General Chemistry

3. STRONG BASES
All the hydroxides of the of the Group 1 and 2
elements (LiOH, NaOH, KOH, RbOH, CsOH,
Ca(OH)2, Ba(OH)2, Sr(OH)2 ) are strong bases,
but only NaOH and KOH are common
laboratory reagents.
• The alkaline earth hydroxides are not very
soluble and are used only when the solubility
factor is not important.
• Low solubility of bases can sometimes be an
advantage.

4. PRACTICE PROBLEM 12
Calculate the pH of a 5.0x10-2 M NaOH solution.
Major species:
Na+, OH-, and H2O
Narrow down contributing species and its
concentration. Write balanced equation.
[OH-] = 5.0x10-2 M
pOH
1.3 pH=12.70

5. BASES
Many types of proton acceptors do not
contain the hydroxide ion. However,
when dissolved in water, these
substances increase the concentration
of hydroxide ion because of their
reaction with water.
NH3(aq) + H2O(l) Û NH+
4(aq) +OH-
(aq)

6. BASES
Kb refers to the equilibrium expression of
the reaction of a base with water to form
the conjugate acid and the hydroxide
ion.
• pH calculations for weak bases are
similar to weak acids.

7. BASES
Other bases that produce the hydroxide
ion by reaction with water:

8. PRACTICE PROBLEM 13
Calculate the pH for a 15.0 M solution of
NH3 (Kb=1.8 x 10-5)
pH=12.20

9. PRACTICE PROBLEM 14
Calculate the pH of a 1.0 M solution of
methylamine (Kb= 4.38 x 10-4)
pH = 12.32

11. POLYPROTIC ACIDS
An acid that can furnish more than one proton
are called polyprotic acids.
• A polyprotic acid dissociates in a stepwise
manner, one proton at a time.
• Each dissociation of a proton, gives its
own Kavalue.
• The conjugate base in the first step is the acid
in the second step.

12. POLYPROTIC EXAMPLE
Triprotic:
H3PO4  H+ + H2PO4
- ; Ka1 = 7.5 x 10-3
H2PO4
-  H+ + HPO4
2- ; Ka2 = 6.2 x 10-8
HPO4
2-  H+ + HPO4
2- ; Ka3 = 4.8 x 10-13
• Ka1 > Ka2 > Ka3
• Each acid involved in dissociation steps
is successively weaker.

13. POLYPROTIC ACIDS

14. POLYPROTIC ACIDS
Most polyprotic acids have very different
successive Ka values.
• Typically, the first dissociation step is the
only step to make an important
contribution to [H+]

15. POLYPROTIC ACIDS
Sulfuric acid is unique among the common
polyprotic acids.
• Sulfuric acid is a strong acid in its first
dissociation step and a weak acid in the
second step.
• When acid concentration is below 1.0M,
both steps contribute to the overall
contribution of [H+]

16. PRACTICE PROBLEM 15
Calculate the pH of a 5.0 M H3PO4 solution
and the equilibrium concentrations of
the species H3PO4, H2PO4
-, HPO4
2-, and
PO4
3-
pH=0.72, [H+]=[H2PO4
-]=0.19M,
[H3PO4]=4.8M,
[HPO4
2-]=6.2x10-8M, [PO4
3-]=1.6x10-19M

17. PRACTICE PROBLEM 16
Calculate the pH of a 1.0 M H2SO4 solution.
[H+]=1.0M, pH=0.00

18. POLYPROTIC SUMMARY
1. Typically, successive Ka values are so much
smaller than the first value that only the first
dissociation step makes a significant
contribution.
2. Sulfuric acid is unique. At 1.0M and higher
the large concentration of H+ from the first
dissociation step represses the second step
and the second step is negligible. For dilute
solutions, the second step does make a
significant contribution and the quadratic
equation must be used to obtain the total H+
concentration.

19. PRACTICE PROBLEM 17
Calculate the pH of a 1.00 x 10-2 H2SO4
solution.
pH = 1.84

21. PROPERTIES OF SALTS
Ionic compounds, also known as salts, can
dissolve in water and under certain
conditions, these ions can behave as
acids or bases.
• Respective partners of strong acids and
strong bases, do not combine with H+ or
OH- and therefore have no effect on pH in
an aqueous solution.
• examples: K+, Na+, Cl-, NO3
-,

22. BASIC SALTS
Ions that are the conjugate base of a weak
acid are strong bases. These ions
produce hydroxide in an aqueous
solution.
• example: C2H3O2
-, F-
C2H3O-
2(aq) + H2O(l) Û HC2H3O(aq) +OH(aq)
-

23. KB AND KA
If Ka is known for a weak acid the Kb for its
conjugate base can be found.
• example: C2H3O2
Ka x Kb =
[H+
][[C2H3O2
-
]
[HC2H3O2 ]
x
[OH-
][[HC2H3O2
-
]
[C2H3O2
-
]
= [H+
][OH-
] = Kw
Ka x Kb = Kw

24. BASIC SALTS SUMMARY
For any salt whose cation has neutral
properties (such as Na+ or K+) and whose
anion is the conjugate base of a weak
acid, the aqueous solution will be basic.

25. PRACTICE PROBLEM 18
Calculate the pH of a 0.30 M NaF solution.
The Ka value for HF is 7.2x10-4.
pH=8.31

26. BASIC SALTS
HCN(aq) + H2O(l) Û H3O+
(aq) +CN(aq)
-
CN-
(aq) + H2O(l) ÛOH-
(aq) + HCN(aq)
In the first reaction CN- is competing with water for
the proton, Ka=6.2x10-10 :
In the second reaction CN- is competing with OH-
for the proton, Kb=1.6x10-5:
Generally, OH- > CN- > H2O

27. ACIDIC SALTS
For any salt whose anion has neutral
properties (such as Cl- or NO3
-) and
whose cation is the conjugate acid of a
weak base, the aqueous solution will be
acidic.
• example: NH4
+, CH3NH3
+

28. ACIDIC SALTS
A second type of salt that produces an acidic
solution is one that contains a highly charged
metal ion.
• Example: Al3+, Al(H2O)6
3+Al(H2O)5OH2++ H+
• The high charge on the metal ion polarizes the
O-H bonds in the water molecule, leaving an
acidic solution.
• Typically, the higher the charge on the metal ion,
the stronger the acidity of the hydrated ion.

29. PRACTICE PROBLEM 19
Calculate the pH of a 0.10 M NH4Cl solution.
The Kb value for NH3 is 1.8x10-5.
pH = 5.13

30. PRACTICE PROBLEM 20
Calculate the pH of a 0.010 M AlCl3
solution. The Ka value for Al(H2O)6
3+ is
1.4x10-5.
pH= 3.43

31. CONFLICTING SALTS
For many salts both ions can affect the pH of
the aqueous solution.
• We can predict whether the solution will
be basic, acidic, or neutral by comparing
the Ka value for the acidic ion with the Kb
value for the basic ion.

32. CONFLICTING SALTS
For many salts both ions can affect the pH of
the aqueous solution.
• The larger of the two constants determines
the characteristic of the solution: acidic,
basic or neutral.
• Ka > Kb =acidic
• Ka < Kb = basic
• Ka = Kb = neutral

33. PRACTICE PROBLEM 21
Predict whether an aqueous solution of each of
the following salts will be acidic, basic or
neutral.
a. NH4C2H3O2
b. NH4CN
c. Al2(SO4)3
a. neutral b. basic c. acidic

34. SALT PROPERTIES

36. STRUCTURE CONSIDERATIONS
Any molecule containing a hydrogen atom is
potentially an acid. Therefore, there are two
main factors that determine if the substance
will act like an acid and if so, what relative
strength it will have:
1. Bond polarities (electronegativity)
2. Number of oxygen atoms in the molecule

37. STRUCTURE CONSIDERATIONS
When looking at relative bond polarities of
binary acids we find:
H – F > H – Cl > H – Br > H – I
Electronegativity goes down the group,
therefore HF is extremely polar and very
strong. HF is the weakest of the acids.

38. STRUCTURE CONSIDERATIONS

39. STRUCTURE CONSIDERATIONS
Generally, oxyacids increase with strength
with an increase in the number of oxygen
atoms attached to the central atom.
• Example: HClO is a weak acid, but HClO4 is
strong.
• This happens because the very
electronegative oxygen atoms are able to
pull electrons away from the O-H bond and
weaken it.

40. STRUCTURE CONSIDERATIONS

41. STRUCTURE CONSIDERATIONS
Oxyacids with the H – O – X grouping, The
higher the electronegativity of X, the greater
the acidity of the molecule.

42. STRUCTURE CONSIDERATIONS

43. STRUCTURE CONSIDERATIONS
Oxyacids behave similarly to hydrated
metal ions.
• The acidity of the water molecules
attached to the metal ion is increased by
the attraction of electrons to the
positive metal ion.
• The greater the charge on the metal ion,
the more acidic the hydrated ion
becomes.

45. OXIDE PROPERTIES
Why is NaOH not an acid?
• The strength of the OH bond is stronger
than Na+ ability to bond to O.

46. OXIDE PROPERTIES
If an oxide with an H – O – X group has a
highly electronegative X, the H is lost
before the OH.
If an oxide with an H – O – X group has a
low electronegative X, OH- can be
formed instead of H+

47. OXIDE PROPERTIES
When a covalent oxide dissolves in water,
an acidic solution forms. These oxides
are called acidic oxides.
• SO3(g) + H2O(l) H2SO4(aq)
• CO2(g) + H2O(l) H2CO3(aq)

48. OXIDE PROPERTIES
When an ionic oxide dissolves in water, a
basic solution results. The most ionic
oxides, such as those of the Group 1 and 2,
produce basic solutions when they are
dissolved in water. These oxides are called
basic oxides.
• CaO(s) + H2O(l) Ca(OH)2(aq)
• K2O(s) + H2O(l) 2KOH(aq)

50. THREE MODELS FOR ACIDS AND BASES
Arrhenius is the most limiting model and was
replaced with a more general (Bronsted-
Lowry) model. An even more general model
was suggested in the 1920’s.

51. LEWIS ACID-BASE MODEL
A Lewis acid is an electron-pair acceptor,
and a Lewis base is an electron-pair
donor. In other words, a Lewis acid has
an empty atomic orbital that can accept
(share) an electron pair from a molecule
with a lone pair.

52. LEWIS THEORY AND COMPLEX IONS
The Al3+ ion accepts one electron pair from
each of six water molecules to form
Al(H2O)6
3+

53. LEWIS THEORY AND COVALENT OXIDES
Sulfur trioxide gains lone pairs from a
water molecule.
SO3(g) + H2O(l) Û H2SO4(aq)

54. PRACTICE PROBLEM 22
For each reaction, identify the Lewis acid and
base.
a. Ni2+
(aq) + 6NH3(aq)  Ni(NH3)6
2+
(aq)
b. H+
(aq) + H2O(aq)  H3O+
(aq)
a. Ni2+ = acid, NH3 = base b. H+ = acid, H2O =
base

55. THE END