Transcript Acids and Bases

Acids and Bases
They’re Everywhere
Definitions
• Arrhenius Acids
–
–
–
–
Acids produce hydrogen ions in solution
Bases produce hydroxide ions in solution
Only have one type of base, hydroxide
First description of what an acid/base was
Definitions
• Bronsted-Lowry Model
– Acid is a proton donor
– Base is a proton acceptor
H---Cl + H---O---H  (H---O---H)+1
H
+

Water is the base (proton acceptor)
HCl is the acid (proton donor)
+
Cl-1
+1
-1
Conjugate Pairs
• The reaction that occurs is really an equilibrium
• The H3O+1 is called the hydronium ion
H---Cl + H---O---H
HCl
Acid

+ H2O
Base
(H---O---H)+1
H
+
Cl-1
H3O+1 + + Cl-1
Acid
Base
These are conjugate acid base pairs.
Water is a base on one side and a acid on the other,
this is called amphoteric
Quick Practice
• Strong acid
HNO3 + H2O ↔ H3O+ + NO3– What are the conjugate pairs?
• Weak acid
HF + H2O ↔ H3O+ + F– What are the conjugate pairs?
Weak Acids
• Weak acids partially dissociate in water:
• HA(aq)  H+(aq) + A- (aq)
– Bronsted-Lowry: HA + H2O  H3O+ + A-
• What is Kc for this dissociation?
• Kc in a weak acid case is better known as Ka, or the
acid-dissociation constant.
• The larger the value of Ka, the stronger the acid.
Acid Dissociation Constant
• HA + H2O  H3O+1 + A• Generalized expression
– The equilibrium expression for this process
Ka = [H3O+1 ] [A-] = acid dissociation constant
[HA]
Note: the [water] is a constant (concentration of a
pure solid or pure liquid is not included in an
equilibrium expression)
Water does play an important part in the dissociation of the acid!
Acid Strength
• HA(aq) + H2O 
H3O+1(aq) + A-(aq)
– A strong acid lies to the right Ka >>1
• [H+1]  [HA]0
– A weak acid lies to the left
Ka<<1
• [H+1] << [HA]0
– Typical Ka (dissociation) constants for weak acids
• HF
• HC2H3O2
7.2 x 10-7
1.8 x 10-5
Conjugate Pairs and Strength
• HA(aq) + H2O 
H3O+1(aq) + A-(aq)
– Strong acids have weak conjugate bases
– Strong bases have weak conjugate acids
http://www.chem.ubc.ca/courseware/pH/index.html
Weak Acid
Acid Strength
• In an acid, the strength of the bond between
the acidic hydrogen and the other atom (H-X)
determines how strong the acid is.
• In general, the strength of an H-X bond
weakens as atoms get bigger.
– So, going down a group, the strength of an acid
increases.
– HF < HCl < HBr < HI
Acid Strength
• Going across a row, bond strengths don’t
change all that much. So, bond polarity is the
major factor – the more polar the bond, the
stronger the acid
• Period 2: CH4 < NH3 < H2O < HF
Acid-Base Behavior and Chemical
Structure
Binary Acids
Acid-Base Behavior and Chemical
Structure
Factors That Affect Acid Strength
Consider H-X. For this substance to be an acid we need:
• H-X bond to be polar with H+ and X- (if X is a metal
then the bond polarity is H-, X+ and the substance is
a base).
• the H-X bond must be weak enough to be broken,
• the conjugate base, X-, must be stable.
Acid-Base Behavior and Chemical Structure
Binary Acids
• Acid strength increases across a period and down a group.
• Conversely, base strength decreases across a period and
down a group.
• What differences in atomic structure account for these
variations?
• The kernel charge increases across a period. Therefore,
the nonmetallic element has a stronger pull on the
shared electron pair and H is more easily ionized.
• As you go down a group, the strength of the H-X bond
weakens as the X has more shells and its size increases.
Therefore, H is more easily ionized
Acid-Base Behavior and Chemical
Structure
Oxyacids
• Oxyacids contain O-H bonds.
• All oxyacids have the general structure Y-O-H.
• The strength of the acid depends on Y and the atoms
attached to Y.
– If Y is a metal (low electronegativity), then the substances
are bases.
– If Y has intermediate electronegativity (e.g. I, EN = 2.5), the
electrons are between Y and O and the substance is a weak
oxyacid.
Acid-Base Behavior and Chemical
Structure
Oxyacids
– If Y has a large electronegativity (e.g. Cl, EN = 3.0), the
electrons are located closer to Y than O and the O-H bond is
polarized to lose H+.
– The number of O atoms attached to Y increase the O-H
bond polarity and the strength of the acid increases (e.g.
HOCl is a weaker acid than HClO2 which is weaker than
HClO3 which is weaker than HClO4 which is a strong acid).
Acid-Base Behavior and Chemical
Structure
Oxyacids
Carboxylic Acids
• On a similar note,
carboxylic acids contain –
OH groups, but are acids,
because of the additional
attached oxygen “aldehyde”
group on the final carbon in
the chain.
• Carboxylic acids are also
stabilized by resonance
once the hydrogen goes
away.
B-l Acids and Bases
• Weak acids and Bases
• Know the difference between a BronstedLowry acid/base and an Arrhenius acid and
base
Strong Acid
• Example 0.10 M HNO3
• H2O  H3O+ + OH• There are two sources of H+
– The nitric acid and water
• Since [H+] >>[OH-] in 0.1M nitric
– Autoionization of water is insignificant
– All the H+ is from HNO3
• [H+] = 0.10M
pH = -log 0.1 = 1.0
Strong Acid
• The acid contributes all the [H+]
• Example 1.0 x 10-10 HCl
– The [H+] from autoionization (1 x 10-7M) is
much higher.
pH = - log 1 x 10-7 = 7
Weak Acids
•
•
•
•
Treat like any equilibrium problem
What is pH of a 1.00M HF solution
Kc = 7.2 x 10 –4
Kw = 1.0 x 10-14
– Since the Kc is so much bigger than the Kw,
– HF is the major source of H+
pH of 1.0 M HF Solution
I
C
E
HF + H2O 
1.0
-x
1.0 – x
H3O+ +
F0(1 x 10-7) 0
+x
+x
x
x
Ka = 7.2 x 10-4 = [F- ] [H+] = x • x = x2
[HF]
1–x
1
Assume x is small (5% rule)
X = 2.7 x 10-2 = [H+]
Check for 5% Rule

x2
[HA]o – x
[HA]
• Ka = x2
Ka • [HA]
(Ka • [HA])1/2
X
[HA]0

x2

x
x 100  5%
• X = 2.7 x 10-2 x 100
1
 5%
Weak Acid Equilibrium
• List the Major species in solution
– Don’t forget water as an acid source!
• Choose the species that can produce H+
– Write balanced equation
•
•
•
•
•
Using the K values, choose dominate source H+
Write the equilibrium expression for dominate H+
Do “ICE” and solve using 5% rule
Verify 5%
Calculate [H+ ] and pH
Polyprotic Acids
• When acids are polyprotic, like the triprotic H3PO4,
where all three protons are weak acids, different Ka
values are used.
• H3PO4  H2PO4- + H+
• H2PO4-  HPO42- + H+
• HPO42-  PO43- + H+
Ka1 = 7.1 x 10-3
Ka2 = 6.3 x 10-8
Ka3 = 4.5 x 10-13
• If Ka1 is more than 103 larger than Ka2, you can
ignore Ka2 and treat it like a monoprotic acid.
Calculate pH of 0.100 M HOCl
• Ka = 3.5 x 10-8
• You Calculate the pH
– Pg 675 if you need book
• pH = 4.23
pH of Weak Acid Mixture
• Calculate the pH of a mixture of 1.00 M
HCN, 5.00 M HNO2 and the equilibrium
concentration of [CN-1]
• HCN
Ka = 6.2 x 10-10
• HNO2 Ka = 4.0 x 10-4
• H20
Kw = 1 x 10-14
How do you approach this?
Calculate pH
• Ka = 4.0 x 10-4 = [H+][NO2-]
[HNO2]
HNO2  H+ + NO2I
5.00
0
0
C
-x
+x
+x
E
5.00 – x
x
x
Calculate pH
• Ka = 4.0 x 10-4 =
=
x2
5.00 - x
x = [H+] = 4.5 x 10-2 M
x2
5
pH = - log [H+] = 1.35
Now calculate [CN-], you now [HCN] and [H+]
Calculate
[CN ]
• Ka = 6.2 x 10 –10
6.2 x 10 –10 = [CN-][H+] = [CN-][4.5 x 10-2]
[CN]
1.00
Solve for [CN-] = 1.4 x 10-8 M
Percent Dissociation
• % dissociation = [amount disassociated] x 100
[initial concentration]
In the HF example
[H+]
=
1.27 x 10-2 M x 100 = 1.27%
[HF]
1.00 M
Calculate the Percent
Dissociation
• 1.00 M HC2H3O2
• Left side of room
• 0.100 M HC2H3O2
• Right side of room
– Write on board
– Write on board
Percent Dissociation
• For solutions of weak acids:
– the more dilute the solution
– The greater the percent dissociation
General Proof
Suppose have acid HA, with [HA]0
Dilute it to 1/10 th initial concentration
Q = (x/10)(x/10) =
x2
= 1/10 Ka
[HA]/10
10 [HA]
Since Q < Ka, the reaction moves to the right
And you get a greater percent dissociation
Ka from % Dissociation
• Lactic acid is 3.7% dissociated @ 0.100 M
• HC3H5O  H+ + C3H5O• Ka = [H+][C3H5O -]
[HC3H5O]
3.7% =
x
x 100 x = 3.7 x 10 -3
[HC3H5O]
[H+][C3H5O -] = (3.7 x 10 –3) (3.7 x 10 –3)
[HC3H5O]
0.100
Ka = 1.4 x 10 -4
Ka =
Weak acid equilibria
Example
Determine the pH of a 0.10 M benzoic acid
solution at 25 oC if Ka = 6.5 x 10-5
HBz(aq) + H2O(l)
H+(aq) + Bz-(aq)
The first step is to write the equilibrium
expression.
Ka =
[H+] [Bz-]
[HBz]
Weak acid equilibria
HBz
H+
Initial conc., M
0.10
0.00 0.00
Change, DM
-x
+x
Eq. Conc., M
0.10 - x
x
Bz-
+x
x
[H+] = [Bz-] = x
We’ll assume that [Bz-] and [H+] are negligible
compared to [HBz], since the value of the Ka<<
[HBz].
(6.5 x 10-5 << 0.10 M)
Weak acid equilibria
Solve the equilibrium equation in terms of x
x2
Ka = 6.5 x 10-5 =
0.10
x
= (6.5 x 10-5 )(0.10)
= 0.00255 M H+
pH = - log (0.0025 M) = 2.6
Weak acid example
Now, lets go for the exact solution
Earlier, we found that for 0.10 M benzoic acid
2
x
Ka = 6.5 x 10-5 =
0.10-x
 X2 + 6.5 x 10-5 X - 6.5 x 10-6 = 0
Use the quadratic equation to solve for x.
X=
-b + b2 - 4ac
2a
Weak acid example
-6.5 x 10-5 + [(6.5 x 10-5)2 +4 x 6.5 x 10-6]1/2
X=
2
X = 0.00252 M H+
versus
pH
= - log (0.00252 M) = 2.599
pH
= - log (0.00255 M) = 2.593
In this case, there is no significant difference between our two
answers. If the Ka value is more than 2 powers of 10 different
than the [acid], you can ignore the change in [acid].
Dissociation of bases, Kb
The ionization of a weak base can also be
expressed as an equilibrium.
B (aq) + H2O(l)
BH+(aq) +OH- (aq)
The strength of a weak base is related to its
equilibrium constant, Kb.
[OH-] [BH+]
Kb =
[B]
Example
Weak base equilibria
The Bz-(aq) formed in the benzoic acid solution is
a weak conjugate base. Determine the pH of a 0.10
M sodium benzoate solution NaBz(aq), at 25 oC
Bz-(aq) + H2O(l)
HBz(aq) + OH-(aq)
The Na+(aq) are spectator ions, and are not
part of the equilibrium expression.
Kb =
[OH-] [HBz]
[Bz-]
Example
Weak base equilibria
The Kb value is related to the Ka value by the
equation Ka x Kb = Kw = 1.0 x 10-14
[H+] [Bz-]
[HBz]
[OH-] [HBz] = [H+] [OH-]
[Bz-]
Kb = Kw / Ka = 1.0 x 10-14 / 6.5 x 10-5
= 1.5 x 10-10
Weak base equilibria
Bz -
OH-
HBz
Initial conc., M
0.10
0.00
0.00
Change, DM
-x
+x
+x
Eq. Conc., M
0.10 - x
x
x
[OH-] = [HBz] = x
We’ll assume that [HBz] and [OH-] are negligible
compared to [Bz -], since the value of the Ka <<
[Bz -].
(1.5 x 10-10 << 0.10 M)
Weak base equilibria
Solve the equilibrium equation in terms of x
x2
Kb = 1.5 x 10-10 =0.10
x
=
= (1.5 x 10-10 )(0.10)
3.9 x 10-6 M
pOH = - log (3.9 x 10-6 M) = 5.4
pH = 14 - pOH = 8.6
Relationship Between Ka and Kb
• What happens when you multiply Ka and Kb
together?
• Ka x Kb = [H+][OH-] = Kw = 1.0 x 10-14
• And, just like pH + pOH = 14.00 for strong
acids/bases at standard temperature…
• pKa + pKb = pKw = 14.00
Acidic and Basic Salts
• Certain ions in solution can exhibit acid/base
properties.
• For example, consider the weak base, ammonia:
– NH3 + H2O  NH4+ + OH-
• What if you dissolve the salt, ammonium sulfate, in
water?
– (NH4)2SO4  2 NH4+ + SO42-
• Because NH4+ is the conjugate acid of NH3, when
this salt dissolves in water, the solution will become
slightly acidic.
Anions and Water
• An anion that is a conjugate base of a weak
acid raises the pH of a solution:
– X- + H2O  HX + OH-
• Example – sodium acetate in water
– Ionic: Na+ + CH3COO- + H2O 
Na+ + CH3COOH + OH– Net ionic: CH3COO- + H2O  CH3COOH + OH-
• The Kb of this reaction can be found using the Ka of
acetic acid (Ka x Kb = Kw)
Cations and Water
• An cation that is a conjugate acid of a weak base
that contains hydrogen lowers the pH of a solution:
– HX+ + H2O  X + H3O+
• Example – ammonium chloride
– Ionic: NH4+ + Cl- + H2O  NH3 + Cl- + H3O+
– Net ionic: NH4+ + H2O  NH3 + H3O+
• The Ka of this reaction can be found using the Kb of
ammonia (Ka x Kb = Kw)
Some Rules
• An anion that is the conjugate base of a
strong acid will not affect the pH of a
solution. (Ex: Br- from HBr)
• An anion that is the conjugate base of a weak
acid will cause an increase in pH (CN- from
HCN)
• A cation that is the conjugate acid of a weak
base will cause a decrease in pH (NH4+ from
NH3)
Some Rules
• Alkali metal cations and Ca2+, Sr2+, and Ba2+
will not affect pH (they are conjugate acids of
strong bases).
• Other metals (Al3+, etc.) will cause a decrease
in pH.
• When a solution contains both a cation and
anion that will affect pH, the ion with the
larger equilibrium constant (Ka or Kb will
have the greater influence on pH).
Lewis Acids and Bases
• There is yet a third definition of acids and
bases – the Lewis definition.
• A Lewis base is defined to be an electron pair
donor.
• A Lewis acid is defined to be an electron pair
receiver.
• Understanding Lewis acids and bases requires
the use of Lewis diagrams.
Lewis Acid-Base Reactions
• Example 1: H+ and NH3
– Empty s orbital on hydrogen
• Example 2: NH3 and BF3
– Empty p orbital on boron
• In order for a Lewis acid to receive an electron
pair, there must be an empty orbital in the
electron configuration.
Coordination Complexes
• Certain transition metal cations can act as
multiple Lewis acids (where the empty orbitals
are is beyond the scope of AP Chemistry)
• Fe3+ has the ability to attract six electron pairs
to itself:
– Reaction of Fe3+ and CN-
• This kind of compound is known as a
coordination complex.
Brief Preview of Organic
Reactions
• Lots of organic chemistry relies on “bonding
sites” – or determining and predicting reaction
mechanisms based on chemical structures.
• Example – Lewis acid/base reaction between
CO2 and H2O to create H2CO3
Hydrolysis of Metal Ions
• When salts dissolve in water, the metal ions
become hydrolyzed, or the water acts like a
Lewis base and forms coordination compounds
with the metal ions.
• This is the mechanism behind metal ions
acting like acids:
• [Fe(H2O)6]3+  Fe(H2O)5(OH)2+ + H+
Lewis Acid/Base Review
• Which of the following compounds can act
as Lewis acids?
• NH3
• H2O
• H+
• SO42• BCl3
Hydride Bases
• We are used to seeing hydrogen ions (H+) as
acids… but there are a class of compounds
called hydrides (H-), which act as bases.
– Ex: NaH (ionic bond w/ H-)