Chapter 20 The Properties of Acids and Bases
Acids and bases play a key role in many areas of chemistry. Often they are part of
equilibrium systems, so now that you understand equilibrium better we can begin to
discuss the intricacies of acid-base systems
Again, I and going to skip around a little bit in the order I present things
20-1 Acids and Bases
Arrhenius Def
Acids make H+ also taste sour
Bases make OH- also taste bitter and make hands slippery
Works only in aqueous solutions
In reality H+ does not exist
Actually H3O+ the hydronium ion
Brønsted- Lowry
Acids = Proton Donors
Bases = Proton Acceptors
Focus on proton-transfer reactions
Works in non-aqueous solutions
Will be focus for rest of chapter
Lewis Theory
acid - electron pair acceptor
Electrophile
base - electron pair donor
Nucleophile
Especially good in understanding organic reactions like NH3 +BF3
20-2 Water
Pure neutral water has some hydronium and hydroxide ions
2 H2O W H3O+ + OHacts as both donor and acceptor
Key Definition:
Kw = [H3O+][OH-] = 1x10-14
X = [H3O+] = [OH-]
1x10-14 =X2
X=SQRT(1X10-14)
X = 1x10-7
Neutral solution when [H3O+] = [OH-]
Basic solution when [H3O+]< [OH-] or [OH-] >1x10-7 & H3O+ > 1x10-7
Acidic solution when [H3O+]> [OH-] or [OH-] <1x10-7 & H3O+ > 1x10-7
2
20-3 Strong Acids and Bases
Key Definition:
Strong acids are acids that undergo 100% dissociation in water
Strong bases are bases that undergo 100% dissociation in water
The common strong acids and bases are listed in table 20.1
YOU NEED TO MEMORIZE THESE COMPOUNDS
Sample calculation
What are the concentrations of H3O+, Cl- and OH- in .03M HCl
100% dissociation so
[H3O+]= [Cl-] = .03M
[OH-] comes from KW
KW = 1x10-14 = [H3O+][OH-]
1x10-14 = .03[OH-]
[OH-] = 1x10-14/.03 = 3.33x10-13
What are the concentrations of H3O+, Ca2+ and OH- in .03M Ca(OH)2
100 % dissociation so
[OH-] = .03M Ca(OH)2 x 2 OH/1Ca(OH)2 = .06M
[Ca2+] = .03M Ca(OH)2 x 1 Ca2+ /1Ca(OH)2 = .03M
[H3O+] comes from KW
KW = 1x10-14 = [H3O+][OH-]
1x10-14 = .06[H3O+]
[H3O+] = 1x10-14/.06 = 1.67x10-13
Neutral Ions
Now let’s look at the back reaction
If HCl + H2O 6 Cl- + H3O+ go to 100% completion
How much of a tendency does Cl- have to do the back reaction:
To Cl- + H2O 6 HCl + OHZip- Zippo - Zilch-Nada
So Cl is a neutral ion
And so on for all the anions of the strong acids
Similarly for Na+
So the cations of strong bases are also neutral
The neutral ions are summarized in table 20.3
YOU NEED TO MEMORIZE THESE IONS
20-4 Carboxylic Acids Skip temporarily
3
20-5 pH and Acidity
You will find that the [H+] and [OH-] concentrations are are important in many
reactions. Even if [H+] and [OH-] are not products or reactants in a reaction, they
often serve as catalysts. In addition, all living organisms carefully try to keep [H+]
carefully regulated because small fluctuations in [H+] or [OH-] can easily kill a
cell
Generally the concentrations of [H+] and [OH-] in water range from 1M to 1x10-14
M. With such a broad range of concentrations it is convenient to switch to a
logarithmic scale. This scale is attributed to Søren Sørenson. The p of pH is
those to mean the ‘power’ or exponent of the hydrogen ion concentration
Key Equation
pH = -log[H3O+] = -log[H+]
Example problems
Just a page ago we calculated that .03M HCl had a [H+] of .03M
What is the pH of this solution?
pH = -log [H+]
=
-log(.03)
= -(-1.52
=1.52
We also determined that a.03M Ca(OH)2 had an [H+] of 1.67x10-13
What is the pH of this solution?
pH = -log [H+]
= -log(1.67x10-13)
= -(-12.778)
=12.778
You can do the same thing with [OH-]
Key Equation
pOH = -log[OH-]
Sample Problem
Continuing with the [OH-] side of the previous problems, the .03M Ca(OH)2 had
an [OH-] .06M
pOH = -log(.06)
= - (-1.222)
4
As we tackle the OH side of the .03M HCl solution I want to show you something
In the original problem we had
[H+] = .03M
then we used
KW = 1x10-14 = [H3O+][OH-]
to find that
[OH-] = 1x10-14/.03 = 3.33x10-13
so pOH = -log (3.33x10-13) = -(-12.48) = 12.48
Bu there is a shortcut
Key Equation
pH + pOH = 14
(I can derive this for you, but you probably don’t want to sit through the math)
Continuing example
we already said that the pH of this solution was 1.52
using 14 = 1.52 + pOH
pOH = 14-1.52 = 12.48
and you get exactly the same number
More on pH scale
Earlier I said that
Neutral solution when [H3O+] = [OH-]
Basic solution when [H3O+]< [OH-] or [OH-] >1x10-7 & H3O+ > 1x10-7
Acidic solution when [H3O+]> [OH-] or [OH-] <1x10-7 & H3O+ > 1x10-7
Making the equivalent statements using the pH scale we have
Neutral pH = pOH = 7
Basic solution is when pH >7 or pOH <7
Acidic solution is when pH <7 or pOH >7
Figure 20.2 pH of some common solutions
Earlier I also said that “Generally the concentrations of [H+] and [OH-] in water
range from 1M to 1x10-14 “ In pH scale this means that generally pH and pOH
values lie between 1 and 14
Emphasis on generally, there are exceptions:
Practice problem
What is the pH and pOH of a 5M HCl solution
[H+] = 5 ! Note >1!
-log(5) = -.7
-.7 !!
14 = pH + pOH; pOH = 14.7!
5
One last trick problem I may throw at you
What is the pH of 1x10-8 HCl?
Well, that should be easy
pH = -log(1x10-8)
= - (-8)
=8
But what is wrong with that answer? HCl is an acid, but you just got a basic pH!
The problem is that you forgot about water. Water is furnishing 1x10-7 H+
so H+ = 1x10-8 from the HCl and 1x10-7 from the water for total of
1.1x10-7
pH = -log(1.1x10-7) = 6.96 an acid pH
Adding the water contribution on top of your original acid or base is only
necessary when the acid (or base) concentration is < or = 1x10-7. It is also a bit
of a quick and dirty approximation, when you get to analytical I will show you the
proper way to handle this problem.
pH6 [H+]
By now you know that I want you to be able to do calculations forward and
backwards. So how do you calculate [H+] from pH?
Example Calculation
If an HNO3 solution has a pH of 4.83, what is the concentration of [HNO3]?
pH = -log [H+]
4.53 = -log[H+]
-4.53 = log [H+]
How do you get rid of the log function? Do 10^ on both sides of the equation
On the left side 10^-4.53 = 10-4.53
On the right side 10^log cancels out so
10-4.53 = [H+]
= 2.95x10-5
Since HNO3 is a strong acid with 100% ionization
[H+] = [HNO3] = 2.95x10-5
Summarizing into a single equation
Key Equation
[H+] = 10-pH
6
20-6 Weak Acids and Bases
If strong acids and bases have 100% dissociation, what do you think happens
with weak acids and bases? You got it, <100% dissociation
I will define % dissociation or % ionization as:
Key Equation
% dissociation or ionization =[A-]/[total acid] x 100% = =[A-]/([HA]+[A-]) x 100%
While my equation doesn’t look like the book equation, they are essentially the
same. The book used [H+] in the numerator I uses [A-] but, since HAWH+ + Athey are the same number!
In the denominator the book uses the term [HA]o and calls this the stoichiometric
concentration. It represents the total you put in, so again we are equivalent.
For a base the equivalent would be
Key Equation
% ionization = [BH+]/total base x100% = [BH+]/([B]+[BH+]) x100%
Sample calculation
If a .01M solution of an acid has a pH of 4.75, what is the % ionization of the
acid?
.1M would represent the total amount of acid we have in the solution
if the pH is 4.75, [H+] = 10-4.75 = 1.78x10-5
if [H+]=[A-] then [A-]= 1.78x10-5
and % ionization = 1.78x10-5/.01 x 100% = 1.78% ionized
If a base is 5% ionized when it has a concentration of 0.0004 M, what is the pH
of the solution?
5% = [BH+]/([B]+[BH+]) x 100%
.0004 = [B]+[BH+]
5% = [BH+]/.0004 x 100%
.05 = [BH+]/.0004
[BH+]=2x10-5
[BH+] = [OH-] = 2x10-5,
pOH = -log(2x10-5) = 4.70
pH = 14-4.70 = 9.30
7
20-7 Ka and Acid Strength
So now you know that weak acids have <100% ionization, how can you tell how
much ionization? How can you tell an acid with a 95% ionization from one with
.005% ionization?
Let’s look at the equilibrium
If the acid dissociation equation is
HA + H2O W H3O+ + AThen the equilibrium equation is:
K =[H3O+][A-]/ [HA]
or
HAW H+ + A-
or
[H+][A-]/[HA]
We give this acid ionization K as special symbol, Ka
And, using the logic from the last chapter a larger K means the reaction goes
more to the right, so
Key Concept
The larger the Ka the more a weak acid ionizes. Or the stronger the acid.
Table 20.4 Ka and pKa of weak acids
Clicker question
Rank the acids, acetic acid HClO2, HIO3 from weakest to strongest acid
Note this table also includes a value for pKa. What is a pKa
Key Equation
pKa = -log Ka
pKa ‘s are handy for certain calculations that we will see in the next chapter
Calculating the pH of a weak acid
You know how to calculate the pH of a strong acid, it is easy because you can
assume 100% ionization. How do you calculate the pH of a weak acid?
Sample calculation
What is the pH of a .006 M solution of acetic acid?
We attack this problem just like any other equilibrium problem
1. Write the balanced reaction CH3COOH + H2O W CH3COO- + H3O+
I will shortcut this to HA W H+ + A2. Write the equilibrium expression that corresponds to the balanced reaction
Ka = [H+][A-]/[HA]
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3. list initial concentrations and start your ICE table
HA
WH+ + AInitial .006
0
0
4. Calculate Q to determine what direction(if any) the reaction will shift
With 0 products we don’t need a Q, you know it will go to products
5. using a single variable(X), define change for each concentration needed to
reach equilibrium using X.
Here, since we want to calculate pH which comes from [H+] we make
[H+]=X
[A-] = X
[HA] = -X
6. Substitute modified concentration terms into the ICE table and the equilibrium
expression
HA
WH+ + AInitial .006
0
0
Change -X
X
X
Eq
.006-X
X
X
Ka = X@X/(.006-X)
Ka = 1.8x10-5 (from table 20.4)
1.8x10-5 = X@X/(.006-X)
7. Solve the equilibrium expression for the unknown
There are two ways to solve this, long and short
Start with the long
(.006-X)1.8x10-5 =X2
1.08x10-7 -1.8x10-5X = X2
X2 + 1.8x10-5X -1.08x10-7 =0
Using the quadratic -b +/- sqrt(b2-4AC) / 2A
-1.8x10-5 +/- sqrt(3.24x10-10 + 4.32x10-7) / 2
-1.8x10-5 +/- sqrt(4.323x10-7) / 2
-1.8x10-5 +/- 6.575x10-4 /2
Ignoring the - root
(6.575x10-4 -1.8x10-5)/2
6.395x10-4/2
3.18x10-4
pH = -log(3.18x10-4) = 3.49
Now the short, quick and dirty
If X si small, the reaction does not go far to the right so
.006-X ~ .006
1.8x10-5 ~ X2/.006
X = sqrt(.006@1.8x10-5 ) = 3.29x10-4 ; pH = 3.48
9
A little off, but not too bad
The quick and dirty method usually works if the Ka is 10-4 or less
if you see a Ka of 10-3 or greater you should use the quadratic
20-8 Successive Approximation
The Q&D answer we got above was a lot faster than the quadratic, but us was
off a bit
3.28x10-4 - 3.18x10-4 = .10x10-4
.10x10-4/3.18x10-4 x 100% = ~3% error
The method of successive approximation is a way to get closer to the real
answer without the quadratic
In the Q&D we did an approximation that .006-X ~ .006
But once we solved the equation and found that X = 3.28x10-4
we can improve our answer
Out improved answer is that .006-3.28x10-4 =.005672
And not solve the problem with is better approximation
1.8x10-5 ~ X2/.005672
X = sqrt(.005672@1.8x10-5 ) = 3.29x10-4 ; pH = 3.48
X = 3.19x10-4
And now our answer is < 1% off.
You can even set up a spread sheet to do this over and over. But most of the
time a single iteration will be just fine for most weak acid problems
Now that you know how to handle weak acids, and you have seen the table with
lots of different weak acids it is time to look at the most common weak acid seen in
organic chemistry the carboxylic acid
20-4 The carboxylic acid (out of order)
The most common acid you see in organic chemistry is the carboxylic acid often
represented with the atoms COOH
The structure is shown above, where R represents H or any C containing group
10
Two of the acids you may be familiar with are acetic acid and formic acid. Acetic
acid is the acid (and the smell) found in vinegar, Formic acid is the irritant found
in and bites ( it was first isolated in 1600's by distilling ants!)
The acidic proton on these structures is the H on the COOH group.
If you think about it you might wonder why. After all H-O-H with OH bonds is not
acidic, and alcohols C-O-H groups are not acidic either. Th answer lies is the
fact that the anion that is formed when the proton leaves is stabilized by
spreading out the negative charge over 2 atoms as shown below:
Is is interesting that you can actually see this in x-ray structures. The C=O bond
is 123 pm, the C-O-H bond is 136 pm and the C-O bond in then resonance stablized
anion is 127 pm
20-9 KB and base strength
Now let’s turn our attention to weak bases
The generic base reaction is:
B + H2O 6BH + OHSo in this case the OH- that you associate with something being basic is the
product of the reaction of a base with water
Notice in the reaction B the base was a proton acceptor which matches our
Bronstead Lowry defintion
11
Or
H
H-N: + H-O-H
H
H
6 H-N-H
H
+ OH-
The pair of electrons are donated to water which agrees with the Lewis model of
a base
Again this is a equilibrium reaction with a K, but we usually use the symbol KB for
the base ionization or base protonation constant. In a similar manner as the
acids
Key Concept
The larger the KB, the more the reaction goes to the right, so the stronger the
base
Table 20.5
Key Equation
pKB ‘s for -log(KB)
Clicker question
Rank the bases ammonia, aniline, and ethylamine from weakest to strongest
base
All of the weak bases found in this table are amines, or nitrogen containing
organic molecules. The lone pair of electrons on the nitrogen that can either
accept protons or donate electrons (depends on the definition of base you use)
is what these bases all have in common
Calculating the [H+] or pH due to a base follows the same kind of logic as
calculation involving acids, you just have to add one more step
Sample Calculation
Let’s calculate the pH of a .04M ethylamine
Again we attack this problem just like any other equilibrium problem
1. Write the balanced reaction CH3CH2NH2 + H2O W CH3CH2NH3 + OHI will shortcut this to B + H2O W BH+ + OH2. Write the equilibrium expression that corresponds to the balanced reaction
KB =[BH+][OH-]/[B]
3. list initial concentrations and start your ICE table
B
H2O
WBH+ + OHInitial .04
Constant
0
0
4. Calculate Q to determine what direction(if any) the reaction will shift
With 0 products we don’t need a Q, you know it will go to products
5. using a single variable(X), define change for each concentration needed to
12
reach equilibrium using X.
Here, since we want to calculate pH which can be calculated form pOH
which comes from [OH-] we make [OH-]=X
[OH-] = X = [BH+]
[B] = -X
6. Substitute modified concentration terms into the ICE table and the equilibrium
expression
B
H2O
WBH+ + OHInitial .04
Const
0
0
Change -X
X
X
Eq
.04-X
X
X
KB = X@X/(.04-X)
KB = 4.5x10-4 (from table 20.5)
4.5x10-4 = X@X/(.04-X)
7. Solve the equilibrium expression for the unknown
There are three ways to solve this, long and short, and successive approximation
Let’s just do the successive approximation
1st approximation
.04-X ~ X
4.5x10-4 =X2/.04
X = sqrt (4.5x10-4 @.04)
x ~ .00424
ND
2 approximation
.04 -X ~.04 - .00424 ~ .03576
4.5x10-4 =X2/.03576
X = sqrt (4.5x10-4 @.03576)
x ~ .0040
But X = [OH-] and we want pH, so
[OH-] =.004, pOH = -log(.004)= 2.40
pH = 14-2.40 = 11.60
20-10 Conjugate Acid-Base Pairs
So far we have concentrated on the forward reaction HA + H2O 6 A- + H3O+
Or B + H2O 6 BH+ + OHBut now let’s look at the reverse reactions
A- + H3O+ 6 HA + H2O Or BH+ + OH- 6 B + H2O
Note that re reverse reactions are also acid base reactions
A- +
H3O+ 6 HA + H2O Or BH+ +
OH- 6 B + H2O
H Acceptor H Donor
H donor H acceptor
Acid
Base
Acid
Base
13
Key Definition
So the acid’s anion is a base in the reverse reaction - we call this a conjugate
base and a base’s cation is an acid in the reverse reaction - we call this a
conjugate acid
Sample problems
What are the conjugates bases of the following acids?
HClO3, H2PO4-, and CH3NH3+
Acids are proton donators, so what ion do you have left after they have donated
their proton - ClO3-, HPO42- and CH3NH2
What are the conjugate acids of the following bases?
CH3NHCH3, HPO4-2 , FBases are proton acceptors so what do you have after you have accepted a
proton? - CH3NH2CH3+, H2PO4-, HF
So the conjugate form of the weak acids are bases, and will make the solution
basic. Also the conjugate form of the weak bases are acids, and will make the
solution acidic. Now a reminder. What is the conjugate form of HCl, and is it an
acids or a base. Similarly, what is the conjugate form of NaOH, and will it make
a solution acid or base? (The anions of strong acids are neutral, the cations of
strong bases are neutral !)
Kb’s of weak acids
So now you know that the acetate anion, CH3COO- should be basic because it is
a conjugate base. But how strong a base is it?
Well consider
CH3COOH + H2OW CH3COO- + H3O+, Ka = 1.75x10-5
For the reverse reaction
CH3COO- + H3O+W CH3COOH + H2O
H2O + H2O WH3O+ + OH-
Reverse so K = 1/Ka
Kw
Now if we add the above two equations together, what do we do with the K’s?
(Multiply them)
The result is
CH3COO- + H2O WCH3COOH + OH- and this is our base reaction for the acetate
ion.
14
KB = Kw/KA = 1x10-14/1.75x10-5 = 5.71x10-10
Key Equation
The Kb for the conjugate base of a weak acid = Kw/KA
Similarly for bases
The KA for the conjugate acid of a weak base = KW /KB
Or, more simply, KAKB=KW
Sample calculation
What is the KA for ammonia (NH4+)
KB for ammonia = 1.8x10-5 (table 20.5)
KA = Kw/KB = 5.7x10-10
20-11 Salt solutions
We started this chapter looking at strong acids and bases, then we worked on
weak acids and bases, and just now we did the conjugate forms of the acids and
bases. Now let’s put all these different pieces together into a compound called a
salt.
Back in chapter 10 you learned that a salt is the product of an acid base
neutralization reaction, so let’s start there.
What do you get when you neutralize an acid like HCl with NaOH?
HCl + NaOH6 NaCl(aq) + H2O
Early on in this chapter you learned that Na+ and Cl- were neutral ions, so the pH
of the neutralized solution is indeed, neutral.
Now what about CH3COOH and NaOH?
CH3COOH + NaOH 6 CH3COONa(aq) + H2O
But as an ionic compound CH3COONaW Na+ + CH3COONa+ neutral CH3COO-basic (C.B of a weak acid)
How about HCl and NH3 6 NH4+ (acidic) and Cl- (neutral)
So neutralization reactions can leave you neutral, or acidic or basic!
What is a person to do? On top of that, sometimes I won’t even give you the
neutralization reaction, I will just give you the salt, say NaHSO3. Are you up the
proverbial creek without a paddle?
Key Procedure:
1. Break each salts into its component anion and cation
2. Ignore the neutral cations and anions
3. Identify the acid/base properties of the remaining cation or anion
15
Table 20.7 is a nice table that summarizes the acid/base properties of many ions
but you shouldn’t have to memorize this table, you should be able to use logic
remember
the cations of strong bases are neutral, so this means basically the first
two column of the periodic table
The anions of strong acids are neutral, and you shoul dknow your straong
acids by now
So you can ignore all of the above
Looking at what remains
The basic anions are the conjugate bases of weak acids
The acidic cations are the conjugate acids of weak bases (and I
think all have a N in them!)
Table 20.7 does have a few that don’t fit above. Let’s examine them more
closely
1. There are no basic cations!
2. Cations with >+2 charge tend to be acidic
Last semester we talked about waters of hydration, that is waters
that are bound to + cations
When you have a highly charge cation, the cation actually grabs
the water so strongly that it splits the water into OH- and H+. The
OH- stays attached to the metal and the H+ is released
Example:
Al3+(aq) is actually [Al(H2O)6]3+ (aq) 6[Al(OH)(H2O)5]2+ (aq) + H+(aq)
Added material NOT in your text
Acid-Base properties of Oxides
There is another class of compounds that have acid/base properties that we can
identify fairly easily and that is the oxides.
In nature you will find oxides of virtually every element. Oxides can be either
acidic or basic. You now have enough chemistry behind you that you can begin
to understand these properties and to predict them of unknown compounds.
Let’s start with the acidic oxides. When a covalent oxide dissolves in water is will
form an acidic solution. Examples:
SO3 (g) + H2O(l)W H2SO4(aq)
SO2 (g) + H2O(l)W H2SO3(aq)
CO2 (g) + H2O(l)W H2CO3(aq)
2NO2 (g) + H2O(l)W HNO3(aq) + HNO2(aq)
16
On the other hand when ionic oxides dissolve in water we get basic solutions
CaO(s) + H2O(l) W Ca(OH)2(aq)
Na2O(s) + H2O(l) W 2NaOH(aq)
K2O(s) + H2O(l) W 2KOH(aq)
and these are called basic oxides
In our covalent oxides the oxygen has a covalent bond to the cental atom and
the H, and the bond between the central atom and the O is stronger than the
bond between the O and the H so the H may be easily removed
In the ionic compound there is no covalent bond between the metal and the O2anion. The oxide anion, in fact, has a high affinity for protons (a strong base)
and will pull the off of water
O2-(aq) + H2O(l) 6 2OH-(aq)
Making most ionic oxides basic
But hold it... In the lab you should have found that CrO3, a metal oxide had a pH
of 2 or less. What gives? (In this compound you have Cr+6 We just talked about
how metal ions with strong + Charge can be acids, here the acidity of the +6
metal ion has overpowered the basicity of the metal-oxide)
Practice problems Maybe Clicker questions
Identify the following compounds as acids or bases
NaCl (N)
CH3COOK (B)
RbF (B)
NH4NO3 (A)
CaO (B)
SO2 (A)
CH3COONH4 (Trick question - N)
20-12 Polyprotic Acids
Let’s talk a bit about H2SO4 and H3PO4
These are examples of polyproitic acids, acid that have more than one proton to
give. There ae several polyprotic acids listed in table 20.9 of your text.
If you look at the K’s of the polyprotics, you will see that form most of them the
first K is relatively strong, but the second is much weaker. Can you figure out
why?
17
After 1 proton remove you have an anion. It is harder to make the second proton
leave a species that is already negatively charged to make it a doubly negatively
charged species
In analytical I will make you do calculations with these guys, and you might
expect these calculations to be difficult because you have multiple states to deal
with, But most of the time the second K is so small compared to the first that
one can be ignored so the calculations -usually- aren’t to bad
Amphoteric salts
Some of the salts of some of the polyprotic acids can act as either an acid or a
base
Key Definition
Species that can act as acid or bases are called amphoteric
The salts of both H2CO3 and H3PO4 do this:
Salt: NaHCO3
Ionizes to Na+ and HCO3-; Na+ is neutral and can be ignored
HCO3 acting as acid
HCO3- + H2O W CO32- + H+
HCO3- acting as base
HCO3- + H2O W H2CO3 + OHThat is why a solution of NaHCO3 can be used in the lab to r=treat with acid or
base burns