Harris: Quantitative Chemical Analysis, Eight Edition
CHAPTER 08:
MONOPROTIC ACID-BASE EQUILIBRIA
CHAPTER 08: Opener A
CHAPTER 08: Opener B
CHAPTER 08: Opener C
CHAPTER 08: Opener D
CHAPTER 08: Opener E
Chapter 8-1. Strong Acids and Bases
“Water almost never produces 10-7M [H+] and 10-7M [OH-]”
Example) In a 10-4M solution of HBr,
14
K w 10 -14
[OH ]     4  10 -10 M
[H ] 10
-
 Water dissociation produces only 10-10M [OH-] and 10-10M [H+]
Box 8-1 Concentrated HNO3 is only slightly Dissociated
- Strong acid in dilute solution are essentially completely dissociated.
- As concentration increases, dissociation decreases because there is not enough
solvent to stabilize the free ions.
Box 8-1 Concentrated HNO3 is only slightly Dissociated
1 049 cm-11  dissociated NO3-11
1,049
*
 undissociated HNO
3
Box 8-1 Concentrated HNO3 is only slightly Dissociated
Temperature dependence of K w
In this table, concentrations in Kw = [H+][OH-] are expressed in molality.
Chapter 8-2. Weak Acids and Bases
HA
֖
H+ + A-
B + H2O
֖
BH+
+
OH-
,
[H  ] [A - ]
Ka 
[HA]
,
[BH  ] [OH - ]
Kb 
[B]
 KaKb = Kw
- The conjugate base of a weak acid is a weak base.
base
The conjugate acid of a weak base is a weak acid.
E
Example)
l )
HA (Ka)
10-4
10-5
A- (Kb)
10-10
10-9
10-7
10-7
As HA becomes a weaker acid
acid, A- becomes a stronger base.
base
(but never a strong base. Recall that KaKb = Kw = 10-14)
Chapter 8-3. Weak-Acid Equilibria
- In a solution of any respectable weak acid, the concentration of H+ due to
acid
id dissociation
di
i i will
ill be
b muchh greater than
h the
h concentration
i due
d to water
dissociation (le chatelier principle).
HA
֖
H1+ + A-
H2O
֖
H2+ + OH-
[H1+] >> [H2+]
[A-] >> [OH-]
Fraction of Dissociation (Ionization Fraction)
- Consider Acid (H3A)
[H 3 A]
[H 3 A]
 

23[H 3A]  [H 2 A ]  [HA ]  [A ] F (formal conc. of H 3 A)
[H 2 A - ]
[HA 2- ]
1 
, 2 
F
F
,
[A 3- ]
3 
F
-  can be determined from the mass balance and equilibrium equations.
Example
HOCl
i)
֖
H+ + OCl-
[H  ] [OCl - ]
Ka 
[HOCl]
FHOCl = [HOCl] + [OCl-]
ii)
 FHOCl
[H  ] [OCl - ]
Ka 
[HOCl]
K a  [[HOCl]
Ka
 [HOCl] 
 [HOCl](1   )

[H ]
[H ]
[HOCl]
[H  ]

 

FHOCl
[H ]  K a
K
[OCl - ]
  a
 1
FHOCl [H ]  K a
- [HOCl] is stronger disinfectant than [OCl-]
- All weak electrolytes dissociate more as they are diluted
(F
(From
the
th above
b
equation
ti for
f 1, when
h the
th solution
l ti is
i diluted,
dil t d [H+] will
ill decrease
d
and thus 1 increases)  Fig. 9-2
Chapter 8-5. Buffers
- Buffer solution is one that resists changes in pH when acids or bases are
added or when dilution occurs.
- The proper functioning of any biological system is critically dependent
on pH
pH.
Fig. 8-3. pH dependence of the rate
of cleavage of an amid bond
byy the enzyme
y chymotrypsin.
y
yp
Henderson-Hasselbach Equation
- The central equation for buffers.
- It is merely a rearranged form of the Ka equilibrium equation.
If a solution is prepared from HA & MA
H+ + AHA
֖
MA
֖
M+ + AA- +H2O ֖ HA + OH
{H  } {A - }  H  [H ] A - [A ]
Ka 

{HA}
 HA [HA]
 HA [HA]

  H  [H ]  K a
 A - [A - ]
 HA [HA]

 log H  [H ]  - logK a - log
 A - [A - ]
 A [A - ]
 pH  pK a  log
 HA [HA]
-
continued
i d
Henderson-Hasselbach Equation
If activities are neglected, s = 1
[A - ]
 pH
H  pK a  log
l
[HA]
- If a solution is prepared from the weak base B and its conjugate acid (BH+)
֖
BH+
 pH  pK a  log
* If [HA] = [A-]
or
H+ + B
[B]
[BH  ]
[BH+] = [B] ,
pKa applies to this acid
pH = pKa
Henderson-Hasselbach Equation
* For every power of 10 change in the ratio [A-]/[HA], the pH changes
by one unit (See Table 9-1)
[A - ]
pH  pK a  log
l
[HA]
Henderson-Hasselbach Equation
[B]
pH  pK a  log
[BH  ]
pK
Ka applies
li to
t this
thi acid
id
[A - ]
pH  pK a  log
[HA]
pK
Ka applies
li to
t this
thi acid
id
If there are 10 different acids and bases in the solution,
the 10 forms of Henderson-Hasselbach equation
q
must all
give the same pH ! , because there can be only one
concentration of H+ in a solution.
Buffer pH
- The most common method of preparing a buffered solution
: Add a mixture of a weak acid (HA) and its conjugate base (MA)
- Weak acid will react with any strong base added.
Conjugate base will react with any strong acid added.
- pKa of the acid should be very close to the desired pH of the buffer solution
for a maximum buffer intensity.
Buffer pH
Let’s examine a buffer solution
(CHA moles/L weak acid + CNaA moles/L conjugate base)
[H  ][A - ]
Ka 
[HA]

K w  [H  ][OH - ]

CT,A = CHA + CNaA = [HA] + [A-]

CT,Na = [Na
[N +] = CNaA

Equilibria ;
Material Balance ;
continued
i d
Buffer pH
Charge Balance ;
[Na+] + [H+] = [A-] + [OH-]

From  & 
[A-] = CNaA + [H+] - [OH-]

From  & 
[HA] = CHA + [OH-] + [H+]

continued
From Henderson-Hasselbach Equation.
[A  ]
pH  pK a  log
[HA]
Putting eqns.  & 
C NaA  [H  ]  [OH  ]
 pH  pK a  log
C HA  [OH - ] - [H  ]
This is original equation and has no approximation except activities.
continued
But for most buffer solutions, and almost invariably for prepared buffers
CNaA >> [H+] - [OH-]
CHA >> [OH-] - [H+]
 pH  pK a  log
C NaA (salt)
C HA (acid)
Approximation;
CNaA = [A-]
CHA = [HA]
Remark;
i) When you mix a weak acid (HA) with a conjugate base (NaA),
you can get what you mix, that is, CHA = [HA] and CNaA = [A-].
Pure HA dissociate very little, and adding extra A- to the solution
will make the HA dissociate even less. Similarly, A- does not
react very much with water, and adding extra HA makes A- react
even less.
HA ֖ H+ + AA- + H2O ֖ HA + OHii) This approximation breaks down for dilute solutions or at extremes
of pH. If CHA or CNaA is small, or if [H+] or [OH-] is large,
the approximation CHA = [HA], and CNaA = [A-] are not good.
iii) IIn acidic
idi solutions,
l i
[H
[H+]] >> [OH
[OH-],
] so [OH-]
[OH ] can bbe iignoredd in
i
eqns.  &  and in basic solution [H+] can be ignored.
Example
What will be the pH?
0 0100 moll off HA
0.0100
A (pK
( a = 2.00)
2 00)
and 0.0100 mol
Solution i))
of A-
pH  p
p
pK a  log
g
C AC HA
 2.00  log
 2.00
0.01
0.01
(
(Wrong!!!)
g )
in 1.00 L of a solution
Example
Solution ii)
pH  pK a  log
 pK a  log
HA
0.01-x
֖
C A -  [H  ]  [OH - ]
C HA  [H  ]  [OH - ]
C A -  [H  ]
C HA  [H  ]
The solution is acidic
[[H+] >> [OH
[ -]
H+ + Ax 0.01+x
[H  ][A - ] x(0.01
(0 01  x)
 Ka 

 10 -2.00
[HA]
0.01  x
 x = [H+] = 4.1410-3
[OH-] = 22.410
410-12 (neglected)
continued
i d
Example
The conc. of HA and A- are not what we mxied
[HA] = CHA – [H+] = 0.00586 M
[A-] = CA- + [H+] = 0.0141 M
 pH  2.00  log
0.0141
 2.38
0.00586
In this example, HA is too strong and conc. are too slow for HA and Ato be equal to their formal concentrations (CHA and CA-, respectively)
Box 8-3 Strong Plus Weak Reacts Completely
A strong acid reacts with a weak base essentially “completely” because the
equilibrium constant is large.
B +
Weak
base
H+
֖
Strong
acid
BH+
1
(for BH  )
K
Ka
If B is tris(hydroxymethyl)aminomethane, then the equilibrium constant for
reaction with HCl is
K
1
1
 -8.075  1.2  10 8
K a 10
Box 8-3 Strong Plus Weak Reacts Completely
A strong base reacts “completely” with a weak acid because the equilibrium
constant is, again, very large.
OH-
+ HA ֖
Strong
base
Weak
acid
A-
+ H2O
K
1
(for A - )
Kb
If HA is acetic acid, then the equilibrium constant for reaction with NaOH is
K
K for HA
1
 a
 1.7  10 9
Kb
Kw
Box 8-3 Strong Plus Weak Reacts Completely
The reaction of a strong acid with a strong base is even more complete than
a strong + weak reaction:
H+
Strong
acid
+
OHStrong
base
֖ H2O
K
1
(for A - )
Kb
* If you mix a strong acid, a strong base, a weak acid, and a weak base, the
strong
g acid and base will neutralized each other until one is used up.
p
The remaining strong acid or base will then react with the weak base or
weak acid.
Preparing a Buffer in Real Life
Tris Buffer [tris(hydroxymethyl)aminomethane]
BH+ (Tris
( i hydrochloride)
h d hl id )
B (Tris)
( i)
pKa = 8.075
If you want to get a tris buffer of 7.80, you can calculate the conc. of BH+ and
B with Henderson Eq.
B (tris)
pH  pK a  log
BH 
7.80
8.075
Preparing a Buffer in Real Life!
If you really wanted to prepare a tris buffer of pH 7.60, you would not do it
b calculating
by
l l ti what
h t tto mix.
i S
Suppose th
thatt you wish
i h tto prepare 11.00
00 L off buffer
b ff
containing 0.100 M tris at a pH of 7.60. You have available solid tris hydro
-chloride
chloride and approximately 1 M NaOH.
NaOH Here
Here’ss how to do it:
1.
Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker
containing about 800 mL of water.
2.
Place a pH electrode in the solution and monitor the pH.
3
3.
Add NaOH until the pH is exactly 7.60.
7 60
4.
Transfer the solution to a volumetric flask and wash the beaker a few
times. Add the washings to the volumetric flask.
5
Dilute to the mark and mix.
You do not mix calculated quantities,
quantities though a quick calculation is helpful
so that you have some idea of how much will be needed.
Reasons why a calculation would be wrong:
1. You might have ignored activity coefficients.
2. The temperature might not be just right.
3. The approximations that [HA] = FHA and [A-] = FAcould be in error.
4. The pKa reported for tris in your favorite table is
probablyy not what you
p
y would measure in your
y
lab.
5. You will probably make an arithmetic error anyway.
Buffer Capacity
The buffer capacity, , also called buffer intensity, is defined as
β
dC b
dC
 a
dpH
dpH
Figure 8-4.
84
a) Cb vs. pH for a solution containing
0.100F HA with pKa =5.00
b) Buffer Capacity vs. pH