-
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
glycolsweet tasteinitial effect drunkenness
metabolized in the liver to glycolic acidHOCH2COOH
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 O2is 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 1,2ethandiol)
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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
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Tro, Chemistry: A Molecular Approach 4
Making an Acid Buffer
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Tro, Chemistry: A Molecular Approach 5
How Acid Buffers WorkHA(aq) + H2O(l) A(aq) + H3O+(aq)
buffers work by applying Le Chteliers Principle to weak acid equilibrium
buffer solutions contain significant amounts of the weak acid molecules, HA these molecules react with added base to neutralize it
you 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
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Tro, Chemistry: A Molecular Approach 6
H2O
How Buffers Work
HA + H3O+AA
AddedH3O+
newHA
HA
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Tro, Chemistry: A Molecular Approach 7
H2O
HA
How Buffers Work
HA + H3O+A
AddedHO
newA
A
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Tro, Chemistry: A Molecular Approach 8
Common Ion EffectHA(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 solution
lowering the H3O+ ion concentration
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Tro, Chemistry: A Molecular Approach 9
Common Ion Effect
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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?
Enter the initial concentrations assuming the [H3O+] from water is 0
Construct an ICE table for the reaction
Write the reaction for the acid with water
HC2H3O2 + H2O C2H3O2 + H3O+
0.100[A-]
equilibriumchange
00.100initial[H3O+][HA]
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Tro, Chemistry: A Molecular Approach 11
0.100[A-]
equilibriumchange
00.100initial[H3O+][HA]
Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
substitute into the equilibrium constant expression
sum the columns to find the equilibrium concentrations in terms of x
represent the change in the concentrations in terms of x
+x+xx0.100 x 0.100 + x x
[ ]( )( )( )x
xxKa +
==+
100.0100.0
OHHC]OH][OH[C
232
3-232
HC2H3O2 + H2O C2H3O2 + H3O+
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Tro, Chemistry: A Molecular Approach 12
[ ]( )( )( )x
xxKa +
==+
100.0100.0
OHHC]OH][OH[C
232
3-232
since Ka is very small, approximate the [HA]eq = [HA]init and [A]eq = [A]init solve for x
determine the value of Ka
x= 5108.1
0.100
+x0.100[A-]
x0.100equilibrium+x-xchange 00.100initial
[H3O+][HA]
0.100 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
[ ]( )( )( )100.0100.0
OHHC]OH][OH[C
232
3-232 xKa ==
+
Ka for HC2H3O2 = 1.8 x 10-5
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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
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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
0.100
+x0.100[A-]
1.8E-50.100equilibrium+x-xchange 00.100initial
[H3O+][HA]
( ) M 100.0108.1100.0100.0]OHC[ 5232 =+=+= x
0.100 + x x 0.100 x
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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
===
+
0.100
+x0.100[A-]
1.8E-50.100equilibrium+x-xchange 00.100initial
[H3O+][HA]
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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
0.100
+x0.100[A-]
1.8E-50.100equilibrium+x-xchange 00.100initial
[H3O+][HA]
[ ]( )( )
( )5
5232
3-232
108.1100.0
108.1100.0OHHC
]OH][OH[C
+
=
=
=aK
Ka for HC2H3O2 = 1.8 x 10-5
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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?
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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?
Enter the initial concentrations assuming the [H3O+] from water is 0
Construct an ICE table for the reaction
Write the reaction for the acid with water
HF + H2O F + H3O+
0.071[A-]
equilibriumchange
00.14initial[H3O+][HA]
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Tro, Chemistry: A Molecular Approach 19
0.071[A-]
equilibriumchange
00.14initial[H3O+][HA]
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute into the equilibrium constant expression
sum the columns to find the equilibrium concentrations in terms of x
represent the change in the concentrations in terms of x
+x+xx0.14 x 0.071 + x x
[ ]( )( )( )x
xxKa +
==+
14.0071.0
HF]OH][[F 3
-
HF + H2O F + H3O+
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Tro, Chemistry: A Molecular Approach 20
[ ]( )( )( )x
xxKa +
==+
14.0071.0
HF]OH][[F 3
-
since Ka is very small, approximate the [HA]eq = [HA]init and [A]eq = [A]init solve for x
determine the value of Ka
x= 3104.1
0.100
+x0.071[A-]
x0.012equilibrium+x-xchange 00.14initial
[H3O+][HA]
0.14 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
( )( )( )14.0071.0100.7 4 xKa ==
Ka for HF = 7.0 x 10-4
4
15.3
100.71010
===
a
pKa
KK a
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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%100104.1104.1
1
3
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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
0.072
+x0.071[A2-]
1.4E-30.14equilibrium+x-xchange 00.14initial
[H3O+][HA]
( ) M 072.0104.1071.0071.0]OHC[ 3232 =+=+= x
0.071 + x x 0.14 x
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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
===
+
0.072
+x0.071[A-]
1.4E-30.14equilibrium+x-xchange 00.14initial
[H3O+][HA]
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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
0.072
+x0.071[A-]
1.4E-30.14equilibrium+x-xchange 00.14initial
[H3O+][HA]
[ ]( )( )
( )4
3
3-
102.714.0
104.1072.0HF
]OH][[F
+
=
=
=aK
Ka for HF = 7.0 x 10-4
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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 base
as long as the x is small approximation is valid
initial
initiala acid][weak
anion] base conjugate[logp pH += K
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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
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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?
Check the x is small approximation
Substitute into the Henderson-HasselbalchEquation
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
HC7H5O2 + H2O C7H5O2 + H3O+
+=
][HA][AlogppH
-
aK( )
+=
050.00.150log781.4pH
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%100050.0102.2 5
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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?
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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?
Check the x is small approximation
Substitute into the Henderson-HasselbalchEquation
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
find the pKa from the given Ka
HF + H2O F + H3O+
+=
][HA][AlogppH
-
aK
( ) 86.214.0
0.071log15.3pH =
+=
32.86-3
-pH3
104.110]OH[
10]OH[+
+
==
=
%5%1%10014.0104.1 3
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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
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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 HAadded base reacts with the HA to make more A
2. an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A]
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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?
Construct a stoichiometry table for the reaction
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.
HC2H3O2 + OH C2H3O2 + H2O
-0.100A-
0.010
0OH
mols After-mols added
0.100mols BeforeHA
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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
0.110
-0.100A-
0
0.010
0OH
0.090mols After-mols added
0.100mols BeforeHA
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Tro, Chemistry: A Molecular Approach 34
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?
Enter the initial number of moles for each
Construct a stoichiometry table for the reaction
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.
HC2H3O2 + OH C2H3O2 + H2O
0.100A-
0.010OH
mols Endmols change
0.100mols BeforeHA
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Tro, Chemistry: A Molecular Approach 35
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?
divide by the liters of solution to find the new molarities
add the change to the initial number of moles to find the moles after reaction
using the added chemical as the limiting reactant, determine how the moles of the other chemicals change
HC2H3O2 + OH C2H3O2 + H2O
new Molarity
0.100A-
0.010OH
mols Endmols change
0.100mols BeforeHA
-0.010-0.010 +0.010
00.1100.090
M 090.0L1.00mol 090.0]HA[ ==
0.090
M 110.0L1.00mol 100.0]A[ - ==
0.110
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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?
Enter the initial concentrations assuming the [H3O+] from water is 0, and using the new molarities of the [HA] and [A]
Construct an ICE table for the reaction
Write the reaction for the acid with water
HC2H3O2 + H2O C2H3O2 + H3O+
0.110[A-]
equilibriumchange
00.090initial[H3O+][HA]
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Tro, Chemistry: A Molecular Approach 37
0.110[A-]
equilibriumchange
00.090initial[H3O+][HA]
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 equilibrium constant expression
sum the columns to find the equilibrium concentrations in terms of x
represent the change in the concentrations in terms of x
+x+xx0.090 x 0.110 + x x
[ ]( )( )( )x
xxKa +
==+
090.0110.0
OHHC]OH][OH[C
232
3-232
HC2H3O2 + H2O C2H3O2 + H3O+
-
Tro, Chemistry: A Molecular Approach 38
[ ]( )( )( )x
xxKa +
==+
090.0110.0
OHHC]OH][OH[C
232
3-232
since Ka is very small, approximate the [HA]eq = [HA]init and [A]eq = [A]init solve for x
determine the value of Ka
x= 51074.1
0.110
+x0.100[A-]
x0.090equilibrium+x-xchange 00.100initial
[H3O+][HA]
0.090 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
[ ]( )( )( )090.0110.0
OHHC]OH][OH[C
232
3-232 xKa ==
+
Ka for HC2H3O2 = 1.8 x 10-5
( )( )( )090.0110.0108.1 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%100100.91074.1
2
5
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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
0.110
+x0.110[A-]
1.5E-50.090equilibrium+x-xchange 00.090initial
[H3O+][HA]
( ) M 110.01074.1110.0110.0]OHC[ 5232 =+=+= x
0.110 + x x 0.090 x
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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
==
=
+
0.110
+x0.110[A-]
1.5E-50.090equilibrium+x-xchange 00.090initial
[H3O+][HA]
-
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
0.110
+x0.110
[A-]
1.5E-50.090equilibrium+x-xchange 00.090initial
[H3O+][HA]
[ ]( )( )
( )5
5232
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?
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
find the pKa from the given Ka HC2H3O2 + H2O C2H3O2 + H3O+
( )547.4
108.1log
logp5
==
=
aa KK
Ka for HC2H3O2 = 1.8 x 10-5
0.110
+x
0.110
[A-]
x0.090equilibrium+x-xchange 00.090initial
[H3O+][HA]
-
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 NaC2H3O2in 1.00 L that has 0.010 mol NaOH added to it?
Check the x is small approximation
Substitute into the Henderson-HasselbalchEquation
HC2H3O2 + H2O C2H3O2 + H3O+
+=
][HA][AlogppH
-
aK
( ) 83.4090.0
0.110log547.4pH =
+=
%5%016.0%100090.0
1074.1 5
-
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
( )
83.4090.0
0.110log547.4pH
=
+=
pKa for HC2H3O2 = 4.745
M 010.0L1.00mol 010.0]OH[ ==
( ) 00.2100.1log]OHlog[pOH
2 ==
=
12.002.00-14.00
pOH -14.00pH00.14pOHpH
==
==+
-
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?
Check the x is small approximation
Substitute into the Henderson-HasselbalchEquation
Assume the [B] and [HB+] equilibrium concentrations are the same as the initial
find the pKa of the conjugate acid (NH4+) from the given Kb
NH3 + H2O NH4+ + OH
+= + ][HB
[B]logppH aK
( )( ) 65.920.00.50log25.9pH =
+=
109.65-3
-pH3
1032.210]OH[
10]OH[+
+
==
=
%5%10020.01032.2 10
-
Tro, Chemistry: A Molecular Approach 48
Ex 16.4 - What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?
Check the x is small approximationCalculate pH from pOH
Substitute into the Henderson-Hasselbalch Equation Base Form, find pOH
Assume the [B] and [HB+] equilibrium concentrations are the same as the initial
find the pKb if given KbNH3 + H2O NH4+ + OH
+=
+
[B]][HBlogppOH bK
( )( ) 35.450.00.20log75.4pOH =
+=
5-4.35-pOH 1074.41010]OH[ ===
%5%0089.0%10050.01074.4 5
-
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
-
0.030-
0.020A-
00.010
0OH
0.17mols After-mols added
0.18mols BeforeHA
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
( ) 09.5090.0
0.110log00.5pH =
+=
after adding 0.010 mol NaOHpH = 5.09
HA + OH A + H2O
0.110
-0.100
A-
0
0.0100
OH
0.090mols After-mols added
0.100mols BeforeHA
( ) 25.417.0
0.030log00.5pH =
+=
after adding 0.010 mol NaOHpH = 4.25( )
%8.1
%1005.00
5.00-5.09 Change %
=
=( )
%0.5
%1004.05
4.05-4.25Change%
=
=
a buffer is most effective with equal concentrations of acid and base
-
0.51
-0.500
A-
0
0.0100
OH
0.49mols After-mols added
0.50mols BeforeHA ( )
%6.3
%1005.00
5.00-5.18Change%
=
=
0.060-
0.050A-
00.010
0OH
0.040mols After-mols added
0.050mols BeforeHA
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
( ) 02.549.0
0.51log00.5pH =
+=
after adding 0.010 mol NaOHpH = 5.02
HA + OH A + H2O
( ) 18.5040.0
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] = 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
-
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
-
aKLowest pH
( )1ppH
10.0logppH=+=
a
a
KK
Highest pH( )
1ppH10logppH
+=+=
a
a
KK
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 = 1.95Nitrous Acid, HNO2 pKa = 3.34Formic Acid, HCHO2 pKa = 3.74Hypochlorous 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 = 1.95Nitrous Acid, HNO2 pKa = 3.34Formic Acid, HCHO2 pKa = 3.74Hypochlorous Acid, HClO pKa = 7.54The 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 : HCHO2would 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 complete
when the reaction is complete we have reached the endpointof 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
NaOHmol100.5L 1
NaOH mol 0.100NaOH L 0.00504=
used HCl moles NaOH mol 1HCl mol 1NaOH mole
=
5.0 x 10-4 mol NaOH
usedHClmol100.5NaOH mol 1HCl mol 1NaOH mol 100.5
4
4
=
excess HCl mol used HCl mol -HCl mol initial
= excessHClmol102.00used HCl mol105.0 -HCl mol 102.50
3-
-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
excessNaOH mol105.0used NaOH mol102.50 -NaOH mol 103.00
4-
-3-3
= NaOHmol1000.3
L 1NaOH 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 Lexcess 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 66
excessNaOH mol105.0used NaOH mol102.50 -NaOH mol 103.00
4-
-3-3
= NaOHmol1000.3
L 1NaOH mol 0.100NaOH L 0.0300
3=
( )11.962.04-14.00pH
04.21009.9-logpOH 3
==== pOH-14.00pH
]-log[OHpOH==
][OHNaOH M NaOH LHCl Lexcess NaOH mol
==+
][OHNaOH M 0.00909 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
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
=
-
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 = 1.37added 15.0 mL NaOH0.00100 mol HClpH = 1.60added 20.0 mL NaOH0.00050 mol HClpH = 1.95
added 25.0 mL NaOHequivalence pointpH = 7.00added 40.0 mL NaOH0.00150 mol NaOHpH = 12.36added 50.0 mL NaOH0.00250 mol NaOHpH = 12.52
-
Tro, Chemistry: A Molecular Approach 68
Titration Curve of Strong Acid with NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50Volume NaOH Added, mL
pH
pH
Derivative
Titration of 25.0 mL of 0.100 M HClwith 0.100 M NaOH
The 1st derivative of the curve is maximum at the equivalence point
Since the solutions are equal concentration, the equivalence point is at equal volumes
-
Tro, Chemistry: A Molecular Approach 69
Species Distribution in a Strong Acid vs pH
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
Frac
tion
of S
peci
es in
Mix
ture
HAA-
After about pH 3, there is practically no HCl left, it has all been reacted and become NaCl + H2O
-
Tro, Chemistry: A Molecular Approach 70
Titration of 25 mL of 0.100 M HCHO2with 0.100 M NaOH
HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq) Initial pH:
x
+x
0.000
[CHO2-]
x0.100 - xequilibrium
+x-xchange
00.100initial
[H3O+][HCHO2]Ka = 1.8 x 10-4
( )( )( )
M 1042.4]O[H100.0100.0
108.1
]HCHO[]O][H[CHO
33
24
2
32
+
+
==
=
=
=
x
xx
xx
Ka
( ) 37.210424.-log]Olog[H- pH
3-3
=== +
%5%2.4%100100.0102.4 3
-
Tro, Chemistry: A Molecular Approach 71
Titration of 25 mL of 0.100 M HCHO2with 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
NaOHmol100.5L 1
NaOH mol 0.100NaOH L 0.00504=
5.0E-4
-
0A-
0
5.0E-4
0OH
2.00E-3mols After-mols added
2.50E-3mols BeforeHA
+=
2
2HCHO molCHO mollogppH aK
14.3pH10.002
10.05log74.3pH 5-4-
=
+=
( ) 74.3108.1-loglog- p
4- === aa KK
-
72
( )( )( )
M 107.1][OH0500.00500.0
106.5
]CHO[]][OH[HCHO
6
211
2
2
==
=
=
=
x
xx
xx
Kb
96
14
3
109.5107.1
101]OH[
]OH[
+
=
=
= wK
114
14
CHO ,
106.5108.1
1012
=
=
=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 HCHO2with 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
NaOHmol1050.2L 1
NaOH mol 0.100NaOH L 0.02503=
2.50E-3
-
0A-
0
2.50E-3
0OH
0mols After-mols added
2.50E-3mols BeforeHA
5.00E-2-x
-x
0.0500
[CHO2-]
xxequilibrium
+x+xchange
00initial
[OH][HCHO2]
CHO2(aq) + H2O(l) HCHO2(aq) + OH(aq)Kb = 5.6 x 10-11
( ) 23.8109.5-log]Olog[H- pH
9-3
=== +
[OH-] = 1.7 x 10-6 M
-
Tro, Chemistry: A Molecular Approach 73
excessNaOH mol105.0used NaOH mol102.50 -NaOH mol 103.00
4-
-3-3
=
NaOHmol1000.3L 1
NaOH mol 0.100NaOH L 0.03003=
][OHNaOH M NaOH LHCl Lexcess 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 74
excessNaOH mol105.0used NaOH mol102.50 -NaOH mol 103.00
4-
-3-3
=
NaOHmol1000.3L 1
NaOH mol 0.100NaOH L 0.03003=
( )11.962.04-14.00pH
04.2101.9-logpOH 3
==== pOH-14.00pH
]-log[OHpOH==
][OHNaOH M NaOH LHCl Lexcess 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
=
-
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 HCHO2pH = 3.74 = pKahalf-neutralization
initial HCHO2 solution0.00250 mol HCHO2pH = 2.37
added 5.0 mL NaOH0.00200 mol HCHO2pH = 3.14
added 10.0 mL NaOH0.00150 mol HCHO2pH = 3.56
added 15.0 mL NaOH0.00100 mol HCHO2pH = 3.92added 20.0 mL NaOH0.00050 mol HCHO2pH = 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 10-6pH = 8.23
added 50.0 mL NaOH0.00250 mol NaOH xspH = 12.52
-
Tro, Chemistry: A Molecular Approach 76
Titration Curve of Weak Acid with NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50Volume NaOH Added, mL
pH
-
Tro, Chemistry: A Molecular Approach 77
Titration Curve of Weak Acid with NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50
Volume NaOH Added, mL
pH
pH
Derivative
Titration of 25.0 mL of 0.100 M HCHO2 with 0.100 M NaOH
The 1st derivative of the curve is maximum at the equivalence point
Since the solutions are equal concentration, the equivalence point is at equal volumes
pH at equivalence = 8.23
-
Tro, Chemistry: A Molecular Approach 78
Species Distribution in a Weak Acid vs pH
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
Frac
tion
of S
peci
es in
Mix
ture
HAA-
at pH 3.74, the [HCHO2] = [CHO2]; the acid is half neutralizedhalf-neutralization occurs when pH = pKa
-
Tro, Chemistry: A Molecular Approach 79
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 Ainit using reaction stoichiometrycalculate pH with Henderson-Hasselbalch using mol HAinit and mol Ainit
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 established
mol A = original mole HAcalculate 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
Use Stoichiometry to determine the volume of added B
Write an equation for the reaction for B with HA.
HNO2 + KOH NO2 + H2O
KOH L .02000 KOH mol 0.200
KOH L 1NO mol 1KOH mol 1
NO L 1NO 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 KOH
Make a stoichiometry table and determine the moles of HA in excess and moles A made
Determine the moles of HAbefore & moles of added B
Write an equation for the reaction for B with HA.
HNO2 + KOH NO2 + H2O
KOH mol 00100.0L 1
KOH mol 200.0mL 1
L 0.001mL 00.5 =
-0
NO2-
00.00100
0OH
mols After-mols added
0.00400mols BeforeHNO2
22 HNO mol 00400.0
L 1HNO mol 100.0
mL 1L 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.
Determine Ka and pKafor HA
Use the Henderson-Hasselbalch Equation to determine the pH
Write an equation for the reaction of HA with H2O
HNO2 + H2O NO2 + H3O+
0.00100
-0
NO2-
00.00100
0OH
0.00300mols After-mols added
0.00400mols BeforeHNO2
Table 15.5 Ka = 4.6 x 10-4
( ) 15.3106.4loglogp 4 === aa KK
+=
2
2HNONOlogppH 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 point
Make a stoichiometry table and determine the moles of HA in excess and moles A made
Determine the moles of HAbefore & moles of added B
Write an equation for the reaction for B with HA.
HNO2 + KOH NO2 + H2O
-0
NO2-
00.00200
0OH
mols After-mols added
0.00400mols BeforeHNO2
22 HNO mol 00400.0
L 1HNO mol 100.0
mL 1L 0.001mL 0.40 =
0.00200 0.00200
at half-equivalence, moles KOH = 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.
Determine Ka and pKafor HA
Use the Henderson-Hasselbalch Equation to determine the pH
Write an equation for the reaction of HA with H2O
HNO2 + H2O NO2 + H3O+
0.00200
-0
NO2-
00.00200
0OH
0.00200mols After-mols added
0.00400mols BeforeHNO2
Table 15.5 Ka = 4.6 x 10-4
( ) 15.3106.4loglogp 4 === aa KK
+=
2
2HNONOlogppH 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 Kas are to each other, the less distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M H2SO3with 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 solutionHInd(aq) + H2O(l) Ind(aq) + H3O+(aq)
the color of the solution depends on the relative concentrationsof Ind:HInd
when Ind:HInd 1, the color will be mix of the colors of Indand HIndwhen 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 pH
pKa 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
mnK
+
= solubilitymolar
-
Tro, Chemistry: A Molecular Approach 99
Ex 16.8 Calculate the molar solubility of PbCl2in pure water at 25C
Create an ICE table defining the change in terms of the solubility of the solid
Write the dissociation reaction and Kspexpression
2SSEquilibrium
+2S+SChange
00Initial
[Cl][Pb2+]
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 PbCl2in pure water at 25C
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Substitute into the Ksp expression
2SSEquilibrium
+2S+SChange
00Initial
[Cl][Pb2+]
Ksp = [Pb2+][Cl]2Ksp = (S)(2S)2
M1043.141017.1
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
Create an ICE table defining the change in terms of the solubility of the solid
Write the dissociation reaction and Kspexpression
(2.10 x 10-2)(1.05 x 10-2)Equilibrium
+2(1.05 x 10-2)+(1.05 x 10-2)Change
00Initial
[Br][Pb2+]
PbBr2(s) Pb2+(aq) + 2 Br(aq)
Ksp = [Pb2+][Br]2
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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
plug into the equation and solve
Substitute into the Ksp expression
Ksp = [Pb2+][Br]2Ksp = (1.05 x 10-2)(2.10 x 10-2)2
(2.10 x 10-2)(1.05 x 10-2)Equilibrium
+2(1.05 x 10-2)+(1.05 x 10-2)Change
00Initial
[Br][Pb2+]( )( )6
222
1063.4
1010.21005.1
=
=
sp
sp
KK
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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
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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
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Tro, Chemistry: A Molecular Approach 106
Ex 16.10 Calculate the molar solubility of CaF2in 0.100 M NaF at 25C
Create an ICE table defining the change in terms of the solubility of the solid
Write the dissociation reaction and Kspexpression
0.100 + 2SSEquilibrium
+2S+SChange
0.1000Initial
[F][Ca2+]
CaF2(s) Ca2+(aq) + 2 F(aq)
Ksp = [Ca2+][F]2
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Tro, Chemistry: A Molecular Approach 107
Ex 16.10 Calculate the molar solubility of CaF2in 0.100 M NaF at 25C
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Substitute into the Ksp expressionassume S is small
0.100 + 2SSEquilibrium
+2S+SChange
0.1000Initial
[F][Ca2+]
Ksp = [Ca2+][F]2Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
( )
( )M1046.1
100.01046.1
100.0
8
2
10
2
=
=
=
S
S
SKsp
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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 solubilityM2(CO3)n(s) 2 Mn+(aq) + nCO32(aq)
H3O+(aq) + CO32 (aq) HCO3 (aq) + H2O(l)
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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 occur
Q = 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
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Tro, Chemistry: A Molecular Approach 110
precipitation occurs if Q > Ksp
a supersaturated solution will precipitate if a seed crystal is added
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Tro, Chemistry: A Molecular Approach 111
Selective Precipitation
a solution containing several different cationscan 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 Kspvalues are significantly different
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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 = Ksp22 ]OH][Mg[ +=Q
( )( )
( )6
13
132
109.1059.0
1006.2]OH[
1006.2]OH][059.0[
=
=
=
= spKQ
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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
( )( )( )
26
62
1060.2011.0
1068.4]OH[
1068.4]OH][011.0[
=
=
=
= spKQ
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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 108.41060.2
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
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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 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
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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
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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
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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
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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)
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Tro, Chemistry: A Molecular Approach 125
Formation Constant the reaction between an ion and ligands to form
a complex ion is called a complex ionformation 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
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Tro, Chemistry: A Molecular Approach 126
Formation Constants
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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?
Determine the concentration of ions in the diluted solutions
Write the formation reaction and Kf expression.Look up Kf value
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)13
43
2
243 107.1
]NH][Cu[])Cu(NH[
== ++
fK
( ) M 107.6L 0.250 L 200.0L 1
mol 101.5 L 200.0]Cu[ 4
-3
2 + =+
=
( ) M 101.1L 0.250 L 200.0L 1
mol 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
0.11
-4(6.7E-4)
0.11
[NH3]
6.7E-4xEquilibrium
+ 6.7E-4-6.7E-4Change
06.7E-4Initial
[Cu(NH3)22+][Cu2+]
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)13
243
43
2107.1
])Cu(NH[]NH][Cu[
== ++
fK
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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)13
43
2
243 107.1
]NH][Cu[])Cu(NH[
== ++
fKconfirm the x is small approximation
Substitute in and solve for x
0.11
-4(6.7E-4)
0.11
[NH3]
6.7E-4xEquilibrium
+ 6.7E-4-6.7E-4Change
06.7E-4Initial
[Cu(NH3)22+][Cu2+]( )( )( )
( )( )( )
13413
4
4
413
107.211.0107.1
107.611.0107.6107.1
=
=
=
x
x
since 2.7 x 10-13
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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 AmphotericMetal 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 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+
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Tro, Chemistry: A Molecular Approach 133
Al3+
Al3+ is hydrated in water to form an acidic solutionAl(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 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