in equilibrium with its
conjugate base.
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)
• A weak base exists in equilibrium with its
conjugate acid.
CH3NH2 (aq) + H2O(l) ⇌
CH3NH3+(aq) + OH-(aq)
Weak Acids and Bases
• HA(aq) + H2O(l) ⇌ H3O (aq) + A (aq)
+

Ka 
[H 3 O ][A

-
]
[HA]
• B(aq) + H2O ⇌ BH (aq) + OH (aq)
+
Kb 
[BH

][OH
[B]

]
-
Competition for Protons
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
-
• Both H2O and A are bases and compete
for the proton.
• If HA is a strong acid, A is an extremely
poor proton acceptor.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
Proton bonded to H3O+
Proton bonded to Cl-
Competition for Protons
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
-
• If HA is a weak acid, A is a weak base.
• The stronger the weak acid, the weaker
the conjugate base.
HF(aq) + H2O ⇌ H3O+(aq) + F-(aq)
Proton bonded to FProton bonded to H3O+
The Influence of the Solvent
• Leveling Effect: In water, all acids
stronger than H3O+ appear to be equally
strong. By using a weaker base than
water as the solvent, we can differentiate
between the strong acids.
weakest
strongest
|
|
HNO3 < H2SO4 < HCl < HBr < HI < HClO4
The Concentration of an Acid
• Analytical Concentration: the
total concentration of all forms of an
acid; both the protonated form (the
acid) and the unprotonated form
(the conjugate base).
Equilibrium Calculations
• You will likely see two kinds of equilibrium
calculations:
1. You measure equilibrium concentrations and
calculate K or
2. you are given starting concentrations and K and
calculate equilibrium concentrations.
• The approach is the same.
1. Write the balanced chemical equation.
2. Calculate equilibrium concentrations/iCe table.
3. Write the algebraic expression for K.
4. Substitute concentrations into expression.
5. Solve.
Calculating Ka for a Weak Acid
• A 0.250 M solution of HF is determined to be
3.7% ionized. Determine Ka for HF.
Determining Ka from pH
• The pH of a 0.100 M solution of HOCl
is 4.26. Calculate Ka for the acid.
HOCl + H2O ⇌ H3O+ + OCl-
Determining Concentrations of Species in Weak
Acid Solution
• Calculate the pH of a 0.50 M
-5
CH3COOH solution. Ka = 1.8 × 10
CH3COOH + H2O ⇌ H3O + CH3COO
+
-
Test Your Skill
• Determine the pH of a 0.025 M solution
-10
of HCN, Ka = 7.2 × 10 .
Calculating pH of a Weak Acid Solution
• Calculate the equilibrium concentrations
and pH of a 0.100 M HF solution,
-4
Ka = 6.3 × 10 . Use approximation.
Check Approximation
• Is 7.9  10-3 << 0.050?
• No, it is not. 5% of 0.050 is 2.5  10-3
and the calculated value is larger. The
approximation fails.
• We need another method to compute the
root.
Successive Approximation
Call the first value y 1  7.9  10
and use it as an approximat
ion to calculate
a new value, y 2. .
6 . 3  10
6 . 3  10
y2
2
4
4


( y 2 )( y 2 )
0 . 100  y 1
(y 2 )
 5 . 80  10
y 2  7 . 6  10
2
0 . 100  7 . 9  10
-3
-5
-3
-3
Check Second Approximation
• Is y2 sufficiently close to y1?
• Restating the question,
is 7.6 10-3 within 5% of 7.9  10-3?
• By within 5%, we mean is 7.6  10-3
within a range varying from 95% to
105% of 7.9  10-3?
• The answer is yes. Since the
second approximation was quite
close to the first, we can accept it.
Fraction Ionized in Solution
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
-
fraction ionized 
[A ]
-
[HA]  [A ]

[H 3 O ]

a nalytical
concentrat
ion of HA
Fraction Ionized in Solution
• Calculate fraction ionized for 0.500 M
-8
HOCl, Ka = 4.0 × 10 .
•
HOCl + H2O ⇌ H3O + OCl
+
-
Solutions of Weak Bases
• A weak base reacts with water to form
hydroxide ions.
B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)
Kb 
[BH

][OH
[B]

]
pH of a Solution of a Weak Base
Calculate the pH of a solution of 0.150 M
-4
methylamine, CH3NH2, Kb = 4.4 × 10 .
CH3NH2 + H2O ⇌ CH3NH3 + OH
+
-
Test Your Skill
• Calculate the pH of a 0.35 M solution of
-8
hydroxylamine, NH2OH, Kb = 1.1 × 10 .
Relating Ka and Kb
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

Ka 
[H 3 O ][A

]
[HA]
A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)
Kb 
[HA][OH
-
[A ]

]
Relating Ka and Kb

K aK b 
[H 3 O ][A


[HA]

K a K b  [H 3 O ][OH
K aK b  K w
]
[HA][OH
-
[A ]

]

]
Strengths of Weak Acid-Base Conjugate Pairs
The conjugate base of a weak acid is a
weak base:
•
CH3COOH + H2O ⇌ H3O+ + CH3COO-
Ka = 1.8 × 10-5
CH3COO- + H2O ⇌ CH3COOH + OH-
Kb = 5.6 × 10-10
HCN + H2O ⇌ H3O+ + CN-
Ka = 6.2 × 10-10
CN- + H2O ⇌ HCN + OH-
Kb = 1.6 × 10-5
Calculating Ka and Kb for Acid-Base Conjugate
Pairs
-4
• Ka for HNO2 is 4.6 x 10 . Calculate Kb for
the NO2 ion.
NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)
Test Your Skill
-9
• Kb for pyridine is 1.8 × 10 . Calculate Ka
for the pyridinium ion.
Ranking Acid-Base Strength
Weaker
acid
HCN
K a = 7.2 x 10 -10
Weaker
base
FK b = 2.8 x 10 -11
Stronger
acid
HF
K a = 3.5 x 10 -4
Stronger
base
CN-5
K b = 1.4 x 10
Conjugate Partners of Strong Acids and Bases
• The stronger the acid, the weaker
the conjugate base.
• strong acid  very weak conjugate
base
• strong base  very weak conjugate
acid
• weak acid  weak conjugate base
pH of Salt Solutions
• The pH of a salt solution is
 =7 (neutral) if it contains the cation of a strong
base (Na+) and the anion of a strong acid (Cl-)
 >7 (basic) if it contains the cation of a strong
base (K+) and the anion of a weak acid (F-)
 <7 (acidic) if it contains the cation of a weak
base (NH4+) and the anion of a strong acid
(NO3-).
The pH of a Salt Solution
• Determine the pH of a 0.59 M solution of
CH3OONa. Ka of CH3OOH = 1.8 × 10-5.
CH3OO (aq) + H2O(l) ⇌ CH3OOH(aq) + OH (aq)
• The acetate ion is a base. Calculate Kb for
the acetate ion:
Mixtures of Acids
• When two or more acids are
present in solution, the effects of
the weaker acids may be ignored if
Ka is smaller by a factor of 100 or
more.
Mixtures of Acids
• To calculate the pH of a mixture of HCl
-8
(strong acid) and HOCl (Ka = 3.0 × 10 ),
ignore the HOCl; treat the solution as a
solution of HCl.
• To calculate the pH of a mixture of
-4
HCOOH (Ka = 1.8 × 10 ) and HCN (Ka =
-10
7.2 × 10 ), ignore the HCN; treat the
solution as a solution of HCOOH.
Binary Hydrides
• Binary hydrides contain hydrogen
and one other element.
• Acidity depends on the HA bond
strength and the stability of A .
Binary Hydrides
• Acidity increases with decreasing H-A
bond dissociation energy
HF << HCl < HBr <
568.2
431.9
366.1
HI
298.3 kJ/mol
Binary Hydrides
• Acidity increases with increasing
electronegativity of A.
CH4
2.5
<
NH3
3.0
<
H 2O
3.5
<
HF
4.0
Oxyacids
• Oxyacids contain H, O, and a third
element X.
X–O–H
• X is usually a non-metal or a
transition metal in a high oxidation
state.
Oxyacids
• As the electronegativity of X
increases, the X-O bond becomes
more covalent and the O-H bond
becomes more polar.
+
X–O–H  X–O + H
Oxyacids
• Two factors increase acidity of the
oxyacid:
 Greater electronegativity of central
atom.
HClO4 > HBrO4
 Greater oxidation number of central
atom (more oxygen atoms).
HClO4 > HClO3 > HClO2 > HClO
Lewis Acids and Bases
• Lewis acid: electron pair acceptor.
• Lewis base: electron pair donor.
• Lewis acid-base reaction: formation
of a coordinate-covalent bond.
AlCl3 + Cl  AlCl4
• Coordinate-covalent bond: covalent
bond in which both electrons come
from one atom.