CHEMICAL EQUILIBRIA--ACID/BASE
Acid/base problems may fall into 4 categories: strong acid/base, weak acid/base, buffers
and hydrolysis. We will go through examples of each of these types of problems one at a time.
Strong Acids and Strong Bases
The strength of the acid is determined by how far the equilibrium lies to the right.
Qualitatively, this may be judged by the Ka of the acid. A large Ka indicates a strong acid; a
small Ka indicates a weak acid. Strong acids, such as HCl, have Ka values in the vicinity of
infinity. This implies that the dissociation of HCl is virtually complete, and the equilibrium lies
completely to the right, therefore, the concentration of the acid equals the concentration of
hydronium ions produced. For instance, a 0.01 M HCl solution will completely dissociate into
0.01 M H3O+ and 0.01 M Cl-. The effective concentration of HCl after "equilibrium" has been
reached will be zero! Analogously, strong bases, such as NaOH, will dissociate completely. The
concentration of OH- in solution will be equal to the concentration of the strong base.
A typical strong acid problem might be: What is the pH of a 0.010 M HCl solution? Since
HCl is a strong acid, the hydronium ion concentration will be equal to the HCl concentration:
[H3O+] = 0.010 M
The pH can be found by taking the negative log of the hydronium ion concentration:
pH = -log[H3O+] = -log(0.010) = 2.00
A typical strong base problem might be: What is the pH of a 0.010 M NaOH solution?
Since NaOH is a strong base, the hydroxide ion concentration will be equal to the NaOH
concentration:
[OH-] = 0.010 M
The pH can be found by first finding the pOH by taking the negative log of the hydroxide
ion concentration, and then converting the pH to pOH. To find the pOH:
pOH = -log[OH-] = -log(0.010) = 2.00
The pH can then be calculated from the equation pH + pOH = 14:
pH = 14.00 - 2.00 = 12.00
Weak Acids and Bases
Weak acids and weak bases do not dissociate completely. An equilibrium exists between
the weak acid, water, H3O+, and the anion of the weak acid. The equilibrium lies to the left hand
side of the equation, indicating that not much H3O+ is being produced. The fact that very little
H3O+ is being produced is the definition of a weak acid. The Ka for a weak acid is small, usually
a number less than 1.
There are three types of problems encountered with weak acids or bases: dissociation,
buffers or hydrolysis. We'll look at each type in detail.
Dissociation of a Weak Acid
In this type of problem, you will be asked to find the hydronium ion concentration and/or
the pH of a weak acid whose initial concentration is known. A typical problem may be:
What is the hydronium ion concentration and pH of a 0.10 M solution of hypochlorous
acid, Ka = 3.5 x 10-8?
In tackling this problem, first note that the Ka is a small number, meaning hypochlorous
acid is a weak acid. To begin the problem, write down the equilibrium involved:
HOCl(aq) + H2O(l)

H3O+(aq) +
OCl¯(aq)
The equilibrium may be expressed mathematically by setting the Ka equal to the mass action
expression:
[H O][OCl-]
Ka  3
[HOCl]
Next, use the equilibrium to establish a table of initial conditions, change in equilibrium and final
equilibrium conditions. Initially, the HOCl concentration is 0.10 M, the concentrations of H3O+
and OCl- are zero:
HOCl(aq) + H2O(l)  H3O+(aq) + OCl¯(aq)
Initial
0.10 M
0
0
Change
Equilibrium
Note that water will not be included in the calculation since is the solvent.
In order for equilibrium to be established, some of HOCl must dissociate and form H3O+
and OCl-. Since it is not known how much will dissociate, we'll call the amount of HOCl lost -x
and the amount of H3O+ and OCl- formed +x:
HOCl(aq) + H2O(l)  H3O+(aq) + OCl¯(aq)
Initial
Change
0.10 M
0
0
-x
+x
+x
Equilibrium
The above operation is justified by LeChatlier's Principle, which states: if a stress is
places on an equilibrium, the equilibrium will shift in the direction which will relieve the stress.
In this case, the stress is the lack of H3O+ and OCl- so the equilibrium will shift to the right to
relieve the stress by forming some H3O+ and OCl-.
By summing the initial concentrations and the change in concentrations, you can obtain
the amount of each species at equilibrium:
HOCl(aq) + H2O(l)  H3O+(aq) + OCl¯(aq)
Initial
0.10 M
0
0
-x
+x
+x
0.10 – x
+x
+x
Change
Equilibrium
These quantities will be used in the mass action expression for the equilibrium of the acid, as
shown below. The Ka for HOCl is 3.5 x 10-8.
Ka 
[H3O][OCl-]
(x)(x)

 3.5 x 10 - 8
[HOCl]
0.10  x
The solution for x becomes simplified because the x shown in bold can be neglected. This x can
be neglected because it will be negligibly small compared to the concentration, 0.10 M. To
determine whether x is negligible, compare the magnitude of the last decimal place of the
concentration of the acid to the magnitude of the equilibrium constant. If the difference in
magnitude is greater than 100, the x may be neglected. In this case, the concentration is known
to the 10-2 place and the equilibrium constant is the magnitude of 10-8. The difference in
magnitude is 106, therefore, x may easily be neglected. This simplifies the equation to:
( x )( x )
 3.5 x 10 - 8
0.10
Multiplying both sides by 0.10 yields:
x2 = 3.5 x 10-9
Taking the square root of both sides yields:
x = [H3O+] = 5.9 x 10-5 = [OCl-] (also)
To find the pH, take the negative log of the hydronium ion concentration:
pH = -log[H3O+] = -log(5.9 x 10-5) = 4.23
Incidentally, the concentration of HOCl at equilibrium would be:
[HOCl] = 0.10 - 5.9 x 10-5 = 0.10 M (2 significant figures)
This shows our assumption that x was negligible was valid.
Dissociation of a Weak Base
A typical weak base problem may read: What is the hydroxide ion concentration and pH
of a 0.10 M solution of NH3, Kb = 1.8 x 10-5?
Again, note that Kb is small. We will follow the same format as we used for weak acid solutions:
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
0.10 M
0
0
Change
Equilibrium
We are assuming that before equilibrium is established, no NH4+ or OH- has formed. To
establish equilibrium, a shift to the right has to occur. Since the amount of NH3 lost, and the
amounts of NH4+ and OH- formed are not known, we assign the value of -x and +x, respectively:
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
Change
0.10 M
0
0
-x
+x
+x
Equilibrium
By summing the initial concentrations and change in concentrations, we have the algebraic
amount of each species in solution at equilibrium:
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
Change
Equilibrium
0.10 M
0
0
-x
+x
+x
0.10 – x
+x
+x
These quantities will be used in the mass action expression:
Kb 
(x)(x)
[NH4][OH-]

 1.8 x 10 - 5
[NH3]
0.10  x
Let's see if we can neglect the x in the denominator. The last decimal place in 0.10 is to the
magnitude of10-2. The magnitude of the constant is 10-5. The difference in magnitude is 103.
Since this difference is greater than 100, the x may be neglected. This simplifies the
mathematical expression to:
(x)(x)
 1.8 x 10 - 5
0.10  x
Multiplying both sides of the equation by the denominator yields:
x2 = 1.8 x 10-6
Taking the square root of both sides yields:
x = [NH4+] = [OH-] = 1.3 x 10-3
To find the pH, first find the pOH by taking the negative log of the hydroxide ion concentration:
pOH = -log[OH-] = -log(1.3 x 10-3) = 2.87
Then subtract the pOH from 14 to find the pH:
pH = 14.00 - pOH = 14.00 - 2.87 = 11.13
Buffer Problems
Buffer solutions consist of a weak acid and its conjugate base (acidic buffer) or a weak
base and its conjugate acid (alkaline buffer). They have the property of resisting pH change even
when strong acid or strong base is introduced into solution. The way to recognize a buffer
problem is to realize that both the concentration of the weak acid or base and their respective
conjugate acid or base concentrations are known. We'll go through an example of each type of
buffer problem below.
Acidic Buffers
Suppose you have a solution which is 0.20 M in acetic acid (CH3CO2H) and 0.10 M in sodium
acetate (NaCH3CO2). What will be the pH of this solution? This is a typical buffer problem.
Notice that the concentrations of both the weak acid and its conjugate base are known. The
equilibrium is still that of acetic acid; the only difference is that we have a common ion (acetate
ion) present.
First, set up a table as we have done before for a weak acid problem:
CH3CO2H(aq) + H2O(l)  CH3CO2¯(aq) + H3O+(aq)
Initial
0.20 M
0.10 M
0
Change
Equilibrium
Notice that the initial concentrations of both CH3CO2H and CH3CO2 ¯ are known. The
acetate ion came from the sodium acetate. The sodium ion is not a part of the equilibrium and is
therefore neglected as a spectator ion. In the next step we apply LeChatelier's principle and see
that in order to establish equilibrium, some CH3CO2H will have to dissociate to produce
hydronium ions:
CH3CO2H(aq) + H2O(l)  CH3CO2¯(aq) + H3O+(aq)
Initial
Change
0.20 M
+x
0.10 M
0
+x
+x
Equilibrium
The concentrations of CH3CO2H, H3O+ and CH3CO2¯ at equilibrium will be the sum of initial
concentrations and the change in concentrations:
CH3CO2H(aq) + H2O(l)  CH3CO2¯(aq) + H3O+(aq)
Initial
0.20 M
Change
+x
Equilibrium
0.20 – x
0.10 M
0
+x
+x
0.10 + x
+x
These algebraic quantities may now be inserted into the mass action equilibrium expression for
acetic acid:
[H3O  ][CH3CO2-]
(x)(0.10  x)
Ka 

 1.8.10 - 5
[CH3CO2H]
(0.20  x)
The values of x shown in bold above may often neglected if they can be shown to be negligibly
small. Compare the concentrations to the Ka value. In this case the concentrations of the weak
acid and its conjugate base are known to the 10-2 place and the Ka is to the 10-5 place. Since the
difference in magnitude is greater than 100 (actually 1000 times different), both of this x
quantities may be neglected. This simplifies the algebraic expression to:
(x)(0.10)
 1.8 x 10 - 5
0.20
Solving for x gives the hydronium ion concentration:
(1.8 x 10 - 5)(0.20)
x  [H3O ] 
 3.6 x 10 - 5
0.10

To find the pH, take the negative log the hydronium ion concentration:
pH = -log[H3O+] = -log[3.6 x 10-5] = 4.44
Alkaline Buffers
Suppose you have a alkaline buffer consisting of 0.20 M aqueous ammonia (NH3) and
0.10 M ammonium chloride (NH4Cl). What is the pH of the solution? Notice that the
concentrations of both the weak base and conjugate acid are known. The ammonium ion is a
common ion to the ammonia equilibrium. Chloride ion is not participating in the equilibrium
(spectator ion) and will be ignored.
Begin by forming the table as we have with other acid\ base problems:
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
0.20 M
0.10 M
0
Change
Equilibrium
Notice that the concentrations of both weak base and conjugate acid are known. Apply
LeChatelier’s Principle and see that the equilibrium will shift to the right to create some OH- to
establish equilibrium. This means some NH3 will be lost and some additional NH4+ will be
formed in addition to some OH-.
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
Change
0.20 M
0.10 M
0
-x
+x
+x
Equilibrium
Sum the initial concentrations and the change in concentrations to find the amount of
NH3, NH4+ and OH- at equilibrium:
NH3(aq) + H2O(l)  NH4+(aq) + OH¯(aq)
Initial
Change
Equilibrium
0.20 M
0.10 M
0
-x
+x
+x
0.20 – x
0.10 + x
+x
Insert the above expressions into the mass action equilibrium expression:
(0.10  x)(x)
[NH4  ][OH-]

 1.8 x 10 - 5
[NH3]
0.20  x
The x quantities shown in bold above often may be neglected. Compare the magnitude of
the concentrations of NH3 and NH4+ to the magnitude of Kb. In this case the magnitudes of the
concentrations are known to the 10-2 magnitude and the Kb is known to the 10-5 magnitude. Since
the difference in magnitude is greater than 100 (difference is 1000), the x quantities shown in
bold may be neglected. This simplifies the expression to the expression shown:
Kb 
(0.10)(x)
 1.8 x 10 - 5
0.20
Solve this expression for x which will provide the hydroxide ion concentration:
(1.8 x 10 - 5)(0.20)
x  [OH ] 
 3.6 x 10 - 5
0.10
To find the pH, first find the pOH by taking the negative log of the hydroxide ion
concentration:
pOH = -log[OH-] = -log(3.6 x 10-5) = 4.44
Next, subtract the pOH from 14.00 to find the pH:
pH = 14.00 - pOH = 14.00 - 4.44 = 9.56
Hydrolysis Problems
Hydrolysis problems deal with the salts of weak acids and weak bases. A salt may be
defined as the product of an acid and a base. The cation of the salt always comes from the base
and the anion always comes from the acid (keep the vowels and consonants together). Not all
salts are neutral. Some salts will hydrolyze in water and produce either acidic or basic solutions.
Let's see how this may occur. The salt, NaCl, will form a neutral solution--no hydrolysis
occurs.
NaCl
/
\
cation
anion
from NaOH from HCl
(strong base)
(strong acid)
Since the sodium ion came from NaOH, a strong base, this ion is a weak conjugate acid
and will not hydrolyze. It will float around in solution as a spectator ion. Likewise, the chloride
ion came from HCl, a strong acid. That makes the chloride ion a weak conjugate base and it will
not hydrolyze. It also will float around in solution as a spectator ion. As a result, the only
hydronium ions present in solution are due to the auto-ionization of water, and the solution will
be neutral in pH.
Next, let us consider the salt, sodium acetate, whose concentration is 0.10 M:
NaCH3CO2
/
cation
from NaOH
(strong base)
\
anion
from CH3CO2H
(weak acid)
The cation, sodium ion, came from a strong base, NaOH, and as discussed above, will not
hydrolyze. However, the anion, acetate ion, came from a weak acid, acetic acid. Acetate ion is
therefore a strong conjugate base and will hydrolyze. This means it will behave as a LowryBronsted base, and accept a proton from water, as shown in the equilibrium below:
C2H3O2¯(aq) + H2O(l)

CH3CO2H(aq) + OH¯(aq)
Notice that hydroxide ion is generated and the salt solution will be alkaline in pH. To
calculate the pH of the solution we need to know the initial concentration of the acetate ion
(which will be given) and the value of the equilibrium constant, Kb (sometimes labeled as Kh, for
hydrolysis constant). The equilibrium constant may be calculated from the Kw of water and the
Ka of the weak acid from which the conjugate base forms:
Kw
1 x 10 - 14
Kb 

 5.6 x 10 - 10
-5
Ka
1.8 x 10
Recall the concentration of the sodium acetate solution was given as 0.10 M. The acetate
ion concentration is also 0.10 M since every mole of sodium acetate contains 1 mole of acetate
ions. One can proceed to solve for the pH as we have done for other equilibrium problems. Begin
by showing the equilibrium and the initial concentrations of species present:
C2H3O2¯(aq) + H2O(l)
Initial
0.10 M
 CH3CO2H(aq) + OH¯(aq)
0
0
Change
Equilibrium
According to LeChatelier’s Principle, the reaction will shift to the right to establish
equilibrium since there is no acetic acid or hydroxide yet present. This means an x amount of
acetate ion will be consumed and an x amount of both acetic acid an hydroxide ion will be
formed, as shown in the table below:
C2H3O2¯(aq) + H2O(l)
Initial
Change
 CH3CO2H(aq) + OH¯(aq)
0.10 M
0
0
-x
+x
+x
Equilibrium
Sum the initial concentrations and the change in concentrations to find the amounts
present at equilibrium:
C2H3O2¯(aq) + H2O(l)
Initial
Change
Equilibrium
 CH3CO2H(aq) + OH¯(aq)
0.10 M
0
0
-x
+x
+x
0.10 – x
+x
+x
These quantities may by substituted into the mass action expression:
Kb 
[CH3CO2H][OH-]
(x)(x)

 5.6 x 10 - 10
[CH3CO2 ]
(0.10 - x)
The x quantity shown in bold in the denominator may be neglected since the difference in
magnitude between the initial concentration of acetate ion (10-2) and the equilibrium constant
(10-10) is 108. This simplifies the above expression to:
Kb 
(x)(x)
 5.6 x 10 - 10
0.10
Multiplying both sides of the equation by the denominator, 0.10, yields:
x2 = 5.6 x 10-11
To find x, take the square root of both sides of the equation:
x = 7.5 x 10-6 = [OH-]
Notice that x corresponds to the hydroxide ion concentration. Now that the hydroxide ion
concentration is known, the pH may be found by first calculating the pOH:
pOH = -log[OH-] = -log(7.5 x 10-6) = 5.12
The pH may now be found by subtracting the pOH from 14.00:
pH = 14.00 - pOH = 14.00 - 5.12 = 8.88
As shown by the equilibrium, the solution is alkaline.
Suppose one had a 0.10 M solution of NH4Cl. What would be the pH of this solution?
First consider if any hydrolysis will take place.
NH4Cl
/
\
cation
anion
from NH3
from HCl
(weak base)
(strong acid)
The chloride anion is the weak conjugate base of a strong acid (HCl) and will not hydrolyze. The
ammonium cation, however, is the strong conjugate acid of a weak base (NH3) and will
hydrolyze--it will act as a Lowry-Bronsted acid and donate a proton to water:
NH4+(aq) + H2O(l)

NH3(aq)
+
H3O+(aq)
Since hydronium ion is being generated, this solution will be acidic in pH. To calculate the pH of
this solution, we need to know the initial concentration of the salt (which will be given), and the
equilibrium constant. The equilibrium constant will be calculated in a similar manner as for
alkaline salts, using the Kw of water in the Kb of the base from which the conjugate acid formed:
Ka 
Kw
1 x 10 - 14

 5.6 x 10 - 10
-5
Kb
1.8 x 10
Using an initial concentration of salt to be 0.10 M, set up a table of initial concentrations,
change in concentrations and equilibrium concentrations:
NH4+(aq) + H2O(l)
Initial

0.10 M
NH3(aq)
+
0
H3O+(aq)
0
Change
Equilbrium
We are assuming that before equilibrium is established, no ammonia or hydronium ion
has been produced. In order to establish equilibrium, an x amount of ammonium ion will be lost
to form an x amount of ammonia and hydronium ion. This follows LeChatlier's Principle--the
reaction will shift to the right to establish equilibrium. The change in concentrations is shown
below:
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Initial
Change
0.10 M
0
0
-x
+x
+x
Equilbrium
Sum the initial concentrations and change in concentrations to find the concentrations
present at equilibrium:
NH4+(aq) + H2O(l)
Initial
NH3(aq)
+
H3O+(aq)
0.10 M
0
0
-x
+x
+x
0.10 – x
+x
+x
Change
Equilbrium

Use the equilibrium concentrations in the mass action expression:
Ka 
[NH3][H3O  ]
(x)(x)

 5.6 x 10 - 10

[NH4 ]
0.10 - x
The x amount shown in bold may be neglected since the difference in magnitude between
the initial concentration (10-2) and the equilibrium constant (10-10) is 108. This simplifies the
above expression to:
(x)(x)
Ka 
 5.6 x 10 - 10
0.10
Multiplying both sides of the equation by the denominator, 0.10, yields:
x2 = 5.6 x 10-11
Solve for x by taking the square root of both sides of the equation:
x = [H3O+] = 7.5 x 10-6
Notice that x is also the hydronium ion concentration. To find the pH take the negative
log of the hydronium ion concentration:
pH = -log[H3O+] = -log(7.5 x 10-6) = 5.12
As shown by the equilibrium, the solution is acidic.
Acid/Base Problem Set
1. What is the pH of a 0.020 M solution of Ba(OH)2?
2. What are the concentration of all species in solution for a solution that is 0.30M in HCN and
0.40M in NaCN?
3. Find the concentration of all species in solution for a 0.45M solution of HNO3.
4. Find the concentration of all species in solution for a 0.45M solution of HNO2.
5. Find the concentration of all species in solution for a 0.45M solution of NaNO3.
6. Find the concentration of all species in solution for a 0.45M solution of KNO2.
7. What is the pH of a 0.35M solution of CH3NH2?
8. Find the concentration of all species in solution for a 0.45M solution of CH3NH2 and 0.30 M
in CH3NH3Br?
9. Calculate the [H3O+], [OH], pH and pOH of the following solutions:
a. 0.010 M HBr
b. 0.30 M Ca(OH)2
c. 0.20 M C6H5COOH (benzoic acid) Ka = 6.2 x 10-5
d. 0.050 M C17H19NO3 (morphine)
Kb = 1.6 x 10-6
10. Predict whether the following salts will be acidic, basic or neutral:
a.
b.
c.
d.
KBr
RbNO2
NH4NO3
NH4C6H5COO (ammonium benzoate – a challenger!)
11. You are titrating 25.0 ml of 0.10 M HCl with 0.10 M NaOH. Calculate the pH of the
resulting solution after addition of (a) 0.0 mL, (b) 10.0 mL, (c) 25.0 mL, (d) 30.0 mL of NaOH
solution.
12. Benzoic acid is often used as an acidulant and preservative in foods. You are titrating 25.0
mL of 0.20 M C6H5COOH (benzoic acid), Ka = 6.2 x 10-5, with 0.10 M NaOH. Calculate the
pH of the resulting solutions after the addition of (a) 0.0 mL, (b) 10.0 mL, (c) 25.0 mL, (d)50.0
mL, (e) 60 mL of NaOH solution.
Answers
1. This is a strong base problem.
Ba(OH)2 (aq) → Ba2+(aq) + 2 OH‾(aq)
0.020 mol Ba(OH) 2
2 mol OH x
 0.040 M OH -  [OH-]
1 L soln
1 mol Ba(OH) 2
pOH = - log[OH‾] = - log(0.040) = 1.40
pH = 14 – pOH = 14 – 1.40 = 12.60
2. This is an acidic buffer problem because a weak acid and a salt of the weak acid are present in
appreciable quantities. From Table 16.3, the Ka of HCN = 4.9 x 10-10.
HCN(aq) + H2O (l)  CN‾(aq) + H3O+(aq)
init
0.30
0.40
0
change -x
x
x
equil
0.30 – x
0.40 + x
x
Ka 
Simplifies to:
[CN-][H3O  ]
(0.40  x)(x)

 4.9 x 10 - 10
HCN
0.30 - x
[CN-][H3O  ]
(0.40)(x)
Ka 

 4.9 x 10 - 10
HCN
0.30
x 
(4.9 x 10 - 10)(0.30)
 3.7 x 10 - 10 M  [H3O]
0.40
[HCN] = 0.30 – (3.7 x 10-10 ) = 0.30 M
[CN‾] = 0.40 + x = 0.40 + (3.7 x 10-10 ) = 0.40 M
[Na+] = 0.40 M
[OH-] 
Kw
1.0 x 10 - 14

 2.7 x 10 - 5

- 10
[H3O ] 3.7 x 10
[H2O] = 55.5 M
3. This is a strong acid problem.
HNO3(aq) + H2O(l)  NO3‾(aq) + H3O+(aq)
[HNO3] = 0 M
[NO3‾] = [H3O+] = 0.45 M
[H2O] = 55.5 M
[OH-] 
Kw
1.0 x 10 - 14

 2.2 x 10 - 14 M

[H3O ]
0.45
4. This is a weak acid problem. The Ka of HNO2 is 4.5 x 10-4.
HNO2(aq) + H2O(l)  NO2‾(aq) + H3O+(aq)
init
0.45
0
0
change -x
+x
+x
equil. 0.45 – x
x
x
Ka 
Simplifies to:
[NO2-][H3O  ]
(x)(x)

 4.5 x 10 - 4
[HNO2]
0.45  x
[NO2-][H3O  ]
(x)(x)
Ka 

 4.5 x 10 - 4
[HNO2]
0.45
x2 = 4.5 x 10-4(0.45) = 2.0 x 10-4
2.0 x 10 - 4  1.4 x 10 - 2 M  [H3O]  [NO2-]
x 
[HNO2] = 0.45 – x = 0.45 – 1.4 x 10-2 = 0.44 M
[OH-] 
Kw
1.0 x 10 - 14

 7.1 x 10 - 13 M

-2
[H3O ]
1.4 x 10
[H2O] = 55.5 M
5. This is a hydrolysis problem. The salt is the salt of a strong acid and strong base, so neither
Na+ (weak conjugate acid) nor NO3‾ (weak conjugate base) will hydrolyze. Both these ions are
spectator ions. Thus, the only [H3O+] or [OH‾] is from the dissociation of water.
[Na+] = [NO3‾] = 0.45 M
[H3O+] = [OH‾] = 1.0 x 10-7 M
[H2O] = 55.5 M
6. This is also a hydrolysis problem. The K+ is a spectator because it is the weak conjugate acid
of a strong base. Nitrite ion will hydrolyze:
NO2‾(aq) + H2O(l)  HNO2(aq) + OH‾(aq)
0.45
0
0
-x
x
x
0.45 – x
x
x
init
change
equil
Need to find Kb:
Kb 
Simplifies to:
Kw
1.0 x 10 - 14
Kb 

 2.2 x 10 - 11
-4
Ka
4.5 x 10
[HNO2][OH-]
(x)(x)

 2.2 x 10 - 11
[NO2 ]
0.45  x
[HNO2][OH-]
(x)(x)
Kb 

 2.2 x 10 - 11
[NO2 ]
0.45
x2 = 2.2 x 10-11(0.45) = 1.0 x 10-11
1.0 x 10 - 11  3.2 x 10 - 6 M  [HNO2]  [OH-]
x 
[NO2ˉ] = 0.45 – x = 0.45 – (3.2 x 10-6) = 0.45 M
[K+] = 0.45 M
[H2O] = 55.5 M
7. This is a weak base problem. The Kb for CH3NH2 is found on Table 16.5 and has a value of
4.4 x 10-4.
CH3NH2 (aq) + H2O(l)  CH3NH3+ (aq) + OHˉ(aq)
init.
0.35
0
0
change
-x
x
x
equil.
0.35 – x
x
x
Kb 
[CH3NH3 ][OH-]
(x)(x)

 4.4 x 10 - 4
CH3NH2
0.35  x
Simplifies to:
Kb 
[CH3NH3 ][OH-]
(x)(x)

 4.4 x 10 - 4
CH3NH2
0.35
x2 = 4.4 x 10-4(0.35) = 1.5 x 10-4
x  1.5 x 10 - 4  0.012 M  [CH3NH3]  [OH-]
[CH3NH2] = 0.35 – x = 0.35 – 0.012 = 0.34 M
Kw
1 x 10 - 14
[H3O ] 

 8.1 x 10 - 13 M
[OH ]
0.012

[H2O] = 55.5 M
8. This is an alkaline buffer problem.
CH3NH2 (aq) + H2O(l)  CH3NH3+ (aq) + OHˉ(aq)
init.
0.45
0.30
0
change
-x
x
x
equil.
0.45 – x
0.30 + x
x
Kb 
[CH3NH3 ][OH-]
(0.30  x)(x)

 4.4 x 10 - 4
CH3NH2
0.45  x
Simplifies to:
Kb 
[CH3NH3 ][OH-]
(0.30)(x)

 4.4 x 10 - 4
CH3NH2
0.45
(4.4 x 10 - 4)(0.45)
x 
 6.6 x 10 - 4 M  [OH-]
0.30
[CH3NH3+] = 0.30 + x = 0.30 - 6.6 x 10-4 = 0.30 M
[CH3NH2] = 0.45 – x = 0.45 - 6.6 x 10-4 = 0.45 M
[H3O] 
Kw
1 x 10 - 14

 1.5 x 10 - 11 M
-4
[OH ]
6.6 x 10
9 a. HBr is a strong acid, thus it dissociates completely:
HBr(aq) + H2O(l) → Br¯(aq)
[H3 O+ ] =
+ H3O+(aq)
0.010 mol HBr 1 mol H3 O+
x
= 0.010 M
1 L soln
1 mol HBr
Using the Kw of water, find the [OH¯]:
Kw = [H3O+][OH¯] = 1.0 x 10-14
[OH − ] =
Kw
1.0 x 10−14
=
= 1.0 x 10−12 M
[H3 O+ ]
0.010
pH = - log[H3O+] = - log(0.010) = 2.00
pOH = - log[OH¯] = - log(1.0 x 10-12) = 12.00
or
pH + pOH = 14, so
pOH = 14 – pH = 14 – 2.00 = 12.00
9 b. Ca(OH)2 is a strong base, thus it dissociates completely:
Ca(OH)2(aq)
[OH − ] =
→ Ca2+(aq) + 2 OH¯(aq)
0.30 mol Ca(OH)2
2 mol OH −
x
= 0.60 M OH −
1 L soln
1 mol Ca(OH)2
[H3 O+ ] =
Kw
1.0 x 10−14
=
= 1.7 x 10−14 M
[OH − ]
0.60
pH = - log[H3O+] = - log(1.7 x 10-14) = 13.78
pOH = - log[OH¯] = - log(0.60) = 0.22
or
pH + pOH = 14, so
pOH = 14 – pH = 14 – 13.78 = 0.22
9 c. Benzoic acid, C6H5COOH, is a weak acid with a Ka = 6.2 x 10-5. Write the equilibrium
reaction and determine the expressions for the equilibrium amounts, starting with an initial
amount of 0.20 M C6H5COOH. We are assuming that initially, there isn’t any acetate or
hydronium ion present. An x amount of benzoic acid has to be lost to form an x amount of both
benzoate ion and hydronium ion. Thus, at equilibrium, an amount of 0.20 – x of benzoic acid is
present, and x amount of both benzoate ion and hydronium ion are present. This can be
conveniently summarized in the table shown below:
C6H5COOH(aq) + H2O (l)  C6H5COO¯(aq) + H3O+(aq)
initial
change
equil.
0.20 M
-x
0.20 – x
0
+x
x
The equilibrium constant expression for the above equilibrium is:
Ka =
[C6 H5 COO− ][H3 O+ ]
= 6.2 x 10−5
[C6 H5 COOH]
Substituting the equilibrium amounts into the above expression gives:
0
+x
x
(x)(x)
= 6.2 x 10−5
0.20 − x
The above expression may be simplified by making an assumption that the “x” in denominator
expression, 0.20 – x is negligibly small. If the magnitude of the constant is more than 100 times
smaller than the magnitude of the known initial concentration, this assumption is valid. In this
problem, the magnitude of the constant (10-5) is 1000 times smaller than the certainty of the
known concentration, 0.20 M (10-2 or second decimal place), so the x in 0.20 – x can be
neglected. One caveat is that a 0.1% error in the result is acceptable when this assumption is
made.
This simplifies the algebraic expression to:
x2
= 6.2 x 10−5
0.20
x2 = 6.2 x 10-5(0.20) = 1.2 x 10-5
Solve for x:
x = √1.2 x 10−5 = 3.5x 10−3 M = [C6 H5 COO− ] = [H3 O+ ]
Now that x is known, solve for [C6H5COOH] = 0.20 – x = 0.20 – 0.0035 = 0.20. Notice that our
assumption that x was negligibly small relative to the benzoic acid concentration was valid.
Use the Kw of water to find the [OH¯]:
Kw
1.0 x 10−14
−
= [OH ] =
= 2.8 x 10−12 M
+
−3
[H3 O ]
3.5 x 10
To find the pH and pOH: pH = -log[H3O+] = -log(3.5 x 10-3) = 2.45
pOH = 14 – pH = 14 – 2.72 = 11.55
9 d. Morphine is a weak organic base, as indicated by the Kb = 1.6 x 10-6. Write the
equilibrium reaction and determine the expressions for the equilibrium amounts, starting with an
initial amount of 0.050 M C17H19NO3. We are assuming that initially, there isn’t any conjugate
acid of morphine (C17H19NO3H+) or any hydroxide (OH¯) present. To establish an equilibrium
and x amount of morphine has be lost to from an x amount of the conjugate acid and x amount of
hydroxide. The resulting equilibrium amounts are then deduced. Once again, this is most easily
shown by the table below:
C17H19NO3(aq) + H2O(l) 
initial
change
equil.
0.050
-x
0.050 – x
C17H19NO3H+(aq)
0
+x
x
The equilibrium constant expression for the above equilibrium is:
+ OH¯(aq)
0
+x
x
Kb =
[C17 H19 NO3 H + ][OH − ]
= 1.6 x 10−6
[C17 H19 NO3 ]
Substituting the equilibrium amounts into the above expression gives:
(x)(x)
= 1.6 x 10−6
0.050 − x
The above expression may be simplified by making an assumption that the “x” in denominator
expression, 0.050 – x is negligibly small. If the magnitude of the constant is more than 100
times smaller than the magnitude of the known initial concentration, this assumption is valid. In
this problem, the magnitude of the constant (10-6) is 1000 times smaller than the certainty of the
known concentration, 0.050M (10-3 or third decimal place), so the x in 0.050 – x can be
neglected. One caveat is that a 0.1% error in the result is acceptable when this assumption is
made.
This simplifies the algebraic expression to:
x2
= 1.6 x 10−6
0.050
Solve for x:
x2 = 1.6 x 10-6(0.050) = 8.0 x 10-8
x = √8.0 x 10−8 = 2.8 x 10-4 = [C17H19NO3H+] = [OH¯]
Now that x is known, solve for [C17H19NO3] = 0.050 – x = 0.050 – (2.8 x 10-4) = 0.050 M
Notice that our assumption that x was negligibly small relative to the benzoic acid concentration
was valid.
Use the Kw of water to find the [H3O+]:
[H3 O+ ] =
To find the pH and pOH:
Kw
1.0 x 10−14
=
= 3.6 x 10−11
[OH − ]
2.8 x 10−4
pH = -log[H3O+] = -log(3.6 x 10-11) = 10.45
pOH = 14 – pH = 14 – 10.45 = 3.55
10. In general, an acid and a base will form a salt and water. The cation of the salt always
comes from the base and the anion of the salt always comes from the acid (as a memory aid,
keep the consonants and vowels together). Not all salts are neutral. The acid or base properties
of a salt are determined by the relative strengths of the acid and base from which the salt
originated. HCl, HBr, HI, HNO3, HClO4 and the first dissociation of H2SO4 are strong acids.
All other acids are weak. Organic acids are always weak. An easy way to recognize and organic
acid is to look for carbon in the formula. For example, acetic acid, CH3CO2H, is a weak organic
acid. The acidic hydrogen is shown in bold font. This organic acid is often referred to, in
general, as a carboxylic acid. The carboxylic acid group is often written in a formula as –CO2H
or COOH. Group IA metal hydroxides, LiOH, NaOH, KOH, RbOH, CsOH, and in Group 2A,
Ca(OH)2, Sr(OH)2 and Ba(OH)2 are typical strong bases. Ammonia, NH3, is a weak base and
organic compounds called amines are weak bases. Amines are derivatives of ammonia. One or
more of the hydrogens on ammonia have been replaced with a carbon group. For example,
methyl amine is CH3NH2. Once again, the presence of carbon in the formula is a clear
indication the base is weak. A strong acid produces a weak conjugate base; a strong base
produces a weak conjugate acid. A weak acid produces a strong conjugate base; a weak base
produces a strong conjugate acid. If a weak conjugate acid or weak conjugate base is produced,
it means that that species is too weak to behave as an acid or base, respectively. On the other
hand, if a strong conjugate acid or strong conjugate base is produced, then that species will
behave as an acid or base, respectively. This process is called hydrolysis.
10 a. KBr is the salt of KOH and HBr. KOH is a strong base. Thus, K+ cation is a weak
conjugate acid because it originated from a strong base (KOH). It will not behave as an acid,
hence it does not hydrolyze and will solely be a spectator ion in solution. Similarly, Br¯ anion is
a weak conjugate base because it originated from a strong acid (HBr). It will not behave as a
base; hence it does not hydrolyze and will solely be a spectator ion in solution. Since neither the
potassium ion nor the bromide ion is contributing to the hydronium ion or hydroxide ion
concentration in solution, the only source for H3O+ or OH¯ in solution is from the dissociation of
water. Thus, the concentration of hydronium ion will be 10-7, and the resulting pH of the
solution will be 7. This is why KBr is a neutral salt.
10 b. RbNO2 is the salt of RbOH and HNO2. RbOH is a strong base. Thus, Rb+ cation is a
weak conjugate acid and will not hydrolyze. It will behave as a spectator ion in solution and not
affect the pH of the solution. Nitrite ion, NO2¯, is the strong conjugate base of the weak acid,
HNO2. Thus it will behave as a base in solution by accepting a proton from water:
NO2¯ (aq) + H2O(l)  HNO2(aq) + OH¯(aq)
Since hydroxide ion is generated in solution, the solution will be alkaline with a pH > 7.
10. c. NH4NO3 is the salt of NH3 and HNO3. The nitrate ion, NO3¯, is the weak conjugate base
of the strong acid, HNO3, thus it will not hydrolyze. It will behave as a spectator ion in solution.
On the other hand, NH3 is a weak base. Thus, NH4+ is the strong conjugate acid of a weak base,
NH3. Thus, the NH4+ cation will undergo hydrolysis and behave as an acid:
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Since hydronium ion is being generated, the solution is acidic with a pH < 7.
10 d. NH4C6H5CO2 is the salt of ammonia, NH3, a weak base, and benzoic acid, C6H5CO2H, a
weak acid. Thus, the ammonium cation, NH4+, is a strong conjugate acid and will hydrolyze
behaving as an acid and the benzoate ion, C6H5CO2¯ is a strong conjugate base and will
hydrolyze behaving as a base:
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
C6H5CO2¯(aq) + H2O(l)  C6H5CO2H(aq) + OH¯(aq)
There are competing equilibria present and the pH will be decided by whichever equilibrium lies
further towards products. This will be determined by comparing the magnitude of each of the
equilibrium constants. The constants may be calculated from the fact that the KaKb = Kw for
any acid/conjugate base pair or base/conjugate acid pair.
Calculate the Ka for NH4+ using the Kw for water and Kb for ammonia:
Ka
Kw
1.0 x 10−14
=
=
= 5.6 x 10−10
−5
Kb
1.8 x 10
Calculate the Kb for C6H5CO2¯ using the Kw of water and the Ka for C6H5CO2H:
Kb =
Kw
1.0 x 10−14
=
= 1.5 x 10−10
Ka
6.5 x 10−5
Comparison of the Ka and the Kb shows that the Ka for ammonium ion is slightly larger than the
Kb for benzoate ion. Thus the hydrolysis of ammonium lies further towards products than the
hydrolysis of benzoate ion. More hydronium ion is produced in solution that hydroxide ion, so
the solution is acidic with a pH < 7.
11. One must do a stoichiometry problem to determine the moles of H3O+ present after each
addition of NaOH and then calculate the concentration of H3O+ based on the total volume of the
solution.
11 a. The initial concentration of H3O+ will be 0.10 M because HCl is a strong acid and
dissociates completely into H3O+ and Cl¯:
initial
change
“equil.”
HCl(aq) + H2O(l) → H3O+(aq)
0.10 M
0M
-0.10 M
+0.10 M
0M
+0.10 M
+ Cl¯(aq)
0M
+0.10 M
+0.10 M
From the [H3O+], calculate the pH:
pH = -log[H3O+] = -log(0.10) = 1.00
To determine the [H3O+] once OH¯ have been introduced, one must use stoichiometry. First find
the total moles of HCl available, calculate the moles of OH¯ introduced and determine which ion
is in excess. Whichever ion is in excess will determine the pH.
First find the total moles of HCl. This will be used repeatedly throughout the problem.
25.0 mL HCl soln x
1 x 10−3 L soln 0.10 mol HCl
x
= 0.0025 mol HCl
1 mL HCl soln
1 L soln
11 b. Next find the moles of hydroxide after the addition of 10.0 mL of NaOH:
10.0 mL soln x
1 x 10−3 L soln 0.10 mol NaOH
x
= 0.0010 mol NaOH
1 mL soln
1 L soln
The moles of NaOH added equals the moles of HCl that reacts. The moles of HCl remaining is
determined by:
0.0025 total moles HCl – 0.0010 mol HCl reacted = 0.0015 mol HCl remaining
The new total volume of the solution is: 10.0 mL soln + 25.0 mL soln = 35.0 mL soln.
This volume in liters is:
35.0 mL soln x
1 x 10−3 L soln
= 0.0350 L soln
1 mL soln
Calculate the new hydronium ion concentration:
0.0015 mol HCl 1 mol H3 O+
x
= [H3 O+ ] = 0.043 M
0.0350 L soln
1 mol HCl
Calculate the new pH: pH = -log[H3O+] = -log(0.043) = 1.34
11 c. Find the moles of hydroxide after the addition of 25.0 mL of NaOH:
25.0 mL soln x
1 x 10−3 L soln 0.10 mol NaOH
x
= 0.0025 mol NaOH
1 mL soln
1 L soln
The moles of NaOH added equals the moles of HCl that reacts. The moles of HCl remaining is
determined by:
0.0025 total moles HCl – 0.0025 mol HCl reacted = 0 mol HCl remaining
The equivalence point of the titration has been reached. The equivalence point is where the
moles of acid equal the moles of base. The salt formed from this reaction is sodium chloride.
Neither the sodium ion nor the chloride ion undergoes hydrolysis and are spectator ions. As a
result, the only source of hydronium ion in solution is from the dissociation of water. Water
produces 1.0 x 10-7 M H3O+ in solution. Use this concentration to calculate the pH:
pH = -log[H3O+] = -log(1.0 x 10-7) = 7.00
11 d. Find the moles of hydroxide after the addition of 30.0 mL of NaOH:
1 x 10−3 L soln 0.10 mol NaOH
30.0 mL soln x
x
= 0.0030 mol NaOH
1 mL soln
1 L soln
Notice the moles of hydroxide is in excess as the equivalence point has been exceeded.
Find the moles of excess hydroxide:
0.0030 mol OH¯ - 0.0025 mol OH¯ reacted = 0.0005 mol OH¯ in excess
The total volume of solution is: 30.0 mL + 25.0 mL = 55.0 mL
Convert this volume into liters:
55.0 mL soln x
1 x 10−3 L soln
= 0.0550 L soln
1 mL soln
Find the concentration of hydroxide ion:
0.0005 mol OH −
= [OH − ] = 9.0 x 10−3 M
0.0550 L soln
Solve for pOH first, then convert to pH:
pOH = -log[OH¯] = -log(9.0 x 10-3) = 2.04
pH = 14 – pOH = 14 – 2.04 = 11.96
12 a.
Initially, the weak acid, benzoic acid is dissociating and we solve the pH as if the
question read: “What is the pH of a 0.10 M benzoic acid solution?” Thus, the problem is
worked similar to Problem 9c:
Benzoic acid, C6H5COOH, is a weak acid with a Ka = 6.2 x 10-5. Write the equilibrium reaction
and determine the expressions for the equilibrium amounts, starting with an initial amount of
0.20 M C6H5COOH. We are assuming that initially, there isn’t any acetate or hydronium ion
present. An x amount of benzoic acid has to be lost to form an x amount of both benzoate ion
and hydronium ion. Thus, at equilibrium, an amount of 0.20 – x of benzoic acid is present, and x
amount of both benzoate ion and hydronium ion are present. This can be conveniently
summarized in the table shown below:
C6H5COOH(aq) + H2O (l)  C6H5COO¯(aq) + H3O+(aq)
initial
change
0.20 M
-x
0
+x
0
+x
equil.
0.20 – x
x
x
The equilibrium constant expression for the above equilibrium is:
Ka
[C6 H5 COO− ][H3 O+ ]
=
= 6.2 x 10−5
[C6 H5 COOH]
Substituting the equilibrium amounts into the above expression gives:
(x)(x)
= 6.2 x 10−5
0.20 − x
The above expression may be simplified by making an assumption that the “x” in denominator
expression, 0.20 – x is negligibly small. If the magnitude of the constant is more than 100 times
smaller than the magnitude of the known initial concentration, this assumption is valid. In this
problem, the magnitude of the constant (10-5) is 1000 times smaller than the certainty of the
known concentration, 0.20 M (10-2 or second decimal place), so the x in 0.20 – x can be
neglected. One caveat is that a 0.1% error in the result is acceptable when this assumption is
made.
This simplifies the algebraic expression to:
x2
= 6.2 x 10−5
0.20
Solve for x:
x2 = 6.2 x 10-5(0.20) = 1.2 x 10-5
x = √1.2 x 10−5 = 3.5 x 10−3 M = [C6 H5 COO− ] = [H3 O+ ]
Now that x is known, solve for [C6H5COOH] = 0.20 – x = 0.20 – 0.0035 = 0.20. Notice that our
assumption that x was negligibly small relative to the benzoic acid concentration was valid.
To find the pH: pH = -log[H3O+] = -log(3.5 x 10-3) = 2.45
12 b. Finding the pH upon the addition of 10.0 mL of strong base is a more complicated
problem. One has to do a stoichiometry problem first. The balanced chemical reaction taking
places is:
C6H5COOH(aq) + OH¯ (aq) → C6H5COO¯(aq) + H2O(l)
Find the total moles of benzoic acid present and the C6H5COOH(aq):
1 x 10−3 L soln 0.20 mol C6 H5 CO2 H
25.0 mL soln x
x
= 5.0 x 10−3 mol C6 H5 CO2 H
1 mL soln
1 L soln
Find the moles of hydroxide added:
1 x 10−3 L soln 0.10 mol NaOH
1 mol OH −
10.0 mL soln x
x
x
= 1.0 x 10−3 mol OH −
1 mL soln
1 L soln
1 mol NaOH
From the balanced reaction, the moles of hydroxide added are equal to the moles of benzoate ion
formed and the moles of benzoic acid that has reacted.
1.0 x 10−3 mol OH − x
1.0 x 10−3 mol OH − x
1 mol C6 H5 CO−
2
= 1.0 x 10−3 mol C6 H5 CO2− formed
−
1 mol OH
1 mol C6 H5 CO2 H
= 1.0 x 10−3 mol C6 H5 CO2 H reacted
1 mol OH −
So far, the moles of hydroxide added are less than the total moles of benzoic acid, thus some of
the benzoic acid remains unreacted. Since some benzoic acid, a weak acid, is present and some
benzoate ion, the conjugate base of the weak acid, is also present, a buffer solution has been
formed.
Find the moles of unreacted benzoic acid:
5.0 x 10-3 mol C6H5CO2H – 1.0 x 10-3 mol C6H5CO2H reacted = 4.0 x 10-3 mol unreacted
C6H5CO2H
The total volume of the solution at this point is the sum of the volumes of benzoic acid solution
and sodium hydroxide solutions:
25.0 mL soln + 10.0 mL soln = 35.0 mL soln
Convert this volume into liters in preparation to calculate the new initial concentrations of
benzoic acid and benzoate ion:
35.0 mL soln x
1 x 10−3 L soln
= 0.0350 L soln
1 mL soln
Find the new initial concentrations of benzoic acid and benzoate ion:
4.0 x 10−3 mol C6 H5 CO2 H
[C6 H5 CO2 H] =
= 0.11 M C6 H5 CO2 H
0.0350 L soln
[C6 H5 CO2− ]
1.0 x 10−3 mol C6 H5 CO−
2
=
= 0.029 M C6 H5 CO−
2
0.0350 L soln
Use these new initial concentrations to set up the calculation for the buffer equilibrium shown
below:
C6H5COOH(aq) + H2O (l)  C6H5COO¯(aq) + H3O+(aq)
initial
change
equil.
0.11 M
-x
0.11 – x
0.029 M
+x
0.029 + x
0
+x
x
The equilibrium constant expression for the above equilibrium is:
Ka =
[C6 H5 COO− ][H3 O+ ]
= 6.2 x 10−5
[C6 H5 COOH]
Substituting the equilibrium amounts into the above expression gives:
(0.029 + x)(x)
= 6.2 x 10−5
0.11 − x
The above expression may be simplified by making an assumption that the “x” in both the
numerator expression, 0.029 + x and the denominator expression, 0.11 – x are negligibly small.
If the magnitude of the constant is more than 100 times smaller than the magnitude of the known
initial concentration, this assumption is valid. In this problem, the magnitude of the constant (105
) is 1000 times smaller than the certainty of the known concentrations, One caveat is that a
0.1% error in the result is acceptable when this assumption is made.
This simplifies the algebraic expression to:
(0.029)(x)
= 6.2 x 10−5
0.11
Rearrange the algebraic expression to solve for x:
x =
(6.2 x 10−5 )(0.11)
= 2.4 x 10−4 = [H3 O+ ]
0.029
Since x = [H3O+], the pH can be calculated as:
pH = -log[H3O+] = -log(2.4 x 10-4) =3.63
12 c. Finding the pH upon the addition of 25.0 mL of strong base is very similar to the previous
problem. One has to do a stoichiometry problem first. The balanced chemical reaction taking
places is:
C6H5COOH(aq) + OH¯ (aq) → C6H5COO¯(aq) + H2O(l)
Find the total moles of benzoic acid present and the C6H5COOH(aq):
25.0 mL soln x
1 x 10−3 L soln 0.20 mol C6 H5 CO2 H
x
= 5.0 x 10−3 mol C6 H5 CO2 H
1 mL soln
1 L soln
Find the moles of hydroxide added:
1 x 10−3 L soln 0.10 mol NaOH
1 mol OH −
25.0 mL soln x
x
x
= 2.5 x 10−3 mol OH −
1 mL soln
1 L soln
1 mol NaOH
From the balanced reaction, the moles of hydroxide added are equal to the moles of benzoate ion
formed and the moles of benzoic acid that has reacted.
2.5 x 10−3 mol OH − x
2.5 x 10−3 mol OH − x
1 mol C6 H5 CO−
2
= 2.5 x 10−3 mol C6 H5 CO2− formed
−
1 mol OH
1 mol C6 H5 CO2 H
= 2.5 x 10−3 mol C6 H5 CO2 H reacted
1 mol OH −
So far, the moles of hydroxide added are still less than the total moles of benzoic acid, thus some
of the benzoic acid remains unreacted. Since some benzoic acid, a weak acid, is present and
some benzoate ion, the conjugate base of the weak acid, is also present, a buffer solution has
been formed.
Find the moles of unreacted benzoic acid:
5.0 x 10-3 mol C6H5CO2H – 2.5 x 10-3 mol C6H5CO2H reacted = 2.5 x 10-3 mol unreacted
C6H5CO2H
The total volume of the solution at this point is the sum of the volumes of benzoic acid solution
and sodium hydroxide solutions:
25.0 mL soln + 25.0 mL soln = 50.0 mL soln
Convert this volume into liters in preparation to calculate the new initial concentrations of
benzoic acid and benzoate ion:
1 x 10−3 L soln
50.0 mL soln x
= 0.0500 L soln
1 mL soln
Find the new initial concentrations of benzoic acid and benzoate ion:
[C6 H5 CO2 H] =
2.5 x 10−3 mol C6 H5 CO2 H
= 0.050 M C6 H5 CO2 H
0.0500 L soln
2.5 x 10−3 mol C6 H5 CO−
2
=
= 0.050 M C6 H5 CO−
2
0.0500 L soln
Use these new initial concentrations to set up the calculation for the buffer equilibrium shown
below:
[C6 H5 CO2− ]
C6H5COOH(aq) + H2O (l)  C6H5COO¯(aq) + H3O+(aq)
initial
change
equil.
0.050 M
-x
0.050 – x
0.050 M
+x
0.050 + x
0
+x
x
The equilibrium constant expression for the above equilibrium is:
Ka =
[C6 H5 COO− ][H3 O+ ]
= 6.2 x 10−5
[C6 H5 COOH]
Substituting the equilibrium amounts into the above expression gives:
(0.050 + x)(x)
= 6.2 x 10−5
0.050 − x
The above expression may be simplified by making an assumption that the “x” in both the
numerator expression, 0.050 + x and the denominator expression, 0.050 – x are negligibly small.
If the magnitude of the constant is more than 100 times smaller than the magnitude of the known
initial concentration, this assumption is valid. In this problem, the magnitude of the constant (105
) is 100 times smaller than the certainty of the known concentrations, One caveat is that a 0.1%
error in the result is acceptable when this assumption is made.
This simplifies the algebraic expression to:
(0.050)(x)
= 6.2 x 10−5
0.050
The above expression simplifies to x = [H3O+] = 6.2 x 10-5
Since x = [H3O+], the pH can be calculated as:
pH = -log[H3O+] = -log(6.2 x 10-5) = 4.21
Notice how [C6H5CO2H] = [C6H5CO2¯]. This unique point is often referred to as the half
equivalence point. At the half equivalence point, pH = pKa.
12 d. Once again, one must do the stoichiometry problem first to solve for the pH upon addition
of 50.0 mL of strong base. The balanced chemical reaction taking places is:
C6H5COOH(aq) + OH¯ (aq) → C6H5COO¯(aq) + H2O(l)
Find the total moles of benzoic acid present and the C6H5COOH(aq):
1 x 10−3 L soln 0.20 mol C6 H5 CO2 H
25.0 mL soln x
x
= 5.0 x 10−3 mol C6 H5 CO2 H
1 mL soln
1 L soln
Find the moles of hydroxide added:
50.0 mL soln x
1 x 10−3 L soln 0.10 mol NaOH
1 mol OH −
x
x
= 5.0 x 10−3 mol OH −
1 mL soln
1 L soln
1 mol NaOH
The moles of acid equal the moles of base. The benzoic acid has been neutralized and only
sodium benzoate and water are present in solution. The moles of benzoate ion equal the moles of
benzoic acid neutralized:
5.0 x 10−3 mol C6 H5 CO2 H x
1 mol C6 H5 CO−
2
= 5.0 x 10−3 mol C6 H5 CO2−
1 mol C6 H5 CO2 H
The total volume of the solution is: 25.0 mL + 50.0 mL = 75.0 mL. Convert this volume to
liters in preparation for finding the concentration of benzoate ion:
75.0 ml soln x
1 x 10−3 L soln
= 0.070 L soln
1 mL soln
Find the concentration of benzoate ion in solution:
[C6 H5 CO2− ]
5.0 x 10−3 mol C6 H5 CO−
2
=
= 0.067 𝑀 C6 H5 CO−
2
0.0750 soln
Recall at the equivalence point, only the salt is moving around in aqueous solution. In this case
the salt is sodium benzoate. The sodium ion is the weak conjugate acid of a strong base, and
thus will not undergo hydrolysis. It will be a spectator ion. On the other hand, the benzoate ion
is the strong conjugate base of a weak acid and will undergo hydrolysis and behave as a base:
C6H5CO2¯(aq) + H2O(l)  C6H5CO2H(aq) + OH¯(aq)
To evaluate the pH, we need to do a hydrolysis equilibrium problem. First, calculate the
equilibrium constant, Kb, for this reaction from the fact that the KaKb = Kw for any
acid/conjugate base pair (or in other instances, for any base/conjugate acid pair).
Kb =
Kw
1.0 x 10−14
=
= 1.6 x 10−10
Ka
6.2 x 10−5
Set up a table to determine the equilibrium concentrations in solution. Initially, 0.067 M
benzoate is present in solution, to establish equilibrium an “x” amount of benzoate must be used
to form an “x” amount of benzoic acid and “x” amount of hydroxide. Combining these
quantities gives us the equilibrium quantities:
initial
change
equil.
C6H5CO2¯(aq) + H2O(l)  C6H5CO2H(aq) + OH¯(aq)
0.067 M
0
0
-x
+x
+x
0.067 – x
x
x
Substitute these quantities into the equilibrium constant expression:
Kb =
[C6H5CO2H][OH − ]
(x)(x)
=
= 1.6 x 10−10
−
[C6 H5 CO2 ]
0.067 − x
The “x” in the denominator term, 0.067 – x can be neglected because the magnitude of the
constant is 107 times smaller than the last certain digit of the known concentration, hence the
expression will simplify to:
x2
= 1.6 x 10−10
0.067
Solve for x:
x2 = (1.6 x 10-10)(0.067) = 1.1 x 10-11
x = √1.1 x 10−11 = 3.3 x 10−6 = [OH − ]
There are two common methods for finding the pH. One method would be to find the pOH and
then convert to pH:
pOH = -log[OH¯] = -log[3.3 x 10-6] = 5.48
pH = 14 – pOH = 14 – 5.48 = 8.52
Alternatively, one could first calculate the [H3O+] using the Kw expression for water and then
find the pH:
Kw
1.0 x 10−14
[H3 O ] =
=
= 3.04 x 10−9 M
[OH − ]
3.3 x 10−6
+
pH = -log[H3O+] = -log(3.04 x 10-9) = 8.52
ACID/BASE “TOOLBOX”
Shown here are the basic tools one needs to be able to calculate hydronium ion, hydroxide ion,
pH, pOH, Ka, and Kb. when dealing with acid/base problems. I call it my “toolbox” and hope
you find it useful!
Fundamental equations:
[H3O+][OH¯] = Kw = 1.0 x 10-14
pX = -logX
KaKb = Kw
All the ways you can find [H3O+] and [OH¯]:
[H3O ] 
Kw
[OH-]
[H3O  ]  10 -
pH
[OH-] 
Kw
[H3O  ]
[OH-]  10 - pOH
All the ways you can find pH and pOH:
pH + pOH = 14
pH = -log[[H3O+]
pH = 14 – pOH
pOH = 14 – pH
pOH = -log[OH¯]
KaKb = Kw is used in hydrolysis problems. Use it to calculate the Kb of the conjugate base of a weak
acid, or conversely, the Ka of the conjugate acid of a weak base:
Kb 
Kw
Ka
Ka 
Kw
Kb
In general, pX is a power function, and it means you take the negative logarithm of the variable or
number, hence, pX = -logX.