### Acids & Bases - Fall River Public Schools

```Weak Acids & Acid Ionization Constant
 Majority of acids are weak. Consider a weak
monoprotic acid, HA:


HA ( aq )  H 2 O ( l )  H 3 O ( aq )  A ( aq )
 The equilibrium constant for the ionization would be:

Ka 

[ H 3 O ][ A ]
[ HA ]

or
Ka 

[ H ][ A ]
[ HA ]
Acid Ionization Constant
 The magnitude of Ka for an acid determines its
strength
+
More [H ] in
Stronger
the Acid
solution
Higher
the Ka
+
Lower [H ]
in solution
Lower
the Ka
Weaker
the
Acid
Determining pH from Ka
 Calculate the pH of a 0.50 M HF solution at 25 C. The
ionization of HF is given by:


HF ( aq )  H ( aq )  F ( aq )

Ka 

[ H ][ F ]
[ HF ]
 7 . 1 x10
4
Determining pH from Ka


0.50
0.00
0.00
-x
+x
+x
0.50 – x
x
x
HF ( aq )  H ( aq )  F ( aq )
Initial (M)
Change (M)
Equilibrium
(M)
Ka 
( x )( x )
( 0 . 50  x )
 7 . 1 x10
4
Determining pH from Ka
0.50 – x ≈ 0.50
 The expression becomes:
x
2
 7 . 1 x10
0 . 50
 x  0 . 019 M
4
Determining pH from Ka
 At equilibrium,
[HF] = (0.50-0.019) = 0.48 M
[H+] = 0.019 M
[F-] = 0.019 M
 pH   log( 0 . 019 )  1 . 72
 This only works if x is less than 5% of 0.50
 Why? Ka are generally only accurate to ±5%
Was the approximation good?
( x)
x100 %
0 . 019 M
( init .concentrat ion )
x100 %  3 . 8 %
0 . 50 M
 Consider if initial concentration of HF is 0.050 M. Same
process above, but x = 6.0 x 10-3 M. This would not be
valid.
6 . 0 x10
3
M
0 . 050 M
x100 %  12 %
Weak Bases & Base Ionization
Constants
 Same treatment as for acids. Given ammonia in water:
NH 3 ( aq )  H 2 O ( l )  NH



4
( aq )  OH
[ NH 4 ][ OH ]
5
K b we solve using [OH-], 1not
. 8 x[H
10+]
 Reminder:
[ NH ]
3

( aq )
Relationship of Ka to their
Conjugate Base
Kw = Ka Kb
Mole Buck
Opportunity!
 This leads us to conclude the stronger the acid (larger
Ka), the weaker its conjugate base (smaller Kb)
Diprotic & Polyprotic Acids
 A more involved process due to stepwise dissociation
of hydrogen ion
 Each step is like a monoprotic acid
 Be sure to think about what is present at each step!
 Note the conjugate base is used as the acid in the next
step
Diprotic Acid Calculation
 Calculate all species present at equilibrium in a 0.10 M
solution of oxalic acid (H2C2O4)
Diprotic Acid Calculation


H 2 C 2 O 4 ( aq )  H ( aq )  HC 2 O 4 ( aq )
Initial (M)
Change
(M)
Equilibriu
m (M)
0.10
0.00
0.00
-x
+x
+x
0.10 – x
x
x
Diprotic Acid Calculation


Ka 
[ H ][ HC 2 O 4 ]
 6 . 5 x10
[ H 2C 2O 4 ]
 Making the approximation
0.10 – x  0.10, we get:
6 . 5 x10
2

x
 x  8 . 1 x10
Approx. good?
8 .1 x 0
2
2
M
0 . 10 M
0 . 10
2
2
M
STOP
x100 %  81 %
Diprotic Acid Calculation
x  6 . 5 x10
2
2
x  6 . 5 x10
3
0
 x = 0.054 M
 After the first stage, we have:
 [H+] = 0.054 M
 [HC2O4-] = 0.054 M
 [H2C2O4] = (0.10 – 0.054) M = 0.046 M
 Next step: Treat conj. base as acid for 2nd step
Diprotic Acid Calculation
 Second dissociation would be:

2

HC 2 O 4 ( aq )  H ( aq )  C 2 O 4 ( aq )
Initial (M)
Change (M)
Equilibriu
m (M)
0.054
0.054
0.00
-y
+y
+y
0.054 – y
0.054 + y
y
Diprotic Acid Calculation

Ka 
[ H ][ C 2 O 4

2
]
5
 6 . 1 x10
[ HC 2 O 4 ]
 Applying the approximation (for both) we obtain:
Approx. good?
( 0 . 054 )( y )
( 0 . 054 )
 y  6 . 5 x10
6 . 1 x10
2
5
M
0 . 054 M
STOP
x100 %  0 . 11 %
Diprotic Acid Calculation!
 Finally, at equilibrium:
 [H2C2O4] = 0.046 M
 [HC2O4-] = (0.054 – 6.1 x 10-5) = 0.054 M
 [H+] = (0.054 + 6.1 x 10-5) = 0.054 M
 [C2O42-] = 6.1 x 10-5 M
Conclusions on Polyprotic Acids
 The past example shows that for diprotic acids, Ka1 >>
Ka2
 From this, we can assume the majority of H+ ions are
produced in the first stage of ionization
 Secondly, concentration of conj. base is numerically
equal to Ka2
Molecular Structure and Strength
of Acids
 Strength of hydrohalic acids (HX) depends on 2
factors:

(called bond enthalpy)

 Strength depends on ease of ionization
 Stronger the bond,
difficult to ionize

the polarity, better chance of ionizing
Strength of Binary Acids
HF << HCl < HBr < HI
 Based on polarity, HF might considered strongest, but
bond strength opposes this separation of charge
 This implies that bond
is the predominant
factor in determining acid strength of binary acids
Strength of Oxoacids
 Divide oxoacids into two groups:
 Oxoacids having different central atoms from the same
group with the same oxidation number

i.e.
and
 Oxoacids have same central atom but different number
of attached groups

i.e.
and
Strength of Oxoacids
• Oxoacids having different central atoms from the same
group with the same oxidation number
 In this group, acid strength increases with increasing
electronegativity of the central atom

Higher polarity means easier to ionize
HClO3 > HBrO3
Strength of Oxoacids
• Oxoacids have same central atom but different number of
attached groups
 In this group, acid strength
number of central atom increases

i.e. the
as oxidation
oxygen, the merrier!
HClO4 > HClO3 > HClO2 > HClO
Acid-Base Properties of Salts
 Salt hydrolyis is the reaction of an anion or a cation of
a salt (or both) with water
 Salts can be neutral, basic, or acidic and follow certain
trends
BEWARE!
 Acids
mixed with bases forms salt + water!
Salts that Produce Neutral Sol’ns
 Alkali metal ion or alkaline earth metal ion (except Be)
do not undergo hydrolysis
 Conjugate bases of strong acids (or bases) do not
undergo hydrolysis
 i.e. Cl-, Br-, NO3-
 So NaNO3 forms a neutral solution
Salts that Produce Basic Sol’ns
 Conjugate bases of a weak acid will react to form OH-
ions
 For example, NaCH3COO forms Na+ and CH3COO- in
solution. Acetate ion is the conjugate base of acetic
acid, and undergoes hydrolysis:
CH 3 COO

( aq )  H 2 O ( l )  CH 3 COOH ( aq )  OH

( aq )
Basic Salt Hydrolysis Calculation
 Calculate the pH of a 0.15 M solution of sodium acetate
(CH3COONa)

CH 3 COONa ( aq )  Na ( aq )  CH 3 COO
 Since the dissocation is 1:1 mole ratio, the
concentration of the ions is the same
 [ CH 3 COO

]  0 . 15 M

( aq )
Basic Salt Hydrolysis Calc.
 Because acetate ion is the conj. base of a weak acid, it
hydrolyzes as:
CH 3 COO
Initial (M)
Change (M)
Equilibriu
m (M)

( aq )  H 2 O ( l )  CH 3 COOH ( aq )  OH

0.15
0.00
0.00
-x
+x
+x
0.15 – x
x
x
( aq )
Basic Salt Hydrolysis Calc.
Kb 
[ CH 3 COOH ][ OH


]
 5 . 6 x10
10
[ CH 3 COO ]
 Applying the approximation,
5 . 6 x10
10
x

Approx. good?
2
9 . 2 x10
0 . 15
 x  9 . 2 x10
6
M
6
M
x100 %  0 . 0061 %
0 . 15 M
STOP
Also known as percent
hydrolysis!
Acidic Salt Hydrolysis
 Conjugate acid of a weak base will react to form H3O+
ions
 Small, highly charged cation will also produce acidic
solutions when hydrated
 Typically
Al ( H 2 O ) 6
3
metals (Al3+, Cr3+, Fe3+)
( aq )  H 2 O ( l )  Al ( OH )( H 2 O ) 5
2
 H 3O

Common Ion Effect Revisited
 Recall that the addition of a common ion causes an
equilibrium to shift
 Earlier we related this to the solubility of a salt
 The idea is just an application of Le Chatelier’s Principle
pH changes due to Common Ion
Effect
 Consider adding sodium acetate (NaCH3COO) to a
solution of acetic acid:


CH 3 COOH ( aq )  CH 3 COO ( aq )  H
 The addition of a common ion here (CH3COO-) will
increase the pH
 By consuming H+ ions
Henderson-Hasselbalch
 Consider:



HA ( aq )  H ( aq )  A ( aq )
 Rearranging Ka for [H+]we get:

[H ] 
K a [ HA ]

[A ]
Ka 

[ H ][ A ]
[ HA ]
Henderson-Hasselbalch
 Take negative log of both sides:


 log[ H ]   log K a  log
pH  pK a  log
or:
pOH  pK b  log
[A ]
[ HA ]
[ Conjugate base ]
[ Acid ]
[ Conjugate acid ]
[ Base ]
Finding pH with common ion present
 Calculate the pH of a solution containing both 0.20 M
CH3COOH and 0.30 M CH3COONa? The Ka of
CH3COOH is 1.8x10-5.
 Sodium acetate fully dissociates in solution
 [ CH 3 COO

]  0 . 30 M
Finding pH with common ion present
 Can use I.C.E. table, or Henderson-Hasselbach
pH  pK a  log
pH  [  log( 1 . 8 x10
[ CH 3 COO

]
[ CH 3 COOH ]
5
)]  log
[ 0 . 30 ]
[ 0 . 20 ]
pH  4 . 92
Effect of Common Ion on pH
 Consider calculating the pH for a 0.20 M acetic acid
solution
 pH = 2.72
 From our last example, its obvious the pH has
increased due to the common ion
Buffer Solutions
 Buffers are a solution of (1) weak acid or weak base and
(2) its salt
 Buffers resist changes in pH upon addition of acid or
base
 Buffer Capacity: refers to the amount of acid or base
a buffer can neutralize
Buffer Solutions
 Consider a solution of acetic acid and sodium acetate
 Upon addition of an acid, H+ is consumed:
 Upon addition of a base, OH- is consumed:
CH 3 COO


( aq )  H ( aq )  CH 3 COOH ( aq )
CH 3 COOH ( aq )  OH

( aq )  CH 3 COO

( aq )  H 2 O ( l )
Buffer Animation
 http://www.mhhe.com/physsci/chemistry/essentialch
emistry/flash/buffer12.swf
Buffer Problem
 (a) Calculate the pH of a buffer system containing
1.0M CH3COOH and 1.0 M CH3COONa. (b) What is
the pH of the buffer system after the addition of 0.10
mole of gaseous HCl to 1.0 L of the solution? Assume
volume of sol’n does not change.
Buffer Problem
 To calculate pH of buffer, I.C.E. or H.H.
pH  [  log( 1 . 8 x10
5
)]  log
pH  pK a  4 . 74
[1 . 0 ]
[1 . 0 ]
 With addition of HCl, we are adding 0.10 M H+, which
will react completely with acetate ion
Buffer Problem
CH 3 COO
Initial (mol)


( aq )  H ( aq )  CH 3 COOH ( aq )
1.0
0.10
1.0
Change
(mol)
-0.10
-0.10
+0.10
Equilibrium
(mol)
0.90
0
1.10
Now acetic acid will still dissociate and amount of H+
formed is the pH of sol’n
Buffer Problem
 Using H.H. (or I.C.E.)
CH 3 COOH ( aq )  CH 3 COO
pH  [  log( 1 . 8 x10
5

)]  log
( aq )  H
[ 0 . 90 ]
[1 . 10 ]
pH  4 . 66

Finding Buffers of Specific pH
 If concentrations of both species are equal this means:

log
[A ]
0
[ HA ]
 Using H.H., to find specific pH, search for pKa  pH
Finding Buffers of Specific pH
 Describe how you would prepare a “phosphate buffer”
with a pH of about 7.40
H 3 PO 4  H

H 2 PO 4  H
HPO
2
4
 H

 H 2 PO 4

2

 HPO
 PO 4
4
3

K a  7 . 5 x10
3
; pK a  2 . 12
K a  6 . 2 x10
8
; pK a  7 . 21
K a  4 . 8 x10
 13
; pK a  12 . 32
Finding Buffers of Specific pH
 Using the HPO42-/H3PO4- buffer:
7 . 40  7 . 21  log
[ HPO
2
4
]

[ H 2 PO 4 ]
log
[ HPO
2
4

]
[ H 2 PO 4 ]
 0 . 19  10
0 . 19
[ HPO
2
4

]
[ H 2 PO 4 ]
This means a mole ratio of 1.5 moles disodium hydrogen phosphate : 1.0 mole
Monosodium dihydrogen phosphate will result in a buffer solution with a 7.4 pH
 10
0 . 19
 1 .5
Finding Buffers of Specific pH
 To obtain this solution, disodium
hydrogen phosphate (Na2HPO4) and
sodium dihydrogen phosphate
(NaH2PO4) is in 1.5:1.0 ratio
 Meaning it is 1.5 M Na2HPO4 and 1.0 M NaH2PO4 are
combined per liter of solution
Acid-Base Titrations
 Three situations will be considered:
 Strong Acid/ Strong Base
 Weak Acid/ Strong Base
 Strong Acid/ Weak Base
 Titrations of weak acid/base are complicated by
hydrolysis
Strong Acid-Strong Base Titration
 Reacting HCl and NaOH, the net ionic equation would
be:

H ( aq )  OH

( aq )  H 2 O ( l )
 For our example, 0.100 M HCl is being titrated by 0.100
M NaOH
 Calculating pH changes depends on the stage of the
titration
Strong Acid-Strong Base Titration
 Scenario #1: After addition of 10.0 mL of 0.100 M
NaOH to 25.0 mL of 0.100 M HCl.
 Total volume = 35.0 mL
Strong Acid-Strong Base Titration
 Scenario #1
0 . 100 mol NaOH
10 . 0 mL x
 1 . 00 x10
3
 2 . 50 x10
3
mol NaOH
1000 mL
25 . 0 mL x
0 . 100 mol HCl
mol HCl
1000 mL
2 . 50 x10
1 . 50 x10
3
3
mol
35 . 0 mL
mol  1 . 00 x10
x
1000 mL
1L
3
mol  1 . 5 x10
3
 0 . 0429 M HCl
mol HCl left
 pH  1 . 37
Strong Acid-Strong Base Titration
 Scenario #2: After addition of 25.0 mL of 0.100 M
NaOH to 25.0 mL of 0.100 M HCl (aka equivalence
point)
 Because equivalent moles of acid/base
Strong Acid-Strong Base Titration
 Scenario #2
 Since no hydrolysis occurs (both are
strong) equivalence point of all
strong acid/base titrations has a

Keep in mind, equivalence point means
Strong Acid-Strong Base Titration
 Scenario #3: After addition of 35.0 mL of 0.100 M
NaOH to 25.0 mL of 0.100 M HCl
 Total volume = 60.0 mL
Strong Acid-Strong Base Titration
 Scenario #3
35 . 0 mL x
0 . 100 mol NaOH
 3 . 50 x10
3
mol NaOH
1000 mL
3 . 50 x10
3
mol NaOH  2 . 50 x10
3
mol HCl  1 . 0 x10
3
mol NaOH
pH  14 . 00  pOH  14 . 00  1 . 78  12 . 22
Weak Acid-Strong Base Titration
 Consider the neutralization between acetic acid and
sodium hydroxide:
CH 3 COOH ( aq )  NaOH ( aq )  CH 3 COONa ( aq )  H 2 O ( l )
 The net ionic is:
CH 3 COOH ( aq )  OH

( aq )  CH 3 COO

( aq )  H 2 O ( l )
 Let’s calculate the pH for this reaction at different
stages
Weak Acid-Strong Base Titration
 Scenario #1: 25.0 mL of 0.100 M acetic acid is titrated
with 10.0 mL of 0.100 M NaOH
 Total volume = 35.0 mL
Weak Acid-Strong Base Titration
 Key things to note:
 Work in
, not molarity in I.C.F.
table

If volume is the same for all species, ratio of
moles is equal to ratio of molar concentration

This means for Ka, no need to convert back to
molarity
 Equivalance point is not 7
 Hydrolysis of salt causes this shift
Weak Acid-Strong Base Titration
 Scenario #1
 Buffer system exists here (CH3COONa/CH3COOH)
 Calculate moles of each
10 . 0 mL x
0 . 100 mol NaOH
 1 . 00 x10
3
mol NaOH
1000 mL
25 . 0 mL x
0 . 100 mol
1000 mL
 2 . 50 x10
3
mol CH 3 COOH
Weak Acid-Strong Base Titration
CH 3 COOH ( aq )  OH
Initial
(mol)

( aq )  CH 3 COO

2.5x10-3
1.00x10-3
0
Change
(mol)
-1.00x10-3
-1.00x10-3
+1.00x10-3
Equilibriu
m (mol)
1.50x10-3
0
1.00x10-3
( aq )  H 2 O ( l )
Buffer System
Wooo!!
Weak Acid-Strong Base Titration
pH  [  log( 1 . 8 x10
5
)]  log
[1 . 00 x10
[1 . 50 x10
pH  4 . 57
3
3
]
]
Weak Acid-Strong Base Titration
 Scenario #2: Adding 25.0 mL of 0.100 M NaOH to 25.0
mL of 0.100 M acetic acid
 Equivalence point is not at 7!
Weak Acid-Strong Base Titration
25 . 0 mL x
0 . 100 mol NaOH
 2 . 50 x10
3
mol NaOH
1000 mL
CH 3 COOH ( aq )  NaOH ( aq )  CH 3 COONa ( aq )  H 2 O ( l )
Initial (mol)
Change
(mol)
Equilibrium
(mol)
2.5x10-3
2.50x10-3
0
-2.50x10-3
-2.50x10-3
+2.50x10-3
0
0
2.50x10-3
Weak Acid-Strong Base Titration
 Acid/Base concentration being zero at the equivalence
point, pH is determined by hydrolysis of salt
CH 3 COO

( aq )  H 2 O ( l )  CH 3 COOH ( aq )  OH

[ CH 3 COO ] 
2 . 50 x10
3
0 . 05 L
mol

( aq )
 0 . 05 M
Weak Acid-Strong Base Titration
 I.C.E. it! Equation becomes:
Kb 
[ CH 3 COOH ][ OH
]

 5 . 6 x10
[ CH 3 COO ]
5 . 6 x10
x  [ OH


10

]  5 . 3 x10
x
2
0 . 05  x
6
M , pH  8 . 72
10
Weak Acid-Strong Base Titration
 Scenario #3: Adding 35.0 mL of 0.100 M NaOH to 25.0
mL of 0.100 M acetic acid
Weak Acid-Strong Base Titration
35 . 0 mL x
0 . 100 mol NaOH
 3 . 50 x10
3
mol NaOH
1000 mL
CH 3 COOH ( aq )  OH
Initial
(mol)
Change
(mol)
Equilibriu
m (mol)

( aq )  CH 3 COO

2.5x10-3
3.50x10-3
0
-2.50x10-3
-2.50x10-3
+2.50x10-3
0
1.00x10-3
2.50x10-3
( aq )  H 2 O ( l )
Weak Acid-Strong Base Titration
 At this point, both OH- and hydrolysis of CH3COO-
affects pH
 Since OH- is much stronger, we can ignore the impact
of the hydrolysis
[ OH

]
1 . 00 x10
3
mol
 0 . 0167 M
0 . 060 L
 pH  12 . 22
Strong Acid-Weak Base Titration
 Consider the titration of HCl with NH3:
 Or
HCl ( aq )  NH 3 ( aq )  NH 4 Cl ( aq )


H ( aq )  NH 3 ( aq )  NH
4
( aq )
 Let’s calculate the pH for this titration at different
stages
Strong Acid-Weak Base Titration
 Key main differences from Weak Acid/ Strong Base:
 Equivalence point is less then 7 due to salt hydrolysis
 Calculation of pH after equivalence point is based on
remaining [H+] only

Even though salt still exists, effect on pH minimal
 First part of titration is the same (buffer!)
Strong Acid-Weak Base Titration
 Calculate the pH at the equivalence point when 25.0
mL of 0.100 M NH3 is titrated by a 0.100 M HCl
solution
Strong Acid-Weak Base Titration
25 . 0 mL x
0 . 100 mol NH
3
 2 . 50 x10
3
mol NH
1000 mL
 At equivalence point, moles of acid equal moles of
base
3
Strong Acid-Weak Base Titration


H ( aq )  NH 3 ( aq )  NH
Initial
(mol)
Change
(mol)
Equilibriu
m (mol)
4
( aq )
2.5x10-3
2.50x10-3
0
-2.50x10-3
-2.50x10-3
+2.50x10-3
0
0
2.50x10-3
Strong Acid-Weak Base Titration
[ NH
NH

4
Initial
(mol)
Change
(mol)
Equilibriu
m (mol)

4
]
2 . 50 x10
3
mol
 0 . 05 M
0 . 05 L

( aq )  H ( aq )  NH 3 ( aq )
0.0500
0
0
-x
+x
+x
0.0500 - x
x
x
Strong Acid-Weak Base Titration

Ka 
5 . 6 x10
10

[ NH 3 ][ H ]
[ NH
x

]
4
2
0 . 0500  x
x

0 . 0500
 Approx. good (0.01%) x  5 . 3 x10  6 M
pH   log( 5 . 3 x10
6
2
)  5 . 28
```