Acids and Bases 2
Reading: Moore chapter 16, sections 16.5-16.10
Questions for Review and Thought: 48, 50, 60, 66, 70, 72, 88, 90, 96, 100
Key Concepts and Skills:
•
definition of Lewis acid, Lewis base, titration, endpoint, common ion effect,
amphoteric compound.
•
Review calculations of Ka, Kb, pH for weak acids and bases; how to calculate the
endpoint pH for titration of weak acids and strong base, and for titration of weak
bases and strong acids.
Lecture Topics:
Calculations involving Ka and Kb for acids and bases
Acid/base equilibria follow the same patterns as all other types of equilibria:
• Write out the balanced chemical equation for the reaction of the acid with water or the
reaction of the base with water
• state the initial conditions in terms of known concentrations
• if given Ka or Kb, calculate Q to determine the direction of reaction; use the variable x
and stoichiometry for the change that occurs upon going to the equilibrium state
• express the equilibrium concentrations in terms of x, and substitute into the expression
for Ka or Kb; solve for x, using quadratic if necessary; if initial concentrations are a factor
of ~103-104 greater than the value of Ka or Kb, assumption that x<<[HA]o (or [A-]o or [B]o,
etc.) can be made.
Examples:
The pH of a 0.1 M solution of aqueous acetic acid has a pH of 2.87. Calculate the Ka for
acetic acid and the percent ionization.
CH3CO2H (aq) + H2O (l)  H3O+ (aq) + CH3CO2-(aq)
Initial
0.1M
1x10-7M
0
(assume 1x10-7~0)
Equilibrium: 0.1-x
x
x
+
Ka= [H3O ]eq[CH3CO2 ]eq/[CH3CO2H]eq; we know that 2.87=-log[H3O+]eq;
[H3O+]eq = x = .0013M; thus [CH3CO2-]eq= .0013M; [CH3CO2H]eq= 0.1-.0013= 0.098M
thus Ka =[.0013]2/[0.098] = 1.72 x 10 –5; %ionization = x/0.1M= .0013/0.1 *100= 1.3%
•Acid/Base ionization constants Kb and Kb are tabulated in Table 16.2
Determine the pH of a solution containing 0.5M HOCl.
HOCl is a weak acid with Ka=3.7 x 10-8
HOCl(aq) + H2O(l)  H3O+ + -OCl(aq)
Initial
0.5 M
1x10-7M 0
Equilibrium
0.5-x
x
x
Ka= 3.7x10-8= [H3O+]eq[-OCl]eq/ [HOCl]eq = x2/0.5-x
Since Ka~10-8, assume x<<0.5, thus 3.7x10-8 (0.5)=x2
x=1.3 x 10-4; thus [H3O+]=1.3 x 10-4 = [-OCl]; [HOCl] = 0.5-1.3x10-4 = 0.49M
pH = -log[H3O+] = -log (1.3x10-4) = 3.88
fraction dissociated = x/0.5M = 1.3x10-4/0.5 *100%= 2.6x10-2%
•Common Ion Effect – the presence of the conjugate base of an acid suppresses the
ionization of the acid
• Determine the pH of a solution containing 0.5M HOCl and 0.01M NaOCl.
HOCl(aq) + H2O(l)  H3O+ + -OCl(aq)
Initial
0.5 M
1x10-7M 0.01M
-7
-9
Q = (1x10 M)(0.01M)/ (0.5M) = 2x10 <Ka reaction proceeds in forward direction
Equilibrium:
0.5-x
x
0.01-x
Ka=3.7x10-8 = (x)(0.01-x)/(0.5-x)
Since Ka~10-8, assume x<<0.01 and x<<0.5, thus 3.7x10-8(0.5)=x(0.1)
x=1.8x10-6=[H3O+]; pH=-log(1.8x10-6)=5.74
fraction dissociated = x/0.5M = 1.8x10-6/0.5 *100%= 3.6x10-4%
thus, the presence of the conjugate base –OCl (the common ion) decreases the amount of
ionization (fraction dissociated) of the acid.
•Some situations when it is NOT safe to make assumptions about the magnitude of x:
when Ka or Kb is within a factor of 103-104 of the concentrations initially present in
solution.
Example: Calculate the pH of a 0.005M solution of dimethylamine (CH3)2NH,
Kb=5.9x10-4
II. Reactions of Acids and Bases; Titration
•The reaction of a strong acid with a strong base gives a salt and water.
HNO3 (aq) + NaOH  NaNO3 + H2O
This is essentially a neutralization reaction when exactly equivalent amounts of HNO3
and NaOH are combined; at this point a pH of 7 is reached.
1. HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq)
2. NaOH(aq)  -OH(aq) + Na+ (aq)
3. HNO3(aq) + H2O(l) + NaOH(aq)  H3O+(aq) + NO3-(aq) + -OH(aq) + Na+ (aq)
since H3O+(aq) + -OH(aq)  2 H2O (l)
and Na+ (aq) + NO3-(aq)  NaNO3
net: HNO3 (aq) + NaOH  NaNO3 + H2O
Net: HNO3 (aq) + NaOH  NaNO3 + H2O
Note: a titration is the slow addition of a known concentration of acid to a base (or the
addition of a known concentration of base to an acid) in the presence of an indicator
(which changes color at the endpoint) or using a pH meter; A titration curve is a plot of
volume of base added (or volume of acid added) versus pH for the titration experiment.
The endpoint is the inflection point of the graph. The endpoint of a strong acid/strong
base titration is always at (about) pH7.
• The reaction of equivalent amounts of a weak acid and a strong base gives a
slightly basic solution (pH>7), the extent of which is determined by Kb of the
conjugate base.
Example: HCOOH(aq) + NaOH (aq)  HCOO-(aq) + H2O (l) +
Weak acid
strong base
weaker base weaker acid
Na+(aq)
HCOO-(aq) + H2O(l)  HCOOH(aq) + -OH(aq) Kb = 5.6x10-11
Weak base weaker acid
weak acid
strong base
The equilibrium lies strongly on the reactant side; yet the presence of the strong base –OH
on the product side makes the endpoint solution slightly basic (pH>7).
However:
• The reaction of equivalent amounts of a weak base and strong acid gives a slightly
acidic solution (pH<7), the extent of which is determined by Ka of the conjugate
acid.
Example:
NH3(aq) + HCl (aq)  NH4+(aq) + Cl-(aq)
Weak base strong acid weak acid extremely weak base
However:
NH4+(aq) + H2O(l)  H3O+(aq) + NH3(aq)
Ka=5.6x10-10
Weak acid very weak base strong acid weak base
The equilibrium lies strongly on the reactant side; yet the presence of the strong acid
H3O+ on the product side makes the endpoint solution slightly acidic (pH<7).
Example: a 0.1M solution of NH3 is exactly neutralized to its endpoint by addition of an
equal volume of 0.1M HCl. What is the pH at the endpoint?
III. Lewis acids and Bases
A Lewis acid is a substance that can accept a pair of electrons to form a new bond. In the
Bronsted definition an acid donates H+, which is a substance capable of accepting a pair
of electrons from H2O to form H3O+
A Lewis base is a substance that can donate a pair of electrons to form a new bond
•Lewis acids are cations or neutral molecules with an available empty orbital
Example: BF3, BBr3, BCl3 all have a vacant p orbital on boron
•Also, metals with electronegative elements attached have a very electropositive metal
centers which behave as Lewis acids: AlCl3, TiCl4, ZnCl2, etc.
•Metal ions having empty obitals can frequently form coordination complexes with bases:
AgCl(s) + 2NH3(aq)  ((NH3)2Ag)+(aq) + Cl-(aq)
FeCl3(s) + H2O (l)  (Fe(H2O)6)3+(aq) + 3Cl-(aq)
Note that coordination of H2O to an electron-deficient metal center such as Fe3+ weakens
some of the OH bonds, making the overall complex a weak acid:
(Fe(H2O)6)3++ H2O (l)  H3O+(aq) + (Fe(H2O)5OH)2+(aq)
Ka= 6.3x10-3
•Lewis acids are electron-deficient centers, whether metals or non metals
Example: O==C==O  O-––C==O + HO-  HCO3Lewis acid
base
•Lewis bases are anions or neutral molecules containing an atom with at least one lone
pair: NH3, PMe3, H2O, H2S, -OH, MeS•You should be able to identify the Lewis acidic/basic atoms within a molecule.
IV. Amphoteric molecules
An amphoteric substance can react both as an acid an as a base
Examples: Al(OH)3, Zn(OH)2, Cr(OH)3
As an acid:
Zn(OH)2 + -OH(aq)  [Zn (OH)3]-(aq) (a zincate complex)
As a base:
Zn(OH)2 + 2H3O+(aq)  Zn2+(aq) + 4 H2O(l)
Problem set #8:
1. Rank the following compounds in order of their acidity (least to greatest)
a. CH3CO2H, Br2CHCO2H, Br3CCO2H, BrCH2CO2H
b. Benzoic acid, fluorobenzoic acid, bromobenzoic acid, iodobenzoic acid
c. Benzoic acid (C6H5-CO2H), acetic acid, HO2C-C6H4-CO2H
2. Arrange the following species according to their strength as bases:
H2O, F-, Cl-, NO2-, CN(use Table 16.2)
3. A 0.100 M solution of a weak acid is 8.1% dissociated. Calculate Ka
4. For each of the following solutions, calculate the pH:
a. HCN, 0.25M (Ka= 6.17x10-10)
b. HBrO, 1.0x10-6M (Ka = 2.0x10-9)
c. Chloroacetic acid, 0.02M (Ka = 1.39 x 10-3)
d. Triethylamine, 0.20M (Kh= 4.0x10-4)
e. Nitric acid (HNO3), 0.25M
5. Calculate the pH of an aqueous solution that is 0.1M in acetic acid and 0.1M in
sodium acetate.
6. The hypobromite ion BrO-, is the conjugate base of the weak acid HOBr,
hypobromous acid. The pH of a 0.1 M NaOBr solution is 10.85 at 25°C. Calculate
the values of Kb for OBr- and Ka for HOBr at 25°C
7. Into 500mL of 0.5M HCl is placed 0.75 mol solid sodium formate (Na+HCOO-).
Assume that the sodium formate dissolves and the volume remains at 500 mL. If
the Ka for formic acid is 1.8x10-4, find [H3O+], [-OH], [HCOOH], and [HCOO-] at
equilibrium.