Review Sheet for Chapter 15
Common ion effect is an application of Le Chatelier’s principle. It is the shift in an equilibrium position
caused by the addition or presence of an ion involved in the equilibrium reaction.
A buffered Solution is one that resists a change in its pH when either hydroxide ions or protons are added.
(Example: blood contains HCO-3 and H2CO3.)
A buffered solution may contain a weak acid & its salts or a weak base & its salt containing a common
ion.
When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the
resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium
calculations.
Below approach can be apply to weak acids & bases and buffered solutions:
1. List the major species in the solution
2. Choose the species that can produce H+ or OH-, and write balanced equation for the reactions
producing H+ or OH3. Using the values of equilibrium constants for the reactions you have written, decide which
equilibrium will dominate in producing H+ or OH4. Write the equilibrium expression for the dominant equilibrium
5. List the initial concentrations of the species participating in the dominant equilibrium.
6. Define the change needed to achieve equilibrium; that is, define x.
7. Write the equilibrium concentrations in terms of x.
8. Substitute the equilibrium concentrations into the equilibrium expression
9. Solve for x the “easy” way; that is, by assuming that [HA]0 – x  [HA]0.
10. Use the 5% rule to verify whether the approximation is valid.
11. Calculate [H+] and pH
How does a buffer work?
Suppose a buffered solution contains large amount of weak acid HA and its conjugate base A-. When
OH- are added to the solution, since weak acid represents the best source of protons, the following
reaction occurs:
OH- + HA A- +H2O
The net result is that OH ions are not allowed to accumulate but are replaced by A- ions.
The equilibrium expression for the dissiociation of HA: K a  HHAA.  
The equilibrium concentration of H+, and the pH is determined by the ratio [HA]/[A-]
The log form of expression for Ka is called the Henderson- Hasselballch equation and is useful for
calculating the pH of solutions when the ration [HA]/[A-] is known.

 
A 
pH  pK a  log HA
  pK a  log

  
base
acid
Buffered solutions also can be formed from weak base and corresponding conjugate acid. (example:
weak base B reacts with any H+ added: B + H+  BH+, and the conjugate acid BH+ + OH-  B + H2O)
When H+ is added to a buffered solution, it reacts essentially to completion with the weak acid present:
H+ + A-  HA or : B + H+  BH+
When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present:
OH- + HA  A- + H2O or OH- + BH+  B + H2O
The buffering capacity of a buffered solution represents the amount of protons or hydroxide ions the
buffer can absorb without signigicant change in pH.
The pH of a buffered solution is determined by the ratio [A-]/[HA]. The capacoty of a buffered solution is
determined bu the mahnitutdes of [HA] and [A-].
[A-]/[HA]  ratio = 1 for most effective buffer
Titration is used to determine the amount of acid or base in a solution. This process involves a solution of
known concentration (the titrant) delivered from a buret into the unknown solution until the substance
being analyzed is just consumed.
The stoichiometric (equivalence) point is often signaled by the color change of an indicator.
pH curve or titration curve is a plotting of the pH of the solution being analyzed as as a function of the
amount of titrant added.
The net ionic reaction for a strong acid- strong base tirtation is H  aq   OH  aq  
 H 2 Ol 
For titration mole is inconveniently large;as a result, use millimole (abbreviated mmol)
# of mmol=Volume (in mL) x molarity
Equivalence (stoichiometric) point: the point in the titration where an amount of base has been added to
exactly react with all the acid originally present.
Strong acid with a strong base has the following charateriststics:
Before the equivalence point, [H+] and hence the pH can be calculated by dividing the # of millimoles of
H+ remaining by the total volume of the solution in millimeters. At equivalence point the pH is 7
After equivalence point, [OH-] can be calculated by dividing the # of millimoles of excess OH- by the
totla volume of the solution. Then [H+] is obtained from Kw.
Weak acids with strong base titration involves a two step procedure:
1. A stoichiometry problem. The reaction of hydroxide ion with the wak acid is assumed to run to
completion, and the concentration of the acid remaining and the conjugate base forned are
determined.
2. An equilibrium problem. The position of the weak acid equilibrium is determined and the pH is
caluclated
Characteristics of weak acids with strong base titration:
The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7.
Near the beginning of the titration of the weak acid, the pH increases more rapidly than it does in the
strong acid case. It levels off near the halfway point and then increases rapidly again. The leveling off
near the halfway point is casued by buffering effects.
Two methods for determining the equivalence point of an acid-base tirtration:
1. Use a pH meter to monitor the pH and plot the titration curve. Center vertical region of the pH
curve indicates the equivalence point
2. Use acid-base indicator, which marks the end point of a titration by changing color.
Phenolphthalein is a commonly used indicator
The end point is defined by the change in color of the indicator. The equivalence point is defined by the
reaction stoichiometry
For most indicators, the ratio of the minor form to the major form must be 1/10 before a color change is
apprarnet.
The principles of equilibrium can be applied when a solid dissolves in water.
The solubility product is Ksp is an equilibrium constant defined by the law of mass action.
Whether a precipitate will form or not can be predicted when two solutions are mixed by using the ion
product Q, which is defined just like Ksp except initial concentrations of the ions are used. If Q is greater
than Ksp , the salt will precipitate.
The equilibrium constant for this reaction is less than 1. What is the strongest base in this system?
 AC 6 H 5 OH aq 
B CN  aq 
C HCN aq 
D C 6 H 5 Oaq 
HOCl aq  
 H aq   OCl aq 
The ionization of hypochlorous acid represented above has K= 3.0x10-8 at 25 C. What is K for this
reaction?
OCl aq  H 2 Ol  
 OH aq  HOCl aq 
 A3.3x10 7
B 3.0 x10 8
C 3.0 x10 6
D 3.3x10 7
What is the pH of a 0.15M solution of formic acid, HCOOH? (Ka for formic acid=1.9x10-4)
 A1.49
B 2.27
C 3.72
D 4.55
Which of these mixtures consittute buffer solutions?
Mixture 1
25ml of .10 M HNO3 and 25mL of .10M NaNO3
25mL of .10M HC2H3O2 and 25mL of .10M NaOH
(A) 1 only
(B) 2 only
(C) both 1 and 2
(D) neither 1 nor 2
The dihydrogen phosphate ion undergoes these reactions in water.

H 2 PO4 aq   H 2 Ol  
 H 3Oaq   HPO42 aq  K=6.2x10-8

H 2 PO4 aq   H 2 Ol  
 OH aq   H 3 PO4 aq  K=1.6x10-7

what is the conjugate base of H 2 PO4  aq  ?
 AHPO42  aq 
B H 2 Ol 
C OH aq 
D H 3 PO4 aq 
HOCl aq  
 H aq   OCl aq 
Hypochlorous acid, HOCl, is a weak acid commonly used as bleaching agent. The acid-dissociation
constant Ka, for the reaction represented above is 3.2x10-8
(a) calculate the H  of a .14 molar solution of HOCl
(b) write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is
dissolved in water and calculate the numerical value of the equilibrium constant for the reaction
(c) calculate the pH of a solution made by combining 4.0 milliliters of 0.14-molar HOCl and 10.0
milliliters of 0.56-molar NaOH.
(d) Household bleach is made by dissolving chlorine gas in water, as represented below. Calculate
the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar
 
Cl2  g   H 2 Ol  
 H aq   Cl  aq   HOCl aq 
Answer Key For Calculation
Mutiple Choices Answers
D
A
OH HOCl   answer
OCl 


H OCl   K  3.0 x10


HOCl 

8
,

H  

1.0 x10 14

,
OH 
1.0 x10 14 OCl 
1.0 x10 14
8

3
.
0
x
10

,
answer
OH  HOCl 
B





1.0 x10 14
OCl 

OH
 3.0 x10 8
HOCl 

answer  3.3x10 7
HCOOH  H   COOH 
i
0.15
0
0

x
x
x
eq
0.15  x
x
x
x2
 1.9 x10  4 ,
.15  x
D
 
x  H ,
 
 log H   answer
A
Free
Re sponse
Answer
a HOCl aq   H aq   OCl aq 
i
0.14
0
0

x
x
x
x
x
eq 0.14  x
K a  3.2 x10 8 
x2
,
.14  x
b NaOCl s   H 2 Ol   HOCl aq   Na aq   OH aq 
Kb 
K w 1.0 x10 14

 3.1x10 7
8
Ka
3.2 x10
 
x  6.7 x10 5 M  H 
c  HOCl aq 
i
OH aq 

(.0400)(0.14 M )
(0.100)(. 0056 M )
 0.0056mmol
 .0056mmol
mo0l
HOCl  mol
OH 
OCl aq 

0.0056mol
i
 0.11
0.050 L

x
indicate eq.
H 2 O l 
0.11  x
eq.
mol ratio !!
   1molHOCl 
H   HOCl  0.065
1mol H
po int

 HOCl aq 
 OH aq 
0
0
x
x
x2
 3.1x10 7
0.11  x
x  1.85 x10  4 M  OH 
x
x
pOH   log OH 
Kb 

 
  3.73
pH  14.00  pOH  10.27
d u sin g
so

 OCl aq   H 2 Ol 


pH   log 0.065  1.19