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

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

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• 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.

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• 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.

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

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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)

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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

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• 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

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Definitions

• Arhenius
• Bronsted Lowry
• Lewis

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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)

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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

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• 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

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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.
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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-
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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

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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.

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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.

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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

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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

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pH Formulas
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• 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

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The pH Scale

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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

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

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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

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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?

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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

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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

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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

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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?

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• YEAR 2

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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.
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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

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Acid Dissociation Constants

• Dissociation constants for some weak acids

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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.

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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

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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

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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

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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

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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

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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

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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

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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-

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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.

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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

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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

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• 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

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• 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)

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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

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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

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• 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

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

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• 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.

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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

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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
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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

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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.

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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

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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
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• 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

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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

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• 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