Section 16
Acid-Base Equilibria
1
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Solutions of a Weak Acid or Base
• The simplest acid-base equilibria are
those in which a single acid or base
solute reacts with water.
– we will first look at solutions of pure weak
acids or bases.
– Next, we will also consider solutions of
salts, which can have acidic or basic
properties as a result of the reactions of
their ions with water. Salts can be neutral,
acidic, or basic.
2
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Ionization Equilibria
• Acid ionization (or acid dissociation) is
the reaction of an acid with water to
produce hydronium ion (hydrogen ion) and
the conjugate base anion (hydrolysis).
HC2 H 3O 2 (aq)  H 2O(l )


H 3O (aq)  C2 H 3O 2 (aq)
– Because acetic acid is a weak electrolyte, it ionizes to a
small extent in water, hence double arrow; not 100% like
strong acid which wrote single arrow.
3
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Ionization Equilibria
• For a weak acid, the equilibrium concentrations
of ions in solution are determined by the acidionization constant (also called the aciddissociation constant, Ka).
– Consider the generic monoprotic acid, HA.
HA(aq )  H 2O(l )


H 3O (aq )  A (aq )
4
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Ionization Equilibria

HA(aq)  H 2O(l )

H 3O (aq)  A (aq)
– The corresponding equilibrium expression is:


[H 3O ][A ]
Kc 
[HA][H 2O]
– Since the concentration of water remains relatively
constant (liquid = 1), we rearrange the equation to get:


[H 3O ][A ]
K a  [H 2O]K c 
[HA]
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
5
Acid-Ionization Equilibria
– Thus, Ka , the acid-ionization constant, equals the
constant [H2O]Kc.


[H 3O ][A ]
Ka 
[HA]
• acid + water is Ka
6
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Experimental Determination of Ka
• The degree of ionization of a weak
electrolyte is the fraction of molecules
that react with water to give ions.
– Electrical conductivity or some other colligative
property can be measured to determine the degree
of ionization.
– With weak acids, the pH can be used to determine
the equilibrium composition of ions in the solution
because it gives the concentration of hydronium ions
at equil.
7
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Nicotinic acid is a weak monoprotic acid with
the formula HC6H4NO2. A 0.012 M solution of
nicotinic acid has a pH of 3.39 at 25°C.
Calculate the acid-ionization constant for this
acid at 25°C.
– It is important to realize that the solution was made
0.012 M in nicotinic acid, however, some molecules
ionize making the equilibrium concentration of
nicotinic acid less than 0.012 M, just a small amount
less.
8
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Let x be the moles per liter of product formed.
HC6 H 4 NO2 (aq)  H 2O(l )
Starting
Change
Equilibrium

H3O (aq)  C6 H 4 NO2 (aq)
0.012
0
0
9
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– We can obtain the value of x from the given
pH.
10
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Substitute this value of x in our equilibrium expression.
– Note first, however, that
(0.012  x)  (0.012  0.00041)  0.01159  0.012
the concentration of unionized acid remains
virtually unchanged. Important point to
remember for later.
– Substitute this value of x in our equilibrium expression.
2
2
x
(0.00041)
5
Ka 

 1.4  10
(0.012  x)
(0.012)
HW 28
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
11
A Problem To Consider
– To obtain the degree of ionization:
% ionization 
[ H  ]acid ionized
[acid ]initial
x 100%
0.00041
% ionization 
x100%  3.4%
0.012
– weak acids typically less than 5%. The lower the %ionized, less H+
formed, the lower Ka, weaker the acid. Ka is measure of strength of
acid or Kb if base. Strength based on ionization not pH directly.
Lower pH more acidic, does not mean stronger acid
12
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Ka is a measure of strength of acid. It indicates the amount ionized to form
hydronium ions. The more hydronium ions formed, naturally the more
acidic and lower pH. However, strength of acid is measure of amount
ionized, not conc of acid. Conc can vary which ultimately will vary the pH
of solution. To determine the strength of an acid you must compare Ka
values not pH of solution. pH is dependent on Ka as well as conc. of
solution. Must compare apples and apples.
10M HC2H3O2 Ka = 1.8 x 10-5
0.1M HF
Ka = 6.9x10-4
pH = 1.87
pH = 2.08
Which stronger acid?
13
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Calculations With Ka
• Once you know the value of Ka, you can
calculate the equilibrium
concentrations of species HA, A-,
and H3O+ for solutions of different
molarities.
– The general method for doing this is same way we
did Kc or Kp except we can simplify the math
because of percent ionize is so small compared to
initial conc of acid; remember 0.012-x=0.012. This
wasn't the case for Kc or Kp problems so can't
neglect in those type problems
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
14
Calculations With Ka
• Note that in our previous example, the degree of
dissociation was so small that “x” was negligible
compared to the concentration of nicotinic acid.
• It is the small value of the degree of ionization that
allowed us to ignore the subtracted x in the
denominator of our equilibrium expression.
• The degree of ionization of a weak acid depends on
both the Ka and the concentration of the acid solution
15
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Calculations With Ka
• How do you know when you can use
this simplifying assumption?
– It can be shown that if the acid concentration, Ca,
divided by the Ka exceeds 100, that is,
if [HA]
Ka
 100
– then this simplifying assumption of ignoring the
subtracted x gives an acceptable error of less than 5%
otherwise must do quadratic formula. Another way
look at it factor of two - three difference in power of 10
is needed between Ka and conc of acid.
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
16
ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2,
for which Ka = 6.0 x 10-4.
HNO2 + H2O
<-->
H3O+ + NO2-
17
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Really two equil: Kw as well but neglect most
of the time as long as [H+] is greater than
10-6 M from the acid.
• Really [H+]tot = [H+]acid +[H+]H2O (water so
small neglect)
18
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted
to 100.0 mL with water for which Ka = 6.0 x 10-4.
HNO2 + H2O
<-->
H3O+ + NO2-
19
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• What is the pH at 25°C of a solution obtained
by dissolving 0.325 g of acetyl salicylic acid
(aspirin), HC9H7O4, in 0.500 L of water? The
acid is monoprotic and Ka=3.3 x 10-4 at 25°C.
– The molar mass of HC9H7O4 is 180.2 g.
20
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• What is the pH at 25°C of a solution obtained
by dissolving 0.325 g of acetylsalicylic acid
(aspirin), HC9H7O4, in 0.500 L of water? The
acid is monoprotic and Ka=3.3 x 10-4 at 25°C.
– These data are summarized below.
HC9 H 7O4 (aq)  H 2O(l )


H3O (aq)  C9 H 7O4 (aq)
Starting[] 0.003607
Change []
Equilibrium[]
0
0
21
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Note that
Ca
Ka
 0.0036
3.3  10
4
 11
which is less than 100 or tell factor 10-3 vs 10-4
not factor 2-3; therefore, won't work, so we
must solve the equilibrium equation exactly .
22
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
– If we substitute the equilibrium concentrations and the
Ka into the equilibrium constant expression, we get
HC9 H 7O4 (aq)  H 2O(l )


H3O (aq)  C9 H 7O4 (aq)
0.003607
Starting[]
Change []
-x
Equilibrium[] 0.003607 –x
0
0
+x
x
+x
x
23
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
– You can solve this equation exactly by using the
quadratic formula.
24
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Taking the upper sign, we get and neglect water
OK
– Now we can calculate the pH.
HW 29
25
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH.
26
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Polyprotic Acids
• Some acids have two or more protons
(hydrogen ions) to donate in aqueous
solution. These are referred to as polyprotic
acids.
– Sulfuric acid, for example, can lose two protons in
aqueous solution.


H 2SO 4 (aq )  H 2O(l )  H 3O (aq )  HSO 4 (aq )

HSO 4 (aq )  H 2O(l )

2
H 3O (aq )  SO 4 (aq )
27
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Polyprotic Acids
– For a weak diprotic acid like carbonic acid, H2CO3,
two simultaneous equilibria must be considered.
H 2CO3 (aq)  H 2O(l )

HCO 3 (aq )  H 2O(l )


H 3O (aq)  HCO3 (aq)

2
H 3O (aq )  CO 3 (aq )
28
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Polyprotic Acids
– Each equilibrium has an associated acid-ionization
constant.
– For the loss of the first proton


[H 3O ][HCO3 ]
K a1 
 4.3  107
[H 2CO 3 ]
– For the loss of the second proton

K a2
2
[H 3O ][CO3 ]
11


4
.
8

10

[HCO3 ]
29
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Polyprotic Acids
– In general, the second ionization constant, Ka2, for
a polyprotic acid is smaller than the first ionization
constant, Ka1. Harder to pull proton off charged
species. Total H+ is sum of all equil H+ but usually
neglect all but the principal rxn.
– In the case of a triprotic acid, such as H3PO4, the
third ionization constant, Ka3, is smaller than the
second one, Ka2.
30
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Polyprotic Acids
– When several equilibria occur at once, it might
appear complicated to calculate equilibrium
compositions.
– However, reasonable assumptions can be made
that simplify these calculations as we show in the
next example.
31
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Ascorbic acid (vitamin C) is a diprotic acid,
H2C6H6O6. What is the pH of a 0.10 M solution?
What is the concentration of the ascorbate ion,
C6H6O62- ?
Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.
– For diprotic acids, Ka2 is so much smaller than Ka1
that the smaller amount of hydronium ion produced in
the second reaction can be neglected.
32
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– The pH can be determined by simply solving the
equilibrium problem posed by the first ionization.
– the first ionization is:
H 2C6 H 6O6 (aq)  H 2O(l )
Starting
Change
Equilibrium
0.10


H3O (aq)  HC6 H 6O6 (aq)
0
0
33
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Assuming that x is much smaller than 0.10 (1265),
you get
34
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Ascorbic acid (vitamin C) is a diprotic acid,
H2C6H6O6. What is the pH of a 0.10 M
solution? What is the concentration of the
ascorbate ion, C6H6O62- ?
The acid ionization constants are Ka1 = 7.9 x
10-5 and Ka2 = 1.6 x 10-12.
– The ascorbate ion, C6H6O62-, which we will call y, is
produced only in the second ionization of H2C6H6O6.

HC6 H 6O6 (aq)  H 2O(l )

2
H3O (aq)  C6 H 6O6 (aq)
35
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
– Assume the starting concentrations for HC6H6O6- and
H3O+ to be those from the first equilibrium.

HC6 H 6O6 (aq)  H 2O(l )
Starting []
0.0028
2
H3O (aq)  C6 H 6O6 (aq)
0.0028
0
Change []
Equilibrium []
36
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
– Substituting into the equilibrium expression
– Assuming y is much smaller than 0.0028 (1.75 x 109 >
100), the equation simplifies to
– Hence,
– typically, the concentration of the anion
ion equals Ka2 if starting with reactants
of first equil.
HW 30
37
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Base-Ionization Equilibria
• Equilibria involving weak bases are
treated similarly to those for weak acids.
– Ammonia, for example, ionizes in water as
follows.

NH 3 (aq )  H 2O(l )

NH 4 (aq )  OH (aq )
– The corresponding equilibrium constant is:


[NH 4 ][OH ]
Kc 
[NH 3 ][H 2O]
38
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Base-Ionization Equilibria
• Equilibria involving weak bases are
treated similarly to those for weak acids.
– Ammonia, for example, ionizes in water as
follows.
NH 3 (aq )  H 2O(l )


NH 4 (aq )  OH (aq )
– The concentration of water is nearly constant.


[NH4 ][OH ]
K b  [H 2O]K c 
[NH3 ]
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
39
Base-Ionization Equilibria
• Equilibria involving weak bases are treated similarly to those for
weak acids. Higher Kb stronger base, ionize more OH-, more
basic.
– In general, a weak base B with the base ionization

B(aq)  H 2O(l )

HB (aq)  OH (aq)
has a base ionization constant equal to

HW 31

[HB ][OH ]
K b
[B ]
40
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• What is the pH of a 0.20 M solution of
pyridine, C5H5N, in aqueous solution?
The Kb for pyridine is 1.4 x 10-9.
– As before, we will follow the three steps in
solving an equilibrium.
1. Write the equation and make a table of
concentrations.
2. Set up the equilibrium constant expression.
3. Solve for x = [OH-].
41
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• What is the pH of a 0.20 M solution of
pyridine, C5H5N, in aqueous solution?
The Kb for pyridine is 1.4 x 10-9.
– Pyridine ionizes by picking up a proton from water
(as ammonia does).
C5 H 5 N(aq)  H 2O(l )
Starting
Change
Equilibrium
C5 H 5 NH  (aq)  OH  (aq)
0.20
0
0
42
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Note that
Cb
Kb
 0.20
1.4 10
9
 1.4 10
8
which is much greater than 100, so we may
use the simplifying assumption that
(0.20-x)  (0.20).
43
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
C5 H 5 N(aq)  H 2O(l )
C5 H 5 NH  (aq)  OH  (aq)
Starting
0.20
Change
-x
Equilibrium 0.20-x
0
+x
x
0
+x
x
– The equilibrium expression is
44
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Using our simplifying assumption that the x
in the denominator is negligible, we get
45
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
– Solving for pOH
– Since pH + pOH = 14.00
HW 32
46
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt Solution
• One of the successes of the BrønstedLowry concept of acids and bases was
in pointing out that some ions can act
as acids or bases.
– Consider a solution of sodium cyanide, NaCN.


NaCN(s) 
 Na (aq)  CN (aq)
H 2O
– A 0.1 M solution has a pH of 11.1 and is therefore
fairly basic.
47
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt Solution
– Sodium ion, Na+, is unreactive with water,
but the cyanide ion, CN-, reacts to produce
HCN and OH- making it a basic solution

CN (aq)  H 2O(l )

HCN(aq)  OH (aq)
– From the Brønsted-Lowry point of view, the
CN- ion acts as a base, because it accepts
a proton from H2O.
48
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt Solution
– You can also see that OH- ion is a product, so you
would expect the solution to have a basic pH. This
explains why NaCN solutions are basic.

CN (aq)  H 2O(l )

HCN(aq)  OH (aq)
– The reaction of the CN- ion with water is referred to
as the hydrolysis of CN-.
49
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt Solution
• The hydrolysis of an ion is the reaction of an
ion with water to produce the conjugate acid
and hydroxide ion or the conjugate base
and hydronium ion.
– The CN- ion hydrolyzes to give the
conjugate acid and hydroxide.

CN (aq)  H 2O(l )

HCN(aq)  OH (aq)
50
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt Solution
– The hydrolysis reaction for CN- has the form of a base
ionization so you write the Kb expression for it.

CN (aq)  H 2O(l )

HCN(aq)  OH (aq)
51
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Acid-Base Properties of a Salt
Solution
• An example of an cation that undergoes hydrolysis is
the ammonium ion


NH 4Cl  NH 4 (aq)  Cl (aq)
– The NH4+ ion hydrolyzes to the conjugate
base (NH3) and hydronium ion.

NH 4 (aq )  H 2O(l )
NH 3 (aq )  H 3O  (aq )
52
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Predicting Whether a Salt is Acidic,
Basic, or Neutral
• How can you predict whether a
particular salt will be acidic, basic, or
neutral and undergoes hydrolysis?
– The Brønsted-Lowry concept illustrates the
inverse relationship in the strengths of conjugate
acid-base pairs.
– Consequently, the anions of weak acids (poor
proton donors) are good proton acceptors.
– Anions of weak acids are basic.
53
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Predicting Whether a Salt is Acidic,
Basic, or Neutral
– On the other hand, the anions of strong acids
(good proton donors) have virtually no basic
character, that is, they do not hydrolyze.
– For example, the Cl- ion, which is conjugate to the
strong acid HCl, shows no appreciable reaction
with water. If did, strong acid breaks up 100% and
hydroxide produced cancelled with hydronium of
strong acid and make neutral solution

HCl (aq )  H 2O(l )  Cl  H 3O

Cl  (aq )  H 2O(l )  HCl  OH 
in essence,
Cl  (aq )  H 2O(l )  no reaction
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
54
Predicting Whether a Salt is Acidic,
Basic, or Neutral
– Small highly charged (attract O more weakens OH bond)
cations have an attraction for the O on water and cause
the production of H+ while large low charged cations do
not. Low charged large ions are group I and II metals
(cations of strong bases) and do not undergo
hydrolysis.
– For example,

Na (aq)  H 2O(l )  no reaction
In summary,
If cation is that of the strong base, it will have neutral
hydrolysis; otherwise, cation will have acidic hydrolysis
If anion of monoprotic strong acid, it will have neutral
hydrolysis; otherwise, anion will have basic hydrolysis
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
55
• Salts may be acidic/basic/neutral and are composed of
cations (positive ions) and anions (negative ions)
cation anion
Acidic or neutral
Cations of strong bases
are neutral; otherwise,
cation contributes acidity
basic or neutral
Anions of monoprotic strong
acids are neutral; otherwise,
anion contributes basicity
Na+, K+, Li+, Ca2+, Sr2+, Ba2+
add neutral; rest of cations add
acidity to salt
Cl-, Br-, I-, NO3-, ClO4- add
neutral; rest of anions add
basicity to salt
Depending on the two components (cation/anion) the overall
salt will be acidic/neutral/basic – nn  n; an  a; nb  b.
56
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Predicting Whether a Salt is Acidic,
Basic, or Neutral
• To predict the acidity or basicity of a
salt, you must examine the acidity or
basicity of the ions composing the salt.
57
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Predicting Whether a Salt is Acidic,
Basic, or Neutral
58
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Predicting Whether a Salt is Acidic,
Basic, or Neutral
•
•
•
•
•
•
•
•
•
AlCl 3
Zn(NO3)2
KClO4
Na3PO4
LiF
NH4F
Ka = 5.6x10-10 > Kb = 1.4x10-11
NH4ClO
Ka = 5.6x10-10 < Kb = 3.6x10-7
NH4C2H3O2
Ka = 5.6x10-10 = Kb = 5.6x10-10
one with larger K will dictate acidity/basicity - stronger
species
HW 33
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
59
The pH of a Salt Solution
• To calculate the pH of a salt solution
would require the Ka of the acidic cation
or the Kb of the basic anion.
– The ionization constants of ions are not listed
directly in tables because the values are easily
related to their conjugate species.
– Thus the Kb for CN- is related to the Ka for HCN.
60
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
The pH of a Salt Solution
• To see the relationship between Ka and Kb for
conjugate acid-base pairs, consider the acid
ionization of HCN and the base ionization of
CN-.
HCN(aq )  H 2O(l )
H 3O  (aq )  CN  (aq )
Ka
CN  (aq)  H 2O(l )
HCN(aq)  OH  (aq)
Kb
H 2O(l )  H 2O(l )


H 3O (aq)  OH (aq) Kw
– When two reactions are added, their equilibrium
constants are multiplied.
61
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
HCN(aq )  H 2O(l )

CN (aq)  H 2O(l )
H 2O(l )  H 2 O(l )


H 3O (aq )  CN (aq )

Ka
HCN(aq)  OH (aq)
Kb
H 3O  (aq)  OH  (aq)
Kw
– Therefore,
K a  K b  K w  1.00 x1014 @ 25o C
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
HW 34
62
The pH of a Salt Solution
• For a solution of a salt in which only
one ion hydrolyzes, the calculation of
equilibrium composition follows that of
weak acids and bases.
– The only difference is first must obtain the
Ka or Kb for the ion that hydrolyzes.
63
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Attack of hydrolysis problem
When attacking pH (hydrolysis problems) always ask
yourself if acid, base, or salt -- if acid or base, is it strong
(straight forward) or weak (Ka or Kb problem) -- If salt, write
dissociation eq and examine cation/anions involved to
determine if acid or base hydrolysis is available then Ka or Kb
problem.
64
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• What is the pH of a 0.10 M NaCN solution
at 25 °C? The Ka for HCN is 4.9 x 10-10.
65
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
66
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
example: What is the pH of 0.10 M NH4I solution?
Kb NH3 = 1.8 x 10-5
HW 35
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
67
The Common Ion Effect
• The common-ion effect is the shift in
an ionic equilibrium caused by the
addition of a solute that provides an ion
common to the equilibrium.
HC2 H 3O 2 (aq)  H 2O(l )

H 3O  (aq)  C2 H 3O 2 (aq)
68
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
The Common Ion Effect

NaC2 H 3O2  Na (aq)  C2 H 3O2 (aq)
HC2 H 3O 2 (aq)  H 2O(l )


H 3O (aq)  C2 H 3O 2 (aq)
– The equilibrium composition would shift to the left
and the degree of ionization of the acetic acid is
decreased, meaning less hydronium formed and
higher pH of solution as compared to pure acid.
– This repression of the ionization of acetic acid by
sodium acetate is an example of the common-ion
effect. Another source of ion coming from an
additional solute added to the solution.
69
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Buffers
• When common ion effect involves weak
acid/conjugate base it is referred to as a buffer.
Buffer is a solution characterized by the ability to
resist changes in pH when limited amounts of acid or
base are added to it.
– Buffers contain either a weak acid and its conjugate
base or a weak base and its conjugate acid.
Typically conj is in form of salt.
– Thus, a buffer contains both an acid species and a base
species in equilibrium which allows it to absorb any
added acid or base and keep pH of solution relatively
constant
70
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
Buffers
– Consider a buffer with equal molar amounts of HA
and its conjugate base A-.
71
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
•
An aqueous solution is 0.025 M in formic acid, HCHO2 and 0.018 M
in sodium formate, NaCHO2. What is the pH of the solution. The Ka
for formic acid is 1.7 x 10-4.
72
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
73
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
• A solution is prepared to be 0.15 M HC2H3O2 and 0.20 M
NaC2H3O2. What is the pH of this solution at 25oC?
Ka HC2H3O2 @ 25oC = 1.8 x 10-5
74
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
• Calculate the pH of 75 mL of a buffer solution (0.15 M
HC2H3O2 / 0.20 M NaC2H3O2 pH = 4.87) to which 9.5 mL of
0.10 M HCl is added. Ka HC2H3O2 @ 25oC = 1.8 x 10-5
75
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
HW 36
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
76
Buffers
– Two important characteristics of a buffer are its
buffer capacity and its pH.
– Buffer capacity depends on the amount of acid
and conjugate base present in the solution.
– The next topic illustrates how to calculate the pH
of a buffer by special eq but can do like we just did
as well to keep with thought process and not
memorize a equation.
77
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
The Henderson-Hasselbalch Equation
• How do you prepare a buffer of given pH?
– A buffer must be prepared from selecting a conjugate
acid-base pair in which the Ka of the acid is
approximately equal to the desired H3O+
concentration.
– Note: you can add extra conjugate base, higher pH, or
less base, lower pH, to get to exact pH wanted.
– To illustrate, consider a buffer of a weak acid HA
and its conjugate base A-.
The acid ionization equilibrium is:
HA(aq )  H 2O(l )


H 3O (aq )  A (aq )
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
78
The Henderson-Hasselbalch Equation
HA(aq )  H 2O(l )


H 3O (aq )  A (aq )
– The acid ionization constant is:


[H 3O ][A ]
Ka 
[HA]
– By rearranging, you get an equation for the H3O+
concentration.
[HA]
[H 3O ]  K a  
[A ]

Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
79
The Henderson-Hasselbalch Equation
– Taking the negative logarithm of both sides of the
equation we obtain:
[ HA]
- log[H 3O ]   log( K a   )
[A ]

[ HA]
- log[H 3O ]   log( K a )  ( log  )
[A ]

– The previous equation can be rewritten

[A ]
pH  pK a  log
[HA]
80
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
The Henderson-Hasselbalch Equation
– More generally, you can write
[base, A ]
pH  pK a  log
[acid , HA]
– This equation relates the pH of a buffer to the
concentrations of the conjugate acid and base. It
is known as the Henderson-Hasselbalch
equation (for buffers only). Also way to work
previous buffer problem. Note pH = pKa when
conc base=acid
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
81
• Let’s prepare a buffer with pH of 4.90
– So to prepare a buffer of a given pH (for example,
pH 4.90) we need a conjugate acid-base pair with
a pKa close to the desired pH or 10-4.90 to find Ka
close to 1.25 x 10-5 (desired hydronium conc.
equivalent to pH desired)
HC2H3O2
Ka = 1.8 x 10-5
H2CO3
Ka = 4.3 x 10-7
HNO2
Ka = 4.5 x 10-4
– The Ka for acetic acid is 1.8 x 10-5, and its pKa is
4.74 or pH of same value if equal conc of acid/salt
are used.
– You could get a buffer of pH 4.90 by increasing the
ratio of [base]/[acid], meaning more basic salt.
82
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
What if I wanted to use 0.100 M HC2H3O2 (Ka = 1.8 x 10-5), what
concentration of NaC2H3O2 would I need to obtain a buffer with pH =4.90.
HW 37
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
83
Acid-Ionization Titration Curves (or mixtures)
• An acid-base titration curve is a plot of the pH of a
solution of acid (or base) against the volume of
added base (or acid).
– Such curves are used to gain insight into the
titration process.
– You can use titration curves to choose an
appropriate indicator that will show when the
titration is complete.
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
84
Figure: Curve for the titration of a
strong acid by a strong base.
• Figure shows a curve for the titration of HCl with NaOH.
– Note that the pH changes slowly until the titration
approaches the equivalence point.
– The equivalence point is the point in a titration
when a stoichiometric amount of reactant has
been added; contrary to end point (observable
point).
– At this equivalence point of SA and SB, the pH of
the solution is 7.0 because it contains a salt,
NaCl, that does not hydrolyze.
– However, the pH changes rapidly from a pH of
about 3 to a pH of about 11.
– To detect the equivalence point, you need an
acid-base indicator that changes color beyond
equivalence point.
– Phenolphthalein can be used because it changes
color in the pH range 8.2-10.
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
86
Previous example equivalent pt was at 7 but not all neutralization reactions
eq pts are at 7. During a titration or just mixing an acid and base together to
make a solution, the resulting solution will contain a salt and water
(Acid + Base --> salt + water). One of the chemical properties of acids and
bases is that they neutralize one another; however, that doesn’t mean that the
product will be neutral.
The misconception is that if the acid and base are in stoich proportions that
the resulting solution is neutral. This is not true. The salt formed may be a
acidic, basic, or neutral salt and will dictate the pH of the solution. Common
sense can help you predict the pH of a mixture.
SA + SB --> neutral salt
WA + SB --> basic salt
SA + WB --> acidic salt
only true neutralization pH =7
original acid and base neutralize but
product has basic properties and basic pH
original acid and base neutralize but
product has acidic properties and acidic pH
87
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
When working acid-base titration problems or mixtures treat
them all the same but must be able to examine the components
remaining after neutralization to determine what affects pH or not
(strong, weak, salts). They are all done the same. The problem will
dictate where you go. We can derive titration curve by doing a
calculation at different points and draw curve or measure with pH
meter.
1.) Calculate mmols initially
2.) write the neutralization reaction and find mmols of
components left after neutralization
3.) find new conc. of species that affect pH
4.) do hydrolysis of main species that affect pH (like earlier in chapter)
88
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Calculate the pH of a solution in which 50.0 mL of 1.000 M
NaOH is added to 50.0 mL of 1.000 M HCl.
89
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Calculate the pH of a solution in which 49.99 mL of 1.000 M
NaOH is added to 50.00 mL of 1.000 M HCl.
90
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Calculate the pH of a solution in which 50.01 mL of 1.000 M
NaOH is added to 50.00 mL of 1.000 M HCl.
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
HW 38
91
A Problem To Consider
• Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is
added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
92
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• continue
93
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Calculate the pH of a solution in which 50.00 mL of 1.000 M NaOH is
added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
94
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• continue
95
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• Calculate the pH of a solution in which 51.00 mL of 1.000 M NaOH is
added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
96
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
A Problem To Consider
• continue
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.
HW 39
97