Download Lec 6 Revision 1

Chapters 6 and 8
Systematic
Treatment of
Equilibrium
Equilibrium constants may be written for dissociations, associations, reactions,
or distributions.
©Gary Christian, Analytical Chemistry,
6th Ed. (Wiley)
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Types of Chemical Equations
• Balanced Chemical Equation
• Charge Balance Equation
• Mass Balance Equation
Charge Balance
The sum of the positive charges in
solution equals the sum of the
negative charges in solution.
n => charge,
n C
i
i
C =>concentration
Cation
i
= n j C anion
j
j
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Mass Balance
The sum of the amounts 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.
EXAMPLE: Write the mass-balance &
charge balance equations for the
system formed when a 0.010 M NH3
solution is saturated with AgCl.
AgCl(s) <=>
Ag+(aq)
+
Cl-1(aq)
Ag+(aq)
+
2 NH3(aq) <=> [Ag(NH3)2+](aq)
NH3(aq)
+
H2O(l) <=> NH4+(aq)
H2O(l) <=> H+(aq)
+
+
OH-(aq)
OH-(aq)
3
AgCl(s) <=> Ag+(aq)
+
Cl-(aq)
Ag+(aq)
+
2 NH3(aq) <=> [Ag(NH3)2+](aq)
NH3(aq)
+
H2O(l) <=> NH4+(aq)
OH-
+
(aq)
Mass Balance Equations:
[Ag+] + [Ag(NH3)2+] = [Cl-]
CNH3= [NH3] + [NH4+] + 2 [Ag(NH3)2+] = 0.010 M
AgCl(s) <=> Ag+(aq)
+
Cl-(aq)
Ag+(aq)
+
2 NH3(aq) <=> [Ag(NH3)2+](aq)
NH3(aq)
+
H2O(l) <=> NH4+(aq)
+
OH-
(aq)
Charge Balance Equation:
[Cl-1] + [OH-1] = [NH4+] + [Ag(NH3)2+] + [Ag+] +[H+]
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Systematic Approach to
Equilibrium Problems
1. Balanced chemical equations
2. What quantity is being sought.
3. Equilibrium-constant expressions
4. Mass-balance expressions for the system
5. Charge balance expression
6. Count equations vs. unknowns. If more unknowns than
equations, seek additional equations, or make
appropriate approximations.
Systematic Approach to
Equilibrium Problems
7. Make suitable approximations
to simplify the algebra.
8. Solve algebraic equations.
9. Check validity of
assumptions.
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EXAMPLE: Write the equation of
mass balance for a 0.100 M
solution of acetic acid.
EXAMPLE: Write the equations of
mass balance for a 1.00X10-5 M
[Ag(NH3)2]Cl solution.
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EXAMPLE: Write the equation of
charge balance for a solution
of H2S.
EXAMPLE: Write the equation of
charge balance for a solution
of 0.1 M Na2HPO4.
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EXAMPLE: Calculate the pH of a
0.100 M solution of acetic
acid in water.
EXAMPLE: Calculate the molar
solubility of Fe(OH)2 in
water.
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EXAMPLE: Calculate the molar
solubility of Fe(OH)2 in water.
Step 1.
Fe(OH)2(s) <=> Fe+2(aq)
2H2O <=> H3O+(aq)
+
+
2 OH-(aq)
OH-(aq)
Step 1.
Fe(OH)2(s) <=> Fe+2(aq)
2H2O <=>
Step 2.
H3O+(aq)
+
+
2 OH-(aq)
OH-(aq)
let x = molar solubility = [Fe+2]
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Step 1.
Fe(OH)2(s) <=> Fe+2(aq)
2H2O <=> H3O+(aq)
Step 2.
2 OH-(aq)
+
+
OH-(aq)
let x = molar solubility = [Fe+2]
Step 3.
Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
Step 1.
Fe(OH)2(s) <=> Fe+2(aq)
2H2O <=> H3O+(aq)
Step 2.
let x = molar solubility = [Fe+2]
+
+
Step 3.
Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
2 OH-(aq)
OH-(aq)
(1)
(2)
Step 4.
Mass Balance Equation
(3)
[OH-] = 2 [Fe+2] + [H3O+]
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Step 1.
Fe(OH)2(s) <=> Fe+2(aq) + 2 OH-(aq)
2H2O <=> H3O+(aq) + OH-(aq)
Step 2.
let x = molar solubility = [Fe+2]
Step 3.
Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
(1)
(2)
Step 4.
Mass Balance Equation
[OH ] = 2 [Fe+2] + [H3O+]
Step 5.
Charge Balance Equation
+2
(4)
2 [Fe ] + [H3O+] = [OH-]
Step 1.
Fe(OH)2(s) <=> Fe+2(aq)
2H2O <=> H3O+(aq) + OH-(aq)
Step 2.
2 OH-(aq)
let x = molar solubility = [Fe+2]
Step 3.
Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
Step 4.
Mass Balance Equation
[OH-] = 2 [Fe+2] + [H3O+]
Step 5.
2 [Fe+2] +
Step 6.
exact
+
(3)
(1)
(2)
(3)
Charge Balance Equation
[H3O+] = [OH-]
(4)
3 unique equations
3 unknowns
solution possible
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Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
(1)
(2)
[OH-] = 2 [Fe+2]
(3)
+
[H3O+]
Step 7.
Simplify Algebra
assume [OH ] ~ 2 [Fe+2]
i.e.
2 [Fe+2] >> [H3O+]
Ksp = [Fe+2][OH-]2 = 8 X 10-16M3
Kw = [H3O+][OH-] = 1 X 10-14M2
[OH-] = 2 [Fe+2]
+
[H3O+]
Assume [OH-] ~ 2 [Fe+2]
(1)
(2)
(3)
i.e.
2 [Fe+2] >> [H3O+]
Step 8.
Solve Algebraic Expressions
Using equation (1)
Ksp = [Fe+2][OH-]2 = [Fe+2](2[Fe+2])2 = 8 X 10-16M3
[Fe+2] =
3
(8 X 10-16/4)M3
=
6 X 10-6M
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Step 8.
[Fe+2] =
Step 9.
Solve Algebraic Expressions
6 X 10-6M
Check Assumption
[OH-]~ 2 [Fe+2] = 2(6 X 10-6M) = 1.2 X 10-5M ~ 1 X 10-5M
[H3O+] = Kw / [OH-]
= (1 X 10-14M2)/(1 X 10-5M) = 1 X 10-9M
2 [Fe+2] >> [H3O+]
Calculation of Chemical
Equilibrium
Concentration calculation
(1) pH of Weak Acid
pH 0.1 M HAc.
A. Reaction equations
HA +H2O
A- +H3O+
2H2O
H3O++OH-
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A. Reaction equations
HA +H2O
A- +H3O+
2H2O
H3O++OH-
B. Equilibrium Constant
Ka =
[A ][H 3O+ ]
[HA]
E. Conversion
Known item: C
[H 3O+ ] = [A ] +
[A ] =
Kw = [H 3O+ ][OH ] = 1014
C. Mass Balance Equations
C = [HA] + [A ]
Kw
[H 3O+ ]
CKa
Ka + [H 3O+ ]
[H 3O+ ] =
CKa
Kw
+
+
Ka + [H 3O ] [H 3O+ ]
D. Charge Balance
[H 3O+ ] = [A ] + [OH ]
[H 3O+ ] =
CKa
Kw
+
Ka + [H 3O+ ] [H 3O+ ]
CKa > 20Kw,
C /Ka > 400
Or is [H+] < 0.05 F
[H 3O+ ] = CKa
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Fraction of Dissociation
For HA = A- + H+,
HA
A [HA]
[HA]
[H + ]
=
=
= +
[H ]+ K a
[HA]+[A ]
F
[A ]
[A ]
K
=
=
= + a
[HA]+[A ]
F
[H ]+ K a
For diprotic weak acid
For H2A = H+ + HA- and HA- = H+ + A2-
H 2A
[H 2 A]
[H + ] 2
=
= + 2
[H 2 A]+ [HA ]+[A 2 ]
[H ] + K a1[H + ]+K a1K a 2
HA
[HA ]
K a1[H + ]
=
= + 2
[H 2 A]+ [HA ]+[A 2 ]
[H ] + K a1[H + ]+K a1K a 2
=
A2
[A 2 ]
K a1K a 2
= + 2
2
[H 2 A]+ [HA ]+[A ]
[H ] + K a1[H + ]+K a1K a 2
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Type
Keq
Formula
Example
Strong acids
Keq=
[H3 O+] = Ca
HCl, HNO3, H2SO4,
HClO4
Strong Bases
Keq=
[OH-]=Cb
NaOH, KOH
Weak Acids
Ka =
[H 3O+ ][A]
HA
[H O ] =
+
3
CKa
Weak Bases
HAc, HCOOH, HClO
NH4OH
[OH ] = CK b
Buffer
HAc-NaAc, NH4+ NH3H2O
Amphiprotic
salt(NaHA)
NaHCO3, NaHPO4,
NaH2PO4,
Polyprotic acids
Same as weak acid
0.1 M H2SO4
Ka2=1.2X10-2
[HSO4-]+[SO4 2- ]=C
[H+]= [HSO4-]+2[SO4 2- ]
[H+]= C+ [SO4 2- ]
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Calculations related to weak acid and base dissociations
Example 1. Calculate the pH of a 0.010 M acetic acid solution.
Ka of acetic acid is 1.75 10-5)
Example 2. What is the pH of a solution of guanidine, (H2N)2CNH?
Ka for (H2N)CNH2 + is 2.9 10-14).
Buffers and pH values in buffer solutions
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.
Henderson-Hasselbalch Equation:
For a buffer solution containing a weak acid and its conjugate base:
pH =
pK a
[A ]
+ log
[HA]
For a buffer solution containing a weak base and its conjugate acid,
pH =
pK a
+ log
[B]
[HB + ]
or
pOH =
pK b
[HB + ]
+ log
[B]
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Effect of addition of some amount of acid to a buffer solution
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.
Example 2. What is the pH if 10.00 mL of 0.0050 M HCl is
added to the above buffer solution?
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