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Unit 12: Acids and Bases
Name:
Acids and Bases
1.
Two types of chemicals which have nearly opposite chemical characteristics are acids and bases.
They have been known of for some time, mainly because of a collection of contrasting observable
properties which made them distinguishable from one another. In the early 1880s, Svante Arrhenius
made the first effort to distinguish between acids and bases in terms of their underlying chemistry.
The Arrhenius definition of an acid was that an acid is a substance which produces hydrogen ions
when it dissolves in water. And for Arrhenius, a base is a substance which produces hydroxide ions
when it dissolves in water. (See Masterton and Hurley, pp. 79-80)
Using Arrhenius' definition, predict whether the following substances are acids (A), bases (B) or
neither (N) based solely upon the chemical formula. Place an A, B or an N in the blank.
2.
HCl
NaOH
Ca(OH)2
CaCl2
HNO3
NH3
NH4 OH
HF
C5H5N
For Arrhenius, acids and bases were electrolytes - substances which dissociated into ions when
dissolved in water. Write the balanced chemical equation for the dissociation of the following acids
and bases.
a. HNO3 (aq)


b. NaOH (aq)
c. HCN (aq)

d. Ca(OH)2 (aq)
3.

Review: Rules for naming acids were presented in Chapter 2 (p. 42). The naming rules depend on
whether the anion part of the acid contains an oxygen atom.
If anion part of the acid does not contain an oxygen atom, then the name takes the form of
hydro
(anion root name) -ic acid. Thus HF is named as hydrofluoric acid and HCN is named
as hydrocyanic acid.
If the anion part of the acid contains an oxygen atom, then the name takes the form of either
(anion root name) -ic acid or
(anion root name)
-ous acid. The -ic suffix is used for
anions like acetate, phosphate, and chlorate and the -ous suffix is used for anions like sulfite,
phosphite, and nitrite. Table 2.2 and Table 2.3 on pp. 39-40 are well worth the look.
4.
Like salts (ionic compounds), acids and bases can be strong electrolytes or weak electrolytes. They
can completely dissociate in water (strong acids and strong bases) or only partially dissociate in
water (weak acids and weak bases). You should know the names and formulas of the following six
strong acids and strong bases. (See Masterton and Hurley, p. 81)
Formula
HCl
Name
Formula
LiOH
hydrobromic acid
NaOH
HI
Name
potassium hydroxide
HNO3
Ca(OH)2
sulfuric acid
strontium hydroxide
perchloric acid
barium hydroxide
Page 1
Unit 12: Acids and Bases
5.
For the following strong acids and strong bases and their given molarity, write the dissociation
reaction and calculate either the hydrogen ion or hydroxide ion concentration. Show your work.
a. A 6.0 M HCl(aq) solution:
[H+] =
b. A 3.0 M NaOH(aq) solution:
[OH-] =
c. A 1.0 M HBr(aq) solution:
[H+] =
d. A 1.0 M Ca(OH)2(aq) solution:
[OH-] =
6.
Recognizing that the Arrhenius definition of acids and bases had several shortcomings, two
scientists (Johannes Bronsted and Thomas Lowry) working independently proposed new
definitions. The Bronsted-Lowry defintions are: an acid is a proton donor and a base is a proton
acceptor. (Masterton and Hurley, pp. 353-354) When dissolved in water, an acid donates a proton
(H+ ion) to another substance. When dissolved in water, a base accepts a proton (H+ ion) from
another substance. Use these definitions to write the balanced chemical equations for the weak acid
and weak base dissociations in water.
+
HCN(aq)
b.
HF(aq)
c.
Dissociation of the weak acid: HNO2 :
+
H2O(l)

a.
H2O(l)

d. Dissociation of the weak base NH3 (ammonia):
e. Dissociation of the weak base C5H5N (pyridine):
Page 2
Unit 12: Acids and Bases
7.
Name:
The Bronsted-Lowry theory of acids and bases perceives every acid-base reaction as being the result
of the deprotonation of an acid and the protonation of a base. That is, there is an acid which gives up a
proton (H+) and a base which receives the proton (H+). When dissolved in water, it is the water
which is doing the accepting (when with an acid) or the donating (when with a base) of the proton.
The general equation for the acid-base reaction is:
HA
acid
+
B

base
A-
conjugate base
+
BH+
conjugate acid
As noted beneath the equation above, the reaction of an acid with a base changes the acid into a
conjugate base and changes the base into a conjugate acid. An acid has one more proton than its
conjugate base; a base has one less proton than its conjugate acid. There are always two sets of
conjugate acid-base pairs in an acid-base reaction. (Masterton and Hurley, pp. 353, 359-360)
Identify the acid (A), base (B), conjugate acid (CA) and conjugate base (CB) in the following acid
dissociation, base dissociation and acid-base reactions. Place an A, B, CA or CB in the blanks.
8.
a.
NH3
_______
+
H2O
_______

NH4+
_______
+
OH_______
b.
HCN
_______
+
H2O
_______

CN_______
+
H3O+
_______
c.
HF
_______
+
H2O
_______

F_______
+
H3O+
_______
d.
C5H5N
_______
+
H2O
_______

C5H5NH+
_______
+
OH_______
e.
NH4+
_______
+
H2O
_______

NH3
_______
+
H3O+
_______
f.
HCO3_______
+
H2O
_______

CO32_______
+
H3O+
_______
g.
HCO3_______
+
HF
_______

H2CO3
_______
+
F_______
f.
HSO4_______
+
H2O
_______

SO42_______
+
H3O+
_______
g.
HSO4_______
+
HCN
_______

H2SO4
_______
+
CN_______
Some substances are amphiprotic or amphoteric. They can act as either an acid or a base. The most
common amphiprotic (amphoteric) substance is water. As seen in the previous question, it acts as
an acid when placed with a base. And it acts as a base when placed with an acid. Identify two other
substances in question 7 which are amphiprotic (amphoteric).
Page 3
Unit 12: Acids and Bases
9.
Unlike the strong acids from question #5, a weak acid only partially dissociates. Its dissociation is
highly reversible when placed in water, with the dominant species present in solution being the
undissociated acid and water. For an acid, the strength of the forward dissociation reaction is
described by an equilibrium constant known as an acid dissociation constant (Ka). The same rules
which apply to any reversible system apply to the weak acid dissociation reactions. Use the
equilibrium constant expression, the Ka value, the dissociation equation and the ICE table to
determine ion concentrations for the following solutions of weak acids. (Reference: Chapter 13.4)
a. 2.0 M HCN
Ka = 6.2 x 10-10
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Ka Expression:
(aq)
+
(aq)
Ka Expression:
(aq)
+
(aq)
Ka Expression:
I
C
E
Algebra/Solution:
b. 1.0 M HC2H3O2
Ka = 1.8 x 10-5
Dissociation Equation:
H2O(l) +
(aq)

I
C
E
Algebra/Solution:
c. 0.50 M HNO2
Ka = 4.0 x 10-4
Dissociation Equation:
H2O(l) +
(aq)

I
C
E
Algebra/Solution:
Page 4
Unit 12: Acids and Bases
10.
Name:
Repeat the procedure from question #9 to determine the ion concentrations for these weak bases:
a. 0.50 M NH3
Kb = 1.8 x 10-5
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Kb Expression:
(aq)
+
(aq)
Kb Expression:
I
C
E
Algebra/Solution:
b. 0.020 M CH3 NH2
Kb = 4.38 x 10-4
Dissociation Equation:
H2O(l) +
(aq)

I
C
E
Algebra/Solution:
11.
As mentioned earlier, water is amphiprotic (amphoteric) - acts as either an acid or a base. But what's
most interesting about water is that it undergoes autoionization as follows
H2O(l)
+
H2O(l)
H3O+(aq)

+
hydronium ion
OH-(aq)
hydroxide ion
The equilibrium constant for this reaction at 25°C is 1.0 x 10-14 . As such,
[H3O+] • [OH-] = 1.0 x 10-14
Since the two product ions always establish an equilibrium in any aqueous solution, the product of
the hydronium ion concentration and the hydroxide ion concentration is always 1.0 x 10-14 at 25°C.
The above mathematical statement is the basis of the so-called pH scale. The pH scale is a
logarithmic scale which relates the hydronium ion concentration to a pH value. Higher pH values
(greater than 7) correspond to basic solutions. Lower pH values (less than 7) correspond to acidic
solutions. A similar scale - the pOH scale - is sometimes used in addition or instead of the pH scale.
The following mathematical equations will have to be used for pH, pOH, and [ ] calculations.
[H3O+] • [OH-] = 1.0 x 10-14
pH = - log [H3O+]
[H3O+] = 10-pH
pOH = - log [OH-]
[OH-] = 10-pOH
pH + pOH = 14.00
Page 5
Unit 12: Acids and Bases
12.
Express your understanding of these mathematical relationships by filling in the table. All
concentrations are in mol/L or M. Assume a temperature of 25 °C.
[H3O+]
0.010 or 1.0 x 10
[OH-]
pH
pOH
Acidic or Basic?
-2
0.0010 or 1.0 x 10-3
5.00
11.00
1.0 x 10-7
0.50 or 5.0 x 10-2
6.0
4.0
13.
Combine your pH and pOH understanding with your strong acid and strong base understanding
(see question #5) to answer the following questions. Assume a temperature of 25 °C. PSAYW
a. Calculate the pH of a 2.0 M HCl aqueous solution.
b. Calculate the pOH and the pH of a 2.0 M NaOH aqueous solution.
c. Calculate the pH of a 0.050 M Ca(OH)2 aqueous solution.
d. An aqueous solution of nitric acid is made and found to have a pH of 2.80. Calculate the
concentration of the nitric acid in the solution (i.e., before its dissociation).
e. Calculate the pH which results when 0.80 g of NaOH are dissolved in 100 mL of water.
Page 6
Unit 12: Acids and Bases
14.
Name:
Historically, acids and bases were distinguished from one another by their observable properties
and not by their chemical formulas or their structures. Associated the following observable
properties with the acids (A), bases (B) or both (AB).
A, B or AB
Statement of an Observable Property
a.
Yields blue on a litmus paper test.
b.
Conducts an electric current when dissolved in water.
c.
An aqueous solution will be pink when phenolphthalein indicator is
added.
d.
Tastes sour (though we would never test its taste in lab).
e.
Tastes bitter (though we would never test its taste in lab).
f.
Yields red on a litmus paper test.
g.
Reacts with metals to produce hydrogen gas.
h.
An aqueous solution will be colorless when phenolphthalein indicator is
added.
i.
May be corrosive.
j.
Reacts with carbonates to produce carbon dioxide gas.
k.
Can turn an indicator a different color.
Combine your pH and pOH understanding with your weak acid and weak base understanding (see
questions #9 and #10) to answer the following questions. Assume a temperature of 25 °C. PSAYW
15.
Calculate the pH of a 0.500 M aqueous solutiion of phenolic acid – HOC6H5 (Ka = 1.6 x 10-10).
16.
Calculate the pH of a 1.3 M aqueous solution of benzoic acid – HC7H5O2 (Ka = 6.5 x 10-5).
Page 7
Unit 12: Acids and Bases
17.
Calculate the pH of a 1.5 M aqueous solution of aniline – C6H5NH2 (Kb = 3.8 x 10-10).
18.
An aqueous solution of acetic acid – HC2H3O2 (Ka =1.8 x 10-5) - has a pH of 3.4. Determine the acetic
acid concentration.
19.
Calculate the pH and the %dissociation of a 0.80 M aqueous solution of acetic acid – HC2H3 O2 (Ka =
1.8 x 10-5).
20.
Approximately 2.1% of a 1.5 M aqueous solution of a monoprotic acid undergoes dissociation.
Calculate the Ka value of the acid.
Page 8
Unit 12: Acids and Bases
21.
Consider the following groups of acids. Identify the strongest acid (or base) in each group. Identify
the weakest acid in each group. By using the word strongest to describe a weak acid, we are refering
to the tendency of the acid to dissociate into ions (which will always be weak compared to a strong
acid but still be stronger than some other weak acids).
a.
b.
c.
22.
Name:
HF
Ka = 6.6 * 10-4
H3PO4
Ka = 7.1 * 10-3
NH4+ ion
Ka = 5.8 * 10-10
Ranking
Ranking
Ranking
HCN
Ka = 6.2 * 10-10
H3BO3
Ka = 5.8 * 10-10
HS- ion
Ka = 1.3 * 10-13
Ranking
Ranking
Ranking
H2C2O4
Ka = 5.4 * 10-2
CH3COOH
Ka = 1.8 * 10-5
CO32- ion
Ka = 4.4 * 10-7
Ranking
Ranking
Ranking
As acid-base problems become more and more complex it becomes crucial to be able to read a
description of an aqueous solution and to determine what major species are present in the solution.
For strong electrolytes (whether salts, acids or bases), it is the dissociated ions which are present in
the solution. For weak electrolytes (whether salts, acids or bases), it is primarily the undissociated
molecular form of the species which is present. For the following description of aqueous solutions,
write the formulas of the major species (present in relatively large concentrations) and of the minor
species (present in relatively small concentrations).
a.
A 2.0 M aqueous solution of NaOH is mixed with a 1.0 M aqueous solution of acetic acid HC2H3 O2 (Ka = 1.8 x 10-5).
Major species:
Minor species:
b.
A 2.0 M aqueous solution of HCl is mixed with a 1.0 M aqueous solution of ammonia - NH3
(Kb = 1.8 x 10-5).
Major species:
Minor species:
c.
A 0.50 M aqueous solution of NaC2H3 O2 is mixed with a 0.50 M aqueous solution of acetic acid
- HC2 H3O2 (Ka = 1.8 x 10-5).
Major species:
Minor species:
d.
A 0.20 M aqueous solution of NH4Cl is mixed with a 0.20 M aqueous solution of ammonia NH3 (Kb = 1.8 x 10-5).
Major species:
Minor species:
Page 9
Unit 12: Acids and Bases
23.
Volumetric analysis is a technique for determining the amount of a certain substance by doing a
titration. In the most common form of a titration, a known volume of a solution of known
concentration (the titrant) is added to a known volume of a solution of unknown concentration (the
analyte). The addition of the titrant continues until the endpoint is reached. The endpoint is
typically signaled by a dramatic change in color (or other observable property) and facilitated by the
use of an indicator. If the titration is correctly performed, the endpoint is also the equivalence point
or the stoichiometric point - the point in the reaction in which the moles of H+ (or H3O+) ions is
equal to the moles of OH- ions.
The analysis of a titration problem begins by writing down all the major species present in solution
and determining the reaction which occurs. Once the reaction is determined and the balanced
chemical equation is written, stoichiometric principles can be used to solve the problem. (Reference:
Chapter 4.8 of Masterton and Hurley; introduction on pp. 82-85; Chapter 14.3 with Examples 14.7,
14.8 and 14.9.)
a.
What volume of 0.250 M NaOH is required to neutralize 50.0 mL of 0.100 M HCl?
Major species:
Net Ionic Eq'n (balanced):
Stoichiometry/Solution:
b.
A chemistry student finds that 42.1 mL of 0.500 M HCl were required to titrate 20 mL of Ba(OH)2
having an unknown concentration. Determine the concentration of the unknown.
Major species:
Net Ionic Eq'n (balanced):
Stoichiometry/Solution:
Page 10
Unit 12: Acids and Bases
c.
Name:
Potassium hydrogen phthalate (KHC8H4 O4, often abbreviated as KHP) has a molar mass of
204.22 g/mol and one acidic proton. In a student titration, 34.67 mL of NaOH with an unknown
concentration are used to neutralize 0.2158 g of KHP. Determine the concentration of the NaOH
solution.
Major species:
Net Ionic Eq'n (balanced):
Stoichiometry/Solution:
d. Baking soda, NaHCO3, is capable of neutralizing stomach acid (mostly HCl). What mass of
baking soda is needed to neutralize 198 mL of stomach acid having an HCl concentration of
0.0531 M?
Major species:
Net Ionic Eq'n (balanced):
Stoichiometry/Solution:
e.
Calculate the mass of aluminum hydroxide required to completely react with 20.0 mL of 0.45M
HCl.
Major species:
Net Ionic Eq'n (balanced):
Stoichiometry/Solution:
Page 11
Unit 12: Acids and Bases
25.
Most acids we’ve studied thus far were monoprotic acids – acids having a single proton to donate.
Some acids are polyprotic acids – acids having more than one proton to donate. Examples of
polyprotic acids include sulfuric acid (H2SO4), phosphoric acid (H3PO4), carbonic acid (H2CO3),
oxylic acid (H2C2 O4), and ascorbic acid (H2C6H6O6). See Table 14.4 on p. 683. Polyprotic acids
dissociate in stepwise fashion – one proton at a time. Each dissociation step has its own Ka value;
these are usually distinguished from each other by labeling them as Ka1, Ka2, … . Thus, for oxalic
acid, the dissociation steps are:
H2C2O4 (aq)
HC2O4-
(aq)
H2O(l) 
+
+
H2O(l) 
H3O+
(aq)
+ HC2O4-
(aq)
H3O+ (aq) + C2O42- (aq)
In general, the first dissociation step occurs more strongly than the second dissociation step. As
such, the first step is more significant in terms of affecting the hydrogen ion concentration and the
pH. There is however one polyprotic acid which falls in a category of its own – sulfuric acid
(H2SO4). Sulfuric acid is a diprotic acid whose first dissociation step occurs completely; that is, the
Ka1 value is very large. The second dissociation step is a partial dissociation step and can make an
appreciable contribution to the overall hydrogen concentration and pH for any solution with a
concentration less than 1.0 M. (Reference: Masterton and Hurley, pp. 366-367)
a.
Calculate the pH and the concentrations of all species at equilibrium for a 2.0 M solution of
carbonic acid (Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11). Show all your work.
Page 12
Unit 12: Acids and Bases
b.
26.
Name:
Calculate the pH and ion concentrations for a 0.250 M solution of sulfuric acid (Ka1 = very large
and Ka2 = 1.2 x 10-2).
Common Ion Problems: Occassionally a weak acid is dissolved in a solution containing a salt
consisting of the conjugate base of the strong acid. That is, the weak acid HA is dissolved in a
solution containing NaA or KA or NH4A or … . In such cases, determining the pH of the solution
demands that an ICE table is used and the initial concentration of the common anion be included in
the ICE table. Solve the following common ion problems using the provided ICE table.
Calculate the pH and the percent dissociation of a solution containing 0.50 M HF (Ka = 7.2 x 10-4)
and 1.5 M NaF.
a.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
I
C
E
Algebra/Solution:
Page 13
+
(aq)
Ka Expression:
Unit 12: Acids and Bases
b.
Calculate the pH and the percent dissociation of a solution containing 0.10 M HC2H3 O2 (Ka = 1.8
x 10-5) and 0.50 M NaC2H3O2.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Ka Expression:
I
C
E
Algebra/Solution:
27.
Question #26 involved analyzing a solution containing a weak acid and the salt of its conjugate base.
The same procedure can be used to analyze a weak base and the salt of its conjugate acid. Use the
ICE table below to solve the following common ion problem.
Calculate the pOH, pH and the percent dissociation of a solution containing 0.25 M NH3 (Kb = 1.8 x
10-5) and 0.10 M NH4Cl.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
I
C
E
Algebra/Solution:
Page 14
+
(aq)
Kb Expression:
Unit 12: Acids and Bases
28.
Name:
Perhaps the most important application of acid-base chemistry involving common ions in solution is
buffering. Buffers are solutions which contain a weak acid and a salt consisting of its conjugate base
(or a weak base and a salt consisting of its conjugate acid). By carefully selecting the right collection
of a weak acid the conjugate base, a solution can be made which offers a high resistance to a pH
change. Even when an acid or base is added to a buffer solution, the solution is very resistant to
changes in its pH.
Suppose that solutions are made by mixing the following sets of compounds. In which case will the
solution act as a buffer?
Solution Compos’n
29.
Buffer?
Solution Compos’n
Buffer?
HNO2 / NaNO2
Yes
No
NH3 / NH4Cl
Yes
No
HNO3 / NaNO3
Yes
No
H2CO3 / NaHCO3
Yes
No
H2S / NH4 HS
Yes
No
H2SO4 / NaHSO4
Yes
No
Ca(OH)2 / HCa(OH)2
Yes
No
NaCl / CaCl2
Yes
No
The pH of a buffered solution can be determined in the same manner as the pH of a common ion
solution is determined – through the use of an ICE table and the equilibrium constant expression.
Use an ICE table to solve the following problems. (Reference: Masterton and Hurley, Chapter 14.1)
a. Calculate the pH of a buffered solution containing 0.10 M H2 CO3 (Ka = 4.4 x 10-7) and 0.25 M
NaHCO3.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Ka Expression:
I
C
E
Algebra/Solution:
b. Calculate the pH of a buffered solution containing 0.40 M NH3 (Kb = 1.8 x 10-5) and 0.25 M NH4Cl.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
I
C
E
Algebra/Solution:
Page 15
+
(aq)
Kb Expression:
Unit 12: Acids and Bases
30.
When a base is added to a weak acid/conjugate base buffer solution, the solution resists the change
in pH because the weak acid neutralizes the base; the weak acid-conjugate base equilibrium shifts in
order to replenish the reacted acid and thus maintain roughly the original acid content in the
solution. This is simply an application of LeChatelier’s principle. The mathematical analysis of such
a process involves two steps: 1) the stoichiometry step in which the amount of weak acid needed to
neutralize the base is determined; 2) the equilibrium step in which the LeChatelier shift of the weak
acid-conjugate base equilibrium is analyzed. (Reference: Masterton and Hurley, pp. 388- 390)
Consider the buffer solution from question #29a: 0.50 M H2 CO3 (Kb = 4.4 x 10-7) and 0.75 M
NaHCO3. Determine the new pH value which would result if 0.05 moles of NaOH is added to 1 liter
of this solution.
As a contrast, determine the pH if 0.05 moles of NaOH is added to 1.0 L of pure water.
Page 16
Unit 12: Acids and Bases
31.
Name:
The concept of a conjugate acid and a conjugate base was discussed in question #7. Now consider
the following truths about the conjugates of acids and bases.
•
•
•
•
The conjugate base of a weak acid tends to form basic solutions in water.
The conjugate acid of a weak base tends to form acidic solutions in water.
The conjugate base of a strong acid has neither acidic or basic properties.
The conjugate acid of a strong base has neither acidic or basic properties.
Identify the following ions as being either acidic, basic or neutral.
32.
Species
Circle
Species
Circle
NO2 -
acidic basic neutral
NO3 -
acidic basic neutral
C2H3O2 -
acidic basic neutral
NH4 +
acidic basic neutral
HS -
acidic basic neutral
OH -
acidic basic neutral
Cl -
acidic basic neutral
F-
acidic basic neutral
Salts are ionic compounds consisting of a cation (+ ion) and an anion (- ion). Salts can actually
produce acidic or basic solutions when dissolved in water, depending on whether the cation or
anion is the conjugate base or acid of a weak acid or weak base. Since an ion which is the conjugate
acid of a weak base is a strong acid, a salt containing this ion will produce an acidic solution.
Similarly, since an ion which is the conjugate base of a weak acid is a strong base, a salt containing
this ion will produce an basic solution. Complete the following statements, thus outlining the logic
which explains whether a salt would form an acidic, basic or neutral solution. (Reference: Masterton
and Hurley, pp. 372-373)
a.
The cation NH4+ is the conjugate _______________ of the ______________ _____________ NH3.
For this reason, the salt NH4Cl would dissolve in water to form a(n) ________________ (acidic,
basic, neutral) solution.
b.
The anion F- is the conjugate _______________ of the ______________ _____________ HF. For
this reason, the salt NH4Cl would dissolve in water to form a(n) ________________ (acidic,
basic, neutral) solution.
c.
The anion NO3- is the conjugate _______________ of the ______________ _____________ HNO3.
For this reason, the salt NaNO3 would dissolve in water to form a(n) ________________ (acidic,
basic, neutral) solution.
d.
The cation Ca2+ is the conjugate _______________ of the ______________ _____________
Ca(OH)2. For this reason, the salt CaCl2 would dissolve in water to form a(n) ________________
(acidic, basic, neutral) solution.
33.
Categorize the following salts as forming acidic, basic or neutral solutions when dissolved in water.
(Reference: Masterton and Hurley, pp. 372-373)
Species
NaCN
Circle
acidic basic neutral
Species
KNO3
Circle
acidic basic neutral
NaCl
acidic basic neutral
NH4NO3
acidic basic neutral
NaC2H3O2
acidic basic neutral
NH4C2H3 O2
acidic basic neutral
Page 17
Unit 12: Acids and Bases
34.
The pH of a salt solution can be determined if the Ka value of its weak acid conjugate is known. The
underlying principle which allows for such a calculation is that the Kb value of the conjugate base of
a weak acid is Kw / Ka (where Kw is 1.0 x 10-14). As an example, if the Ka for HCN is 6.2 x 10-10, then
the Kb of CN- (the conjugate base of HCN) is (1.0 x 10-14 / 6.2 x 10-10) = 1.6 x 10-5. Use this information
to solve the following problems.
a. Calculate the pH of a 0.50 M NaCN solution. The Ka HCN is 6.2 x 10-10.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Kb Expression:
I
C
E
Algebra/Solution:
b. Calculate the pH of a 2.0 M NaC2H3O2 solution. The Ka HC2H3O2 is 1.8 x 10-5.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
I
C
E
Algebra/Solution:
Page 18
+
(aq)
Kb Expression:
Unit 12: Acids and Bases
35.
Name:
The procedure used in question #34 for the conjugate of a weak acid can also be used for the
conjugate of a weak base.
Calculate the pH of a 0.25 M solution of ammonium nitrate. The Kb of ammonia (NH3) is 1.8 x 10-5.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
+
(aq)
Ka Expression:
I
C
E
Algebra/Solution:
36.
As seen in question #35, a salt containing a cation which is the conjugate acid of a weak base
produces an acidic solution. A second type of salt which produces an acidic solution is one which
contains a highly charged metal ion such as Al3+ or Fe3+ or Cr3+. When present in water, the cation
becomes quickly hydrated and its hydrated form is slightly acidic.
Determine the pH of a 2.5 M iron(III) chloride solution. The Ka value of Fe(H2O)63+ is 6.0 x 10-3.
Dissociation Equation:
H2O(l) +
(aq)

(aq)
I
C
E
Algebra/Solution:
Page 19
+
(aq)
Ka Expression:
Unit 12: Acids and Bases
37.
Strong Acid-Strong Base Titration: Consider the titration of 20.0 mL of 0.500 M HNO3 with 0.250 M
NaOH. Determine the pH value of the resulting solution after the addition of the following amounts
of base to the acid. (Reference: Masterton and Hurley, pp. 394-395; Example 14.7)
Volume NaOH
(mL)
pH
Please show your work.
0.00
10.0
20.0
30.0
35.0
38.0
40.0
42.0
45.0
50.0
60.0
Construct a plot of pH vs. volume of acid used.
Be sure to label axes values.
Identify the endpoint with a large • on the
diagram.
What is the pH value at the endpoint of a strong
acid-strong base titration?
Use Figure 14.4 and Table 14.2 to identify a
suitable indicator for the detection of this
endpoint.
Page 20
Unit 12: Acids and Bases
38.
Name:
Weak Acid-Strong Base Titration: Consider the titration of 20.0 mL of 0.500 M HC2H3O2 (Ka = 1.8 x
10-5) with 0.250 M NaOH. Determine the pH value of the resulting solution after the addition of the
following amounts of base to the acid. (Reference: Masterton and Hurley, pp. 396-397; Example
14.8)
Volume NaOH
(mL)
pH
Please show your work.
0.00
10.0
20.0
30.0
35.0
38.0
40.0
42.0
45.0
50.0
60.0
Construct a plot of pH vs. volume of acid used.
Be sure to label axes values.
Identify the endpoint with a large • on the
diagram.
What is the pH value at the endpoint of a weak
acid-strong base titration?
Use Figure 14.4 and Table 14.2 to identify a
suitable indicator for the detection of this
endpoint.
Page 21
Unit 12: Acids and Bases
39.
Weak Base-Strong Acid Titration: Consider the titration of 20.0 mL of 0.500 M NH3 (Kb = 1.8 x 10-5)
with 0.250 M HCl. Determine the pH value of the resulting solution after the addition of the
following amounts of acid to the base. (Reference: Masterton and Hurley, pp. 397-398)
Volume HCl
(mL)
pH
Please show your work.
0.00
10.0
20.0
30.0
35.0
38.0
40.0
42.0
45.0
50.0
60.0
Construct a plot of pH vs. volume of acid used.
Be sure to label axes values.
Identify the endpoint with a large • on the
diagram.
What is the pH value at the endpoint of a weak
base-strong acid titration?
Use Figure 14.4 and Table 14.2 to identify a
suitable indicator for the detection of this
endpoint.
Page 22
Unit 12: Acids and Bases
Name:
26.
Hydrolysis of Salts
27.
Structure and Acidity and Acidic properties of Oxides (Chapter 14.9-10)
29.
Introduce buffer problems and repeat the common ion; do some conceptual Qs on preparing buffers
30.
pH curves (might save this for last - Chapter 15:4-5)
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Unit 12: Acids and Bases
References:
1.
Wikipedia on Definitions of Acids and Bases:
http://en.wikipedia.org/wiki/Acid-base_reaction_theories
2.
Identifying conjugate acid-base pairs (practice with answers):
http://chemistryandphysics.astate.edu/draganjac/AcidBasePairs.html
3.
Acid-Base Titration:
http://preparatorychemistry.com/Bishop_Titration.htm
Page 24