Chapter 15: Applications of
Aqueous Equilibria - Titrations
GENERAL CHEMISTRY: ATOMS FIRST
John E. McMurray – Robert C. Fay
Prentice Hall
Titration
• in an acid-base titration, a solution of known
concentration (titrant) is slowly added from a burette
to a solution of unknown concentration in a flask until
the reaction is complete
 when the reaction is complete we have reached the endpoint
of the titration
• an indicator may be added to determine the endpoint
 an indicator is a chemical that changes color when the pH
changes
• when the moles of H3O+ = moles of OH−, the titration
has reached its equivalence point
Titration
The Titration Curve
• is a plot of pH vs. amount of added titrant
• the inflection point of the curve is the equivalence
•
•
point of the titration
prior to the equivalence point, the unknown solution in
the flask is in excess, so the pH is closest to its pH
the pH of the equivalence point depends on the pH of
the salt solution
 equivalence point of neutral salt, pH = 7
 equivalence point of acidic salt, pH < 7
 equivalence point of basic salt, pH > 7
• beyond the equivalence point, the known solution in
the flask (from the burette) is in excess, so the pH
approaches its pH
Titration Curve:
Known Strong Base Added to a Strong Acid
Strong Acid/Strong Base
Titration of 25 mL of 0.100 M HCl with
0.100 M NaOH
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq)
• initial pH = -log(0.100) = 1.00
• initial moles HCl = (0.0250 L)(0.100 mol/L) = 2.50 x 10-3 mol HCl
• now add 5.0 mL (0.0050L) NaOH
0.100 mol NaOH
1L
 5.0  10 4 mol NaOH
0.0050 L NaOH 
1 mol HCl
1 mol NaOH
 5.0  10 4 mol HCl used
5.0  10 4 mol NaOH 
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
2.50  10 initial mol HCl - 5.0 10 mol HCl used
-3
-4
 2.00  10 -3 mol HCl excess
2.00  10 -3 mol HCl excess
0.0250 L HCl  0.0050 L NaOH
 0.0667 M HCl  [H 3 O  ]
pH  -log[H 3O  ]
pH  -log 0.0667   1.18
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• when 25 mL NaOH added you reach the equivalence point
• no HCl and no NaOH present in flask, pH = 7.00
• after 30 mL (0.0300L) NaOH added, after the equivalence point:
0.100 mol NaOH
0.0300 L NaOH 
1L
 3.00  103 mol NaOH
3.00 10-3 mol NaOH - 2.50 10-3 mol NaOH
(amount of NaOH used to reach equivalenc e pt)
 5.0 10-4 mol NaOH excess
5.0  10 -4 mol NaOH excess
0.0250 L HCl  0.0300 L NaOH
 0.00909 M NaOH  [OH  ]
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
[OH-] = 9.09 x 10-3
[H3O  ][OH  ]  K w
Kw
1 10 14
12
[H 3O ] 


1
.
1
0

10
[OH  ] 9.09  10 3

pH  - log[H 3O ]


pH  - log 1.10  10-9  11.96
pOH  -log[OH  ]
pH  14.00 - pOH


pOH  -log 9.09  103  2.04
pH  14.00 - 2.04  11.96
Adding NaOH to HCl
added
5.0
mL
25.0 mL
30.0
0.100
mLNaOH
M
NaOH
HCl
0.00200
0.00250 mol HCl
0.00050
NaOH
1.18
pH = 11.96
1.00
added
added35.0
10.0mL
mLNaOH
NaOH
0.00100
0.00150mol
molNaOH
HCl
pH
pH==12.22
1.37
added40.0
15.0mL
mLNaOH
NaOH
added
0.00100mol
molNaOH
HCl
0.00150
pH==12.36
1.60
pH
added
added20.0
50.0mL
mLNaOH
NaOH
0.00050
0.00250mol
molHCl
NaOH
pH
pH==1.95
12.52
added 25.0 mL NaOH
equivalence point
pH = 7.00
Titration of 25 mL of 0.100 M HCl with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(l)
• after equivalence point
mol NaOH excess
L HCl  L NaOH
 M NaOH  [OH  ]


[H3O ][OH ]  K w
5.0  10-4 mol NaOH
0.0250 L HCl  0.0300 L NaOH
 0.0091 M NaOH  [OH ]
[H 3O  ] 
Kw
[OH ]
1  10 14
12


1
.
1
0

10
9.1  10 3
pH  - log[H 3O ]
pH  - log 1.10 10-12   11.96
Titrating Weak Acid with a Strong Base
• the initial pH is that of the weak acid solution
calculate like a weak acid equilibrium problem
e.g., 15.5 and 15.6
• before the equivalence point, the solution
becomes a buffer
calculate mol HAinit and mol A−init using reaction
stoichiometry
calculate pH with Henderson-Hasselbalch using mol
HAinit and mol A−init
• half-neutralization pH = pKa
Titrating Weak Acid with a Strong Base
• at the equivalence point, the mole HA = mol Base,
so the resulting solution has only the conjugate
base anion in it before equilibrium is established
mol A− = original mole HA
calculate the volume of added base like Ex 4.8
[A−]init = mol A−/total liters
calculate like a weak base equilibrium problem
e.g., 15.14
• beyond equivalence point, the OH is in excess
[OH−] = mol MOH xs/total liters
[H3O+][OH−]=1 x 10-14
Titration of a weak acid, 25 mL of 0.100 M
HCHO2,with a strong base, 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(l)
Ka = 1.8 x 10-4 = [CHO2-] [H3O+]
[HCHO2]
Titration of a weak acid, 25 mL of 0.100 M
HCHO2,with a strong base, 0.100 M NaOH
HCHO2(aq) + H2O(l)  CHO2-(aq) + H3O+(aq)
[HCHO2] [CHO2-] [H3O+]
initial
0.100
0.000
≈0
change
-x
+x
+x
x
x
equilibrium 0.100 - x
4.2  10 3
 100 %  4.2%  5%
0.100
Ka = 1.8 x 10-4
[CHO 2  ][H 3O  ]
Ka 
[HCHO 2 ]
2



x
x
x
1.8  10  4 

0.100  x  0.100
x  [H 3O  ]  4.24  10 3 M
pH  - log[H 3O  ]


 -log 4.24  10-3  2.37
Titration of a weak acid, 25 mL of 0.100 M
HCHO2,with a strong base, 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(l)
• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3
mols start
HA
A-
OH−
2.50E-3
0
0
-
-
5.0E-4
mols added
mols after
2.00E-3 5.0E-4
given Ka = 1.8 x 10-4
pK a  - logK a


 -log 1.8  10-4  3.74
≈0
add 5.0 mL NaOH
0.100 mol NaOH
1L
 5.0  10 4 mol NaOH
0.0050 L NaOH 
 mol CHO2 

pH  pK a  log
 mol HCHO 
2

 5.0 10-4 

pH  3.74  log 
-3 
 2.00 10 
pH  3.14
Titration of a weak acid, 25 mL of 0.100 M
HCHO2,with a strong base, 0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3
• at equivalence
CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq)
A-
OH−
mols Before 2.50E-3
0
0
mols added
-
-
2.50E-3
mols After
0
2.50E-3
≈0
HA
added 25.0 mL NaOH
0.100 mol NaOH
1L
 2.50  103 mol NaOH
0.0250 L NaOH 
2.50  10-3 mol CHO2
2.50  10-2 L HCHO 2  2.50  10-2 L NaOH
 5.00  10-2 M CHO2
Titration of a weak acid, 25 mL of 0.100 M
HCHO2,with a strong base, 0.100 M NaOH**
• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3
Kw
• at equivalence
K


CHO2−(aq)
+ H2O(l)  HCHO2(aq) +
OH−
b , CHO 2
(aq)
[HCHO2] [CHO2-] [OH−] K = 5.6 x 10-11
b
initial
0
≈0
0.0500
change
+x
-x
+x
equilibrium
x
5.00E-2-x
x
[OH-] =
1.7 x 10-6 M
[H 3O  ] 
Kw
[OH ]
Kb 
Ka
1  10 14
11


5
.
6

10
1.8  10  4
[HCHO 2 ][OH  ]
[CHO 2  ]
2



x
x
x
5.6  10 11 

0.0500  x  0.0500
x  [OH ]  1.7  10  6 M
1  10 14
9


5
.
9

10
1.7  10  6
pH  - log[H 3O  ]


 -log 5.9  10-9  8.23
Adding NaOH to HCHO2
initial HCHO
added
30.0mL
35.0
5.0
10.0
25.0
mL
solution
NaOH
2NaOH
0.00050 molpoint
0.00100
0.00250
0.00200
0.00150
equivalence
NaOH 2xs
HCHO
pH = 3.56
0.00250
11.96
12.22
2.37
3.14
mol CHO2−
−]
[CHO
= 0.0500
M
added
NaOH
212.5
init mL
added
mL NaOH
−] 40.0
[OH
=
1.7
x 10-6 2
0.00125
eq mol HCHO
0.00150 mol NaOH xs
pH = 8.23
pH = 3.74
12.36= pKa
half-neutralization
added 50.0
15.0 mL NaOH
0.00100 mol NaOH
0.00250
HCHO2xs
pH = 12.52
3.92
added 20.0 mL NaOH
0.00050 mol HCHO2
pH = 4.34
A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the volume of KOH at the
equivalence point
Write an equation for
the reaction for B with
HA.
Use Stoichiometry to
determine the volume
of added B
0.0400 L HNO 2 
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L
 0.0400 L
1 mL
0.100 mol HNO 2 1 mol KOH
1 L KOH


1 L NO2
1 mol HNO 2 0.200 mol KOH
 0.0200 L KOH
0.0200 L 
1 mL
 20.0 mL
0.001 L
A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH after adding 5.00 mL
KOH
Write an equation
for the reaction for
B with HA.
Determine the
moles of HAbefore &
moles of added B
Make a
stoichiometry table
and determine the
moles of HA in
excess and moles
A made
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L 0.100 mol HNO 2

 0.00400 mol HNO 2
1 mL
1L
0.001 L 0.200 mol KOH
5.00 mL 

 0.00100 mol KOH
1 mL
1L
HNO2
NO2-
OH−
0
≈0
mols Before 0.00400
0.00100
mols added
mols After 0.00300 0.00100 ≈ 0
A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH after adding 5.00 mL
KOH.
Write an equation for
the reaction of HA
with H2O
Determine Ka and pKa
for HA
Use the HendersonHasselbalch Equation
to determine the pH
HNO2 + H2O  NO2 + H3O+
Table 15.5 Ka = 4.6 x 10-4


pK a   log K a   log 4.6 104  3.15
 NO2  

pH  pK a  log 
 HNO 2 


HNO
NO2OH−
 02.00100
pH  3.15  log 
  2.67
≈0
mols Before 0.00400
 0.003000 
0.00100
mols added
mols After 0.00300 0.00100 ≈ 0
A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH at
the half-equivalence point
Write an equation
for the reaction for
B with HA.
Determine the
moles of HAbefore &
moles of added B
Make a
stoichiometry table
and determine the
moles of HA in
excess and moles
A made
HNO2 + KOH  NO2 + H2O
40.0 mL 
0.001 L 0.100 mol HNO 2

 0.00400 mol HNO 2
1 mL
1L
at half-equivalence, moles KOH = ½ mole HNO2
HNO2
NO2-
OH−
0
≈0
mols Before 0.00400
0.00200
mols added
mols After 0.00200 0.00200 ≈ 0
A 40.0 mL sample of 0.100 M HNO2 is titrated with
0.200 M KOH. Calculate the pH at
the half-equivalence point.
Write an equation for
the reaction of HA
with H2O
Determine Ka and pKa
for HA
Use the HendersonHasselbalch Equation
to determine the pH
HNO2 + H2O  NO2 + H3O+
Table 15.5 Ka = 4.6 x 10-4


pK a   log K a   log 4.6 104  3.15
 NO2  

pH  pK a  log 
 HNO 2 


HNO
NO2OH−
 02.00200
pH  3.15  log 
  3.15
0
≈0
mols Before 0.00400
 0.00200
0.00200
mols added
mols After 0.00200 0.00200 ≈ 0
Titration Curve of a Weak Base with
a Strong Acid
Titration of a Polyprotic Acid
• if Ka1 >> Ka2, there will be two equivalence
points in the titration
the closer the Ka’s are to each other, the less
distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M H2SO3
with 0.100 M NaOH
Monitoring pH During a Titration
• the general method for monitoring the pH during the
course of a titration is to measure the conductivity of
the solution due to the [H3O+]
 using a probe that specifically measures just H3O+
• the endpoint of the titration is reached at the
•
equivalence point in the titration – at the inflection
point of the titration curve
if you just need to know the amount of titrant added to
reach the endpoint, we often monitor the titration with
an indicator
Monitoring pH During a Titration
Monitoring a Titration with
an Indicator
• for most titrations, the titration curve shows a
very large change in pH for very small additions
of base near the equivalence point
• an indicator can therefore be used to determine
the endpoint of the titration if it changes color
within the same range as the rapid change in pH
pKa of H-Indicator ≈ pH at equivalence point
Indicators
• many dyes change color depending on the pH of the solution
• these dyes are weak acids, establishing an equilibrium with
the H2O and H3O+ in the solution
H-Ind(aq) + H2O(l)  Ind(aq) + H3O+(aq)
• the color of the solution depends on the relative
concentrations of Ind:HInd
 when Ind:H-Ind ≈ 1, the color will be mix of the colors of
Ind and HInd
 when Ind:H-Ind > 10, the color will be mix of the colors of
Ind
 when Ind:H-Ind < 0.1, the color will be mix of the colors of
H-Ind
Phenolphthalein
Methyl Red
H
C
(CH3)2N
H
C
C
H
C
C
C
H
N
(CH3)2N
C
OH-
C
N
N
CH
C
C
H
H
C
H
C
NaOOC
H
C
C
C
H
N
H
C
H
H3O+
H
C
N
H
C
N
C
C
H
CH
C
NaOOC
C
H
Acid-Base Indicators
Chapter 15: Applications of Aqueous
Equilibria-Buffers
GENERAL CHEMISTRY: ATOMS FIRST
John E. McMurray – Robert C. Fay
Prentice Hall
Buffers
• buffers are solutions that resist changes in pH
when an acid or base is added
• they act by neutralizing the added acid or base
• there is a limit to their neutralizing ability,
eventually the pH changes
• Solution made by mixing a weak acid with a
soluble salt containing its conjugate base anion
Formation of an Acid Buffer
How Acid Buffers Work
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
• buffers follow Le Châtelier’s Principle
• buffers contain significant amount of weak acid, HA
 The HA molecules react with added base to neutralize it
 the H3O+ combines with OH− to make H2O
 H3O+ is then replaced by the shifting equilibrium
• buffer solutions also contain significant amount of
conjugate base anion, A−
 The A− molecules react with added acid to make more HA
and keep H3O+ constant
How Buffers Work
H2O
new
HA
HA
HA

Added
H3O+
A−−
+
H3O+
How Buffers Work
H2O
new
A−
HA
HA

Added
HO−
A−−
+
H3O+
Common Ion Effect
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
Le
Châtelier’s Principle
• adding a salt, NaA, containing the acid anion,
shifts the position of equilibrium to the left (A−
is the conjugate base of the acid)
• this lowers the H3O+ ion concentration and
causes the pH to be higher
Common Ion Effect
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Write the reaction for
the acid with water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Ka for HC2H3O2 = 1.8 x 10-5
H2O + HC2H3O2  C2H3O2 + H3O+
[HA]
initial
change
equilibrium
0.100
[A-] [H3O+]
0.100
≈0
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
H2O + HC2H3O2  C2H3O2 + H3O+
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
initial
change
[HA]
[A-]
0.100
0.100
x
+x
equilibrium 0.100 x 0.100 + x
[H3O+]
0
+x
x
[C2H3O-2 ][H3O ] 0.100  x x 
Ka 

HC2H3O2 
0.100  x 
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
[H3O+]
since Ka is very small,
0.100
0.100
initial
≈0
approximate the
change
-x
+x
+x
[HA]eq = [HA]init and
[A−]eq = [A−]init solve for x equilibrium 0.100
0.100x 0.100
0.100+x
x
determine the value of Ka
[HA]
[C2H3O-2 ][H3O ] 0.100  x x 
Ka 

HC2H3O2 
0.100  x 
[C2H3O-2 ][H3O ] 0.100  x 
Ka 

HC2H3O2 
0.100 
1.8  10 5  x
[A-]
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
check if the
approximation is
valid by seeing if
x < 5% of
[HC2H3O2]init
[HA]
initial
change
equilibrium
0.100
-x
0.100
[A-]
[H3O+]
0.100
≈0
+x
+x
0.100
x
x = 1.8 x 10-5
5
1.8  10
 100 %  0.018 %  5%
1
1.00  10
the approximation is valid
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
substitute x into the
equilibrium
concentration
definitions and solve
[HA]
[A-]
[H3O+]
≈0
+x
0.100
0.100
initial
change
-x
+x
0.100x 0.100
0.100
+ x 1.8E-5
x
equilibrium 0.100
x = 1.8 x 10-5
HC 2H3O2   0.100  x  0.100  1.8  105   0.100 M


[C2H3O2 ]  0.100  x  0.100  1.8 10
[H3O ]  x  1.8  10 5 M
5
  0.100 M
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
substitute [H3O+]
into the formula for
pH and solve
[HA]
[A-]
0.100
initial
change
-x
equilibrium 0.100
0.100

pH  -log H3O

  log 1.8  10

+x
[H3O+]
≈0
+x
0.100
1.8E-5

  4.74
5
What is the pH of a buffer that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
check by substituting
[HA] [A-]
[H3O+]
the equilibrium
0.100
0.100
initial
≈0
concentrations back into
-x
+x
+x
the equilibrium constant change
expression and
1.8E-5
equilibrium 0.100 0.100
comparing the calculated

Ka to the given Ka
[C H O ][ H O ]
Ka 
the values match
2
3
2
3
HC 2H3O 2 

0.100 1.8  10 5 
5

 1.8  10
0.100 
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Write the reaction for
the acid with water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
Ka for HF = 7.0 x 10-4
H2O + HF  F + H3O+
[HA]
initial
change
equilibrium
0.14
[A-]
[H3O+]
0.071
≈0
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
Ka for HF = 7.0 x 10-4
H2O + HF  F + H3O+
[HA]
[A-] [H3O+]
0.14
0.071
initial
0
x
change
+x
+x
equilibrium 0.14 x 0.071 + x
x
[F ][H3O ] 0.071  x x 
Ka 

HF
0.14  x 
-

What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
[H3O+]
since Ka is very small,
0.14
0.071
initial
≈0
approximate the
change
-x
+x
+x
[HA]eq = [HA]init and
[A−]eq = [A−]init solve for x equilibrium 0.14
0.100+x
x 0.071
0.012
x
determine the value of Ka
K a  10  pK a  10 3.15
K a  7.0  10  4
[HA]
[A-]
[F- ][H3O ] 0.071  x x 
Ka 

HF
0.14  x 
K a  7.0  10 4 
0.071x 
0.14
1.4  10 3  x
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
check if the
approximation is
valid by seeing if
x < 5% of
[HC2H3O2]init
[HA]
initial
change
equilibrium
0.14
-x
0.14
[A2-] [H3O+]
0.071
≈0
+x
+x
0.071
x
x = 1.4 x 10-3
3
1.4  10
 100 %  1%  5%
1
1.4  10
the approximation is valid
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
substitute x into the
equilibrium
concentration
definitions and solve
[HA]
[A2-]
[H3O+]
≈0
+x
0.14
0.071
initial
change
-x
+x
0.14x 0.071
0.072+ x 1.4E-3
x
equilibrium 0.14
x = 1.4 x 10-3
HF  0.14  x  0.14  1.4 103   0.14 M


[C2H3O2 ]  0.071  x  0.071  1.4 10
[H3O ]  x  1.4  10 3 M
3
  0.072 M
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium

pH  -log H3O

  log 1.4  10

[HA]
[A-]
0.14
0.071
-x
+x
[H3O+]
≈0
+x
0.14
0.072
1.4E-3

  2.85
3
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
check by substituting
[HA] [A-]
the equilibrium
0.14
0.071
initial
concentrations back into
-x
+x
the equilibrium constant change
expression and
0.072
equilibrium 0.14
comparing the calculated
Ka to the given Ka

the values are
close enough
[H3O+]
≈0
+x
1.4E-3
[F ][ H3O ]
Ka 
HF

0.072  1.4  10 3
4

 7.2  10
0.14 


Henderson-Hasselbalch Equation
• calculating the pH of a buffer solution can be
simplified by using an equation derived from
the Ka expression called the HendersonHasselbalch Equation
• the equation calculates the pH of a buffer from
the Ka and initial concentrations of the weak
acid and salt of the conjugate base
as long as the “x is small” approximation is valid
[conjugate base anion] initial
pH  pK a  log
[weak acid] initial
Deriving the Henderson-Hasselbalch Equation
[A - ][ H 3O  ]
Ka 
HA 
 [HA] 

[H 3O ]  K a  - 
 [A ] 
Take the log of both sides and multiply by -1

 [HA]  
pH   log K a    log -  
 [A ]  

pK a  - log K a

 [HA]  
pH  pK a    log -  
 [A ]  

  [HA]  
 log[ H3O ]   log K a  -  
Remember:
[A
]




[HA]
[A  ]
 log   log




[HA]
[HA]
[A ]
 log[ H3O ]   log K a    log -  
 [A ]  

Therefore
pH  - log[H 3O ]
 [A - ] 

pH  pK a  log

[HA]



What is the pH of a buffer that is 0.050 M
HC7H5O2 and 0.150 M NaC7H5O2?
Assume the [HA] and
[A-] equilibrium
concentrations are the
same as the initial
Substitute into the
Henderson-Hasselbalch
Equation
Check the “x is small”
approximation
2.2 10 5
100%  0.044%
0.050
which is  5%
HC7H5O2 + H2O  C7H5O2 + H3O+
Ka for HC7H5O2 = 6.5 x 10-5
pK a   log K a 
  log 6.5  10   4.187
5
 [A - ] 

pH  pK a  log

[HA
]


 0.150  
pH  4.187  log

 0.050 
pH  4.66

[H3O ]  10
-pH
[H3O ]  10-4.66  2.2  10 5
What is the pH of a buffer that is 0.14 M HF
(pKa = 3.15) and 0.071 M KF?
find the pKa from the
given Ka
Assume the [HA] and
[A-] equilibrium
concentrations are the
same as the initial
Substitute into the
Henderson-Hasselbalch
Equation
Check the “x is small”
approximation
HF + H2O  F + H3O+
 [A - ] 

pH  pK a  log

[HA
]


 0.071 
pH  3.15  log
  2.86
 0.14 
[H3O  ]  10-pH
[H3O  ]  10-2.86  1.4  103
1.4  103
 100%  1%  5%
0.14
Do I Use the Full Equilibrium Analysis or
the Henderson-Hasselbalch Equation?
•
•
a)
b)
•
the Henderson-Hasselbalch equation is generally
good enough when the “x is small” approximation is
applicable
generally, the “x is small” approximation will work
when both of the following are true:
the initial concentrations of acid and salt are not very
dilute
the Ka is fairly small
for most problems, this means that the initial acid and
salt concentrations should be over 1000x larger than
the value of Ka
How Much Does the pH of a Buffer
Change When an Acid or Base Is Added?
•
•
1.
although buffers do resist change in pH when acid or
base are added to them, their pH does change
calculating the new pH after adding acid or base
requires breaking the problem into 2 parts
a stoichiometry calculation for the reaction of the
added chemical with one of the ingredients of the
buffer to reduce its initial concentration and increase
the concentration of the other
 added acid reacts with the A− to make more HA
 added base reacts with the HA to make more A−
2. an equilibrium calculation of [H3O+] using the new
initial values of [HA] and [A−]
What is the pH of a buffer that has 0.100 mol HC2H3O2
and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol
NaOH added to it?
If the added chemical is
a base, write a reaction
for OH− with HA. If the
added chemical is an
acid, write a reaction
for it with A−.
Construct a
stoichiometry table for
the reaction
HC2H3O2 + OH−  C2H3O2 + H2O
HA
A-
mols Before 0.100 0.100
mols added
mols After
OH−
0
0.010
What is the pH of a buffer that has 0.100 mol HC2H3O2
and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol
NaOH added to it?
Fill in the table –
tracking the changes in
the number of moles
for each component
HC2H3O2 + OH−  C2H3O2 + H2O
HA
A-
OH−
mols Before 0.100 0.100 ≈ 0
0.010
mols added
0.090 0.110
≈0
mols After
What is the pH of a buffer that has 0.100 mol HC2H3O2
and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol
NaOH added to it?
Write the reaction for
the acid with water
Construct an ICE
table for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0,
and using the new
molarities of the
[HA] and [A−]
HC2H3O2 + H2O  C2H3O2 + H3O+
[HA]
initial
change
equilibrium
0.090
[A-]
[H3O+]
0.110
≈0
What is the pH of a buffer that has 0.100 mol HC2H3O2
and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol
NaOH added to it?
HC2H3O2 + H2O  C2H3O2 + H3O+
represent the change
in the concentrations
in terms of x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression
initial
change
[HA]
[A-]
0.090
0.110
x
+x
equilibrium 0.090 x 0.110 + x
[H3O+]
0
+x
x
[C 2 H 3O -2 ][H 3O  ] 0.110  x  x 
Ka 

HC 2 H 3O 2 
0.090  x 
What is the pH of a buffer that has 0.100 mol HC2H3O2
and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol
NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
[H3O+]
since Ka is very small,
0.100
0.100
initial
≈0
approximate the
change
-x
+x
+x
[HA]eq = [HA]init and
[A−]eq = [A−]init solve for x equilibrium 0.090
0.090x 0.110
0.110+x
x
determine the value of Ka
[A-]
[HA]
[C 2 H 3O -2 ][H 3O  ] 0.110  x  x 
Ka 

HC 2 H 3O 2 
0.090  x 
[C 2 H 3O -2 ][H 3O  ] 0.110 x 
Ka 

HC 2 H 3O 2 
0.090
1.8  10
5

0.110 x 

0.090
1.47  105  x
What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has
0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check if the
approximation is
valid by seeing if
x < 5% of
[HC2H3O2]init
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.090
0.110
≈0
-x
+x
0.090
0.110
+x
x
x = 1.47 x 10-5
5
1.47  10
 100%  0.016%  5%
2
9.0  10
the approximation is valid
What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has
0.010 mol NaOH added to it?
substitute x into the
equilibrium
concentration
definitions and solve
[A-]
[HA]
[H3O+]
≈0
+x
0.090
0.110
initial
change
-x
+x
0.090x 0.110
0.110
x
equilibrium 0.090
+ x 1.5E-5
x = 1.47 x 10-5
HC2H3O2   0.090  x  0.090  1.47 105   0.090 M


[C 2 H 3O 2 ]  0.110  x  0.110  1.47  10
5
[H3O ]  x  1.47 105 M
  0.110 M
What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has
0.010 mol NaOH added to it?
substitute [H3O+]
into the formula for
pH and solve
[HA]
[A-]
0.090
initial
change
-x
equilibrium 0.090
0.110

pH  -log H 3O


  log 1.47  10

5
+x
[H3O+]
≈0
+x
0.110
1.5E-5
  4.83
What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has
0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check by substituting
the equilibrium
concentrations back into initial
the equilibrium constant change
expression and
equilibrium
comparing the calculated
Ka to the given Ka
[HA]
[A-]
0.090
0.110
-x
+x
[H3O+]
≈0
+x
0.090
0.110
1.5E-5
[C 2 H 3O -2 ][H 3O  ]
Ka 
HC 2 H 3O 2 
the values match

0.1101.47  105 

 1.8  105
0.090
What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has
0.010 mol NaOH added to it?
find the pKa from the
given Ka
HC2H3O2 + H2O  C2H3O2 + H3O+
Assume the [HA] and
[A-] equilibrium
concentrations are the
same as the initial
Ka for HC2H3O2 = 1.8 x 10-5
pK a   log K a
[HA]
[A-]
initial
0.090
0.110
≈0
change
-x
+x
+x
0.090
0.110
x
equilibrium
[H3
O+]

  log 1.8  105
 4.7 45

What is the pH of a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L
that has 0.010 mol NaOH added to it?
Substitute into the
HC2H3O2 + H2O  C2H3O2 + H3O+
Henderson-Hasselbalch
pKa for HC2H3O2 = 4.745
Equation
Check the “x is small”
approximation
[H3O  ]  10-pH
[H3O  ]  10-4.83  1.47  105
 [A - ] 

pH  pK a  log

[HA
]


 0.110 
pH  4.745  log
  4.83
 0.090 
1.47  105
 100%  0.016%  5%
0.090
Compare the effect on pH of adding 0.010 mol NaOH to
a buffer that has 0.100 mol HC2H3O2 and 0.100 mol
NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00
L of pure water?
HC2H3O2 + H2O  C2H3O2 + H3O+
pKa for HC2H3O2 = 4.745
 [A - ] 

pH  pK a  log

[HA
]


 0.110 
pH  4.7 45  log

 0.090 
 4.83
0.010 mol
[OH ] 
 0.010 M
1.00 L

pOH   log[OH  ]

2

  log 1.0  10  2.00
pH  pOH  14.00
pH  14.00 - pOH
 14.00 - 2.00
 12.00
What is the pH of a buffer that is 0.50 M NH3
(pKb = 4.75) and 0.20 M NH4Cl?
find the pKa of the
conjugate acid (NH4+)
from the given Kb
Assume the [B] and
[HB+] equilibrium
concentrations are the
same as the initial
Substitute into the
Henderson-Hasselbalch
Equation
Check the “x is small”
approximation
NH3 + H2O  NH4+ + OH−
pK a  pK b  14
pK a  14 - pK b  14 - 4.75  9.25
 [B] 
pH  pK a  log

 
 [HB ] 
 0.50 
pH  9.25  log
  9.65
 0.20 
[H3O  ]  10-pH
[H3O  ]  10-9.65  2.23  1010
2.23  1010
 100%  5%
0.20
Buffering Effectiveness
• a good buffer should be able to neutralize moderate
•
•
•
•
amounts of added acid or base
however, there is a limit to how much can be added
before the pH changes significantly
the buffering capacity is the amount of acid or base a
buffer can neutralize
the buffering range is the pH range the buffer can be
effective
the effectiveness of a buffer depends on two factors (1)
the relative amounts of acid and base, and (2) the
absolute concentrations of acid and base
Effect of Relative Amounts of Acid & Conjugate Base
Buffers are most effective when [acid] = [base]
Buffer 1
0.100 mol HA & 0.100 mol AInitial pH = 5.00
HA + OH−  A + H2O
HA
A-
OH−
mols Before
0.100
0.100
0
mols added
-
-
0.010
mols After
0.090
0.110
≈0
 [A - ] 

pH  pK a  log

 [HA ] 
pKa (HA) = 5.00
 0.110 
pH  5.00  log 
  5.09
 0.090 
Effect of Relative Amounts of Acid & Conjugate Base
Buffers are most effective when [acid] = [base]
Buffer 12
0.18 mol HA & 0.020 mol AInitial pH = 4.05
HA
mols
Before
mols added
mols After
A-
0.18 0.020
-
-
0.17 0.030
OH−
0
0.01
0
≈0
HA + OH−  A + H2O
pKa (HA) = 5.00
 [A - ] 

pH  pK a  log

 [HA ] 
 0.030 
pH  5.00  log 
  4.25
 0.17 
Effect of Relative Amounts of Acid and
Conjugate Base
Buffers are most effective when [acid] = [base]
pKa (HA) = 5.00
Buffer 1
0.100 mol HA & 0.100 mol AInitial pH = 5.00
 [A - ] 

pH  pK a  log

 [HA ] 
HA + OH−  A + H2O
 0.110 
pH  5.00  log 
  5.09
 0.090 
after adding 0.010 mol NaOH
pH = 5.09
HA
A-
OH−
mols Before
0.100
0.100
0
mols added
-
-
0.010
mols After
0.090
0.110
≈0
% Change

5.09 - 5.00 100%
5.00
 1.8%
Effect of Relative Amounts of Acid and
Conjugate Base
Buffers are most effective when [acid] = [base]
Buffer 12
0.18 mol HA & 0.020 mol AInitial pH = 4.05
pKa (HA) = 5.00
after adding 0.010 mol NaOH
 0.030 
pH  5.00  log 
  4.25
 0.17 
 [A - ] 

pH  pK a  log

 [HA ] 
after adding 0.010 mol NaOH
pH = 4.25
% Change
HA + OH−  A + H2O
HA
A-
OH−
mols Before
0.18
0.020
0
mols added
mols After
-
-
0.010
0.17
0.030
≈0

4.25 - 4.05

100%
4.05
 5.0%
Start: Effect of Absolute Concentrations
of Acid and Conjugate Base
a buffer is most effective when the concentrations
of acid and base are largest (0.1M vs. 0.5M)
pKa (HA) = 5.00
Buffer 1
0.50 mol HA & 0.50 mol AInitial pH = 5.00
 [A - ] 

pH  pK a  log

 [HA ] 
HA + OH−  A + H2O
 0.51 
pH  5.00  log 
  5.02
 0.49 
after adding 0.010 mol NaOH
pH = 5.02
HA
A-
OH−
mols Before
0.50
0.500
0
mols added
-
-
0.010
mols After
0.49
0.51
≈0
% Change
vs. 5.09
for 0.1M

5.02 - 5.00 

100%
5.00
 0.4%
vs. 1.8%
for 0.1M
Effect of Absolute Concentrations of
Acid and Conjugate Base
a buffer is most effective when the
concentrations of acid and base are largest
Buffer 12
0.050 mol HA & 0.050 mol AInitial pH = 5.00
pKa (HA) = 5.00
 [A - ] 

pH  pK a  log

 [HA ] 
HA + OH−  A + H2O
after adding 0.010 mol NaOH
pH = 5.18
HA
A-
OH−
mols Before
0.050
0.050
0
mols added
-
-
0.010
mols After
0.040
0.060
≈0
Add 0.01 mol NaOH
 0.060 
pH  5.00  log 
  5.18
 0.040 
% Change

5.18 - 5.00 

100%
5.00
 3.6%
Effectiveness of Buffers
• a buffer will be most effective when the
[base]:[acid] = 1
equal concentrations of acid and base
• effective when 0.1 < [base]:[acid] < 10
• a buffer will be most effective when the [acid]
and the [base] are large
Buffering Range
• we have said that a buffer will be effective when
•
0.1 < [base]:[acid] < 10
substituting into the Henderson-Hasselbalch we can
calculate the maximum and minimum pH at which
the buffer will be effective
 [A - ] 

pH  pK a  log

[HA
]

 Highest pH
Lowest pH
pH  pK a  log 0.10
pH  pK a  log 10
pH  pK a  1
pH  pK a  1
therefore, the effective pH range of a buffer is pKa ± 1
when choosing an acid to make a buffer, choose
one whose is pKa is closest to the pH of the buffer
Which of the following acids would be the
best choice to combine with its sodium salt to
make a buffer with pH 4.25?
Chlorous Acid, HClO2
Nitrous Acid, HNO2
Formic Acid, HCHO2
Hypochlorous Acid, HClO
pKa = 1.95
pKa = 3.34
pKa = 3.74
pKa = 7.54
Which of the following acids would be the
best choice to combine with its sodium salt to
make a buffer with pH 4.25?
Chlorous Acid, HClO2
Nitrous Acid, HNO2
Formic Acid, HCHO2
Hypochlorous Acid, HClO
pKa = 1.95
pKa = 3.34
pKa = 3.74
pKa = 7.54
The pKa of HCHO2 is closest to the desired pH of
the buffer, so it would give the most effective
buffering range.
What ratio of NaCHO2 : HCHO2 would be
required to make a buffer with pH 4.25?
Formic Acid, HCHO2, pKa = 3.74
 [A - ] 

pH  pK a  log 
 [HA ] 
 [CHO 2  ] 

4.25  3.74  log 

[HCHO
]
2 

 [CHO 2  ] 

0.51  log 

[HCHO
]
2 

10
 [CHO 2  ] 

log
 [HCHO2 ] 


 10
0.51

[CHO 2 ]
 3.24
[HCHO 2 ]
to make the buffer with pH
4.25, you would use 3.24 times
as much NaCHO2 as HCHO2
Buffering Capacity
• buffering capacity is the amount of acid or base that
•
•
•
•
can be added to a buffer without destroying its
effectiveness
the buffering capacity increases with increasing
absolute concentration of the buffer components
as the [base]:[acid] ratio approaches 1, the ability of the
buffer to neutralize both added acid and base improves
buffers that need to work mainly with added acid
generally have [base] > [acid]
buffers that need to work mainly with added base
generally have [acid] > [base]
Chapter 15: Applications of Aqueous
Equilibria - Solubility Equilibria
GENERAL CHEMISTRY: ATOMS FIRST
John E. McMurray – Robert C. Fay
Prentice Hall
Solubility Equilibria
• all ionic compounds dissolve in water to some
degree
however, many compounds have such low solubility
in water that we classify them as insoluble
• we can apply the concepts of equilibrium to
salts dissolving, and use the equilibrium
constant for the process to measure relative
solubilities in water
Solubility Product
• the equilibrium constant for the dissociation of a solid
•
•
•
•
salt into its aqueous ions is called the solubility
product, Ksp
for an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
the solubility product would be
Ksp = [Mm+]n[Xn−]m
for example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
Molar Solubility
• solubility is the amount of solute that will dissolve in a
given amount of solution
 at a particular temperature
• the molar solubility is the number of moles of solute that
will dissolve in a liter of solution
 the molarity of the dissolved solute in a saturated solution
• for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
molar solubility  n  m 
K sp
n
n
m
m
Ex 16.8 – Calculate the molar solubility of PbCl2
in pure water at 25C
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Initial
Ex 16.8 – Calculate the molar solubility of PbCl2
in pure water at 25C
Substitute into the
Ksp expression
Find the value of
Ksp from Table
16.2, plug into the
equation and
solve for S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
3
2+]
[Pb
K sp  4 S
[Cl−]
Initial K
0
05
1.17  10
sp
3
3
S
Change
+S
+2S
4
4
Equilibrium
S  1.43S 10  2 M 2S
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Initial
[Pb2+]
[Br−]
0
0
Change
+(1.05 x 10-2) +2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25C is 1.05 x 10-2 M
Substitute into the
Ksp expression
plug into the
equation and
solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10-2)(2.10 x 10-2)2


[Pb2+]

[Br−2]
2
2
K

1
.
05

10
2
.
10

10
sp
Initial
0
0
6
K

4
.
63

10
Change
+(1.05 x 10-2) +2(1.05 x 10-2)
sp
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Ksp and Relative Solubility
• molar solubility is related to Ksp
• but you cannot always compare solubilities of
compounds by comparing their Ksp values
• in order to compare Ksp values, the compounds
must have the same dissociation stoichiometry
The Effect of Common Ion on Solubility
• addition of a soluble salt that contains one of the
ions of the “insoluble” salt, decreases the
solubility of the “insoluble” salt
• for example, addition of NaCl to the solubility
equilibrium of solid PbCl2 decreases the
solubility of PbCl2
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the equilibrium to the left
Ex 16.10 – Calculate the molar solubility of CaF2
in 0.100 M NaF at 25C
Write the
dissociation
reaction and Ksp
expression
Create an ICE
table defining the
change in terms of
the solubility of
the solid
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
[F−]
0
0.100
Change
+S
+2S
Equilibrium
S
0.100 + 2S
Initial
Ex 16.10 – Calculate the molar solubility of CaF2
in 0.100 M NaF at 25C
Substitute into the
Ksp expression
assume S is small
Find the value of
Ksp from Table
16.2, plug into the
equation and
solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
K sp
Initial
ChangeS

2+] 2
[Ca
 S 0.100 
0 10
1.46  10
+S 2
0.100 
Equilibrium
S 8
S  1.46  10
[F−]
0.100
+2S
M
0.100 + 2S
The Effect of pH on Solubility
• for insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxide
 and the lower the pH, the higher the solubility
 higher pH = increased [OH−]
•
M(OH)n(s)  Mn+(aq) + nOH−(aq)
for insoluble ionic compounds that contain anions of
weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l)
Precipitation
• precipitation will occur when the concentrations of the
•
ions exceed the solubility of the ionic compound
if we compare the reaction quotient, Q, for the current
solution concentrations to the value of Ksp, we can
determine if precipitation will occur
 Q = Ksp, the solution is saturated, no precipitation
 Q < Ksp, the solution is unsaturated, no precipitation
 Q > Ksp, the solution would be above saturation, the salt
above saturation will precipitate
• some solutions with Q > Ksp will not precipitate unless
disturbed – these are called supersaturated solutions
precipitation occurs
if Q > Ksp
a supersaturated solution will
precipitate if a seed crystal is added
Selective Precipitation
• a solution containing several different cations
can often be separated by addition of a reagent
that will form an insoluble salt with one of the
ions, but not the others
• a successful reagent can precipitate with more
than one of the cations, as long as their Ksp
values are significantly different
Ex 16.13 What is the minimum [OH−] necessary to just
begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
From the Ksp table: Ksp = 2.06 x 10-13
2
 2
Q  [Mg ][OH ]
Q  K sp
[0.059 ][OH ]  2.06  10
 2

[OH ] 
13
2.06 10   1.9 10
13
0.059 
6
Ex 16.14 What is the [Mg2+] when Ca2+ (with
[0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
2
 2
Q  [Ca ][OH ]
Q  K sp
[0.011][OH ]  4.68  10
 2

[OH ] 
6
4.68 10   2.06 10
6
0.011
2
Ex 16.14 What is the [Mg2+] when Ca2+ (with
[0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
2
 2
Q  [Mg ][OH ]



when Q  K sp

[Mg 2  ][ 2.06  10  2 ]2  2.06  10 13
13
2
.
06

10
10
[Mg 2  ] 

4
.
8

10
M
2
2.06  10  2


when Ca2+ just
begins to
precipitate out, the
[Mg2+] has dropped
from 0.059 M to
4.8 x 10-10 M
Qualitative Analysis
• an analytical scheme that utilizes selective
precipitation to identify the ions present in a
solution is called a qualitative analysis scheme
wet chemistry
• a sample containing several ions is subjected to
the addition of several precipitating agents
• addition of each reagent causes one of the ions
present to precipitate out
Qualitative Analysis
Group 1 Insoluble Chlorides
• group one cations are Ag+, Pb2+, and Hg22+
• they form water insoluble compounds with Cl−
 AgCl, PbCl2, Hg2Cl2
as long as the concentration is large enough
PbCl2 may be borderline
molar solubility of PbCl2 = 1.43 x 10-2 M
• precipitated by the addition of HCl, which acts
to decrease the solubility due to LeChatelier’s
principle.
Group 2 Acid Insoluble Sulfides
• group two cations are Cd2+, Cu2+, Bi3+, Sn4+,
As3+, Pb2+, Sb3+, and Hg2+
• HgS, CdS, CuS, SnS2 ,etc.
• they form water insoluble compounds with HS−
and S2− at low pH
• They are precipitated by the
addition of H2S in HCl
Group 3 Base Insoluble Sulfides &
Hydroxides
• group three cations are Fe2+, Co2+, Zn2+, Mn2+,
Ni2+ precipitated as sulfides
 FeS, CoS, ZnS, MnS, NiS
• Cr3+, Fe3+, and Al3+ precipitated as hydroxides
 Al(OH3), Fe(OH3), Cr(OH3 )
• all these cations form compounds with S2− that
are insoluble in water at high pH
• precipitated by the addition of H2S in NaOH
Group 4 Insoluble Phosphates
• group four cations are Mg2+, Ca2+, Ba2+
• all these cations form compounds with PO43−
that are insoluble in water at high pH
• precipitated by the addition of (NH4)2HPO4
Group 5
• group five cations are Na+, K+, NH4+
• all these cations form compounds that are
soluble in water – they do not precipitate
• They are identified by the color of their flame
Complex Ion Formation
• transition metals tend to be good Lewis acids
• they often bond to one or more H2O molecules to form
a hydrated ion
 H2O is the Lewis base, donating electron pairs to form
coordinate covalent bonds
•
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
ions that form by combining a cation with several
anions or neutral molecules are called complex ions
 e.g., Ag(H2O)2+
• the attached ions or molecules are called ligands
 e.g., H2O
Complex Ion Equilibria
• if a ligand is added to a solution and it forms a
stronger bond than the original ligand, it will
replace the original ligand
Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l)
generally H2O is not included, since its complex ion is
always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Formation Constant
• the reaction between an ion and ligands to form
a complex ion is called a complex ion
formation reaction
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
• the equilibrium constant for the formation
reaction is called the formation constant, Kf

[Ag(NH 3 )2 ]
Kf 

2
[Ag ][ NH 3 ]
Formation Constants
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Write the
2+(aq) + 4 NH (aq)  Cu(NH ) 2+(aq)
Cu
3
3 2
formation reaction
2
[
Cu(NH
)
and Kf expression.
13
3 4 ]
Kf 

1
.
7

10
2
4
[
Cu
][
NH
]
Look up Kf value
3
Determine the
concentration of
ions in the diluted
solutions
1.5  10-3 mol
0.200 L 
1L
[Cu 2  ] 
 6.7  10  4 M
0.200 L  0.250 L
2.0  10-1 mol
0.250 L 
1L
[ NH 3 ] 
 1.1  10 1 M
0.200 L  0.250 L
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Kf 
[Cu 2 ][ NH 3 ]4
[Cu(NH 3 )42 ]
 1.7  1013
Create an ICE
[Cu2+]
[NH3] [Cu(NH3)22+]
table. Since Kf is
large, assume all
Initial
6.7E-4
0.11
0
2+
the Cu is
Change
-≈6.7E-4 -4(6.7E-4)
+ 6.7E-4
converted into
complex ion, then Equilibrium
x
0.11
6.7E-4
the system returns
X= the small amt of Cu2+ left in solution
to equilibrium
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is
mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Substitute in
and solve for x
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq)
[Cu(NH 3 )42 ]
13
Kf 

1
.
7

10
[Cu 2 ][ NH 3 ]4
confirm the
“x is small”
approximation

6.7  10 

4
13
1.7  10[Cu
2+]
x 0.11
[NH43]
6.7E-4
0.11

6.7  10 
x
 2.7  10
Change
-≈6.7E-4 -4(6.7E-4)
1.7 10 0.11
4
Initial
13
4
[Cu(NH3)22+]
13
0
+ 6.7E-4
Equilibrium
x -4
0.11
6.7E-4
-13
since 2.7 x 10 << 6.7 x 10 , the approximation is valid
The Effect of Complex Ion Formation
on Solubility
• the solubility of an ionic compound containing a
metal cation that forms a complex ion increases
in the presence of aqueous ligands
AgCl(s)  Ag+(aq) + Cl−(aq)
Ksp = 1.77 x 10-10
The silver chloride solubility increases due to consumption of the Ag+ ion by NH3
according to LeChatelier’s principle as seen by the large formation constant:
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107
Solubility of Amphoteric
Metal Hydroxides
• many metal hydroxides are insoluble
• all metal hydroxides become more soluble in acidic
solution
 shifting the equilibrium to the right by removing OH−
• some metal hydroxides also become more soluble in
basic solution
 acting as a Lewis base forming a complex ion
• substances that behave as both an acid and base are said
•
to be amphoteric
some cations that form amphoteric hydroxides include
Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
3+
Al
• Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
• addition of OH− drives the equilibrium to the right
and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l)