Acids and Bases

1
Acids and Bases

• L. Scheffler
• Lincoln High School

2
Acids and Bases

• The concepts acids and bases were loosely defined
  as substances that change some properties of
  water.
• One of the criteria that was often used was
  taste.
• Substances were classified
• salty-tasting
• sour-tasting
• sweet-tasting
• bitter-tasting
• Sour-tasting substances would give rise to the
  word 'acid', which is derived from the Greek word
  oxein, which mutated into the Latin verb acere,
  which means 'to make sour'

3
Acids

• React with certain metals to produce hydrogen
  gas.
• React with carbonates and bicarbonates to produce
  carbon
• dioxide gas
• Vinegar is a solution of acetic acid. Citrus
  fruits contain citric acid.

Bases

• Have a bitter taste
• Feel slippery.
• Many soaps contain bases.

4
Properties of Acids

• Produce H (as H3O) ions in water (the
  hydronium ion is a hydrogen ion attached to a
  water molecule)
• Taste sour
• Corrode metals
• Electrolytes
• React with bases to form a salt and water
• pH is less than 7
• Turns blue litmus paper to red Blue to Red
  A-CID

5
Properties of Bases

• Generally produce OH- ions in water
• Taste bitter, chalky
• Are electrolytes
• Feel soapy, slippery
• React with acids to form salts and water
• pH greater than 7
• Turns red litmus paper to blue Basic Blue

6
Arrhenius Definition

• Arrhenius
• Acid - Substances in water that increase the
  concentration of hydrogen ions (H).
• Base - Substances in water that increase
  concentration of hydroxide ions (OH-).
• Categorical definition easy to sort substances
  into acids and bases
• Problem many bases do not actually contain
  hydroxides

7
Bronsted-Lowry Definition

• Acid - neutral molecule, anion, or cation
  that donates a proton.
• Base - neutral molecule, anion, or cation
  that accepts a proton. HA B ?
  HB A-
• Ex HCl H2O ? H3O
  Cl-
• Acid Base Conj Acid Conj
  Base
• Operational definition - The classification
  depends on how the substance behaves in a
  chemical reaction

8
Conjugate Acid Base Pairs
Conjugate Base - The species remaining after an
acid has transferred its proton. Conjugate
Acid - The species produced after base
has accepted a proton. HA
A- - conjugate acid/base pair
A- - conjugate base of acid HA B
HB - conjugate acid/base pair
HB - conjugate acid of base B
9
Examples of Bronsted-Lowry Acid Base Systems
Note Water can act as acid or base Acid Base
Conjugate Acid Conjugate Base HCl H2O
? H3O Cl- H2PO4- H2O ?? H3O
HPO42- NH4 H2O ?? H3O
NH3 Base Acid Conjugate Acid
Conjugate Base NH3 H2O ?? NH4
OH- PO43- H2O ?? HPO42- OH-
10
G.N. Lewis Definition
Lewis Acid - an electron pair acceptor Base - an
electron pair donor
11
The pH Scale
pH H3O OH-
pOH
12
pH and acidity

• Acidity or Acid Strength depends on Hydronium Ion
  Concentration H3O
• The pH system is a logarithmic representation of
  the Hydrogen Ion concentration (or OH-) as a
  means of avoiding using large numbers and powers.
• pH - log H3O log(1 /
  H3O)
• pOH - log OH- log(1 /
  OH-)
• In pure water H3O 1 x 10-7 mol / L (at
  20oC)
• ? pH - log(1 x 10-7) - (0 - 7)
  7
• pH range of solutions 0 - 14
• pH lt 7 (Acidic) H3O gt 1 x 10-7 m
  / L
• pH gt 7 (Basic) H3O lt 1 x 10-7 m / L

13
pH and acidity
Kw H3O OH- 1.0 x10-14 In
water H3O OH- 1.0 x10-7 pH pOH
14
14
Calculating the pH

• pH - log H3O
• Example 1 If H3O 1 X 10-10pH - log 1 X
  10-10
• pH - (- 10)
• pH 10
• Example 2 If H3O 1.8 X 10-5pH - log 1.8
  X 10-5
• pH - (- 4.74)
• pH 4.74

15
Indicators
16
pH and acidity
The pH values of several common substances are
shown at the right. Many common foods are weak
acids Some medicines and many household cleaners
are bases.
17
Neutralization

• An acid will neutralize a base, giving a salt
  and water as products
• Examples
• HCl NaOH ? NaCl
  H2O
• H2SO4 2 NaOH ? Na2SO4 2 H2O
• H3PO4 3 KOH ? K3PO4 3 H2O
• 2 HCl Ca(OH) 2 ? CaCl2 2 H2O

18
Neutralization Calculations

• If the concentration of acid or base is
  expressed in Molarity or mol dm-3
• then
• --The volume in dm3 multiplied by the
  concentration yields moles.
• -- If the volume is expressed in cm3
• the same product yields millimoles

19
Neutralization Problems

• If an acid and a base combine in a 1 to 1
  ratio, then the volume of the acid multiplied by
  the concentration of the acid is equal to the
  volume of the base multiplied by the
  concentration of the base
• Vacid C acid V base C base
• If any three of the variables are known it is
  possible to determine the fourth

20
Neutralization Problems

• Example 1 Hydrochloric acid reacts with
  potassium hydroxide according to the following
  reaction
• HCl KOH ? KCl H2O
• If 15.00 cm3 of 0.500 M HCl exactly
  neutralizes 24.00 cm3 of KOH solution, what is
  the concentration of the KOH solution?
• Solution
• Vacid Cacid Vbase Cbase
• (15.00 cm3 )(0.500 M) (24.00
  cm3 ) Cbase
• Cbase (15.00 cm3 )(0.500
  M)
• (24.00
  cm3 )
• Cbase 0.313 M

21
Neutralization Problems

• Whenever an acid and a base do not combine in
  a 1 to 1 ratio, a mole factor must be added to
  the neutralization equation
• n Vacid C acid V base C base
• The mole factor (n) is the number of times
  the moles the acid side of the above equation
  must be multiplied so as to equal the base side.
  (or vice versa)
• Example
• H2SO4 2 NaOH ? Na2SO4 2 H2O
• The mole factor is 2 and goes on the acid
  side of the equation. The number of moles of
  H2SO4 is one half that of NaOH. Therefore the
  moles of H2SO4 are multiplied by 2 to equal the
  moles of NaOH.

22
Neutralization Problems

• Example 3 Phosphoric acid reacts with
  potassium hydroxide according to the following
  reaction
• H3PO4 3 KOH ? K3PO4 3 H2O
• If 30.00 cm3 of 0.300 M KOH exactly
  neutralizes 15.00 cm3 of H3PO4 solution, what
  is the concentration of the H3PO4 solution?
• Solution
• In this case the mole factor is 3 and it goes on
  the acid side, since the mole ratio of acid to
  base is 1 to 2. Therefore
• 3 Vacid Cacid Vbase Cbase
• 3 (15.00 cm3 )(Cacid) (30.00 cm3 )
  (0.300 M)
• Cacid (30.00 cm3 )(0.300 M)
• (3) (15.00 cm3 )
• Cacid 0.200 M

23
Neutralization Problems

• Example 2 Sulfuric acid reacts with sodium
  hydroxide according to the following reaction
• H2SO4 2 NaOH ? Na2SO4 2 H2O
• If 20.00 cm3 of 0.400 M H2SO4 exactly
  neutralizes 32.00 cm3 of NaOH solution, what is
  the concentration of the NaOH solution?
• Solution
• In this case the mole factor is 2 and it goes on
  the acid side, since the mole ratio of acid to
  base is 1 to 2. Therefore
• 2 Vacid Cacid Vbase Cbase
• 2 (20.00 cm3 )(0.400 M) (32.00 cm3 )
  Cbase
• Cbase (2) (20.00 cm3 )(0.400 M)
• (32.00 cm3 )
• Cbase 0.500 M

24
Neutralization Problems

• Example 4 Hydrochloric acid reacts with
  calcium hydroxide according to the following
  reaction
• 2 HCl Ca(OH)2 ? CaCl2 2 H2O
• If 25.00 cm3 of 0.400 M HCl exactly
  neutralizes 20.00 cm3 of Ca(OH)2 solution, what
  is the concentration of the Ca(OH)2 solution?
• Solution
• In this case the mole factor is 2 and it goes on
  the base side, since the mole ratio of acid to
  base is 2 to 1. Therefore
• Vacid Cacid 2 Vbase Cbase
• (25.00 cm3) (0.400) (2) (20.00 cm3)
  (Cbase)
• Cbase (25.00 cm3 ) (0.400 M)
• (2) (20.00 cm3 )
• Cbase 0.250 M

25
Acid Base Dissociation
Acid-base reactions are equilibrium
processes. The relationship between the relative
concentrations of the reactants and products is a
constant for a given temperature. It is known as
the Acid or Base Dissociation Constant. The
stronger the acid or base, the larger the value
of the dissociation constant.
26
Acid Strength

• Strong Acid - Transfers all of its protons to
  water - Completely
  ionized - Strong
  electrolyte -
  The conjugate base is weaker and has a
  negligible tendency to be
  protonated.
• Weak Acid - Transfers only a fraction of its
  protons to water
• - Partly ionized - Weak
  electrolyte - The conjugate
  base is stronger, readily
  accepting protons from water
• As acid strength decreases, base strength
  increases.
• The stronger the acid, the weaker its conjugate
  base
• The weaker the acid, the stronger its conjugate
  base

27
Acid Dissociation Constants

• Dissociation constants for some weak acids

28
Base Strength

• Strong Base - all molecules accept a proton
  - completely ionizes
  - strong electrolyte
  - conjugate acid is very weak,
  negligible tendency
  to donate protons.
• Weak Base - fraction of molecules accept proton
  - partly ionized
  - weak electrolyte - the conjugate acid
  is stronger. It more readily donates
  protons.
• As base strength decreases, acid strength
  increases.
• The stronger the base, the weaker its conjugate
  acid.
• The weaker the base the stronger its conjugate
  acid.

29
Common Strong Acids/Bases
Strong Acids Hydrochloric Acid Nitric
Acid Sulfuric Acid Perchloric Acid
Strong Bases Sodium Hydroxide Potassium
Hydroxide Barium Hydroxide Calcium
Hydroxide While strong bases they are not very
soluble
30
Water as an Equilibrium System
Water has the ability to act as either a
Bronsted- Lowry acid or
base. Autoionization spontaneous formation of
low concentrations of H and OH- ions by
proton transfer from one molecule to another.
Equilibrium Constant for Water
31
Weak Acid Equilibria

• A weak acid is only partially ionized.
• Both the ion form and the unionized form exist at
  equilibrium
• HA H2O ?? H3O A-
• The acid equilibrium constant is
• Ka H3O A-
• HA
• Ka values are relatively small for most weak
  acids. The greatest part of the weak acid is in
  the unionized form

32
Weak Acid Equilibrium Constants

• Sample problem . A certain weak acid dissociates
  in water as follows
• HA H2O ?? H3O A-
• If the initial concentration of HA is 1.5 M and
  the equilibrium concentration of H3O is 0.0014
  M. Calculate Ka for this acid
• Solution
• Ka H3O A-
• HA
• I C E Substituting
• HA 1.5 -x 1.5-x Ka
  (0.0014)2 1.31 x 10-6
• A- 0 x x 1.4986
• H3O 0 x x
• x 0.0014
• 1.5-x 1.4986

33
Weak Base Equilibria

• Weak bases, like weak acids, are partially
  ionized. The degree to which ionization occurs
  depends on the value of the base dissociation
  constant
• General form B H2O ? BH OH-
• Kb BHOH-
• B
• Example
• NH3 H2O ? NH4 OH-
• Kb NH4 OH-
• NH3

34
Weak Base Equilibrium Constants

• Sample problem . A certain weak base dissociates
  in water as follows
• B H2O ?? BH OH-
• If the initial concentration of B is 1.2 M and
  the equilibrium concentration of OH- is 0.0011
  M. Calculate Kb for this base
• Solution
• Kb BH OH-
• B
• I C E Substituting
• B 1.2 -x 1.2-x Kb
  (0.0011)2 1.01 x 10-6
• OH- 0 x x 1.1989
• BH 0 x x
• x 0.0011
• 1.2-x 1.1989

35
Weak Acid Equilibria Concentration Problems

• Problem 1. A certain weak acid dissociates in
  water as follows HA H2O ?? H3O A-
• The Ka for this acid is 2.0 x 10-6. Calculate
  the HA A-, H3O and pH of a 2.0 M
  solution
• Solution
• Ka H3O A- 2.0 x 10-6
• HA
• I C E Substituting
• HA 2.0 -x 2.0-x Ka x2
  2.0 x 10-6
• A- 0 x x 2.0-x
• H3O 0 x x If x ltltlt 2.0 it
  can be dropped from the
  denominator
• The x2 (2.0 x10-6)(2.0) 4.0 x10-6 x 2.0
  x 10-3
• A- H3O 2.0 x10-3 HA 2.0 - 0.002
  1.998
• pH - log H3O -log (2.0 x 10-3) 2.7

36
Weak Acid Equilibria Concentration Problems

• Problem 2. Acetic acid is a weak acid that
  dissociates in water as follows CH3COOH
  H2O ?? H3O CH3COO-
• The Ka for this acid is 1.8 x 10-5. Calculate
  the CH3COOH,CH3COO- H3O and pH of a
  0.100 M solution
• Solution
• Ka H3O CH3COO- 1.8 x 10-5
• CH3COOH
• I C E
  Substituting
• CH3COOH 0.100 -x 0.100-x Ka
  x2 1.8 x 10-5
• CH3COO- 0 x x
  0.100-x
• H3O 0 x x
  If x ltltlt 0.100 it can be dropped
  from the denominator
• The x2 (1.8 x10-5)(0.100) 1.8 x10-6 x
  1.3 x 10-3
• CH3COO-- H3O 1.3 x10-3 CH3COOH
  0.100 - 0.0013 0.0987
• pH - log H3O -log (1.3 x10-3) 2.88

37
Weak Base Equilibria

• Example1. Ammonia dissociates in water according
  to the following equilibrium
• NH3 H2O ?? NH4 OH-
• Kb NH4 OH- 1.8 x 10-5
• NH3
• Calculate the concentration of NH4 OH- NH3
  and the pH of a 2.0M solution.
• I C E Substituting
• NH3 2.0 -x 2.0-x Kb x2
  1.8x 10-5
• OH- 0 x x 2.0-x
• NH4 0 x x If x ltltlt
  2.0 it can be dropped from the
  denominator
• The x2 (1.8 x10-5)(2.0) 3.6 x10-5 x 6.0
  x 10-3
• OH- NH4 6.0 x10-3 NH3 2.0- 0.006
  1.994
• pOH - log OH- -log (6.0 x10-3) 2.22
• pH 14-pOH 14-2.22 11.78

38
Amphoteric Solutions

• A chemical compound able to react with both an
  acid or a base is amphoteric.   
• Water is amphoteric. The two acid-base couples of
  water are H3O/H2O and H2O/OH-It behaves
  sometimes like an acid, for example
• And sometimes like a base
• Hydrogen carbonate ion HCO3- is also amphoteric,
  it belongs to the two acid-base couples
  H2CO3/HCO3- and HCO3-/CO32-

39
Common Ion Effect

• The common ion effect is a consequence of Le
  Chateliers Principle
• When the salt with the anion (i.e. the conjugate
  base) of a weak acid is added to that acid,
• It reverses the dissociation of the acid.
• Lowers the percent dissociation of the acid.
• A similar process happens when the salt with the
  cation (i.e, conjugate acid) is added to a weak
  base.
• These solutions are known as Buffer Solutions.

40
Buffer Solutions - Characteristics

• A solution that resists a change in pH. It is pH
  stable.
• A weak acid and its conjugate base form an acid
  buffer.
• A weak base and its conjugate acid form a base
  buffer.
• We can make a buffer of any pH by varying the
  concentrations of the acid/base and its conjugate.

41
Buffer Solution Calculations

• Calculate the pH of a solution that is 0.50 M
  CH3COOH and 0.25 M NaCH3COO.
• CH3COOH H2O ?? H3O CH3COO- (Ka 1.8
  x 10-5)
• Solution
• Ka H3O CH3COO- 1.8 x 10-5
• CH3COOH
• I C E .
  Substituting
• CH3COOH 0.50 -x 0.50-x Ka
  x (0.25x) 1.8 x 10-5
• CH3COO- 0.25 x 0.25x
  (0.50-x)
• H3O 0 x x
  If x ltltlt 0.25 it can be dropped from both
  expressions in ( ) since adding
  or subtracting a small
  amount will not
  significantly change the value of the ratio
• Then the expression becomes x(0.25)/(0.50)
  1.8 x 10-5
• x 3.6 x 10-5 H3O
• pH - log H3O -log(3.6 x 10-5 )
  4.44

42
Buffer Solutions - Equations

• 1. Ka H3O A-
  HA
• H3O Ka HA
• A-
• The H3O depends on the ratio HA/A-
• Taking the negative log of both sides of
  equation 2 above
• pH -log(Ka HA/A-)
• pH -log(Ka) - log(HA/A-)
• pH pKa log(A-/HA)

43
Henderson Hasselbach Equation

• pH pKa log(A-/HA)
• pH pKa log(Base/Acid)
• This expression is known as the
  Henderson-Hasselbach equation. It provides a
  shortcut from using the I.C.E. model for buffer
  solutions where the concentration of both A-
  and HA are significantly greater than zero.

44
Using the Henderson -Hasselbach Equation

• pH pKa log(A-/HA)
• Example
• Calculate the pH of the following of a
  mixture that contains 0.75 M lactic acid
  (HC3H5O3) and 0.25 M sodium lactate (Ka 1.4 x
  10-4)
• HC3H5O3 H2O ?? H3O C3H5O3-
• Solution
• Using the Henderson-Hasselbach equation
• pH - log (1.4 x 10-4) log (
  0.25/0.75 )
• 3.85 (-0.477) 3.37

45
Henderson-Hasselbach Equation and Base Buffers

• For a base a similar expression can be written
• pOH pKb log (BH / B)
• pOH pKb log (Acid / Base)
• Example Calculate the pH of a solution that
  contains 0.25 M NH3 and 0.40 M NH4Cl
• (Kb 1.8 x 10-5)
• Solution
• pOH - log(1.8 x 10-5) log (0.40/0.25)
• 4.74 0.204 4.94
• pH 14 - pOH 14 - 4.94 9.06

46
Henderson-Hasselbach Equation Base Buffers

• Methyl amine is a weak base with a Kb or 4.38 x
  10-4
• CH3NH2 H2O ?? CH3NH3 OH-
• Calculate the pH of a solution that is 0.10 M
  in methyl amine and 0.20 M in methylamine
  hydrochloride.
• pOH pKb log (BH / B)
• Solution
• pOH -log (4.38 x 10-4) log (0.20 /
  0.10)
• 3.36 0.30 3.66
• pH 14- 3.66 10.34

47
Additional Buffer Problems

• How many grams of sodium formate, NaCHOO, would
  have to be dissolved in 1.0 dm3 of 0.12 M formic
  acid, CHOOH, to make the solution a buffer of pH
  3.80? Ka 1.78 x 10-4

pH pKa Log (A-/HA) Solution
3.80 -log (1.78 x 10-4) Log A- -
Log 0.12 3.80 3.75 Log A- -
(-0.92) Log A- 3.80 - 3.75 - 0.92 -
0.87 A- 10-0.87 0.135 mol
dm-3 The molar mass of NaCHOO 231212(16)
58.0 gmol-1 So (0.135 mol dm-3)(58.0 gmol-1 )
7.8 grams per dm-3
48
Relationship of Ka, Kb Kw

• HA weak acid. Its acid ionization is

• A- is the conjugate base Its base ionization
  is

• Multiplying Ka and Kb and canceling like terms

49
Titration Curves

• A graph showing pH vs volume of acid or base
  added
• The pH shows a sudden change near the equivalence
  point
• The Equivalence point is the point at which the
  moles of OH- are equal to the moles of H3O

50
Strong acid-strong base Titration Curve

• At equivalence point, Veq
• Moles of H3O Moles of OH-
• There is a sharp rise in the pH as one approaches
  the equivalence point
• With a strong acid and a strong base, the
  equivalence point is at pH 7
• Neither the conjugate base or conjugate acid is
  strong enough to affect the pH

51
Weak acid-strong base Titration Curve

• The increase in pH is more gradual as one
  approaches the equivalence point
• With a weak acid and a strong base, the
  equivalence point is higher than pH 7
• The strength of the conjugate base of the weak
  acid is strong enough to affect the pH of the
  equivalence point

52
Buffered Weak Acid-Strong Base Titration Curve

• The initial pH is higher than the unbuffered acid
• As with a weak acid and a strong base, the
  equivalence point for a buffered weak acid is
  higher than pH 7
• The conjugate base is strong enough to affect the
  pH

53
Polyprotic Weak Acids

• Polyprotic acids have more than one hydrogen
  that can be neutralized
• Phosphoric Acid has three hydrogen ions.
• H3PO4 H2O ?? H3O H2PO4-
• H2PO4- H2O ?? H3O HPO42-
• HPO42- H2O ?? H3O PO43-
• At given pH only one acid form and one conjugate
  base predominate
• pH 0-4.7 H3PO4 and H2PO4-
• pH 4.7-9.7 H2PO4- and HPO42-
• pH 9.7-14 HPO42- and PO43-

54
Polyprotic Weak Acid-Strong Base Titration Curve

• Phosphoric Acid has three hydrogen ions.
• There are three equivalence points
• H3P04 H2O ?? H3O H2PO4-
• H2PO4- H2O ?? H3O HPO42-
• HPO42- H2O ?? H3O PO43-

55
Salts

• A salt is the neutralization product of an acid
  and a base.
• The anion comes from the acid and the cation from
  the base.
• Examples
• HCl NaOH ? NaCl H2O.
• H2SO4 2 KOH ? K2SO4 H2O.

56
Salts

• If a salt is the result of a
• -- Strong acid and a strong base, the pH is near
  neutral.
• HCl NaOH ? NaCl
  H2O.
• -- Weak acid and a strong base, the pH will be
  greater than 7.
• CH3COOH NaOH ? NaCH3COO H2O
• -- Strong acid and a weak base, the pH will be
  lower than 7.
• NH4OH HCl ? NH4Cl H2O
• -- Weak acid and a weak base, the pH depends on
  whether the acid or the base is stronger.