Strong acids and bases
Ch.8 & 9
Systematic Treatment of Equilibrium
& Monoprotic Acid-base Equilibrium
The sum of the positive charges in solution equals the
sum of the negative charges in solution.
pH
1.0
3.0
5.0
8.0?
ni CiCation =
i
n =>
Kw
[OH-]
[Cl-]
Charge balance.
=
+
Mass balance. [Cl-] = 1.0 x 10-8 M
H 2O
Kw
[H+] = 1.05 x 10-7 M, -9.51 x 10-8 M
pH = 6.98
C =>concentration
-
When the (strong) acid or base concentration is around ~10-6
- 10-8 M, we have to be concerned with autoprotolysis.
OH + H +
[OH-] = [H+] - (1.0 x 10-8 M)
Kw = [H+ ][OH-] = [H+ ]([H+] − (1.0 x 10-8 M))= 1.0× 10-14
[H+]2 − (1.0×10−8 )[H+] −Kw=0
j
Material balance. The quantity of all species in a
solution containing a particular atom (or group of
atoms) must equal the amount of that atom (or group)
delivered to the solution.
-
OH + H +
Systematic Treatment of Equilibrium Strong acids and bases
[H+]
charge,
n j C anion
j
Mass Balance
We have to consider autoprotolysis of water:
H 2O
pH
Charge Balance
Strong acids and bases
Conc. (M)
0.10
1.0×10-3
1.0×10-5
1.0×10-8
Conc. (M)
0.10
1.0×10-3
1.0×10-5
1.0×10-8
For extremely low strong acid or base concentration
(<10-8), pH = 7.00
Problem:
What is the pOH, and pH of a 2.48 x 10-7 M solution of
NaOH?
ANS: 6.55; 7.45
1
Systematic Treatment of Equilibrium Weak acids and bases
A typical weak-acid problem
Example: Calculate the pH of a 1 L solution prepared by
dissolving 0.10 moles of acetic acid?
• Weak Acid: acid dissociation constant (Ka)
Ka
HA +H2O
Ka
HA
A- +H3O+
-
A
Ka =
+H+
[ A ][ H O ] [ A ][ H ]
=
[ HA]
[ HA]
−
+
−
+
3
HOAc
• Weak Base, hydrolysis reaction: base hydrolysis
constant (Kb)
[ BH ][OH ]
Kb
Kb =
B + H2O
BH ++OH[ B]
+
Ka
H
Kw
H O
OH
2
−
+
+ OAc
-
−
+
+
−
−
−5
Mass Balance:
[HOAc] + [OAc-] = C = 0.10
−14
A typical weak-acid problem
A typical weak-acid problem
Example: Calculate the pH of a 1 L solution prepared by dissolving
0.10 moles of acetic acid?
Example: Calculate the pH of a 1 L solution prepared by dissolving
0.10 moles of acetic acid?
Simplifying assumptions
(1)
+
Charge Balance:
[H+] = [OH-] + [OAc-]
3
Simplifying assumptions
[ A ][ H ] [H ] ≈ [A ]
[H ]
x
x = [H + ]
Ka =
  →
  →
[ HA]
C − [H ]
C−x
−
−
Kw = [H+ ][OH-] = 1.0× 10-14
+H +
• Relationship between Ka and Kb for conjugate
acid/base pairs:
Kw = Ka × Kb = [ H O ][OH ] = [ H ][OH ] = 10
[OAc ][ H ]
= 1.75 × 10
[ HOAc]
Ka =
+
+
−
+
2
2
+
(1)
C = 0.10 M and Ka = 1.75×10-5, x = 1.3×10-3 M
Ka =
[ A ][ H ] [H + ] ≈ [A − ]
[H ]
x
x = [H + ]
  →
  →
[ HA]
C − [H ]
C−x
−
+
+
2
2
+
C = 0.10 M and Ka = 1.75×10-5, x = 1.3×10-3 M
(2)
Ka =
[ A ][ H ]
x
− x
~
C→ x
=
C
[ HA]
C
C−x
−
+
2
2
x = Ka ⋅ C = 1.75 × 10 × 0.10 = 1.3 × 10 M
−5
A typical weak-acid problem
Example: Calculate the pH of a 1 L solution prepared by dissolving
0.10 moles of acetic acid?
• Fraction of dissociation ( )
• For a monoprotic weak acid:
α
[ A− ]
=
−
A typical weak-base problem
Example: Find the pH of 0.10 M ammonia.
NH 3 + H 2 O
[A ]
[A ]
x
x
=
=
=
[ A ] + [ HA] [ A ] + (C − [ A ]) x + (C − x) C
−
−
−
α=
−
x 1.3 × 10
=
= 0.013
C
0.10
−3
−3
H O
2
Kb
Kw
NH
OH
-
+
4
+ OH
+H +
−
Kb =
+
[OH − ][ NH 4 ]
= 1.75 ×10 −5
[ NH 3 ]
Kw = [H+ ][OH-] = 1.0× 10-14
Charge Balance:
[OH-] = [H+] + [NH4+]
Mass Balance:
[NH3] + [NH4+] = C = 0.10
2
A typical weak-base problem
A typical weak-base problem
Example: Find the pH of 0.10 M ammonia.
Example: Find the pH of 0.10 M ammonia.
NH 3 + H 2 O
Kb
Kw
H O
+
NH
-
OH
2
4
+ OH
−
Kb =
+
[OH − ][ NH 4 ]
= 1.75 ×10 −5
[ NH 3 ]
Kw = [H+ ][OH-] = 1.0× 10-14
+H +
Simplifying assumptions
(1)
Kb =
−
+
−
[OH ][ NH ] [OH ] ≈ [NH 4 ]
[OH ]
x
x = [ OH ]
   
→
 →
[ NH ]
C − [OH ]
C−x
+
−
−
2
2
4
−
3
C = 0.10 M and Kb = 1.76×10-5, x = 1.3×10-3 M
Charge Balance:
[OH-] = [H+] + [NH4+]
Mass Balance:
[NH3] + [NH4+] = C = 0.10
A typical weak-base problem
A typical weak-base problem
Example: Find the pH of 0.10 M ammonia.
Example: Find the pH of 0.10 M ammonia.
Simplifying assumptions
(1)
Kb =
−
+
−
[OH ][ NH ] [OH ] ≈ [NH 4 ]
[OH ]
x
x = [ OH ]
   
→
 →
[ NH ]
C − [OH ]
C−x
+
−
−
2
2
4
−
3
C = 0.10 M and Kb = 1.76×10-5, x = 1.3×10-3 M
(2)
• Fraction of dissociation ( )
• For a monoprotic weak base:
+
2
2
4
+
[ NH ]
[ NH ]
x
x
=
=
=
[ NH ] + [ NH ] [ NH ] + (C − x ) x + (C − x) C
4
4
+
+
4
3
α=
[OH ][ NH ]
x
x
~C
Kb =
=
C− x
→
[ NH ]
C−x
C
−
+
α=
4
x 1.3 × 10
=
= 0.013
C
0.10
−3
3
x = Kb ⋅ C = 1.76 × 10 × 0.10 = 1.3 × 10 M
−5
−3
Type
Keq
Formula
Example
Strong acids
Keq=∞
[H3O+] = Ca
HCl, HNO3, HClO4
Strong Bases
Keq=∞
[OH-]=Cb
NaOH, KOH
Weak Acids
Ka =
Weak Bases
Kb =
+
[ H ][ A]
HA
[OH − ][ BH + ]
[ B]
[H ] =
+
Ka ⋅ C
Ka =
Amphiprotic
salt(NaHA)
Polyprotic acids
Ka1 =
Ka 2 =
[ H 3O ][ A]
HA
−
b
[ H 3O + ] = Ka
C HA
C NaA
[
Ka1Ka 2
]
[ H 3O + ][ HA]
H 3O + =
H2 A
[ H 3O + ][ A2− ]
HA−
NH3
[OH ] = K ⋅ C
Buffer
+
HAc, HCOOH, HClO
HAc-NaAc, NH4+ NH3H2O
Buffers
• A buffer solution resists changes in pH
when acids or bases are added or when
dilution occurs.
• A buffer solution generally consists of a
mixture of an acid and its conjugate base.
NaHCO3, NaHPO4,
NaH2PO4,
Same as weak acid
3
Henderson-Hasselbalch Equation
Ka =
+
−
[ H ][ A ]
[ HA]
Kb =
−
[OH ][ HB ]
[ B]
For a buffer solution containing a weak acid and its conjugate base:
pH =
[ A− ]
+ log
[ HA]
pK a
For a buffer solution containing a weak base and its conjugate acid,
pH =
+ log
pK a
[ B]
[ HB + ]
or
pOH =
pK b
Henderson-Hasselbalch Equation
+
+ log
[ HB + ]
[ B]
• pH = pKa
+ log
[ A− ]
[ HA]
= 4.76 + log
pK a
+ log
[ A− ]
[ HA]
= 4.76 + log
+ log
[ A− ]
[ HA]
Buffer Capacity
β=
dCb
dCa
=−
dpH
dpH
20.00 * 0.10 / 30.00
= 5.06
10.00 * 0.10 / 30.00
Example 2. What is the pH if 10.00 mL of 0.0050 M HCl is
added to the above buffer solution?
pH =
pK a
• As the acid increases, the
pH goes down.
Example 1. Calculate the pH of a buffer prepared by adding 10.00
mL of 0.10 M acetic acid and 20.00 mL of 0.10 M sodium acetate.
pK a
pH =
• For every power of 10
change in the ratio
[A-]/[HA], the pH changes
by one unit.
Effect of adding acid to a buffer
pH =
[A-] = [HA]
(20.00 * 0.10 − 0.0050 *10.00) / 40.00
= 5.03
(10.00 * 0.10 + 0.0050 *10.00) / 40.00
•Buffer capacity is a maximum when pH = pKa
•Best to chose a buffer system whose pKa is within
±1 pH unit of your desired pH
Diprotic Buffers
Diprotic Buffers
• For H2A, we can write two Henderson-Hasselbalch
equations:
[ A2 − ]
[ HA− ]
pH = pK 2 + log
pH = pK1 + log
[ HA − ]
[ H 2 A]
• For H2A, we can write two Henderson-Hasselbalch
equations:
[ A2 − ]
[ HA− ]
pH = pK 2 + log
pH = pK1 + log
[ HA − ]
[ H 2 A]
Ex: What is the pH of a solution prepared by dissolving 1.0 g KHP
and 1.20 g Na2P in 50.0 mL water? pK2=5.408
Ex: What is the pH of a solution prepared by dissolving 1.0 g KHP
and 1.20 g Na2P in 50.0 mL water? pK2=5.408
pH =
pK 2
+ log
[ P 2− ]
[ HP − ]
= 5.408 + log
1.20 g / 210.094 g / mol
= 5.47
1.00 g / 204.221g / mol
4
Diprotic acids/Dibasic bases
Polyprotic acids/Polybasic bases
Ch.10
Polyprotic Acid-base Equilibrium
H+
H2A
HA-
Ka =
HA-
+
H+ + A2-
1
[ H ][ HA ]
[ H A]
+
−
2
[ H ][ A ]
Ka =
[ HA ]
+
2
Ka1, Ka2 …
H3PO4
H+ + H2PO4H2PO4
H+ + HPO422HPO4
H+ + PO43-
2−
−
Ka1 = 7.11 x 10-3
Ka2 = 6.32 x 10-8
Ka3 = 7.1 x 10-13
H2L+ (Acidic form)
Amino Acid
What’s the pH of a 0.0500 M solution of H2L+
A zwitterion: a molecule with both positive and negative charges
Leucine
Assume that H2L+ acts as a monoprotic acid
[ H ][ HL ]
x
Ka =
=
[H L ]
C−x
+
2
1
+
2
x=1.32×10-2 M, pH=1.88
H2L+ (Acidic form)
L- (Basic form)
What’s the pH of a 0.0500 M solution of H2L+ ?
What’s the pH of a 0.0500 M solution of L- ?
The relations between acid and base equilibrium constants
for a diprotic acid/base:
Ka Kb = Kw
1
Assume that H2 acts as a monoprotic acid
[ H ][ HL ]
x
Ka =
=
[H L ]
C−x
+
2
2
1
H2L+
HL + H+
HL L- + H+
L− + H2O
HL + OH−
HL+ H2O
H2L+ + OH−
+
2
x=1.32×10-2
M, pH=1.88
Check our assumption by calculating [L-]
Ka =
2
[ H ][ L ]
[ HL]
+
−
[L ] =
−
2
Ka Kb = Kw
L+
1
Ka1
Ka2
Kb1=Kw/Ka2
Kb2=Kw/Ka1
Ka [ HL]
= Ka = 1.80 × 10 M
[H ]
−10
2
+
2
5
L- (Basic form)
What’s the pH of a 0.0500 M solution of L- ?
L− + H2O
HL+ H2O
HL + OH−
H2L+ + OH−
Kb1=Kw/Ka2=5.59×10-5
Kb2=Kw/Ka1=2.13×10-12
assume that L- acts as a monobasic species with Kb=Kb1
Kb =
1
[OH ][ HL]
x
=
[L ]
C−x
−
2
L- (Basic form)
What’s the pH of a 0.0500 M solution of L- ?
L− + H2O
HL+ H2O
1
x=1.64×10-3 M, [H+]= Kw/x = 6.08×10-12 M, pH = 11.22
Kb1=Kw/Ka2=5.59×10-5
Kb2=Kw/Ka1=2.13×10-12
assume that L- acts as a monobasic species with Kb=Kb1
Kb =
−
HL + OH−
H2L+ + OH−
[OH ][ HL]
x
=
[L ]
C−x
−
2
−
x=1.64×10-3 M, [H+]= Kw/x = 6.08×10-12 M, pH = 11.22
Check our assumption by calculating [H2L+]
Kb =
2
HL (Intermediate Form)
What’s the pH of a 0.0500 M solution of HL ?
HL is amphiprotic, meaning it can both donate and accept a proton
HL L- + H+
HL+ H2O
H2L+ + OH−
[ H L ][OH ]
[ HL ]
+
−
[H L ] =
+
2
2
Kb [ HL ]
= Kb = 2.13 × 10 M
[OH ]
−12
2
−
2
HL (Intermediate Form)
What’s the pH of a 0.0500 M solution of HL ?
HL L- + H+
HL+ H2O
H2L+ + OH−
Ka2=1.80×10-10
Kb2=Kw/Ka1=2.13×10-12
Ka2=1.80×10-10
Kb2=Kw/Ka1=2.13×10-12
Charge balance:
[H+]+[H2L+] = [L-]+ [OH-]
We have to use the systematic treatment to determine pH.
HL (Intermediate Form)
What’s the pH of a 0.0500 M solution of HL ?
HL L- + H+
HL+ H2O
H2L+ + OH−
[HL]=C
Ka2=1.80×10-10
Kb2=Kw/Ka1=2.13×10-12
Equilibrium Constants:
HL (Intermediate Form)
What’s the pH of a 0.0500 M solution of HL ?
HL L- + H+
HL+ H2O
H2L+ + OH−
Ka2=1.80×10-10
Kb2=Kw/Ka1=2.13×10-12
[HL]=C
Check our assumptions:
6
HL (Intermediate Form)
For diprotic weak acid
What’s the pH of a 0.0500 M solution of HL ?
HL L- + H+
HL+ H2O
H2L+ + OH−
For H2A = H+ + HA- and HA- = H+ + A2-
Ka2=1.80×10-10
Kb2=Kw/Ka1=2.13×10-12
αH A =
[ H 2 A]
[H + ]
=
−
2−
2
+
[ H 2 A] + [ HA ] + [ A ]
[ H ] + K a1[ H + ] + K a1K a 2
α HA =
[ HA− ]
K a1[ H + ]
=
+ 2
[ H 2 A] + [ HA− ] + [ A2− ]
[ H ] + K a1[ H + ] + K a1K a 2
α
[ A2 − ]
K a1 K a 2
=
[ H 2 A] + [ HA− ] + [ A2 − ]
[ H + ] 2 + K a1[ H + ] + K a1K a 2
2
2
If Ka2C>>Kw, than the 2nd term in the numerator can be dropped
and if Ka1<<C, the 1st term in the denominator can also be
neglected:
• Isoionic Point: the pH obtained when the
pure, neutral polyprotic acid HA (the
natural zwitterion) is dissolved in water
A2
−
=
Type
Keq
Formula
Example
Strong acids
Keq=∞
[H3O+] = Ca
HCl, HNO3, HClO4
Strong Bases
Keq=∞
Weak Acids
Weak Bases
Kb =
[OH − ][ B ]
[ BOH ]
Ka =
[ H 3O + ][ A]
HA
Amphiprotic
salt(NaHA)
Ka1 =
Ka 2 =
Polyprotic acids
3
+
CKa
HAc, HCOOH, HClO
NH4OH
[OH − ] = CK b
[ H 3O + ] = Ka
[ H 3O + ][ HA]
H2 A
H 3O + =
[ H 3O + ][ A2− ]
HA−
NaOH, KOH
[H O ] =
[ H O ][ A]
Ka = 3
HA
Buffer
• Isoelectric Point: the pH at which the
average charge of the polyprotic acid is zero
[OH-]=Cb
+
[
]
C HA
C NaA
Ka1Ka 2
HAc-NaAc, NH4+ NH3H2O
NaHCO3, NaHPO4,
NaH2PO4, amino acid
Treated as
monoprotic weak
acids
7