AcidBase Chemistry

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Acid-Base Chemistry
Chapter 16/17
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Acids Bases
Arrhenius Definition
Acids release H in solution Bases release OH- in
solution. Examples HCl ? H Cl- NaOH ?
Na OH-
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• Arrhenius definition of acids and bases has
  limitations.
• It only applies to aqueous solutions.
• Bronsted and Lowry proposed a definition based on
  acid base reactions transferring H ions from one
  substance to another.

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H Ion in Water

• H ions are bare protons
• H will react strongly with the nonbonding pair
  of electrons in a water molecule to form
• Hydronium ion (H3O).

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H and H3O are used interchangeably to represent
acidic solutions

• HCl ? H Cl-
• HCl(g) H2O(l) ?H3O(aq) Cl-(aq)

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The Bronsted Definition
Acids are proton donors. (bad) Bases are
proton acceptors. (bba) Example HBr
H2O ? H3O Br-
A
B
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• An acid and a base always work together to
  transfer a proton.
• Bronsted acids must have an H to transfer.
• Bronsted bases must have a nonbonding pair of
  electrons that can bind with H.

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Water may act as an acid or a base

• HCl(g) H2O(l) ?H3O(aq) Cl-(aq)
• NH3(aq)H2O(l)?NH4(aq) OH-(aq)
• In the 1st reaction above, water accepts a proton
  (Bronsted base).
• In the 2nd reaction above, water donates a proton
  (Bronsted acid)
• Amphoteric substances
• a substance that is capable of acting as an
  acid or a base.
• Know this term and definition.

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Conjugate Acid-Base Pairs

• Acid base equilibrium
• both the forward and reverse reactions involve
  proton transfer.
• HX H2O ?X- H3O
• Forward HX is acid/ H2O is the base
• Reverse H3O acid/ X- is base

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HX H2O ?X- H3O

• An acid and a base that differ only in the
  presence or absence of H are called a Conjugate
  acid-base Pair.
• Every acid has a conjugate base.
• Every base has a conjugate acid.
• HX is the conjugate acid of X-
• H3O is the conjugate acid of H2O

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Example Identify Bronsted acid base also the
conjugate acid and base. NH3 H2O ? NH4
OH-
A
B
CA
CB
NH3 acts as a Bronsted base by accepting a
proton. Water acts as a Bronsted acid by donating
a proton.
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Example Write the formula of the conjugate
base of H2SO4 in aqueous solution.
HSO4-
H2SO4 H2O ?HSO4- H3O
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Example
Consider the reaction HSO4- HCO3- ? SO42-
H2CO3 Identify the acids and bases for the
forward and reverse reactions. Identify the
conjugate acid-base pairs.
B
A
CA
CB
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Lewis Acids and Bases
Bases are electron pair donors (LBD) Acids are
electron pair acceptors (LAA)
Example H2O NH3 ? OH- NH4 Is
really H2O NH3 ? OH- HNH3
electron pair acceptor(H)
Electron pair donor(NH3)
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Identify the Lewis Acid and Lewis Base in the
following reaction. (start by writing the Lewis
Structures of each component.)
BF3 NH3 ? BF3NH3
Lewis Base (Electron Donor)
H
F
H
F

B
N
H
F
B
N
H
F
H
F
H
F
Lewis Acid (Electron Acceptor)
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The main use of the Lewis Acid/Base theory is to
provide an explanation for the formation of
complex compounds.
AgCl 2NH3 ? Ag(NH3)2
The silver ion has the electron
configurationKr4d105s1. It loses the 1
electron to form AgCl. This means it will have
vacant s and p orbitals in its 5th energy level.
These vacant orbitals can accept electron pairs
from Lewis bases.
Common Trend Metal ions tend to form Lewis Acids
because of vacant orbitals in highest energy
level.
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Cu2 4NH3 ? Cu(NH3)42
Lewis Base
Lewis Acid
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• Identify the Lewis acid and Lewis base in each of
  the following reactions
• SnCl4(s) 2Cl-(aq) ?SnCl62-(aq)
• b) Hg2(aq) 4CN- ?Hg(CN)42-(aq)

• SnCl4 accepts 2 pair of e- from Cl- so SnCl4 is
  the Lewis acid) /Cl- is Lewis base.
• Hg2 accepts 4 pair of e- (Lewis Acid)

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Summary of Definitions
Arrhenius defines acids and bases in
water. Bronsted defines acids and bases in
solvents other than water. Lewis defines acids
and bases without a solvent explains complex
compound formation (metal ions are Lewis acids).
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Acid and Base Strength

• Strong acids completely transfer their protons to
  water (completely ionized).
• HCl ? H OH- (single arrow)
• Weak acids are those that only partially
  dissociate in aqueous solution.
• HC2H3O2 H C2H3O2- (double arrow)

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Strong Acid
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Weak Acid
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Flash Animations of Weak and Strong
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Strong Acids Hydrochloric Acid Hydrobromic
Acid Hydroiodic Acid Nitric Acid Sulfuric
Acid Chloric Acid Perchloric Acid
HCl
HBr
HI
HNO3
H2SO4
HClO3
HClO4
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Binary Acids

• When comparing the strength of binary acids, the
  strength depends on two factors
• 1) The size of the anion in the acid.
• 2) The polarity of the molecule.

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Binary Acids

• When comparing two acids that both have an anion
  in the same group on the periodic table the
  strength of the acid increases as the anion
  increases in size.
• A larger anion indicates a weaker bond between H
  and anion.
• More ionization is possible making acids with
  large anions stronger acids.

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Acid Strength

• Polarity of the bond (?Polarity ?acidity)
• H X (only used to compare same row acids)
• Strength of the bond
• Strong bonds are more difficult to break than
  weak bonds. (?strength ?acidity)
• This explains why the very polar HF is a stronger
  acid than H2O. HF is still weaker than HCl and
  HBr because going down a group has a bigger
  influence on acid strength than across a period.
• Stability of conjugate base (X-). If X- is more
  stable, it is more likely to form as the result
  of the loss of H (will learn more about this
  soon).

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• Consider the acid HA, Binary Acid
• When comparing binary acids with only H and a
  nonmetallic element, the larger A is, the
  stronger the acid.
• Which is stronger?
• HCl or HF
• HBr or HCl

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• Bond strength changes much less across a period.
• Bond polarity becomes the determining factor for
  binary acids in the same row.
• Which is stronger?
• HCl or H2S

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Consider the acid HXO, an Oxyacid 2. For acids
containing oxygen, the more electronegative X,
the stronger the acid. Which is stronger? HClO3
or HIO3 HBrO2 or HClO2
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Consider the acid HXO, and oxoacid 2. For
acids containing oxygen, the more oxygen atoms
present, the stronger the acid. Which is
stronger? HClO4 or HClO3 HBrO or HBrO2
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Binary versus Oxoacids
Example HCl HClO3
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Oxoacids (sometimes called oxyacids)

• Contain 1 or more O-H bonds
• One or more OH groups or additional oxygen atoms
  are bonded to the central atom.

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Carboxylic Acids

• Contain carboxyl groups
• COOH
• Strength of carboxylic acids increase as the
  number of electronegative atoms in the acid
  increases.

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• Carboxylic acids are the largest group of organic
  acids.
• Acetic acid is a carboxylic acid
• HC2H3O2 (CH3COOH)
• Which is the stronger acid
• CF3COOH or HC2H3O2

CF3COOH
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Predict the relative strengths of the oxyacids in
each of the following groups
HClO, HBrO, and HIO
HClO gt HBrO gt HIO
HNO3 and HNO2
HNO3gtHNO2
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Carboxylic acids have similar structures as
hydrocarbons, except a CH3 in the hydrocarbon is
replaced with COOH group (organic acid
group). Naming carboxylic acids follows same
rules as hydrocarbons, except we change the
ending to oic and add the word acid. HC2H3O2 or
CH3COOH acetic acid or ethanoic acid. CHOOH
methanoic acidCH3COOH ethanoic acidCH3CH2COOH
Propanoic acid CH3CH2CH2COOH Butanoic
acidCH3CH2CH2CH2COOH Pentanoic acid Etc.
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Strong Bases All group 1 Metal Hydroxides LiOH,
NaOH, KOH, RbOH, CsOH Heavy Group 2 Metal
Hydroxides Ca(OH)2, Sr(OH)2, Ba(OH)2)
Know Strong Acids and Bases! All others are weak!
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NH3 is a commonly discussed weak base. Know this!
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Strong acids and bases dissociate 100
Example NaOH ? Na OH-
1 M
1 M
1 M
Na
OH-
Na
OH-
OH-
Na
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1 M
1 M
1 M
Example HCl ? H Cl-
H
Cl-
H
Cl-
H
Cl-
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Weak acids and bases dissociate X
X
Example HF ? H F-
small
small
1 M
HF
H
HF
F-
HF
HF
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Acetic acid is only 1 ionized. It is a weak
acid. Because it only ionizes 1, more HC2H3O2
is present, and equilibrium lies to
left. HC2H3O2 H2O ? H3O C2H3O2-
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Weak makes strong. Strong makes weak.
The stronger the acid, the weaker its conjugate
base.
The stronger the base, the weaker its conjugate
acid.
Note These are relative terms. The conjugate
bases of weak acids are still weak.
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See figure 16.12 on page 645.

• Substance with negligible acidity (CH4) contain
  hydrogen but show no acidic behavior in water.
• The conjugate bases (Ex CH3-) are strong bases
  reacting with water completely and forming OH-
  ions.

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Handout
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• When the acid is a stronger acid than H3O, the
  equilibrium lies to the right.
• HCl H2O ? H3O Cl-
• When H3O is the stronger acid, the equilibrium
  lies to the left.
• HC2H3O2 H2O ? H3O C2H3O2-

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In all Acid-base reactions, the equilibrium
favors the transfer of the proton from the
stronger acid to the stronger base.This means
that the side of the reaction with a weak acid
and base is favored.

• Which direction is favored(HPO42- is a stronger
  acid than H2O).
• PO43-(aq) H2O(l) ? HPO42-(aq) OH-(aq)

Since HPO42- is the stronger acid, it wants to
transfer its proton to the stronger base (OH-).
This means right to left is favored.
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Which direction is favored NH4(aq) OH- ?
NH3(aq) H2O(l)
OH- is the stronger base, so we know NH4 must be
a strong acid and will want to transfer its
proton to OH-. This means left to right is
favored.
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Example The strengths of the following acids
increase in the order HCN lt HF lt HNO3. Arrange
the conjugate bases of these acids in order of
increasing base strength.
NO3- lt F- lt CN-
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The Autoionization of Water

• Water can act as a Bronsted acid (donate proton)
  or base (accept proton) depending on the
  environment.
• One water can donate a proton to another water
  molecule, this is the autoionization of water.

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H2O H2O ? H3O OH-

• The equilibrium constant expression

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Autoionization is a rapid equilibrium process.
We usually write the equation H2O H
OH-
The ions only last for a brief moment until they
are recaptured by a water molecule. At any given
moment, only a very small proportion exists as
ions (about 1 in a billion particles).
HOH-
We can represent the equilibrium constant
expression of water as
K
H2O
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• Remember, just like solids, we dont include pure
  liquids in equilibrium expressions.
• This is because the concentration of H2O is large
  and constant.
• The equilibrium expressions for the ions in pure
  water can therefore be represented as
• Kw H3OOH- HOH-
• Kw (ion equilibrium constant of H2O)
• Kw 1.0 x 10-14 at 25ºC

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• When HOH- the solution is neutral.
• When the concentration of one increases the other
  decreases, the product of the 2 must 1.0 x
  10-14 at 25ºC.
• HgtOH-the solution is acidic
• HltOH-the solution is basic

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The Amphoteric Property of Water
Arrhenius H2O(l) ? H(aq) OH-(aq) Kw
HOH-
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The Amphoteric Property of Water
Bronsted H2O(l) H2O(l) ? H3O(aq)
OH-(aq) Kw H3OOH-
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The Ion Product Constant of Water
Kw HOH-
1.0 x 10-14
At equilibrium H OH- 1.0x10-7 M.
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pH

• Common way to express H.
• Means power of hydrogen.
• A solution with a pH of 2 is 100 times more
• acidic than a solution with a pH of 4.
• The negative logarithm of the hydrogen ion
• concentration in mol/L (M).

pH -logH
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What is the pH of a neutral solution?
pH -log (1.0x10-7) -(-7.0) 7.0
An unknown solution has a hydrogen ion conc. of
5.40 x 10-6. Calculate the pH of the solution
and determine if the solution is acidic or basic.
pH -log (5.40 x 10-6) 5.27
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• pH decreases as H increases.
• We can determine if a solution is acidic or basic
  simply by looking at the hydrogen ion
  concentration.
• If the H concentration is greater than 1.0 x
  10-7, the solution is acidic. (Ex 1.0 x 10-6
  pH of 6)
• If the H concentration is less than 1.0 x 10-7,
  the solution is basic. (Ex 1.0 x 10-8 pH of 8)
• We must simply look at the exponent to determine
  if it is an acid or a base.

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Calculating the Hydrogen Ion Concentration from pH
(The reverse of calculating the pH)
pH -log H
Consider a neutral solution pH -log 1.0 x
10-7 7 To find H given a pH of 7 H 10
pH So, we type 10 -7 H 10-7 1.0 x
10-7 Typing inverse log (10x) and then -7 would
be the same. Some calculators have an antilog
button.
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What is the hydrogen ion concentration of a
solution with a pH of 2.4?
H 10 pH H 10-2.4 4.0 x 10-3
An unknown base solution has a pH of 10.6.
Calculate the H concentration.
H 10 pH H 10-10.6 2.5 x 10-11
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• What is the pH of a solution with OH- 3.0 x
  10-4?

Kw HOH-
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Orange juice has a hydroxide ion concentration of
6.3 x 10-12 M. Determine the pH of orange juice.
1.0 x 10-14
1.6 x 10-3
6.3 x 10-12
pH -log (1.6 x 10-14) 2.8
You will see that there is another way to
calculate this perhaps an easier way.
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Other "p" Scales

• The negative log of a quantity is the p, or
  potential of that quantity.

We can calculate the pOH of a solution as we did
the pH of a solution.
pOH - log OH-
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pH pOH -log Kw 14
If given the OH- concentration of a solution, we
can calculate the pOH of the solution, and then
substract it from 14 to find the pH. Lets see
an example.
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Example
A particular NaOH solution has an OH-
concentration of 2.9 x 10-4M. Calculate the pH.
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?H means that H were released into the
solution making it acidic. ?OH- means that OH-
were released into the solution making it basic.
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The concentration of OH- ion in a certain
household ammonia cleaning solution is 0.0025 M.
Calculate the concentration of the H ions.
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Example The H ion concentration in a certain
solution is 5.0 x 10-5 M. What is the OH- ion
concentration?
Kw H3OOH-
OH- Kw H
1.0 x 10-14 5.0 x 10-5M
2.0 x 10-10 M
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Measuring pH

• pH Meters
• Acid-base indicators
• These have an acid form and a base form that
  differ in color.
• Litmus
• Red indicate a pH of less than 5
• Blue indicates pH above 8

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Other pH Indicators

• Bromthymol blue
• Phenophthalein
• Alizarin yellow R
• Universal

• Methyl violet
• Thymol Blue
• Methyl orange
• Methyl red

See page 604.
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Strong Acids and Bases

• Both are strong electrolytes
• Completely ionize in aqueous solution.
• Strong acids
• 6 of the 7 are monoprotic (HX)
• -Exist as only H ions and X- ions
• Since they completely ionize, calculating the pH
  of a strong acid or base is a straight forward
  process.

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The molarity of the solution tells us the H
concentration. Calculate the pH of a .100 M
solution of HCl (aq) HCl (aq) ? H Cl-
H .100 M Cl- .100 M pH
-log H 1
Calculate the pH of a .0100 M solution of HCl
(aq) H .0100 M NO3- .0100 M pH -log
H 2 As concentration decreases, pH
increases.
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The same method is used to calculate pOH or pH of
a base. Calculate the pH of a .200 M solution of
Ca(OH)2 . Ca(OH)2 ? Ca2 2OH-
OH- .400 M (Notice how we doubled the
concentration) pOH -log OH- pOH -log
.400 pOH .398 pH 14 - .398 13.6
Note Many strong bases are not very soluble in
water. Do not confuse this with pH. The small
of ions that do dissolve still leads to the
corresponding pH as calculated above.
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Strong Bases

• Besides the common strong bases, strong basic
  solutions can be formed when other substances
  react with water to form OH-.
• Metal oxides (Na2O,CaO)
• Each mole of O2- forms 2 moles OH-
• Na2O(s) H2O(l)??2Na(aq)2OH-(aq)

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• Ionic hydrides and nitrides also react with water
  in this way.
• N3- 3H2O ? NH3 3OH-
• H- H2O ? H2 OH-
• Because the anions O2-, H-, and N3-, are all
  stronger bases than OH-(the conjugate base of
  H2O) they are able to remove a proton from water.

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Weak Acids

• Partially ionize in aqueous solution.
• HA(aq) H2O(l) ? H3O(aq) A-(aq)
• HA(aq) ? H(aq) A-(aq)
• Equilibrium constant expression

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• Ka is the acid-dissociation constant or The Acid
  Ionization Constant
• Subscript a indicates this is the equilibrium
  constant of an acid.
• The magnitude of Ka indicates the tendency of
  Hydrogen to ionize.

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The Acid Ionization Constant (Ka)

HA ? H A-
The larger the Ka, the stronger the acid.
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• To calculate Ka given pH and initial molarity.
• Similar to Keq problems from last chapter.
• If given pH and initial molarity, 10-pH gives us
  the concentration of H and A-.
• Once you know H and A-, you must simply
  subtract this from initial molarity to find
  acid.

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A 0.085 M solution of HC8H7O2 is found to have a
pH of 2.68. Calculate Ka for this acid.
HC8H7O2 ? H C8H7O2-
H and C8H7O2- 10-2.68 2.09 x 10-3
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A 0.085 M solution of HC8H7O2 is found to have a
pH of 2.68. Calculate Ka for this acid.
HC8H7O2 ? H C8H7O2-
H and C8H7O2- 2.09 x 10-3
We were given the initial concentration of
HC8H7O2 to start. 2.09 x 10-3 of this ionized
into H and C8H7O2- To find new concentration
(at equilibrium) of HC8H7O2, we simply subtract
this from initial concentration. 0.085 2.09 x
10-3 .0829
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So, at this point, we have H 0.00209
C8H7O2- 0.00209 HC8H7O2 0.0829 Plug into
equilibrium expression for Ka
Ka 5.3 x 10-5
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Determining the pH of a weak Acid
pH - logH
As usual, we use the equation above. However,
since only some of a weak acid dissociates, we
must first calculate the concentration of H at
equilibrium.
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The Key to these types of problems is to
determine what you are given and use this
information to find other values which can be
plugged into equilibrium constant
expression. (May use table for these as well if
it helps organize your work).
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Dont forget If you are given an initial
conc. and ionization, you can simply multiply
to find H. Ex A .400 M Weak acid is 3.5
ionized H .400 x .035 .014Can now calculate
Ka using same method as previously described.
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Calculations with Ka pH

• A given pH of a solution will always represents
  an equilibrium condition.
• Proton transfer reactions are usually very rapid.

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Using Kato calculate pH

• Calculate the pH of 0.30M acetic acid at 25ºC if
  the Ka 1.8 x 10-5.

Step 1
HC2H3O2 ? H C2H3O2
Step 2
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HC2H3O2 ? H C2H3O2
Step 3
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Step 4
At this point, we would end up needing to use the
quadratic equation. Dont do it! Waste of
time! Because Ka is very small and the
equilibrium lies far to the left, we are able to
make the assumption that 0.30-x is equal to
0.30. We can therefore, eliminate x in the
denominator. (unless ionization is over 5).
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Solving for x
100
What is the ionization of the solution in the
previous problem?
Hequilibrium
ionization
HC2H3O2 initial
H was equal to 0.0023 and the molarity of
HC2H3O2 was 0.30 mol/L.
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• The Ka for niacin is 1.5 x10-5. A) What is the
  pH of a 0.010M solution of niacin (HC6H4O2N)? B)
  What is the ionization of the solution?
• Step 1
• HC6H4O2N ? H C6H4O2N-

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HC6H4O2N ?H C6H4O2N-

• Step 2

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HC6H4O2N ? H C6H4O2N-
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HC6H4O2N ?H C6H4O2N-
Solve for x
X3.9 x10-4
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Polyprotic Acids

• More than 1 ionizable H atom.
• These H atoms ionize in separate steps
• The acid dissociation constants for each H atom
  are designated Ka1, Ka2, and so on.

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H3PO4 ?H H2PO4- (Ka1) H2PO4- ?H
HPO42- (Ka2)

• It is always easier to remove the first proton.
• Thus Ka1 is always larger than Ka2
• Ka2 is larger than Ka3

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Ionizations Occur in Steps
H2SO3
? H HSO3-
Ka1 1.3 x 10-2
HSO3-
? H SO3-2
Ka2 6.3 x 10-8
SO3-2
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The Ka values are so small for the second and
third ionizations that we will only consider the
1st ionization when calculating the H
concentration. Just know 3 Kas do exist, and
that they get very small after the first.
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Weak Bases

• In water many substances act as weak bases.
• Weak base H2O ?Conj. Acid OH-
• Remember that ammonia (NH3) is a weak base.
• NH3(aq) H2O(l) ? NH4 OH-(aq)

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NH3(aq) H2O(l) ? NH4 OH-(aq)

• The equilibrium constant expression

• Because water concentration is constant, it is
  isnt included in the equilibrium expression for
  Kb.

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Kb (base dissociation constant)

• Kb always refers to the equilibrium in which the
  base reacts with water to form the conjugate acid
  and OH-.
• A lone pair of electrons is needed to bond with
  the H. (Example N in NH3)

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The Base Ionization Constant Kb
B H2O ? HB OH-
Indicates strength of the base (as Kb goes up
more ionization occurs)
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What is the pH of a 0.400M solution of
ammonia?Kb 1.8 x 10-5 (always start by writing
the equation of a weak base dissociating in
water).

• H2O(l) NH3(aq) ?NH4(aq)OH-(aq)

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• H2O(l) NH3(aq) ?NH4(aq)OH-(aq

pOH -log(2.7x10-3) 2.57 pH 14 2.57 11.4
115
Types of Weak Bases

• Divided into 2 categories
• Neutral substances that have a nonbonding pair of
  electrons.
• Most of these bases contain N, these are called
  amines.

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• Second category is composed of anions of weak
  acids.
• NaClO in aqueous solution Na ClO-
• In an acid base reaction, the Na is a spectator
  ion.
• The ClO- is the conjugate base of a weak acid,
  hypochlorous acid. (ClO- acts as a weak base)
• ClO-(aq) H2O(l) ? HClO(aq) OH-(aq)

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Relationship between Ka Kb

• Consider NH4NH3 (conjugate acid-base pair)
• NH4(aq) ? NH3(aq) H(aq)
• NH3(aq) H2O(l) ? NH4(aq)OH-(aq)

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When 2 reactions are added together to give a
3rd, the equilibrium constant for the 3rd is
equal to the product of the 2 reactions.

• NH4(aq) ?NH3(aq) H(aq)
• NH3(aq) H2O(l) ? NH4(aq)OH-(aq)

H2O(l) ? H(aq) OH-(aq)
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Reaction 1 Reaction 2 Reaction 3K1 x K2 K3

• NH4(aq) ?NH3(aq) H(aq)
• NH3(aq) H2O(l) ? NH4(aq)OH-(aq)

H2O(l) ? H(aq) OH-(aq)
120

• Ka x Kb Kw
• As Ka gets larger, Kb gets smaller.
• It is possible to calculate Kb for any weak base
  if the Ka for the conjugate acid is known.

Given on Exam Kw Ka X Kb 1.0 x 10-14
_at_25ºC
121
The Ka of acetic acid is 1.7 x 10-5. What is the
conjugate base, and what is the Kb of its
conjugate base?
Kw Ka X Kb 1.0 x 10-14 _at_25ºC
Conjugate base is the acetate ion C2H3O2-
122
Ka
HC2H3O2 ? H C2H3O2-
Kb
C2H3O2- H2O ? HC2H3O2 OH-
Ka x Kb Kw
H2O ? H OH-
123
Kw
HOH- 1.0 x 10-14
Kw Ka X Kb
Kw
1.0 x 10 -14
Kb
Kb
Ka
1.7 x 10-5
Kb 5.6 x 10-10 Shows it is a
weak base.
124
The Ka of hydrofluoric acid is 6.8 x 10-4. What
is the conjugate base, and what is the Kb of its
conjugate base?
Kw Ka X Kb 1.0 x 10-14 _at_25ºC
Conjugate base is the fluoride ion F-. Kb 1.5 x
10-11
125
Acid-Base Properties of Salt Solutions

• Salt solutions are strong electrolytes and ionize
  in water
• Acid-base properties of salt solutions are a
  result of the anions and cations formed.

126
Salt - an ionic compound formed by the reaction
of an acid and a base.
strong acid
strong base
NaCl
127
strong acid
strong base
NaNO3
128
weak acid
strong base
NaC2H3O2
Remember! Weak ? strong conjugate
129
Hydrolysis

• Many ions react with water to form H or OH-.
• This reaction is hydrolysis.
• The anions from weak acids react with water to
  form OH-.
• Conjugate base is strong in relative terms.
• C2H3O2-H2O(l) ? HC2H3O2 OH-

130

• The anion of a strong acid (NO3-) forms a weak
  conjugate base.
• It does not react with water and does not affect
  pH.
• Anions with an ionizable proton (HPO42-) are
  amphoteric.
• Reaction with water will be determined by
  relative Ka and Kb.

131
Will a solution of the salt Na2HPO4 be acidic or
basic? Turn to table on page 635.
In order to determine this, we need to first
consider the hydrolysis of the salt.
Acid
HPO42- H2O ? H3O PO43- or HPO42- H2O ?
H2PO4- OH-
Base
The one that dominates in water depends on which
has the largest equilibrium constant, Ka or Kb.
132
From a list of equilibrium constants, we know
that

• HPO42- H2O ? H3O PO43- Ka3 4.2 x
  10-13
• We must now determine the Kb value of HPO42-
• Since HPO42- acts as a base in the following
  reaction
• HPO42- H2O ? H2PO4-(aq) OH- Ka2 6.2 x
  10-8
• We must calculate Kb from the Ka of the conjugate
  acid (H2PO4-).

133
HPO42-(aq) ? H2PO4-(aq) OH-

• Ka x Kb Kw
• Given in data table Ka2 (H2PO4-) 6.2 x 10-8
• Kb x 6.2 x 10-8 1.0x10-14
• Kb 1.6 x 10-7

134
HPO42- H2O ? H3O PO43- ka3 4.2 x
10-13 HPO42- H2O ? H2PO4- OH- kb 1.6
x 10-7

• Solution is basic because Kb is about 5x larger
  than Ka.

135

• Cations of alkali metal and alkaline earth metals
  do not hydrolyze in water.
• They do not influence pH.
• The pH of a salt solution can be qualitatively
  predicted by considering the cation and anion
  composing the salt.

136
Consider the relative strengths of the acids and
bases that the salt is derived from.

• Salts derived from strong acid and strong base.
• NaCl and NaNO3
• Neither the cation or anion hydrolyses.
• No change in pH (pH7)

137
strong acid
strong base
NaNO3
Strong ? weak conjugates
NO HYDROLYSIS Neutral Salt
138

• Salts of strong base and weak acid.
• NaClO and Ba(C2H3O2)
• Anion hydrolyses/ Cation does not
• Basic (pH above 7)

139
weak acid
strong base
NaC2H3O2
Weak ? strong conjugate
C2H3O2- H2O ? HC2H3O2 OH-
NaC2H3O2 is a basic salt
140

• Salt of a weak base and strong acid.
• NH4Cl and Al(NO3)3
• The cation hydrolyses and the anion does not.
• Acid (pH less than 7)

141
strong acid
weak base
NH4Cl
Weak ? strong conjugate
NH4 H2O ? NH3 H3O
NH4Cl is an acidic salt
142

• Salt of a weak base and weak acid.
• NH4C2H3O2 and FeCO3
• Both cation and anion exhibit hydrolyses.
• pH depends on the extent of each hydrolysis.
• Relative Ka and Kb of ions.

143
Salts formed by weak acid weak base can be
either acidic OR basic
Kb lt Ka
acidic
basic
Kb gt Ka
nearly neutral
Kb ? Ka
144
Example Write net ionic equations to show which
of the following ions hydrolyze in aqueous
solution a. NO3-
NO3- H2O ? NO RXN
145
b. NO2-
NO2- H2O ? HNO2 OH-
BASIC
146
c. NH4
NH4 H2O ? NH3 H3O
ACIDIC
147
Example Predict whether the following aqueous
solutions will be acidic, basic, or
neutral KI NH4I KC2H3O2
(answers neutral, acid, base)
148
Chemical Structure and behavior

• When substances are placed in water some act as
  acids, some as bases, and others are neutral.
• Structure of the compound plays a role in the
  behaviors.

149
Acid Base Reactions
Write the net ionic equation and decide whether
solution is acidic, basis, or neutral
Case 1 strong acid strong base
NaOH HCl ?? NaCl HOH
Na OH- H Cl- ? Na Cl- H2O
Net H OH- ? H2O
Equimolar solution will be NEUTRAL
150
Case 2 weak acid strong base
HC2H3O2 NaOH ? NaC2H3O2 HOH
HC2H3O2
?
Na
OH-
Na
C2H3O2-
H2O
Net HC2H3O2 OH- ? C2H3O2- H2O
solution will be BASIC
151
HC2H3O2 OH- ? C2H3O2- H2O
WEAK ACID ? STRONG
CONJUGATE BASE
Acetate ion hydrolyzes in water to form OH-
C2H3O2- H2O ? HC2H3O2 OH-
152
Case 3 strong acid weak base
HCl NH3 ? NH4Cl
H Cl-
?
NH3
NH4 Cl-
Net H NH3 ? NH4
solution will be ACIDIC
153
H NH3 ? NH4
WEAK BASE ? STRONG CONJUGATE ACID
NH4 H2O ? NH3 H3O
154
Case 4 weak acid weak base
?
Both form strong conjugates. It depends on the
the reaction
155
Hydrolysis of Metal Ions

• The Lewis concept helps explain why some metal
  ions display acidic properties in solution.
• Positively charged metal ions attract unshared
  electrons pairs in water molecules.
• This attraction is what causes hydrolysis,
  dissolving of salts.

156

• Water molecules bound to the metal ion are more
  acidic than those in the solution.
• This effect usually has increasing acidity (Ka
  hydrolysis) with increasing charge of the metal
  ion and decreasing radius of the ion.

157
Start of Chapter 17
158
Common-Ion Effect
(related to Le Chateliers principle) Adding a
solute with an ion that is common to a system at
equilibrium will cause the equilibrium to shift.
Qualitative Aspect
HC2H3O2(aq) H(aq) C2H3O2-(aq)
Add sodium acetate (C2H3O2- is common ion) ?
shift to left
Add hydrochloric acid (H is common ion) ? shift
to left
159
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160
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161
Common-Ion Effect
Quantitative Aspect (similar to weak acid
problems)
Lets compare the pH of two solutions.
1) solution made up of 0.30 mol of HC2H3O2 and
0.30 mol of NaC2H3O2. in 1.00 Liters.
Vs.
2) solution made up of only 0.30 mol of HC2H3O2
in 1.00 Liters.
162
Spectator
Less H
C2H3O2-(aq)
Na
HC2H3O2(aq) H(aq) C2H3O2-(aq)
Based on our previous discussion, the pH of the
solution with NaC2H3O2 should be higher (less
acidic) because of the shift to the left.
Lets calculate to verify!
163
HC2H3O2 ? H C2H3O2-
Must consider added solute at start
164
Ka 1.8 x 10-5
x(0.30 x)
1.8 x 10-5
(0.30 - x)
Again, x is small, so adding or subtracting it
will be insignificant.
(0.30x)
x 1.8 x 10-5
1.8 x 10-5
(0.30)
Substituting into table tells us H 1.8 x 10-5
165
pH -log 1.8 x 10-5 pH 4.74
Solution without Acetate Common Ion Added
HC2H3O2 ? H C2H3O2
166
x2
1.8 x 10-5
(0.30 - x)
x2
1.8 x 10-5
(0.30)
x 2.32 x 10-3
pH -log 2.32 x 10-5 pH 2.63
pH of solution with acetate added was indeed
higher (4.74 vs. 2.63)
167
Same process can be seen when a common ion is
added to a weak base. NH3(aq) H2O(l)
NH4(aq) OH-(aq)
The addition of NH4Cl(aq) would cause the
solution to become less basic (shift to left
reducing OH-).
168
Buffers --A buffer is a solution that resists a
change in pH. --created by the addition of a
common ion to a weak acid or base. --excess
conjugate acid or base absorbs the addition of an
acid or base. --examples include blood (pH 7.4),
buffered aspirin, antacid buffers. Buffer
Capacity --The amount of acid or base a buffered
solution can handle before a drastic change in pH
is noticeable. --depends on amount of conjugate
acid or base, gt amount gt resistance. (You
couldnt expect a buffered solution to resist a
change in pH upon the addition of 2.0 L of HCl).
169

• Quantitative Calculations of Buffers
• Three types you must understand
• How to calculate pH of buffer solution.
• (already did thiscommon ion calculationwill
  review on next few slides).
• 2) Determine effect of adding strong acid or base
  to a buffered solution.
• 3) Determine how to prepare a buffer with a
  certain pH.

170
Calculating the pH of a Buffer Solution (remember,
a buffer solution is a solution created by
adding a commonion to a weak acid or base).
171
1) Calculate the pH of a buffer solution made up
of 0.30 M HC2H3O2 and 0.30 M NaC2H3O2. in 1.00
Liters.
HC2H3O2 ? H C2H3O2-
172
Ka 1.8 x 10-5
x(0.30 x)
1.8 x 10-5
(0.30 - x)
x is small, so adding or subtracting it will be
insignificant.
(0.30x)
x 1.8 x 10-5
1.8 x 10-5
(0.30)
Substituting into table tells us H 1.8 x 10-5
173
pH -log 1.8 x 10-5 pH 4.74
pH of this particular buffer solution is 4.74.
174
We will now see the benefit of a buffer solution
by looking at how this type of solution resists a
change in pH when a small amount of strong acid
or strong base is added to the solution.
175
Consider adding .01 mol of HCl (g) to pure
water. The H would completely ionize to form H
ions. Since the pH - log H The pH of this
solution would then be -log 0.01 2
176
Now we will consider adding .01 mol of HCl (g) to
the acetic acid/sodium acetate buffer solution we
made earlier. Calculation
177
Again our buffered solution is made up of 0.30 M
HC2H3O2 and 0.30 M NaC2H3O2. in 1.00 Liters.
HC2H3O2 ? H C2H3O2-
178
Adding 0.01 mol of HCl to the buffer will cause
the equilibrium to shift to the left (common ion
H).
HC2H3O2 ? H C2H3O2-
179
This causes a change in our starting amounts.
0.01 H
HC2H3O2 ? H C2H3O2-
0.0
0.30
0.29
0.30
0.31
180
Our new table
HC2H3O2 ? H C2H3O2-
181
Ka 1.8 x 10-5
x(0.29 x)
1.8 x 10-5
(0.31 - x)
x is small, so adding or subtracting it will be
insignificant.
(0.29x)
x 1.9 x 10-5
1.8 x 10-5
(0.31)
Substituting into table tells us H 1.9 x 10-5
182
pH -log 1.9 x 10-5 pH 4.72
pH of the buffer solution before addition of HCl
was 4.74. pH after adding 0.01 HCl to pure water
is 2 pH after adding 0.01 mol HCl to buffered
solution is 4.72(proves resistance to pH
change)
183
Consider adding .01 mol of NaOH (s) to pure
water. The NaOH would completely dissociate to
form OH- ions. Since the pOH - log OH- The
pOH of this solution would then be -log 0.01
2 pH 14-2 12
184
Now we will consider adding .01 mol of NaOH (s)
to the acetic acid/sodium acetate buffer solution
we made earlier. Calculation
185
Again, our buffered solution is made up of 0.30 M
HC2H3O2 and 0.30 M NaC2H3O2. in 1.00 Liters.
HC2H3O2 ? H C2H3O2-
186
Adding 0.01 mol of NaOH to the buffer will
neutralize some of the H to form water and at
the same time, more C2H3O2- will be created to
restore equilibrium.
HC2H3O2 ? H C2H3O2-
187
Again, this causes a change in our starting
amounts.
0.01 OH-
HC2H3O2 ? H C2H3O2-
0.0
0.30
0.31
0.30
0.29
188
Our new table
HC2H3O2 ? H C2H3O2-
189
Ka 1.8 x 10-5
x(0.31 x)
1.8 x 10-5
(0.29 - x)
x is small, so adding or subtracting it will be
insignificant.
(0.31x)
x 1.7 x 10-5
1.8 x 10-5
(0.29)
Substituting into table tells us H 1.7 x 10-5
190
pH -log 1.7 x 10-5 pH 4.77
pH of the buffer solution before addition of NaOH
was 4.74. pH after adding 0.01 NaOH to pure
water is 12 pH after adding 0.01 mol NaOH to
buffered solution is 4.77 (small change from
4.74)
191
Summary
pH of buffer solution before adding a strong acid
or base was 4.74.pH after adding 0.01 mol of
HCl to the buffered solution is 4.72pH after
adding 0.01 mol of NaOH to the buffered solution
is 4.77
Strong Acid Buffer
Strong Base
4.72 4.74 4.77
Proves resistance to change.
192
Example of How To Prepare a Buffer
193
How many moles of NH4Cl must be added to 2.0 L of
0.10 M NH3 to form a buffer whose pH is 9.00?
Assume that the addition of NH4Cl does not change
the volume of the solution.
As always, first 2 steps are to write equation
and equilibrium expression.
NH3(aq) H2O(l) NH4(aq) OH-(aq)
Kb 1.8 x10-5
We must use given information to calculate
concentration of NH4.
We know Kb and the Initial concentration of NH3
0.10 M
We obtain OH- from the pH given in the
problem. pOH 14.00 9.00 5.00 10-pOH
1 x 10-5
Now we can solve for NH4
194
NH3
NH4 Kb
OH-
(0.10 M)
NH4 (1.8 x 10-5)
0.18 M
(1.0 x 10-5 M)
0.18 M x 2.0 L 0.36 mol NH4Cl
L
Conclusion 0.36 mol of NH4Cl must be added to 2.0
L of 0.10 M NH3 to form a buffer solution with a
pH of 9.
195
Now that you have learned the concepts behind the
calculations of buffers, its time to learn the
shortcut!
196
Henderson Hasselbalch Equation
(Given on AP Exam)
A-
Acid
pH pKa log
HA
HB
Base
pOH pKb log
B
pKa -log Ka pKb - log Kb
197
pKa -log Ka pKb - log Kb
pKa and pKb are positive whole numbers of Ka and
Kb
Explanation of how the equation is derived is
explained in your textbook.
198
A-
Conjugate base
pH pKa log
HA
Acid
HB
Conjugate acid
pOH pKb log
Base
B
When doing equilibrium calculations, we normally
neglect the amounts of the acid and base of the
buffer that ionize. (because they are weak)
This allows us to use the initial concentrations
directly in the equation above.
Example
199
Solution made up of 0.30 mol of HC2H3O2 and 0.30
mol of NaC2H3O2. in 1.00 Liters. We previously
found that this buffered solution has a pH of
4.74.
A-
pH pKa log
HA
pKa - log Ka
pKa - log 1.8 x 10-5 4.74
.30
.60
pH 4.74 log
.30
pH 4.74
Note the change in pH if .60 mol of NaC2H3O2 is
added.
pH 5.04
200
What is the pH of a buffer that is 0.10 M NH3 and
0.18 M NH4Cl? Kb 1.8 x 10-5
HB
pOH pKb log
B
.18
pOH -log 1.8 x 10-5 log
.10
pOH 5.00 pH 14 5 9
201
Turn to page 667 and calculate the practice
exercise using the Henderson-Hasselbach
equation. pKa of benzoic acid 6.3 x 10-5
A-
pH pKa log
HA
pKa -log Ka
202
Titration the gradual addition of one solution to
another solution to reach an equivalence point,
usually done to determine the concentration of an
unknown.
203
Standard Solutiona solution in which the
concentration is known which is used to determine
the unknown concentration of a second solution.
Equivalence pointpoint at which equivalent
quantities of acid and base have been brought
together (the completion of a neutralization
reaction).
Titrantthe solution being added to cause
neutralization. Can be acid or base, depending
on which is known and which is unknown.
Buretlong graduated piece of glassware used in a
titration. Usually consists of a stop valve
which allows the addition of a titrant on a drop
by drop basis.
Indicator solutionsolution added to flask which
indicates the change of a solution in terms of
its acidic or basic properties.
Titration rangerange of colors that occur as a
result of the indicator being partly in its
acidic form and partly in its basic form
(indicator with proper range is needed).
End pointinstant at which the indicator changes
color. Although the end point should be close to
the equivalence point, they are not the same.
204
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205
Titration Curve A curve plotted during a
titration which allows us to determine the
equivalence point in the titration. Created by
measuring the pH with a pH meter as drops or mLs
of titrant are added from buret. Computer
software and probes are often used because
software draws curve for you. The shape of the
curve differs based on type of titration being
performed.
206
Titration Curve titration of a strong acid with a
strong base (notice it takes 25.0 mL of 0.100 M
NaOH to reach equilavence point)
Remember
207
Titration Curve (titration of strong base with
strong acid)
208
Titration Curve (titration of weak acid with
strong base)
209
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210
Titration Curve (titration of polyprotic acid
with strong base)
211
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212

• Must be able to calculate concentration and pH of
  solution after specific volume of acid or base is
  added (allows us to determine equivalence point).
• Must be able to calculate amount of titrant
  needed to neutralize acid or base.
• Must be able to calculate the above in previously
  mentioned titration conditions.

Because of a lack of time, you will need to learn
some of the these calculations on your own.
213
Calculating pH when concentrations and volumes of
all species are known.
Calculate the pH when 49.00 mL of 0.100 M NaOH
solution is added to 50 mLs of 0.100 HCl
solution.
Calculate the pH when 51.00 mL of 0.100 M NaOH
solution is added to 50 mLs of 0.100 HCl
solution.
214
Calculating unknown concentration
A student titrates 40.0 mLs of an HCl solution
of unknown concentration with 0.5500 M NaOH
solution. The equilvence point is reached after
adding 24.64 mL of the NaOH solution. What is
the concentration of the HCl solution (H)?
MHVH MOH-VOH-
215
Solubility and Solubility Products
216
Types of Solutions

• Saturated
• Unsaturated
• Supersaturated

217
Solubility Equilibria

• Acid and base equilibria involve homogenous
  systems (same phase).
• We will now consider heterogeneous equilbria
  systems (different phase).
• Precipitation of Ionic Compounds
• Saturated solution
• Contains maximum amount of solute and solution is
  in contact with undissolved solute in a state of
  equilibrium.

218

• Saturated solution
• The solution is in contact with undissolved
  solute.

Solution Equilibrium
Na
Cl-
Na
Na
Cl-
Cl-
Cl-
Na
Cl-
Na
NaCl(s)
219
Degrees of Solubility
NaCl(s) ? Na(aq) Cl-(aq)
soluble
AgCl(s) ? Ag(aq) Cl-(aq)
insoluble
Ca(OH)2(s) ? Ca2(aq) 2OH-(aq)
slightly soluble
220

• To write an equilibrium product constant for a
  heterogeneous system, ignore the concentrations
  of pure solids or pure liquids.
• BaSO4(s) ?Ba2(aq) SO42-(aq)
• The Kc depends only on the molar concentration of
  the species in solution.
• Solubility-Product constant (Ksp)

221
BaSO4(s) ?Ba2(aq) SO42-(aq)

• Even though BaSO4 is not included in the Ksp,
  some must be present for the system to be at
  equilibrium.
• Ksp Ba2SO42-

222

• Ksp is the equilibrium constant for the
  equilibrium that exists between an ionic solid
  and its ions in a saturated solution.
• A very small Ksp indicates that only a small
  amount of solid will dissolve in water.

223
Rules for Writing Ksp

• Ksp is equal to the product of the concentration
  of the ions in the equilibrium, each raised to
  the power of its coefficient in the equation.

224
The Solubility Product Constant Ksp
Example AgBr(s) ? Ag(aq) Br-(aq)
Ksp AgBr-
225
Example Ca(OH)2(s) ? Ca2(aq) 2OH-(aq)
Ksp Ca2OH-2
Example Ag2CrO4(s) ? 2Ag(aq) CrO42-(aq)
Ksp Ag2CrO42 -
226

• Give the Ksp expressions for
• Barium carbonate
• Silver sulfate
• Calcium flouride

a)Ba2CO32- b)Ag2 SO42- c)Ca2F-2
227
Solubility and Ksp
Solubility grams of solute
liter of solution
Molar moles of solute solubility liter of
saturated solution
228

• The solubility of a substance can change as
  concentrations of other solutes change.
• Mg(OH)2 solubility is dependent on pH as well as
  concentration of Mg2
• Ksp for each substance has only one value at any
  specific temperature.

229
Calculating Ksp from Solubility Data
Step 1 Use data to determine molar
solubility. Step 2 Use molar solubility and
stoichiometry of dissociation to determine
concentrations of cations and anions. Step 3
Write equilibrium expression, calculate Ksp.
230
A saturated solution of AgCl in contact with
undissolved solid is prepared at 25C. The
concentration of Ag ions in the solution is
found to be 1.35 x 10-5 M. Assuming that AgCl
dissociates completely in water and that there
are no other simultaneous equilibria involving
the Ag or Cl- ions in the solution, calculate
Ksp for this compound.
AgCl(s) ? Ag(aq) Cl-(aq)
Ksp AgCl-
1.35 x 10-51.35 x 10-5
Ksp 1.82 x 10-10
231

• Solid silver sulfate is added to water at a
  specific temperature. It is allowed to sit until
  an equilibrium of the two phases exist. Analysis
  shows that the Ag concentration in solution is
  1.5 x 10-2 M. Calculate the solubility product of
  the salt.

Ag2SO4(s) 2Ag(aq) SO42-(aq)
Ksp Ag2SO42-
1.5 x 10-2 mol Ag
1 mol SO42-
.0075 mol SO42-
2 mol Ag
L
Ksp 1.5 x 10-22.0075 1.7 x 10-6
232
Calculating solubility from Ksp
Step 1 Use equilibrium expression Ksp value
to determine concentrations of cations and
anions. Step 2 Use ion concentrations to
determine molar solubility. Step 3 convert
molar solubility to solubility (grams
solute/liter).
233
Calculate the solubility of copper (II) hydroxide
in g/L. Ksp 2.2 x 10-20 Molar mass Cu(OH)2
97.57g/mol
Cu(OH)2 ? Cu2(aq) 2OH-(aq)
234

• Ksp Cu2OH-2
• 2.2 x 10-20 (x)(2x)2

235
Solubility Factors

• Common-ion effect
• pH
• Complex ion formation
• Amphoterism

236
Common-ion effect Solubility

• In general, the solubility of a slightly soluble
  salt is decreased by the presence of a second
  solute that supplies a common ion.
• Ksp is unchanged by additional solutes.

237
The common ion effect decreases the solubility of
the salt.
AgBr(s) ? Ag(aq) Br-(aq)
Adding Ag, shift left Adding Br-, shift left
238

• The Ksp for manganese(II) hydroxide is 1.6 x
  10-13. Calculate the molar solubility of
  manganese (II) hydroxide in a solution that
  contains 0.020 M NaOH.

Mn(OH)2(s) ? Mn2(aq) 2OH-(aq)
239
Mn(OH)2(s) ? Mn2(aq)
2OH-(aq)

• Ksp Mn2OH-2

240
Ksp Mn2OH-2

• The solubility of Mn(OH)2 is very small compared
  to the 0.020M NaOH.

241
You try one.
CaF2 is much less soluble than Ca(NO3)2.
Calculate the molar solubility of CaF2 in a
solution that is 0.010 M in Ca(NO3)2. Ksp CaF2
3.9 x10-11
Answer 3.1 x 10-5 M Explanation to Follow
242
CaF2(s) ? Ca2(aq)
2F-(aq)

• Ksp Ca2F-2

243
Ksp Ca2F-2

• The solubility of CaF2 is very small compared to
  the 0.010M Ca(NO3)2.

244
Solubility pH

• The solubility of almost any ionic compound is
  affected if the solution is made sufficiently
  acidic or basic.
• The solubility of slightly soluble salts
  containing basic anions increases as H
  increases (pH?).

245
pH Can Change Solubility
Mg(OH)2(s) ? Mg2(aq) 2OH-(aq)
Add OH-, raises pH, shift left
Add acid, lowers pH, more Mg2 must be
made shift to right
246
Consider CaF2
CaF2(s) ? Ca2(aq) 2F-(aq)
2F-(aq) 2H(aq) ? 2HF(aq)
CaF2(s) 2H (aq) ? Ca2(aq) 2HF(aq)
All substances with basic anions are more soluble
in acidic solution.
247

• Is Ni(OH)2(s) more soluble in acidic or basic
  solution?
• Write the equations that show how this reaction
  occurs.

Ni(OH)2(s) ? Ni2(aq) 2OH-(aq)
2OH-(aq) 2H(aq) ? 2H2O(l)
Ni(OH)2(s) 2H (aq) ? Ca2(aq) 2H2O(l)
248
Which of the following are more soluble in acidic
solution than in basic solution?
CuCN CaCO3 BaF2 AgCl
All basic anions
249
Homework Read review packet on Additional
Equilibrium concepts. Read through questions and
answers at end of packet.
250
At 25ºC the molar solubility of Mg(OH)2 in pure
water is 1.4 x 10-4M. Calculate its molar
solubility in a buffer medium whose pH is Ksp
1.2x10-11 a.) 12.00 b.) 11.00
Mg(OH)2(s) ? Mg2(aq) 2OH-(aq)
251

• Calculate the concentration of aqueous ammonia
  necessary to initiate the precipitation of
  iron(II) hydroxide from a 0.0030 M solution of
  FeCl2.
• Ksp of Fe(OH)2 1.6 x 10-14
• Kb of NH3 1.8 x 10-5

NH3(aq) H2O?NH4(aq) OH-(aq)
Fe2(aq) 2 OH-(aq) ?Fe(OH)2(s)
252
Formation of Complex Ions

• Metal ions have the ability to act as Lewis acids
  (electron pair acceptors) toward water molecules,
  which act as Lewis bases.
• Lewis bases other than water also interact with
  metal ions (particularly transition metals)

253
AgCl Ksp1.8x10-10

• In the presence of aqueous ammonia AgCl will
  dissolve.
• AgCl(s) ? Ag(aq) Cl-(aq)
• Ag(aq) 2 NH3(aq) ? Ag(NH3)2(aq)
• AgCl(s) 2NH3(aq)?Ag(NH3)2(aq)Cl-(aq)
• Presence of NH3 drives the top reaction to the
  right (?solubility of AgCl).

254
Complex Ions
an ion made up of the metal ion with one or more
molecules or ions (Lewis bases) bonded to it.
Examples Ag(NH3)2 Cu(CN)4-2
255

• The stability of a complex ion in aqueous
  solution is dependent upon the size of the
  equilibrium constant for its formation (Kf).

Ag(aq) 2 NH3(aq) ? Ag(NH3)2(aq)
256
Stable or Unstable?
Ag (aq) 2CN- (aq) ? Ag(CN)2- (aq)
1.0 x 1021
The larger the Kf the more stable the ion.
257
The formation of a complex ion has a strong
affect on the solubility of a metal salt.
AgI(s) ? Ag(aq) I-(aq)
Adding CN- shift right
Ag