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Title: Acids and Bases http://www.shodor.org/unchem/basic/ab/ http://www.chemtutor.com/acid.htm


1
Acids and Bases http//www.shodor.org/unche
m/basic/ab/ http//www.chemtutor.com/acid.htm

2
  • Strong Acids
  • Strong acids completely dissociate in water,
  • HCl(aq) gtH1 (aq) Cl-1 (aq)
  • 1moldm-3 1moldm-3 1moldm-3
  • HCl - hydrochloric acid
  • HNO3 - nitric acid
  • H2SO4 - sulfuric acid
  • HBr - hydrobromic acid
  • HI - hydroiodic acid
  • HClO4 - perchloric acid
  • Weak Acids (Carboxylic Acids)
  • A weak acid only partially dissociates in
    water
  • CH3COOH(aq) gtH1 (aq) CH3COO- (aq)
  • 1moldm-3 x moldm-3
    x moldm-3
  • Examples of weak acids include hydrofluoric
    acid, HF, and
  • acetic acid, CH3COOH.

3
  • Strong Bases
  • The hydroxides of the Group I and Group II metals
    usually are considered to be strong bases.
  • Ionize completely
  • NaOH(aq) gt Na(aq) OH- (aq)
  • LiOH - lithium hydroxide
  • NaOH - sodium hydroxide
  • KOH - potassium hydroxide
  • RbOH - rubidium hydroxide
  • Ca(OH)2 - calcium hydroxide
  • Ba(OH)2 - barium hydroxide
  • Weak Bases ( AMINES)
  • Examples of weak bases include ammonia, NH3,
  • Ionize partially
  • NH3 H2 Ogt NH41 OH-
  • Methylamine and diethylamine, (CH3CH2)2NH.

4
  • Which is not a strong acid?
  • A. Nitric acid
  • B. Sulfuric acid
  • C. Carbonic acid
  • D. Hydrochloric acid

5
Properties of Acids (page 145)
  • 1.Produce H (as H3O) ions in water
  • HCl (aq) gt H (aq) Cl (aq)
  • 2.Taste sour
  • 3.Corrode metals
  • Zn(s) HCl(aq)gt ZnCl2 (aq) H2 (g)
  • 4.Electrolytes
  • 5.React with bases to form a salt and water
  • HCl NaOH gt NaCl H2 O
  • 6.pH is less than 7
  • 7.Turns blue litmus paper to red
  • 8. React with carbonates and bicarbonates to
    produce carbon dioxide.
  • CaCO3 (s) 2HCl(aq) gt CaCl2 (aq) H2
    O(l) CO2 (g)

6
Properties of Bases (page 146)
  • Generally produce OH- ions in water
  • NaOH gt Na OH-
  • Taste bitter, chalky,soapy,slippery
  • Are electrolytes
  • Displacement of ammonia from ammonium salts
  • React with acids to form salts and water
  • pH greater than 7
  • Turns red litmus paper to blue

7
  • Which substance, when dissolved in water, to give
    a 0.1 mol dm- solution, has the highest pH?
  • A. HCl
  • B. NaCl
  • C. NH3
  • D. NaOH

8
Definitions
  • Arhenius
  • Bronsted Lowry
  • Lewis

9
Arrhenius Definition
  • Arrhenius
  • Acid - Substances in water that increase the
    concentration of hydrogen ions (H).
  • HCl(g) gt H1 (aq) Cl-1 (aq)
  • Base - Substances in water that increase
    concentration of hydroxide ions (OH-).
  • NaOH(s) gt Na1 (aq) OH-1 (aq)

10
Bronsted-Lowry Definition
  • Acids are species that donate a proton (H).
  • HNO3 (aq) H2O(l) gt NO3-(aq) H3O(aq)
  • NO3- is called the conjugate base of the acid
    HNO3, and H3O is the conjugate acid of the base
    H2O

11
  • Bases are species that accept a proton.
  • NH3 (aq) H2O(l) gt NH4(aq) OH-(aq)
  • NH3 is a base and H2O is acting as an acid. NH4
    is the conjugate acid of the base NH3, and OH- is
    the conjugate base of the acid H2O.
  • A compound that can act as either an acid or a
    base, such as the H2O in the above examples, is
    called amphoteric

12
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.
13
Bronsted-Lowry Acid Base Systems
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-
14
Lewis Definition
  • A Lewis acid is defined to be any species that
    accepts lone pair electrons.
  • A Lewis base is any species that donates lone
    pair electrons.
  • H OH-1 gt H2O

15
Auto Ionization of Water http//www.wwnorton.
com/college/chemistry/gilbert2/tutorials/interface
.asp?chapterchapter_16folderself_ionization
  • self-ionization of water
  • H2O (l) gt H (aq) OH- (aq)
  • or
  • 2 H2O (l) gt H3O (aq) OH- (aq)
  • hydronium ion
  • It is an example of autoprotolysis, and relies on
    the amphoteric nature of water.

16
Kw
  • The autoionization of water in equilibrium
  • H2O(l) ? H(aq) OH-(aq)
    ?Hgt 0
  • The equilibrium constant expression for this
    reaction is given by
  • Keq H OH-/H2 O Kc x H2O Kw
    1.0 x 10-14

  • ion product for water
  • The value for Kw is for room temperature, 25 C,
    and 1.0 atm
  • The equilibrium constant expression applies not
    only to pure (distilled) water but to any aqueous
    solution. It can be used to calculate either H
    or OH- provided one of them is known.

17
Kw and T http//mmsphyschem.com/autoIon.htm
  • H2O(l) ? H(aq) OH-(aq) ?Hgt 0
  • T Kw
  • 0 1.14 x 10-15
  • 5 1.85 x 10-15
  • 10 2.92 x 10-15
  • 15 4.53 x 10-15
  • 20 6.81 x 10-15
  • 25 1.01 x 10-14
  • The ionization of water is endo so,
  • as the temperature increases,so does the Kw

18
Formulas
  • pH -log H
    pKa -log Ka
  • pOH -log OH
    pKb -log Kb
  • pH pOH 14
    pKb pKa 14
  • H OH 1 x 10 -14 Ka x Kb
    Kw
  • Kw 1.0 x 10-14 pKw 14 pKw
    pH pOH

19
pH Formulas
20
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21
  • Lime(calcium hydroxide) is added to a lake to
    neutralize the effects of acid rain. The pH value
    of the lake water rises from 4 to 7. What is the
    change in concentration of H in the lake
    water?
  • A. An increase by a factor of 3
  • B. An increase by a factor of 1000
  • C. A decrease by a factor of 3
  • D. A decrease by a factor of 1000

22
The pH Scale

23
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

24
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.
25
Indicators Substances that change color when the
concentration of hydrogen changes.
26
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

27
Neutralization Problems
  • moles acid moles base
  • 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 M acid V base M base

28
Neutralization Problems
  • Example 1
  • 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?

29
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
  • moles base 2 x moles acid

30
Neutralization Problems
  • Example 3
  • 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

31
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

32
Neutralization Problems
  • Example 4
  • 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?

33
  • YEAR 2

34
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 Ka and
Kb The stronger the acid or base, the larger the
value of the dissociation constant.
35
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36
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

37
Acid Dissociation Constants
  • Dissociation constants for some weak acids

38
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.

39
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

40
Weak Acid Equilibrium Constants
  • Sample problem . A certain weak acid dissociates
    in water as follows
  • HF H2O ?? H3O F-
  • If the initial concentration of HF is 1.5 M and
    the equilibrium concentration of H3O is 0.0014
    M. Calculate Ka for this acid

41
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

42
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

43
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

44
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

45
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

46
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-

47
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.

48
Buffers
  • A buffer solution is na aqueous solution that
    resists a change in pH when a small amount of
    acid, base or water is added to it.
  • Acidic Buffers Weak acid and its salt
  • Can be prepared by using excess of a weak acid
    and a strong base .
  • NaOH(aq) CH3 COOH(aq) gt CH3 COONa(aq) CH3
    COOH(aq) H2 O (l)

  • salt excess weak acid
  • CH3COOH (aq) gt CH3COO- (aq) H (aq)
    Ka 1.74 x 10-5
  • A CB
  • pH pKa log CB / A Henderson
    Hasselbach Equation

49
Formulas
  • pH -log H pKa -log Ka
  • pOH -log OH pKb -log Kb
  • pH pOH 14 pKb pKa 14
  • H OH 1 x 10 -14 Ka x Kb
  • Kw 1.0 x 10-14 pKw 14

50
  • 1. Given 30 cm3 of 0.1 M ethanoic acid and 10
    cm3 of 0.1 M NaOH, find the pH of the buffer
    solution.
  • CH3COOH (aq) gt CH3COO- (aq) H (aq) Ka
    1.74 x 10-5
  • A CB
    pKa -log (1.74 x 10-5
    )4.76
  • n CH3COOH 0.1 x 0.03 0.003
  • n NaOH 0.1 x 0.01 0.001
  • excess CH3COOH after titration 0.002
  • pH pKa log CB / A
  • pH 4.76 log(0.001/0.002) 4.5

51
  • 2. 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)

52
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

53
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

54
  • Basic Buffers Weak base and its salt
  • Can be prepared by using excess of a weak base
    and a strong acid .
  • HCl(aq) CH3 NH2(aq) gt CH3 NH2Cl(aq) CH3
    NH3 (aq) H2 O (l)
  • salt
    excess base
  • CH3NH2 (aq) H2 O gt CH3NH3 (aq) OH- (aq)
    Kb 4.37 x 10-4
  • B CA
  • pOH pKb log CA / B
    Henderson Hasselbach Equation

55
  • Find the pH of a buffer made with 0.025 M
    methylamine and 0.010 M HCl.

56
  • 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.

57
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

58
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
59
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

60
Titration Curves
  • Simulations http//chem-ilp.net/labTechniques/Aci
    dBaseIdicatorSimulation.htm
  • 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
  • The end point is when the indicatorchanges
    color acurate to the addition of one dro.

61
Equivalence Point x End Pointhttp//www.chem.ubc.
ca/courseware/pH/section15/index.html
  • The endpoint of a titration is NOT the same thing
    as the equivalence point.
  • The equivalence point is a single point defined
    by the reaction stoichiometry as the point at
    which the base (or acid) added exactly
    neutralizes the acid (or base) being titrated.
  • The endpoint is defined by the choice of
    indicator as the point at which the colour
    changes. Depending on how quickly the colour
    changes, the endpoint can occur almost
    instantaneously or be quite wide.
  • Intuition may suggest that the endpoint of the
    titration will occur at the equivalence point if
    we choose an indicator whose pKa is equal to the
    pH of the equivalence point

62
Indicators Substances that change color when
the concentration of hydrogen changes. The choice
of an indicator is determined by the pH of the
solution at the equivalence point
http//www.avogadro.co.uk/chemeqm/acidbase/titrati
on/phcurves.htm
63
  • An indicator is a solution of a weak acid in
    which the the conjugate base has a different
    color from that of the undissociated acid
  • Given na aqueous solution of na indicator HIn(aq)
  • HIn(aq) ? H(aq) In-(aq)
  • Color A Color
    B
  • (in acid solution) ( in basic
    solution)
  • The end point of the indicator will be when pH
    pkIN
  • At low pH we see color A
  • At high pH we see color B

64
Strong acid-strong base
  • 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
  • Indicators methyl red and phenophtalein
  • Change between 4 and 10

65
  • a strong acid-strong base has a very sharp
    equivalence point, meaning that a very large
    change in pH occurs due to the addition of a very
    small amount of titrant, often a single drop.
  • . This means that any indicator that starts to
    change colour in this range will signal equally
    well that the equivalence point has been reached.
  • Any acid-base indicator that changes colour
    between pH 4 and pH 10 is suitable to detect the
    end-point for a strong acid - strong base
    titration. Both methyl orange and phenolphthalein
    could be used. Just one drop of the added base
    will bring about a change in colour of the
    indicator.

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Weak acid-strong base
  • 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
  • CH3COOH NaOH
  • Ka
  • Indicator phenophtalein

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WEAK ACID - STRONG BASE
NaOH(aq) HF(aq) ? NaF(aq) H2O(l)
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Strong Acid X Weak Base
  • The equivalence point is below 7 because the salt
    (NH4Cl) formed at the neutralization reacts with
    water to give H ions. The equivalence point lies
    at about pH 5.3. It is, therefore necessary to
    use an indicator with pH range slightly on the
    acidic side. Methyl orange can be used.
    Phenolphthalein is not suitable because its
    colour change occurs away from the equivalence
    point.

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Weak Acid Weak Base
  • There will hardly be a pH change around the
    equivalence point

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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

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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-

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SALTS Hydrolysis
  • Salts are ionic and already completely
    dissociated. They are also strong electrolytes.
  • I. When a salt comes from a strong acid and a
    strong base, they form neutral solutions when
    they dissolve in water.
  • NaCl(aq) gt Na(aq) Cl-(aq)

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  • II. When a salt comes from a weak acid and a
    strong base, it forms na alkaline solution when
    it dissolves in water.
  • CH3COOH(aq) NaOH(aq) gt CH3COONa(aq)
    H2O(l)
  • CH3COOH(aq) gt Na(aq) CH3COO-(aq)

  • H2O(l) gt OH-(aq) H(aq)

  • ?

  • CH3COOH(aq)

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  • III. When a salt comes from a strong acid and a
    weak base, it will be acidic in solution.
  • HCl(aq) NH3(aq) gt NH4Cl(aq)
  • NH4Cl(aq) gt NH4(aq)
    Cl-(aq)
  • H2O(l) gt OH-(aq)
    H(aq)
  • ?
  • NH3(aq)

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http//home.clara.net/rod.beavon/AlCl3_and_water
.htm
  • The acidity of a salt also depends on the size
    and charge of the anion.
  • Aluminum chloride reacts vigorously with water to
    produce a strong acidic solution.
  • AlCl3(s) 3H2O(l) ? Al(OH )3 (s) 3HCl(g)
    amphoteric nature
  • The charge of aluminum is spread over the small
    ion giving it a high charge density. It will then
    attract six pair of electrons forming the
    hexahydrated ion CC page 155
  • http//en.wikipedia.org/wiki/Hydrolysis
  • The greater the charge of nthe ion or
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