Chapter 16Aqueous
IonicEquilibrium
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
2
The Danger of Antifreeze• each year, thousands of pets and wildlife die
from consuming antifreeze• most brands of antifreeze contain ethylene
glycol sweet taste initial effect drunkenness
• metabolized in the liver to glycolic acid HOCH2COOH
• if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO3
−, causing the blood pH to drop• when the blood pH is low, it ability to carry O2
is compromised acidosis
• the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol
ethylene glycol(aka 1,2–ethandiol)
Tro, Chemistry: A Molecular Approach 3
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• but just like everything else, there is a limit to
what they can do, eventually the pH changes• many buffers are made by mixing a solution of
a weak acid with a solution of soluble salt containing its conjugate base anion
Tro, Chemistry: A Molecular Approach 4
Making an Acid Buffer
Tro, Chemistry: A Molecular Approach 5
How Acid Buffers WorkHA(aq) + H2O(l) A−
(aq) + H3O+(aq)
• buffers work by applying Le Châtelier’s Principle to weak acid equilibrium
• buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize ityou can also think of the H3O+ combining with the OH− to
make H2O; the H3O+ is then replaced by the shifting equilibrium
• the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant
Tro, Chemistry: A Molecular Approach 6
H2O
How Buffers Work
HA + H3O+A−A−
AddedH3O+
newHA
HA
Tro, Chemistry: A Molecular Approach 7
H2O
HA
How Buffers Work
HA + H3O+
A−
AddedHO−
newA−
A−
Tro, Chemistry: A Molecular Approach 8
Common Ion Effect HA(aq) + H2O(l) A−
(aq) + H3O+(aq)
• adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left
• this causes the pH to be higher than the pH of the acid solutionlowering the H3O+ ion concentration
Tro, Chemistry: A Molecular Approach 9
Common Ion Effect
Tro, Chemistry: A Molecular Approach 10
Ex 16.1 - 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
HC2H3O2 + H2O C2H3O2 + H3O+
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach 11
[HA] [A-] [H3O+]
initial 0.100 0.100 0
change
equilibrium
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
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
+x+xx
0.100 x 0.100 + x x
x
xxKa
100.0
100.0
OHHC
]OH][OH[C
232
3-232
HC2H3O2 + H2O C2H3O2 + H3O+
Tro, Chemistry: A Molecular Approach 12
x
xxKa
100.0
100.0
OHHC
]OH][OH[C
232
3-232
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq
= [A−]init solve for x
x 5108.1
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.100 0.100 x0.100 x
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
0.100 +x
100.0
100.0
OHHC
]OH][OH[C
232
3-232 x
Ka
Ka for HC2H3O2 = 1.8 x 10-5
Tro, Chemistry: A Molecular Approach 13
Ex 16.1 - 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
%5%018.0%1001000.1
108.11
5
the approximation is valid
x = 1.8 x 10-5
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.100 0.100 x
Tro, Chemistry: A Molecular Approach 14
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
x = 1.8 x 10-5
substitute x into the equilibrium concentration definitions and solve
M 100.0108.1100.0100.0OHHC 5232 x
M 108.1]OH[ 53
x
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.100 0.100 1.8E-5
M 100.0108.1100.0100.0]OHC[ 5232 x
0.100 + x x 0.100 x
Tro, Chemistry: A Molecular Approach 15
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
substitute [H3O+] into the formula for pH and solve
74.4108.1log
OH-logpH5
3
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.100 0.100 1.8E-5
Tro, Chemistry: A Molecular Approach 16
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the values match
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.100 0.100 1.8E-5
5
5232
3-232
108.1100.0
108.1100.0
OHHC
]OH][OH[C
aK
Ka for HC2H3O2 = 1.8 x 10-5
Tro, Chemistry: A Molecular Approach 17
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
Tro, Chemistry: A Molecular Approach 18
Practice - 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
HF + H2O F + H3O+
[HA] [A-] [H3O+]
initial 0.14 0.071 ≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach 19
[HA] [A-] [H3O+]
initial 0.14 0.071 0
change
equilibrium
Practice - 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
+x+xx
0.14 x 0.071 + x x
x
xxKa
14.0
071.0
HF
]OH][[F 3-
HF + H2O F + H3O+
Tro, Chemistry: A Molecular Approach 20
x
xxKa
14.0
071.0
HF
]OH][[F 3-
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq
= [A−]init solve for x
x 3104.1
[HA] [A-] [H3O+]
initial 0.14 0.071 ≈ 0
change -x +x +x
equilibrium 0.012 0.100 x0.14 x
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
0.071 +x
14.0
071.0100.7 4 x
Ka
Ka for HF = 7.0 x 10-4
4
15.3
100.7
1010
a
pKa
K
K a
Tro, Chemistry: A Molecular Approach 21
Practice - 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
%5%1%100104.1
104.11
3
the approximation is valid
x = 1.4 x 10-3
[HA] [A2-] [H3O+]
initial 0.14 0.071 ≈ 0
change -x +x +x
equilibrium 0.14 0.071 x
Tro, Chemistry: A Molecular Approach 22
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
x = 1.4 x 10-3
substitute x into the equilibrium concentration definitions and solve
M 14.0104.114.014.0HF 3 x
M 104.1]OH[ 33
x
[HA] [A2-] [H3O+]
initial 0.14 0.071 ≈ 0
change -x +x +x
equilibrium 0.14 0.072 1.4E-3
M 072.0104.1071.0071.0]OHC[ 3232 x
0.071 + x x 0.14 x
Tro, Chemistry: A Molecular Approach 23
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute [H3O+] into the formula for pH and solve
85.2104.1log
OH-logpH3
3
[HA] [A-] [H3O+]
initial 0.14 0.071 ≈ 0
change -x +x +x
equilibrium 0.14 0.072 1.4E-3
Tro, Chemistry: A Molecular Approach 24
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the values are close enough
[HA] [A-] [H3O+]
initial 0.14 0.071 ≈ 0
change -x +x +x
equilibrium 0.14 0.072 1.4E-3
4
3
3-
102.714.0
104.1072.0
HF
]OH][[F
aK
Ka for HF = 7.0 x 10-4
Tro, Chemistry: A Molecular Approach 25
Henderson-Hasselbalch Equation• calculating the pH of a buffer solution can be
simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation
• the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate baseas long as the “x is small” approximation is valid
initial
initiala acid][weak
anion] base conjugate[logp pH K
Tro, Chemistry: A Molecular Approach 26
Deriving the Henderson-Hasselbalch Equation
][A
[HA]]OH[
HA
]OH][[A
-3
3-
a
a
K
K
][A
[HA]log]OHlog[
-3 aK
]Olog[H- pH 3
][A
[HA]loglog]OHlog[
-3 aK
][A
[HA]loglogpH
-aK
aK log- pKa
][A
[HA]logppH
-aK
[HA]
][Alog
][A
[HA]log
[HA]
][AlogppH
-
aK
Tro, Chemistry: A Molecular Approach 27
Ex 16.2 - 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
HC7H5O2 + H2O C7H5O2 + H3O+
][HA
][AlogppH
-
aK
050.0
0.150log781.4pH
Ka for HC7H5O2 = 6.5 x 10-5
781.4105.6log
logp5
aa KK
4.66pH
54.66-3
-pH3
102.210]OH[
10]OH[
%5%044.0%100050.0
102.2 5
Tro, Chemistry: A Molecular Approach 28
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
Tro, Chemistry: A Molecular Approach 29
Practice - 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+
][HA
][AlogppH
-
aK
86.2
14.0
0.071log15.3pH
32.86-3
-pH3
104.110]OH[
10]OH[
%5%1%10014.0
104.1 3
Tro, Chemistry: A Molecular Approach 30
Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?
• 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:
a) the initial concentrations of acid and salt are not very dilute
b) 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
Tro, Chemistry: A Molecular Approach 31
How Much Does the pH of a Buffer Change When an Acid or Base Is Added?• though 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 parts1. 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−]
Tro, Chemistry: A Molecular Approach 32
Ex 16.3 - 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- OH−
mols Before 0.100 0.100 0
mols added - - 0.010
mols After
Tro, Chemistry: A Molecular Approach 33
Ex 16.3 - 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
mols added - - 0.010
mols After 0.090 0.110 ≈ 0
Tro, Chemistry: A Molecular Approach 36
Ex 16.3 - 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] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach 37
[HA] [A-] [H3O+]
initial 0.090 0.110 0
change
equilibrium
Ex 16.3 - 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?
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
+x+xx
0.090 x 0.110 + x x
x
xxKa
090.0
110.0
OHHC
]OH][OH[C
232
3-232
HC2H3O2 + H2O C2H3O2 + H3O+
Tro, Chemistry: A Molecular Approach 38
x
xxKa
090.0
110.0
OHHC
]OH][OH[C
232
3-232
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq
= [A−]init solve for x
x 51074.1
[HA] [A-] [H3O+]
initial 0.100 0.100 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 x0.090 x
Ex 16.3 - 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?
0.110 +x
090.0
110.0
OHHC
]OH][OH[C
232
3-232 x
Ka
Ka for HC2H3O2 = 1.8 x 10-5
090.0
110.0108.1 5 x
Tro, Chemistry: A Molecular Approach 39
Ex 16.3 - 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
%5%016.0%100100.9
1074.12
5
the approximation is valid
x = 1.47 x 10-5
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 x
Tro, Chemistry: A Molecular Approach 40
Ex 16.3 - 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?
x = 1.47 x 10-5
substitute x into the equilibrium concentration definitions and solve
M 090.01074.1090.0090.0OHHC 5232 x
M 1074.1]OH[ 53
x
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 1.5E-5
M 110.01074.1110.0110.0]OHC[ 5232 x
0.110 + x x 0.090 x
Tro, Chemistry: A Molecular Approach 41
Ex 16.3 - 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
83.41074.1log
OH-logpH5
3
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 1.5E-5
Tro, Chemistry: A Molecular Approach 42
Ex 16.3 - 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?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the values match
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 1.5E-5
5
5
232
3-232
108.1090.0
1074.1110.0
OHHC
]OH][OH[C
aK
Ka for HC2H3O2 = 1.8 x 10-5
Tro, Chemistry: A Molecular Approach 43
Ex 16.3 - 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
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
HC2H3O2 + H2O C2H3O2 + H3O+
547.4
108.1log
logp5
aa KK
Ka for HC2H3O2 = 1.8 x 10-5
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change -x +x +x
equilibrium 0.090 0.110 x
Tro, Chemistry: A Molecular Approach 44
Ex 16.3 - 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 Henderson-Hasselbalch Equation
Check the “x is small” approximation
HC2H3O2 + H2O C2H3O2 + H3O+
][HA
][AlogppH
-
aK
83.4
090.0
0.110log547.4pH
%5%016.0%100090.0
1074.1 5
pKa for HC2H3O2 = 4.745
54.83-3
-pH3
1074.110]OH[
10]OH[
Tro, Chemistry: A Molecular Approach 45
Ex 16.3 – 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+
][HA
][AlogppH
-
aK
83.4
090.0
0.110log547.4pH
pKa for HC2H3O2 = 4.745
M 010.0L 1.00
mol 010.0]OH[
00.2100.1log
]OHlog[pOH2
12.00
2.00-14.00
pOH -14.00pH
00.14pOHpH
Tro, Chemistry: A Molecular Approach 46
Basic BuffersB:(aq) + H2O(l) H:B+
(aq) + OH−(aq)
• buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq) NH4+
(aq) + OH−(aq)
Tro, Chemistry: A Molecular Approach 47
Ex 16.4 - 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−
][HB
[B]logppH aK
65.9
20.0
0.50log25.9pH
109.65-3
-pH3
1032.210]OH[
10]OH[
%5%10020.0
1032.2 10
9.254.75-14p -14p
14pp
ba
ba
KK
KK
Tro, Chemistry: A Molecular Approach 49
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
HA A- OH−
mols Before 0.18 0.020 0
mols added - - 0.010
mols After 0.17 0.030 ≈ 0
Effect of Relative Amounts of Acid and Conjugate Base
Buffer 10.100 mol HA & 0.100 mol A-
Initial pH = 5.00
Buffer 120.18 mol HA & 0.020 mol A-
Initial pH = 4.05pKa (HA) = 5.00
][HA
][AlogppH
-
aK
09.5
090.0
0.110log00.5pH
after adding 0.010 mol NaOHpH = 5.09
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
25.4
17.0
0.030log00.5pH
after adding 0.010 mol NaOHpH = 4.25
%8.1
%1005.00
5.00-5.09
Change %
%0.5
%1004.05
4.05-4.25
Change %
a buffer is most effective with equal concentrations of acid and base
HA A- OH−
mols Before 0.50 0.500 0
mols added - - 0.010
mols After 0.49 0.51 ≈ 0
%6.3
%1005.00
5.00-5.18
Change %
HA A- OH−
mols Before 0.050 0.050 0
mols added - - 0.010
mols After 0.040 0.060 ≈ 0
Effect of Absolute Concentrations of Acid and Conjugate Base
Buffer 10.50 mol HA & 0.50 mol A-
Initial pH = 5.00
Buffer 120.050 mol HA & 0.050 mol A-
Initial pH = 5.00pKa (HA) = 5.00
][HA
][AlogppH
-
aK
02.5
49.0
0.51log00.5pH
after adding 0.010 mol NaOHpH = 5.02
HA + OH− A + H2O
18.5
040.0
0.060log00.5pH
after adding 0.010 mol NaOHpH = 5.18
%4.0
%1005.00
5.00-5.02
Change %
a buffer is most effective when the concentrations of acid and base are largest
Tro, Chemistry: A Molecular Approach 52
Effectiveness of Buffers
• a buffer will be most effective when the [base]:[acid] = 1equal 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
53
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
][HA
][AlogppH
-
aK
Lowest pH
1ppH
10.0logppH
a
a
K
KHighest pH
1ppH
10logppH
a
a
K
K
therefore, the effective pH range of a buffer is pKa ± 1when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer
Tro, Chemistry: A Molecular Approach 54
Ex. 16.5a – 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 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
Tro, Chemistry: A Molecular Approach 55
Ex. 16.5a – 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 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO 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.
Tro, Chemistry: A Molecular Approach 56
Ex. 16.5b – What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?
Formic Acid, HCHO2, pKa = 3.74
][HA
][AlogppH
-
aK
][HCHO
][CHOlog51.0
][HCHO
][CHOlog74.325.4
2
2
2
2 24.3][HCHO
][CHO
1010
2
2
51.0][HCHO
][CHOlog
2
2
to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2
Tro, Chemistry: A Molecular Approach 57
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]
Tro, Chemistry: A Molecular Approach 58
Buffering Capacity
a concentrated buffer can neutralize more added acid or base than a dilute buffer
Tro, Chemistry: A Molecular Approach 59
Titration• in an acid-base titration, a solution of unknown
concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is completewhen the reaction is complete we have reached the endpoint
of the titration
• an indicator may be added to determine the endpointan 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
Tro, Chemistry: A Molecular Approach 60
Titration
Tro, Chemistry: A Molecular Approach 61
Titration Curve• 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 known 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 solutionequivalence 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 unknown solution in the burette is in excess, so the pH approaches its pH
Tro, Chemistry: A Molecular Approach 62
Titration Curve:Unknown Strong Base Added to
Strong Acid
Tro, Chemistry: A Molecular Approach 63
Tro, Chemistry: A Molecular Approach 64
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 mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10-3
• before equivalence point
NaOH added molesL 1
NaOH mol 0.100NaOH added of L
added 5.0 mL NaOH
NaOH mol 100.5L 1
NaOH mol 0.100NaOH L 0.0050
4
used HCl moles NaOH mol 1
HCl mol 1NaOH mole
5.0 x 10-4 mol NaOH
used HCl mol 100.5NaOH mol 1
HCl mol 1NaOH mol 100.5
4
4
excess HCl mol used HCl mol -HCl mol initial
excess HCl mol102.00
used HCl mol105.0 -HCl mol 102.503-
-4-3
]O[HHCl M NaOH LHCl L
excess HCl mol
3
]O[HHCl M 0.0667 NaOH L 0050.0HCl L 0.0250
HCl mol102.00
3
-3
2.00 x 10-3 mol HCl
]O-log[HpH 3 18.10667.0-logpH
Tro, Chemistry: A Molecular Approach 65
excess NaOH mol105.0
used NaOH mol102.50 -NaOH mol 103.004-
-3-3
NaOH mol 1000.3
L 1
NaOH mol 0.100NaOH L 0.0300
3
][OHNaOH M 0.00909 NaOH L 0300.0HCl L 0.0250
NaOH mol100.5 -4
123
14
3
1001.11009.9
101
]OH[]OH[
wK
][OHNaOH M NaOH LHCl L
excess NaOH mol
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)
• at equivalence, 0.00 mol HCl and 0.00 mol NaOH
• pH at equivalence = 7.00
• after equivalence point
NaOH added molesL 1
NaOH mol 0.100NaOH added of L
added 30.0 mL NaOH5.0 x 10-4 mol NaOH xs
excess NaOH mol HCl mol initial -added NaOH mol
wK ]][OHOH[ 3
96.111001.1log- pH 9- ]Olog[H- pH 3
Tro, Chemistry: A Molecular Approach 67
added 30.0 mL NaOH0.00050 mol NaOHpH = 11.96
added 35.0 mL NaOH0.00100 mol NaOHpH = 12.22
Adding NaOH to HCl
25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00
added 5.0 mL NaOH0.00200 mol HClpH = 1.18
added 10.0 mL NaOH0.00150 mol HClpH = 1.37
added 15.0 mL NaOH0.00100 mol HClpH = 1.60
added 20.0 mL NaOH0.00050 mol HClpH = 1.95
added 25.0 mL NaOHequivalence pointpH = 7.00
added 40.0 mL NaOH0.00150 mol NaOHpH = 12.36
added 50.0 mL NaOH0.00250 mol NaOHpH = 12.52
Tro, Chemistry: A Molecular Approach 70
Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)
• Initial pH:[HCHO2] [CHO2
-] [H3O+]
initial 0.100 0.000 ≈ 0
change -x +x +x
equilibrium 0.100 - x x x
Ka = 1.8 x 10-4
M 1042.4]O[H
100.0100.0108.1
]HCHO[
]O][H[CHO
33
24
2
32
x
x
x
xx
Ka
37.210424.-log
]Olog[H- pH3-
3
%5%2.4%100100.0
102.4 3
Tro, Chemistry: A Molecular Approach 71
Titration of 25 mL of 0.100 M HCHO2 with 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
• before equivalence added 5.0 mL NaOH
NaOH mol 100.5L 1
NaOH mol 0.100NaOH L 0.0050
4
HA A- OH−
mols Before 2.50E-3 0 0
mols added - - 5.0E-4
mols After 2.00E-3 5.0E-4 ≈ 0
2
2
HCHO mol
CHO mollogppH aK
14.3pH10.002
10.05log74.3pH
5-
4-
74.3108.1-log
log- p4-
aa KK
72
M 107.1][OH
0500.00500.0106.5
]CHO[
]][OH[HCHO
6
211
2
2
x
x
x
xx
Kb
96
14
3
109.5107.1
101
]OH[]OH[
wK
114
14
CHO ,
106.5108.1
101
2
a
wb K
KK
22-
2-2
2-2
-3
CHO M105.00
NaOH L102.50HCHO L102.50
CHO mol102.50
2
2
2
CHO M
NaOH LHCHO L
CHO mol
Titration of 25 mL of 0.100 M HCHO2 with 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 added 25.0 mL NaOH
NaOH mol 1050.2L 1
NaOH mol 0.100NaOH L 0.0250
3
HA A- OH−
mols Before 2.50E-3 0 0
mols added - - 2.50E-3
mols After 0 2.50E-3 ≈ 0
[HCHO2] [CHO2-] [OH−]
initial 0 0.0500 ≈ 0
change +x -x +x
equilibrium x 5.00E-2-x x
CHO2−
(aq) + H2O(l) HCHO2(aq) + OH−(aq)
Kb = 5.6 x 10-11
23.8109.5-log
]Olog[H- pH9-
3
[OH-] = 1.7 x 10-6 M
Tro, Chemistry: A Molecular Approach 73
excess NaOH mol105.0
used NaOH mol102.50 -NaOH mol 103.004-
-3-3
NaOH mol 1000.3L 1
NaOH mol 0.100NaOH L 0.0300
3
][OHNaOH M NaOH LHCl L
excess NaOH mol
][OHNaOH M 0.0091 NaOH L 0300.0HCl L 0.0250
NaOH mol100.5 -4
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)
• after equivalence point
NaOH added molesL 1
NaOH mol 0.100NaOH added of L
added 30.0 mL NaOH
5.0 x 10-4 mol NaOH xs
excess NaOH mol HCHO mol initial -added NaOH mol 2
12
3
14
3
1001.1101.9
101
]OH[]OH[
wK
96.111001.1log- pH 9- ]Olog[H- pH 3
wK ]][OHOH[ 3
Tro, Chemistry: A Molecular Approach 75
added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96
added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22
Adding NaOH to HCHO2
added 12.5 mL NaOH0.00125 mol HCHO2
pH = 3.74 = pKa
half-neutralization
initial HCHO2 solution0.00250 mol HCHO2
pH = 2.37
added 5.0 mL NaOH0.00200 mol HCHO2
pH = 3.14
added 10.0 mL NaOH0.00150 mol HCHO2
pH = 3.56
added 15.0 mL NaOH0.00100 mol HCHO2
pH = 3.92
added 20.0 mL NaOH0.00050 mol HCHO2
pH = 4.34
added 40.0 mL NaOH0.00150 mol NaOH xspH = 12.36
added 25.0 mL NaOHequivalence point0.00250 mol CHO2
−
[CHO2−]init = 0.0500 M
[OH−]eq = 1.7 x 10-6
pH = 8.23
added 50.0 mL NaOH0.00250 mol NaOH xspH = 12.52
Tro, Chemistry: A Molecular Approach 79
Titrating Weak Acid with a Strong Base
• the initial pH is that of the weak acid solutioncalculate like a weak acid equilibrium problem
e.g., 15.5 and 15.6
• before the equivalence point, the solution becomes a buffercalculate 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
Tro, Chemistry: A Molecular Approach 80
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 establishedmol A− = original mole HA
calculate the volume of added base like Ex 4.8
[A−]init = mol A−/total literscalculate 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
Tro, Chemistry: A Molecular Approach 81
Ex 16.7a – 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
HNO2 + KOH NO2 + H2O
KOH L .02000
KOH mol 0.200
KOH L 1
NO mol 1
KOH mol 1
NO L 1
NO mol 0.100NO L .04000
22
22
L 0400.0mL 1
L 0.001mL 0.40
mL 0.20L 0.001
mL 1L 0200.0
Tro, Chemistry: A Molecular Approach 82
Ex 16.7b – 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 KOHWrite 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
KOH mol 00100.0L 1
KOH mol 200.0
mL 1
L 0.001mL 00.5
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00100
mols After ≈ 0
22 HNO mol 00400.0
L 1
HNO mol 100.0
mL 1
L 0.001mL 0.40
0.00300 0.00100
Tro, Chemistry: A Molecular Approach 83
Ex 16.7b – 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 Henderson-Hasselbalch Equation to determine the pH
HNO2 + H2O NO2 + H3O+
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00100
mols After 0.00300 0.00100 ≈ 0
Table 15.5 Ka = 4.6 x 10-4
15.3106.4loglogp 4 aa KK
2
2
HNO
NOlogppH aK
67.20.00300
00100.0log15.3pH
Tro, Chemistry: A Molecular Approach 84
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence pointWrite 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
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00200
mols After ≈ 0
22 HNO mol 00400.0
L 1
HNO mol 100.0
mL 1
L 0.001mL 0.40
0.00200 0.00200
at half-equivalence, moles KOH = ½ mole HNO2
Tro, Chemistry: A Molecular Approach 85
Ex 16.7b – 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 Henderson-Hasselbalch Equation to determine the pH
HNO2 + H2O NO2 + H3O+
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00200
mols After 0.00200 0.00200 ≈ 0
Table 15.5 Ka = 4.6 x 10-4
15.3106.4loglogp 4 aa KK
2
2
HNO
NOlogppH aK
15.30.00200
00200.0log15.3pH
Tro, Chemistry: A Molecular Approach 86
Titration Curve of a Weak Base with a Strong Acid
Tro, Chemistry: A Molecular Approach 87
Titration of a Polyprotic Acid• if Ka1 >> Ka2, there will be two equivalence
points in the titrationthe 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
Tro, Chemistry: A Molecular Approach 88
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
Tro, Chemistry: A Molecular Approach 89
Monitoring pH During a Titration
Tro, Chemistry: A Molecular Approach 90
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
HInd(aq) + H2O(l) Ind(aq) + H3O+
(aq)
• the color of the solution depends on the relative concentrations of Ind:HIndwhen Ind:HInd ≈ 1, the color will be mix of the colors of Ind
and HInd when Ind:HInd > 10, the color will be mix of the colors of Ind
when Ind:HInd < 0.1, the color will be mix of the colors of HInd
91
Phenolphthalein
Tro, Chemistry: A Molecular Approach 92
Methyl Red
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N NH
NaOOC
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N N
NaOOC
H3O+ OH-
Tro, Chemistry: A Molecular Approach 93
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 pHpKa of HInd ≈ pH at equivalence point
94
Acid-Base Indicators
Tro, Chemistry: A Molecular Approach 95
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
Tro, Chemistry: A Molecular Approach 96
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 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)
• and its equilibrium constant is Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach 97
Tro, Chemistry: A Molecular Approach 98
Molar Solubility• solubility is the amount of solute that will dissolve in a
given amount of solutionat 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)
mnmn
sp
mn
K
solubilitymolar
Tro, Chemistry: A Molecular Approach 99
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach 100
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Substitute into the Ksp expression
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
M 1043.14
1017.1
4
4
2
35
3
3
S
SK
SK
sp
sp
Tro, Chemistry: A Molecular Approach 101
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Tro, Chemistry: A Molecular Approach 102
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C 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
[Pb2+] [Br−]
Initial 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)
PbBr2(s) Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Tro, Chemistry: A Molecular Approach 103
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C 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−]
Initial 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)
6
222
1063.4
1010.21005.1
sp
sp
K
K
Tro, Chemistry: A Molecular Approach 104
Ksp and Relative Solubility
• molar solubility is related to Ksp
• but you cannot always compare solubilities of compounds by comparing their Ksps
• in order to compare Ksps, the compounds must have the same dissociation stoichiometry
Tro, Chemistry: A Molecular Approach 105
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
Tro, Chemistry: A Molecular Approach 106
Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
[Ca2+] [F−]
Initial 0 0.100
Change +S +2S
Equilibrium S 0.100 + 2S
CaF2(s) Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
Tro, Chemistry: A Molecular Approach 107
Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C
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
[Ca2+] [F−]
Initial 0 0.100
Change +S +2S
Equilibrium S 0.100 + 2S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
M 1046.1
100.0
1046.1
100.0
8
2
10
2
S
S
SKsp
Tro, Chemistry: A Molecular Approach 108
The Effect of pH on Solubility• for insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxideand the lower the pH, the higher the solubilityhigher 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)
Tro, Chemistry: A Molecular Approach 109
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 occurQ = Ksp, the solution is saturated, no precipitationQ < Ksp, the solution is unsaturated, no precipitationQ > 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
Tro, Chemistry: A Molecular Approach 110
precipitation occurs if Q > Ksp
a supersaturated solution will precipitate if a seed crystal is added
Tro, Chemistry: A Molecular Approach 111
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
Tro, Chemistry: A Molecular Approach 112
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
22 ]OH][Mg[ Q
6
13
132
109.1059.0
1006.2]OH[
1006.2]OH][059.0[
spKQ
Tro, Chemistry: A Molecular Approach 113
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 22 ]OH][Ca[ Q
2
6
62
1060.2011.0
1068.4]OH[
1068.4]OH][011.0[
spKQ
Tro, Chemistry: A Molecular Approach 114
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
M 108.4
1060.2
1006.2]Mg[
1006.2]1060.2][Mg[
when
1022
132
13222
spKQ
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M 22 ]OH][Mg[ Q when Ca2+ just
begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M
Tro, Chemistry: A Molecular Approach 115
Qualitative Analysis
• an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis schemewet 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
Tro, Chemistry: A Molecular Approach 116
Qualitative Analysis
117
Tro, Chemistry: A Molecular Approach 118
Group 1
• group one cations are Ag+, Pb2+, and Hg22+
• all these cations form compounds with Cl− that are insoluble in wateras long as the concentration is large enoughPbCl2 may be borderline
molar solubility of PbCl2 = 1.43 x 10-2 M
• precipitated by the addition of HCl
119
Group 2
• group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+
• all these cations form compounds with HS− and S2− that are insoluble in water at low pH
• precipitated by the addition of H2S in HCl
120
Group 3
• group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides
• all these cations form compounds with S2− that are insoluble in water at high pH
• precipitated by the addition of H2S in NaOH
Tro, Chemistry: A Molecular Approach 121
Group 4
• 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
122
Group 5• group five cations are Na+, K+, NH4
+
• all these cations form compounds that are soluble in water – they do not precipitate
• identified by the color of their flame
Tro, Chemistry: A Molecular Approach 123
Complex Ion Formation• transition metals tend to be good Lewis acids• they often bond to one or more H2O molecules to form
a hydrated ionH2O 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 ionse.g., Ag(H2O)2
+
• the attached ions or molecules are called ligandse.g., H2O
Tro, Chemistry: A Molecular Approach 124
Complex Ion Equilibria• if a ligand is added to a solution that forms a
stronger bond than the current ligand, it will replace the current 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)
Tro, Chemistry: A Molecular Approach 125
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
23
23
]NH][[Ag
])[Ag(NH
fK
Tro, Chemistry: A Molecular Approach 126
Formation Constants
Tro, Chemistry: A Molecular Approach 127
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 formation reaction and Kf expression.
Look up Kf value
Determine the concentration of ions in the diluted solutions
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
134
32
243 107.1
]NH][Cu[
])Cu(NH[
fK
M 107.6L 0.250 L 200.0
L 1mol 101.5
L 200.0]Cu[ 4
-3
2
M 101.1L 0.250 L 200.0
L 1mol 100.2
L 250.0]NH[ 1
-1
3
128
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?
Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium
[Cu2+] [NH3] [Cu(NH3)22+]
Initial 6.7E-4 0.11 0
Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4
Equilibrium x 0.11 6.7E-4
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
132
43
43
2
107.1])Cu(NH[
]NH][Cu[
fK
Tro, Chemistry: A Molecular Approach 129
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)
134
32
243 107.1
]NH][Cu[
])Cu(NH[
fK
Substitute in and solve for x
confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)2
2+]
Initial 6.7E-4 0.11 0
Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4
Equilibrium x 0.11 6.7E-4
13413
4
4
413
107.211.0107.1
107.6
11.0
107.6107.1
x
x
since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid
Tro, Chemistry: A Molecular Approach 130
The Effect of Complex Ion Formation on Solubility
• the solubility of an ionic compound that contains 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
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq) Kf = 1.7 x 107
• adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+
131
Tro, Chemistry: A Molecular Approach 132
Solubility of Amphoteric Metal Hydroxides
• many metal hydroxides are insoluble• all metal hydroxides become more soluble in acidic
solutionshifting the equilibrium to the right by removing OH−
• some metal hydroxides also become more soluble in basic solutionacting 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+
Tro, Chemistry: A Molecular Approach 133
Al3+
• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6
3+(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 moleculesAl(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)
Tro, Chemistry: A Molecular Approach 134