Acid and Base Equilibrium

1
Acid and Base Equilibrium
2
Arrhenius Definition

• Acids produce hydrogen ions in aqueous solution.
  (Arhennius definition)
• Bases produce hydroxide ions when dissolved in
  water. (Arhennius definition)
• Limits to aqueous solutions.
• Only one kind of base.
• NH3 ammonia could not be an Arrhenius base.

3
Brønsted-Lowry Definitions

• And acid is an proton (H1) donor and a base is a
  proton acceptor.
• Acids and bases always come in pairs.
• HCl is an acid.
• When it dissolves in water it gives its proton to
  water.
• HCl(g) H2O(l) ? H3O1 Cl-1 OR
• HCl ? H1 Cl-1
• Water is a base makes hydronium ion.

4
General Acid/Base Equation

• The general equation is
• HA(aq) H2O(l) ? H3O1(aq) A-1(aq)
• Acid Base ? Conjugate acid Conjugate
  base
• This is an equilibrium situation.
• Competition for H1 between H2O and A-1
• The stronger base controls direction.
• If H2O is a stronger base it takes the H1
• Equilibrium moves to right.

5
Acid dissociation constant Ka

• The equilibrium constant for the general
  equation.
• HA(aq) H2O(l) ? H3O1(aq) A-1(aq)
• Ka H3O1A-1 HA
• H3O1 is often written H1 ignoring the water in
  equation (it is implied). Either way of
  expressing it is fine.

6
Acid dissociation constant Ka

• HA(aq) ? H1(aq) A-1(aq)
• Ka H1A-1 HA
• We can write the expression for any acid.
• Strong acids dissociate completely and their
  equilibrium is far to the right.
• Conjugate base must be weak.
• Dont forget stoichiometry if you have an acid
  with 2 H1 ions like H2SO4

7
A Way to Describe Acids

• Strong acids
• Ka is large
• H is equal to HA
• A- is a weaker base than water

• Weak acids
• Ka is small
• H ltltlt HA
• A- is a stronger base than water

8
More Ways to Describe Acids

• Polyprotic Acids- more than 1 acidic hydrogen
  (diprotic, triprotic). Sulfuric Acid H2SO4
• Oxyacids - Proton is attached to the oxygen of an
  ion. An example is H3PO4 Phosphoric Acid
• Organic acids contain the Carboxyl group -COOH
  with the H attached to O. An example of this is
  Acetic Acid which we call vinegar.
• Organic Acids are generally very weak.

9
Amphoteric Nature of Water

• Behaves as both an acid and a base.
• Water autoionizes as shown below.
• 2 H2O(l) ? H3O1(aq) OH-1(aq)
• KW H3O1OH-1 H1OH-1
• At 25ºC KW 1.0 x 10-14
• In EVERY aqueous solution.
• Neutral solution H1 OH-1 1.0 x 10-7
• Acidic solution H1 gt OH-1
• Basic solution H1 lt OH-1

10
pH

• pH -logH1
• Used because H1 is usually very small.
• As pH decreases, H1 increases exponentially.
• Log is based on a factor of 10.
• Only the digits after the decimal place of a pH
  are significant.
• H1 1.0 x 10-8 pH 8.0
• pOH -logOH-1 AND pKa -log Ka

11
Relationships

• KW H1OH-1 1.0 x10-14
• -log KW -log(HOH-) OR
• -log KW -logH -logOH-
• pKW pH pOH
• pH pOH 14.00
• H,OH-, pH and pOH
• Given any one of these we can find the other
  three variables.

12
Basic
Acidic
Neutral
13
Calculating pH of Solutions

• Always write down the major species in solution.
• Strong Acids /or Bases.
• Water
• Known Conjugates.
• Remember these are equilibria.
• Remember the stoichiometry.
• Dont try to memorize there is no one way to do
  this.

14
Strong Acids

• HCl, HNO3, H2SO4, HClO4 (HBr, HI)
• Completely dissociated.
• Ka is said to be infinite, since there is no
  reactant left.
• H1 HA
• OH-1 is going to be small because of
  equilibrium and is only present due to the
  autoionization of water.

15
Weak Acids

• Ka will be small, often, very small.
• Determine whether most of the H1 will come from
  the acid or the water. You do this by comparing
  Ka to Kw and see which one has a higher
  dissociation.
• The rest of the problem is just like last
  previous problems.
• SET UP THE ICE BOX
• DETERMINE THE SHIFT
• If HAlt 10-7 water contributes more H1

16
Example of a Ka Problem

• Calculate the pH of 2.0 M acetic acid HC2H3O2
  with a Ka of 1.8 x 10-5
• Calculate pOH, OH-1, H1
• HC2H3O2 ? H1 C2H3O2-1
• Initial 2.0 0 0
• Change -x x x
• Equilibrium 2.0 x x x
• 1.8 x 10-5 x2 / 2 x .006 5
  rule works

17
The Complete Solution

• We found that x .006 and that this was the
  amount of H1. Solving for pH gives 2.22
• Solving for pOH ? 14 2.22 11.78
• Solving for OH-1 ? H1 OH-1 10-14
• so OH-1 1.66 x 10-12
• It wasnt asked in this problem, but .006 is also
  the amount of the acetate ion in solution.

18
A mixture of Weak Acids

• Determine the major species.
• The stronger will dominate.
• Bigger Ka if concentrations are comparable
• Calculate the pH of a mixture of
• 1.20 M HF (Ka 7.2 x 10-4) and
• 3.4 M HOC6H5 (Ka 1.6 x 10-10)
• The HF is clearly the dominant species as its Ka
  is 6 factors of 10 higher. Note Remember that
  the Ka is a negative exponent.

19
Upon Further Review

• We can apply some LeChateliers Principal Ideas
  to this problem.
• 1.20 M HF (Ka 7.2 x 10-4) and
• 3.4 M HOC6H5 (Ka 1.6 x 10-10)
• From the problem, we see that HOC6H5 is a minor
  contributor, but if LeChateliers Principal is
  taken into account we see that it is even less
  than originally thought. The addition of H1
  ions from the HF (the major species) further
  shift the HOC6H5 dissociation to the left
  restricting its H1 contribution.

20
Percent dissociation

• amount dissociated x 100 initial
  concentration
• For a weak acid percent dissociation increases as
  acid becomes more dilute.
• Calculate the dissociation of 1.00 M and
  .00100 M Acetic acid (Ka 1.8 x 10-5)
• As HA0 decreases H1 decreases but
  dissociation increases.

21
If you know the dissociation

• What is the Ka of a weak acid that is 8.1
  dissociated as 0.100 M solution?
• Still set up the ICE BOX and you know that 8.1
  of the .1 M solution is going to dissociate (x)
  and that 91.9 of the solution will remain as a
  molecule.

22
Bases

• The OH-1 is a strong base. (Arrhenius)
• Hydroxides of the alkali metals are strong bases
  because they dissociate completely when
  dissolved.
• The hydroxides of alkaline earths Ca(OH)2 etc.
  are strong dibasic bases, but they dont
  dissolve well in water.
• Used as antacids because OH-1 will combine with
  H1 to form water.

23
Bases without OH-

• Bases are proton acceptors. (Brønsted-Lowry)
• NH3 H2O ? NH41 OH-1
• It is the lone pair on nitrogen that accepts the
  proton.
• Many weak bases contain a Nitrogen atom.
• B(aq) H2O(l) ? BH1(aq) OH-1(aq)
• Kb BHOH- B
• The Law of Mass Action doesnt change.

24
Strength of Bases

• Group 1 and 2 hydroxides are strong.
• Eg. NaOH, KOH, Ca(OH)2
• Most others are weak.
• The smaller the Kb ? the weaker the base.
• Calculate the pH of a solution of 4.0 M pyridine
  (Kb 1.7 x 10-9)

N
25
Polyprotic acids

• Always dissociate stepwise, one H1 at a time.
• The first H1 comes off much easier than the
  second. Its Ka will always be greater.
• Ka1 for the first step is much bigger than Ka2
  for the second.
• Denoted Ka1, Ka2, Ka3 to clarify which step is
  being referred to.

26
Polyprotic acid 2

• H2CO3 ? H1 HCO3-1 Ka1 4.3 x 10-7
• HCO3-1 ? H1 CO3-2 Ka2 4.3 x 10-10
• The base in the first step is often the acid in
  second. Your Ka values will tell you this.
• In calculations we can normally ignore the second
  dissociation when determining pH.
• However, we need to solve the second when we are
  looking for the anion CO3-2 in this instance.

27
Calculate the Concentration

• of all the ions in a solution of 1.00 M Arsenic
  acid H3AsO4
• Ka1 5.0 x 10-3
• Ka2 8.0 x 10-8
• Ka3 6.0 x 10-10
• Each one is done individually.
• Set up an ICE Box. The math is easy, the initial
  set up is the tricky part.

28
Sulfuric acid is special

• For starters, remember that it is a strong acid.
• (??? What does that tell you about Ka1 ???)
• Ka2 1.2 x 10-2
• Calculate the concentrations of H2SO4, HSO4-1 and
  SO4-2 in a 2.0 M solution of H2SO4
• Do the same calculations in a solution of H2SO4
  at 2.0 x 10-3 M

29
Salts as acids/bases - Strong

• Salts are ionic compounds.
• Solutions of salts that are made up of the
  conjugates of strong acids and strong bases are
  neutral. For example NaCl KNO3
• NaOH (strong base) HCl (strong acid)? NaCl
• KOH (strong base) HNO3 (strong acid) ? KNO3
• There is no equilibrium for strong acids and
  bases. For teaching purposes, you can think of
  these as strong salts. (This is not general
  chemistry terminology, but it fits.)
• We ignore the reverse reaction because K is so
  large in the forward reaction.

30
Salts as acids/bases - Weak

• Some salts do have a pH when in solution
• NaNO2 will be basic in aqueous solution.
• WHY, YOU ASK ? OK I will tell you.
• Here is whats going on, the major species in
  solution are Na1, NO2-1, and from water H1, and
  OH-1.
• When the Na1 and the OH-1 merge, they are
  soluble, dissociating and releasing OH-1 ions
  into solution.
• However, when H1 and NO2-1 merge, the HNO2
  molecule is favored the H1 ions are not in
  solution.
• Therefore the solution has more OH-1 ions than
  H1 ions and is therefore basic.

31
Conjugate Acids and Bases

• When an acid or base dissociates, it leaves
  behind an ion. For example.
• HNO2 ? H1 NO2-1
• Fe(OH)3 ? OH-1 Fe(OH)21
• The H1 or the OH-1 have already been discussed
  thoroughly (see Arhennius definition of acids and
  bases) but in each equation there is also an ion
  that is formed.
• These are called conjugates.
• In the example above the NO2-1 is called the
  conjugate base of HNO2
• The Fe(OH)21 is the conjugate acid of Fe(OH)3

32
More on Basic Salts

• If the anion of a salt is the conjugate base of a
  weak acid - basic solution.
• In an aqueous solution of NaF
• The major species are Na1, F-1, and H2O
• F-1 H2O ? HF OH-1
• Kb HFOH-1 F-1
• but Ka H1F-1 HF
• So there is a relationship between Ka and Kb

33
Ka tells us Kb

• The anion of a weak acid is a weak base.
• Calculate the pH of a solution of 1.00 M NaCN.
  Ka of HCN is 6.2 x 10-10
• The CN-1 ion competes with OH- for the H1
• How do we know that this is going to be a basic
  solution ?
• Because the Ka of HCN is low so the molecule is
  preferred and it doesnt dissociate and liberate
  any H1 ions.
• If any NaOH forms, it will dissociate immediately
  releasing OH-1 ions.

34
Acidic salts

• A salt with the cation of a weak base and the
  anion of a strong acid will be acidic.
• Calculate the pH of a solution of 0.40 M NH4Cl
  (the Kb of NH3 1.8 x 10-5).
• H1 (from water) Cl-1 ? forms HCl which will
  immediately dissociate to form and acid soln.
• Some acidic salts are those of highly charged
  metal ions.
• More on this later.

35
Anion of weak acid, cation of weak base

• IF
• Ka gt Kb acidic
• Ka lt Kb basic
• Ka Kb Neutral

36
Structure and Acid/Base Properties

• Any molecule with an H in it is a potential acid.
• The stronger the X-H bond the less acidic
  (compare bond dissociation energies) because the
  H1 ion isnt liberated as easily.
• The more polar the X-H bond the stronger the acid
  (use electronegativities).
• The more polar H-O-X bond - stronger acid.

37
Strength of oxyacids

• The more oxygen hooked to the central atom, the
  more acidic the hydrogen.
• HClO4 gt HClO3 gt HClO2 gt HClO
• Remember that the H1 is attached to an oxygen
  atom.
• The oxygen atoms are electronegative and so they
  pull electrons away from hydrogen atom making it
  an Hydrogen ion (H1).

38
Hydrated metals

• Highly charged metal ions pull the electrons of
  surrounding water molecules toward them.
• Make it easier for H1 to come off.

H
Al3
O
H
39
Acid-Base Properties of Oxides

• Non-metal oxides dissolved in water can make
  acids.
• SO3 H2O H2SO4
• Metal oxides dissolve in water to produce bases.
• CaO H2O Ca(OH)2

40
Lewis Acids and Bases

• Most general definition of all.
• Acids are electron pair acceptors.
• Bases are electron pair donors.

F
H
B
F
N
H
F
H
41
Lewis Acids and Bases

• Boron triflouride wants more electrons.

F
H
B
F
N
H
F
H
42
Lewis Acids and Bases

• Boron triflouride wants more electrons.
• BF3 is Lewis base NH3 is a Lewis Acid.

F
H
F
H
B
N
F
H
43
Lewis Acids and Bases
(
H
Al3
6
O
H
3
(
H
Al
O
H
44
Summary of Definitions of Acids and Bases

• Arrhenius Acid Hydrogen Ion donor
• Arrhenius Base Hydroxide Ion donor
• Brønsted-Lowry Acid Proton donor
• Brønsted-Lowry Base Proton acceptor
• Lewis Acid Electron Pair acceptor
• Lewis Base Electron Pair donor

45
Titration - Introduction

• A titration is a chemistry lab procedure where an
  acid is added to a base (or vice-versa) to
  determine its exact concentration.
• Exact volumes are needed so a buret is usually
  the tool of choice.
• Accurate measurement is needed to get a good
  result.
• The formula M1V1 M2V2 comes in handy.
• The central idea is that the number of moles of
  acid will the number of moles of base.

46
Titration - Procedure

• An exact amount of acid with the unknown
  concentration is put in a flask.
• A small amount of an indicator is added to the
  base.
• A color change might take place depending on the
  indicator.
• A Standard solution of base with known
  concentration is put in a buret.
• The acid is carefully added until the solution
  shows the desired color change which depends on
  the indicator used.

47
Indicators
Indicators are certain chemicals that have a
peculiar property of having one color when they
are in an acid solution and another when they are
in a basic solution. The most common indicator
is phenolphthalein, because it happens to change
close to pH 7, which is neutral.
Phenolphthalein is clear when it is in acidic
solution and a light pink when it just turns
basic. If more base is accidently added, the
solution turns a dark purple.