CHEMISTRY
Chapter 15
Acid-Base Equilibria
Acids and Bases
• Arrhenius’ Definition:
• Acids - are substances that produce hydrogen ions
(protons or H+) in solution.
Bases - are substances that produce hydroxide ions in
solution.
Strong Acids and Strong Bases – totally ionize in
solution
Weak Acids and Weak Bases – partially ionize in
solution
Bronsted-Lowry
• Bronsted-Lowry Acid – any compound that donates H+
• Bronsted-Lowry Base – any compound that accepts H+
• - does not need to have OH in chemical formula
• - ex. NH3
Acid Dissociation in Water
• General Rxn. when Acid dissolves in H2O
• HCl +
acid
H2O
base
H3O+
+ Clconj. Acid conj. base
Properties of H2O @ 25 oC
• H2O (l) D H+ (aq)
+
OH- (aq)
• Neutral, can act as an acid and a base
• Kw = [H+][OH-] = 1.0 x 10-14 only @ 25 oC
• Kw = water dissociation constant
Acidity vs. Basicity
• If [H+] >[OH-] , solution is acidic
• If [H+] <[OH-] , solution is basic
• The term pX = -log [concentration of X]
• So:
pH = -log [concentration of H+]
•
pOH = -log [concentration of OH-]
• pH = power of hydrogen; the power of H to which 10 is
raised
pH
• Kw = [H+][OH-] = 1.0 x 10-14 @ 25 oC
• pKw = [-log H+] + [-log OH-]= - [log 1.0 x 10-14]
= 14
• pH = -log [H+]
• pOH = -log [OH-]
Properties of H2O @ 25 oC
• pKw = - [log 1.0 x 10-14] = 14
• pH + pOH = 14
• pH = 7
• pH = pOH
pOH = 7
Things to Remember
• pKw = - [log 1.0 x 10-14] = 14 @ 25 oC
• pH + pOH = 14
• pH <7 ; acidic
• pH > 7; basic
• pH is between 0 - 14
Broensted-Lowry’s
Definition
•
• Acid – is a proton (H+) donor.
• Base – is a proton (H+) acceptor.
• * Broensted-Lowry Definition is more general
– It even applies to bases that have no –OH such as NH3.
Terminologies
• H+
• OH• H3O+
=
=
=
proton
hydroxide ion
hydronium ion
• Conjugate base –acid minus proton
• Conjugate acid – base plus proton
More Terminologies
• Conjugate acid-base pair
– Consists of 2 substances related to each other by the
donation and acceptance of a single proton (H+).
• Acid Dissociation Constant (Ka)
Equations
•
•
•
•
•
•
•
•
pH = - log [H+]
pOH = - log [OH-]
[H+] = 10 – pH
[OH-] = 10 - pOH
Kw = 10 - pKw
pKw = pH + pOH
pKw = - log [Kw]
Kw = [H+][OH-]
Sample Problem
• At 40 oC, a solution has Kw = 2.916 x 10-14; pH = 7.51
•
•
•
•
•
•
Calculate the following:
A. pOH of the solution
B. hydrogen ion concentration [H+]
C. hydroxide ion concentration [OH-]
D. pKw
E. Is the solution acidic basic or neutral?
Equilibrium
• K
=
[H3O+ ][Cl- ]
[HCl][H2O]
=
[H+][Cl-]
[HCl]
– H2O removed from top and bottom since H3O+ is simply
H+ dissolved in water.
• Remember:
Keq
=
[products]
[reactants]
Acid Strength
• Strength of acid is given by the equilibrium position of the
dissociation reaction:
• HA (aq) + H2O (l)
H3O+ + A-
• Strong acid – totally ionized and equilibrium
lies far to the right
• Weak acid – only partially ionized and
equilibrium lies far to the left
Strong Acid vs. Weak Acid
• Strong Acid – yields a weak conjugate base (one that
has weak affinity for proton; weaker than H2O)
• Weak Acid – yields a strong conjugate base (one that
has strong affinity for proton; stronger than H2O)
Comparison
Property
Strong Acid
Weak Acid
Ka value
Large Ka
Small Ka
Equil. Position
Far to the right
Far to the Left
Equil. Concn
[H+] = [HA]0
[H+] << [HA]0
Conj. Base
Strength vs H2O
A- much weaker base
than H2O
A- much stronger base
than H2O
Please Note!
• Tuesday’s experiment is Experiment 29:
Choice I.
Sample Problems
• Given [OH-] = 1.0 x 10-12 M, calculate pH. Is the
solution basic, acidic or neutral?
• Given [H+] = 4.30 x 10-6 M, calculate pH. Is the solution
basic, acidic or neutral?
Strong Acids and Bases
• If the molarity of the acid or base is less than 10-6 M
then the autoionization of water needs to be taken into
account. In other words, water is the primary source
of H+ and OH-, so the pH would be neutral.
Strong Acids and Bases
Strong Acids
• The strongest common acids are HCl, HBr, HI, HNO3,
HClO3, HClO4, and H2SO4.
• are strong electrolytes.
• All strong acids ionize completely in solution:
Strong Acids and Bases
• If the molarity of the acid or base is less than 10-6 M
then the autoionization of water needs to be taken into
account. In other words, water is the primary source
of H+ and OH-, so the pH would be neutral.
pH of Strong Acids and Bases
• The pH (and hence pOH) of a strong acid is given by the
initial molarity of the acid.
• The pOH (and hence pH) of a strong base is given by the
initial molarity of the base.
• Be careful of stoichiometric ratios!
Please Note!
• Tuesday’s experiment is Experiment 29:
Choice I.
Bronsted-Lowry Acids and
Bases
• Bronsted-Lowry acids – compounds that
donate a proton (H+)
• Bronsted-Lowry Bases – compounds that
accept a proton (H+)
• Note that Bronsted-Lowry bases need not have
the –OH group on the formula
Weak Acids
• Weak acids are only partially ionized in solution.
• There is a mixture of ions and unionized acid in solution.
• Therefore, weak acids are in equilibrium:
HA(aq) + H2O(l)
HA(aq)
H3O+(aq) + A-(aq)
H+(aq) + A-(aq)
[H3O  ][ A - ]
Ka 
[HA]
[H  ][ A - ]
Ka 
[HA]
NOTE
• For Weak Acids and Weak Bases:
• USE ICE to determine
pH and pOH.!
+
H,
OH ,
Sample A Problem
• A solution of 0.10 M formic acid (HCOOH) has a pH
of 2.38 at 25 oC.
• A. Calculate Ka for formic acid at this temperature.
• B. What percent of this solution is ionized?
Sample Problem
• The Ka of acetic acid is 1.8 x 10-5.
• A. Calculate the pH of a 0.30 M solution of
CH3COOH.
• B. Calculate OH- and pOH.
• C. Calculate Kb.
• Calculate % ionization.
A Simple Trick
• Use of approximation: eliminates the difficulty of
quadratic equations.
• Approximation is Valid if:
X_______ x 100 < 5 %
[Initial Concn.]
Relationship between Ka and
Kb
• Ka x Kb = 1.0 x 10 -14
only at 25 oC.
pH of polyprotic acids
• Treat polyprotic acids as separate steps!
• #1. H2A (aq) D H+ (aq) + HA- (aq) Ka1
• # 2. HA- (aq) D
H+ (aq) + A-2 (aq) Ka2
• Initial [H+] in Step 2 is Equil. [H+] from Step 1.
• Total [H+] = SUM from Steps 1 & 2
HOMEWORK
• What is the pH of a 1.00 M solution of
tartaric acid, H2C4H4O6 (aq.) at 25.0 oC?
• Answer: pH = 1.49
Sample Problem
• The Ka of acetic acid is 1.8 x 10-5. Calculate the Kb of
of CH3COOH.
Sample Problem
• The Kb of ammonia is 1.8 x 10-5. Calculate the pH of
a 0.15 M solution of NH3.
Sample Problem
• Calculate the concentration of an
aqueous solution of NaOH that has a pH
of 11.50.
HOMEWORK
• What is the pH of a 1.00 M solution of
tartaric acid, H2C4H4O6 (aq.) at 25.0 oC?
• Answer: pH = 1.49
Weak Acids
Calculating Ka from pH
• Weak acids are simply equilibrium calculations.
• The pH gives the equilibrium concentration of H+.
• Using Ka, the concentration of H+ (and hence the pH) can
be calculated.
– Write the balanced chemical equation clearly showing the
equilibrium.
– Write the equilibrium expression. Find the value for Ka.
– Write down the initial and equilibrium concentrations for
everything except pure water. We usually assume that the
change in concentration of H+ is x.
Weak Acids
Calculating Ka from pH
• Substitute into the equilibrium constant expression and
solve. Remember to turn x into pH if necessary.
Using Ka to Calculate pH
• Percent ionization is another method to assess acid
strength.
• For the reaction
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
% ionization 
[H3O ]eqm
[HA]0
 100
Sample Problem
• A solution of NH3 in water has a pH of 10.50. What is
the initial molarity of the solution?
Other Weak Bases
• Amines ex. Methylamine (CH3NH2)
• carbonate ion (CO32-)
• hypochlorite ion (ClO-1)
Weak Bases
• Also use ICE!
• Calculation is the same as for weak
acids!
• Main difference is that you get [OH-]
and pOH first.
Effects of Salts on pH
•
•
•
•
•
Conjugate bases of strong acids have no effect on pH.
Conjugate acids of strong bases have no effect on pH.
Conjugate bases of weak acids increase pH (more basic).
Ex. F- (aq) + H2O(l) D HF (aq) + OH- (aq)
Conjugate acids of weak bases decrease pH (more
acidic).
• NH4+(aq) + H2O (l) D NH3 (aq) + H3O+ (aq)
Relationship Between Ka
and Kb
Acid-Base Properties of
Salt Solutions
Combined Effect of Cation and Anion in Solution
• A cation that is the conjugate acid of a weak base will
cause a decrease in the pH of the solution.
• Metal ions will cause a decrease in pH except for the
alkali metals (Grp. I) and alkaline earth
metals.(Grp.II)
• When a solution contains both cations and anions from
weak acids and bases, use Ka and Kb to determine the
final pH of the solution.
Sample Problem
• Determine whether the resulting solution in water will be
acidic, basic or neutral.
•
•
•
•
A. K+ClO3B. Na+CH3COOC. Na2HPO4
D. NH4+Cl-
Ka for HPO4- = 4.2 x 10-13
Sample Problem
• Predict whether the potassium salt of citric acid
(K2+HC6H5O7-) will form an acidic, basic or neutral
solution in water.
Weak Acids
Polyprotic Acids
• Polyprotic acids have more than one ionizable proton.
• The protons are removed in steps not all at once:
H2SO3(aq)
H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2
HSO3-(aq)
H+(aq) + SO32-(aq)
Ka2 = 6.4 x 10-8
• It is always easier to remove the first proton in a
polyprotic acid than the second.
• Therefore, Ka1 > Ka2 > Ka3 etc.
Weak Acids
Polyprotic Acids
Sample Problem
• The solubility of CO2 in pure water at 25 oC and 0.1 atm
is 0.0037 M. The common practice is to assume that all
of the dissolved CO2 is in the form of carbonic acid
(H2CO3), which is produced by the reaction between the
CO2 and H2O.
• What is the pH of a 0.0037 M solution of H2CO3?
• Ka1 = 4.3 x 10-7
• Ka2 = 5.6 x 10 -11
Answer
• pH
=
4.4
• x1 = 4.0 x 10-5 M
• [CO3-] = 5.6 x 10-11 M
Sample Problem
• Calculate the pH and concentration of oxalate ion
(C2O42-), in a 0.020 M solution of oxalic acid (H2C2O4)
Acid-Base Behavior and
Chemical Structure
Binary Acids
Lewis Acids and Bases
• Brønsted-Lowry acid is a proton donor.
• Focusing on electrons: a Brønsted-Lowry acid can be
considered as an electron pair acceptor.
• Lewis acid: electron pair acceptor.
• Lewis base: electron pair donor.
• Note: Lewis acids and bases do not need to contain
protons.
• Therefore, the Lewis definition is the most general
definition of acids and bases.
Lewis Acids and Bases
• Lewis acids generally have an incomplete octet (e.g.
BF3).
• Transition metal ions are generally Lewis acids.
• Lewis acids must have a vacant orbital (into which the
electron pairs can be donated).
• Compounds with p-bonds can act as Lewis acids:
H2O(l) + CO2(g)  H2CO3(aq)
End of Chapter 16
Acid-Base Equilibria
Problem 1
• Give the conjugate base of the following BronstedLowry acids:
• H2SO3
• H2AsO4• NH4+
Problem 2
• By what factor does [H+] change for a pH change of:
• A. 2.00 units
• B. 0.50 units
Problem 3
• Calculate [OH-] and pH for:
• A.) 1.5 x 10-3 M Sr(OH)2. Sr(OH)2 is a strong base.
• B.) a solution formed by adding 10 mL of 0.100 M
HBr to 20.0 mL of 0.200 M HCl
Problem
•
• Calculate the pH of a solution made by adding 15.00
grams of NaH in enough water to make 2.5 L of
solution
Problem
• Write the ionization and equilibrium expressions for
HBrO2.
Problem
• A particular sample of vinegar has a pH of 2.9.
Assuming acetic acid is the only acid in the vinegar,
find the initial concentration of acetic acid in the
vinegar.
Problem
• The acid dissociation constant for benzoic acid
(HC7H5O2) is 6.3 x 10-5. Calculate the equilibrium
concentrations of H3O+, C7H5O2- and HC7H5O2 if the
initial concentration of HC7H5O2 is 0.050 M.
Problem
•
Calculate the pH of 0.120 M pyridine (C5H5N). Kb
for pyridine is 1.7 x 10-9.
Problem
• A 0.200 M solution of a weak acid, HA is 9.4%
ionized. Using this information, calculate [H+], [A-],
[HA] and Ka for HA.
Problem
• An unknown salt is either NaF, NaCl, or NaOCl.
When 0.05 mole of the salt is dissolved in water to
form 0.500 L of solution, the pH of the solution is
8.08. What is the identity of the salt?
Problem
• Write the chemical equation and the Kb expression
for the ionization of the following bases in aqueous
solution:
• A. Dimethylamine (CH3)2NH
• B. Formate ion
(HCOO-)
• C. Carbonate ion (CO32-)
Problem
• Calculate the molar concentration of OH- ions in a
0.075M solution of ethylamine.
• Kb of C2H5NH2 = 6.4 x 10-4.
• Calculate the pH of this solution.
Problem
• Ka for acetic acid (CH3COOH) is 1.8 x 10-5 while Ka
for hypochlorous (HClO) ion is 3.0 x 10-8.
• A. Which is the stronger acid?
• B. Which is the stronger conjugate base? Acetate ion
(CH3COO-) or chlorous (ClO-) ion?
• C. Calculate kb values for CH3COO- and ClO-.
Solubility vs. Ksp
• Solubility – refers to the quantity that dissolves to
form a saturated solution. Unit is gm/liter or
moles/liter for molar solubility.
• - solubility if affected by temperature
• Solubility product constant – is the equilibrium
constant for the equilibrium that exists between the
ionic solute and its saturated aqueous solution
Ksp
• Solubility product constant – the equilibrium
constant indicating how soluble the product is in
water.
• Example:
CaF2 (s) D Ca2+ (aq) + 2F- (aq)
• Ksp = [Ca2+][F-]2
Problem 1
• Give the ionization equation and Ksp expression for
the reaction:
• Ag2CrO4 (s) D
?
+
?
Problem 2
• The Ksp for CaF2 is 3.9 x 10-11 at 25 oC. Assuming
that CaF2 dissociates completely upon dissolving and
that there are no other important equilibria affecting
its solubility:
•
a. calculate the solubility of CaF2 in moles per liter.
• b. calculate the solubility of CaF2 in grams per liter.
Problem 3
• The Ksp for LaF3 is 2.0 x in 10-19. What is the
solubility of LaF3 in water in moles per liter?
• What is the solubility of LaF3 in water in grams per
liter?
Answer
• A. 9.28 x 10-6 M
Factors Affecting Solubility
• Common-Ion Effect
• Concentration
Problem 4
• Calculate the molar solubility of CaF2 at 25 oC in a
solution that is:
• A. 0.010 M in Ca(NO3)2
• B. 0.025 M in NaF
Precipitation of Ions
• Remember Q, the reaction quotient?
• If: Q > Ksp, prepitations occurs until Q = Ksp
Q = Ksp, equilibrium exists (saturated solution)
Q < Ksp, solid dissolves until Q = Ksp.
Problem 1
• A solution contains 1.0 x 10-12 M Ag+ and 2.0 x 10-2 M
Pb2+. When Cl- is added, both AgCl and Ksp
precipitate from the solution.
• What concentration of Cl- is necessary to begin the
precipitation of each salt?
• Which salt precipitates first?
Insoluble Chlorides
• Of the common metals ions, only Ag+, Hg2 2+, Pb 2+
form insoluble chlorides.
Qualitative Analysis
• Order of separation of ions:
Cl-  S2-  (OH)-  (PO4) 3-  NH4+
1st step:
2nd step:
3rd step:
4th step:
add 6M HCl
add H2S and 0.20 M HCl
add (NH4)2S
add (NH4)2HPO4 and NH3