A generalized balanced chemical reaction:
aA  bB 

 cC  dD
Is this chemical system at equilibrium?
Chemical Equilibrium
A chemical system is at equilibrium
a. when none of it’s thermodynamic properties change with
time (macroscopic)
b. when the rate of the forward reaction exactly equals the
rate of the reverse reaction.
Why is chemical equilibrium important?
All ‘chemical transformations’ occur because they tend to
reach a state of ‘chemical equilibrium’.
aA + bB
aA + bB
cC + dD
At equilibrium – microscopically, transformation occur in both
directions at equal rates. (dynamic equilibrium).
No changes of macroscopic concentration values occur, if
the reaction system is unperturbed from equilibrium ‘state’.
At equilibrium all species shown in the equation co-exist,
however small/large the concentrations are.
Every system (reaction or otherwise) strives to reach stability,
which is the state of equilibrium.
Reactions with v.large K are nearly complete (in the forward
direction); forward reaction much favored. Amount of product
is determined by quantity of limiting reactant only.
cC + dD
At equilibrium the concentration of species are related
by way of the equilibrium constant, defined as;
K
[C]c [D]d
[A]a [B]b
[ ]  equilibrium concentration value
K = equilibrium constant (is a function of temperature)
K measures the equilibrium position of the reaction
Large K more products than reactants AT EQUILIBRIUM,
and vice versa.
(Expression for K is associated with a balanced equation
as written).
Systems not at equilibrium - Reaction Quotient Q:
Such reactions are ideal for quantitative (gravimetric and
volumetric) analysis.
A chemical reaction system that is not at equilibrium
would have the concentrations not conforming to the
equilibrium constant expression.
Reactions with very small K have much less tendency to go
forward and the reverse reaction is favored!!
_________________________________________________
[C]cne [D]dne
K
[A]cne [B]bne
A chemical system at equilibrium can be perturbed to a nonequilibrium state. Then the system would readjust to reach a
‘new’ equilibrium state, in order to offset the disturbance
(Le Chaterlier’s Principle).
[C]ne , [D]ne , ...  non-equilibrium concentration values
1
Any reaction strives to reach the equilibrium state.
aA  bB 

 cC  dD
At equilibrium the concentrations of the species would
satisfy the equilibrium expression.
K
[C]c [D]d
[A]a [B]b
[ ]  equilibrium concentration value
Systems not at equilibrium - Reaction Quotient Q:
Q
[C]c [D]d
[A]a [B]b
[X]  non-equilibrium (ne) concentration values
aA + bB
cC + dD
Q = conc. ‘ratio’ of non. eq. system  K
Q<K ; Q>K ; non equilibrium: so that Q=K i.e. @eqlm.
Reaction species mixed with their concentrations not
satisfying the equilibrium concentration expression (Q  K)
would change their concentrations so as to reach the
equilibrium expression values (Q  K).
For K and Q expressions use the numerical value of the
concentrations expressed in units of ;
solutions – value of molarity (mol/L) M {std.state - 1M};
gases – value of partial pressure (atm) {std.state - 1atm};
for solvents, solids and pure liquids set [ ] =1.
Examples of K expression:
K and Q are dimensionless quantities.
Q ‘captures’ the state of the reaction system on it’s
way to equilibrium.
Q is a measure of the how far the reaction state
is away from the equilibrium state.
H2CO3(aq) = HCO3-(aq) + H+(aq)
K
[ HCO3 ][ H  ]
[ H 2CO3 ]
Bi2S3(s) + 6H+(aq) = 2Bi3+ (aq)+ 3H2S(aq)
K
Manipulating equilibrium constants:
aA  bB  cC  dD
K 
[ A]a [ B ]b
1
K' 

[C ]c [ D]d K
1
K'
K
If a reaction is a summation of two or more ‘reactions’; then
their equilibrium constants are related;
A B  C  D
[C ]c [ D]d
[ A]a [ B ]b
cC  dD  aA  bB;
[ Bi 3 ]2 [ H 2 S ]3
[ H  ]6
C  D E  F G
for the reverse reaction
A B  E  F G
[C ][ D]
[ A][ B]
[ F ][G ]
K2 
[C ][ D][ E ]
[ F ][G ]
K
[ A][ B][ E ]
K1 
(1)
(2)
(3)
Because (3)  (1)  (2)
[ F ][G ]
[C ][ D]
[ F ][G ]
K  K1 K 2 


[ A][ B] [C ][ D][ E ] [ A][ B][ E ]
K  K1 K 2
2
Aqueous - Solution Chemistry
A B  C  D
2 A  2 B  2C  2 D
nA  nB  nC  nD
K1 
K
K
[C ][ D ]
[ A][ B]
2
2
[C ] [ D]
= K12
[ A]2 [ B]2
[C ][ D ]
= K1n
[ A][ B]
(Expression for K is associated with a balanced equation
as written).
Examples of equilibria:
Most reactions in quantitative analysis are performed
in aqueous solutions. Reactants that are generally water
soluble are; electrolytes, polar molecules or small molecules.
dissociation: ~100% strong electrolyte
 1% fairly weak electrolyte
The ions of dissolved weak electrolytes exist in equilibrium
with it’s un-dissociated molecules.
AB(aq)
A+ (aq) + B- (aq)
Structure of hydronium ion H3O+
Acid dissociation Ka:
HA + H2O
acid
K a,HA 
A- + H3O+ (note: 3 species + solvent)
conjugate base
[A  ][H3O+ ]
[HA]
note: [H2O] – solvent; set to 1.
stronger HA (larger Ka) ~ weaker base, Aweaker HA (smaller Ka) ~ stronger base, A- log Ka = pKa
Bronsted Acid is a substance which increase [H3O+] in water.
Strong acids dissociate nearly completely in solution.
Strong acids consumes/reacts with basic groups when
encountered, almost completely.
Know all strong acids and
bases. Ka >>0.
All other acids and bases
are weak; Ka small.
Strengths of acids/bases expressed by their K values.
(only a few strong acids are encountered in aqueous
phase)
3
Bronsted Base is a substance which decreases [H3O+] in water
or increases the [OH-] in water.
Base constant Kb for a base A:- ;
Strong bases consume acidic hydrogens nearly completely,
vice versa.
Bases other than simple hydroxides render aqueous
solutions basic due to hydrolysis, (breakup of water
molecules)
A salt is any ionic compound which is neither a Bronsted acid
nor a Bronsted base.
A:- + H2O
K b,A- 
Strengths of acids/bases expressed by their K values.
(only a few strong acids are encountered in aqueous phase)
acid-base reaction Keq:
A+ H3O+
Base Acid
K eq 
HA
+
H2O
weaker acid
[HA]
1

[A ][H3O+ ] K a,HA
-
note: inverse relationship
HA + OH- hydrolysis
[OH- ][HA]
[A - ]
- log Kb = pKb
All solvents with active hydrogens (i.e. protic solvents)
undergo auto-protolysis (self ionization).
2NH3  NH4  NH2
2CH3COOH  CH3COOH2  CH3COO
2H2O  H3O  HO
Any acid in water (for example) generates hydronium ions.
Acidic species actually available in aqueous solutions is H3O+.
Hydration
shell.
H3O+ ions in solution
is hydrated.
Ideal hydrated
hydronium ion.
4
Auto-protolysis of solvents (water, Kw):
self dissociation of water: 2 H2O
OH- + H3O+
Kw = [OH-][H3O+]
The strength of an acid or base character of a solute is
highly dependent on the solvent in which it is dissolved.
It is the solvent that determines the acid or base strength
of a solute.
Leveling effect: All acids stronger than H3O+ is leveled
down to the strength of H3O+ in water.
All bases stronger than HO- is leveled down to HO- in water.
For aqueous solutions;
Weak acids and bases:
acid dissociation Ka of a weak acid HA:
OH- + H3O+
2 H2O
HA + H2O
WA
A- + H3O+
Conjugate pairs shown in
color.
Conjugate base
A
,
K
A
,
b
A
- log Ka = pKa; smaller Ka  larger pKa
w
A
H
,
a
[H+ ][A  ]
[HA]
-
+
-14
pK a  pK b  pK w
K
K a,HA 
+
K aK b  K w
,
K K bK w H
a
K
=
=
Weaker acid  stronger conj. base
Stronger acid  weaker conj. base
always exist in water
Kw = [OH ][H3O ] =[OH ][H ]=10 @ 25oC
Pure water:
K a,HAK b, A   K w
[H+] = [OH-] = 10-7
-
stronger conj. Base  weaker acid
weaker conj. Base  stronger acid
Weaker acid  stronger conj. base
Stronger acid  weaker conj. base
Solubility of Ionic Compounds
Solubility is an equilibrium process between the ionic
compound (solid) and it’s dissolved ions.
Solubility is expressed as concentrations of solutes
in solution at saturation point of dissolution.
5
Solubility product Ksp:
BaSO4(s)
Ba++(aq) + SO4-- (aq) KspBaSO4 = 1.1×10-10
-2
Ksp = [Ba+2][SO4 ] “saturated solution” concentrations!!!
no [BaSO4] in Ksp expression – solid, set to 1
Compounds with very small Ksp are ideal for gravimetric
analysis.
Eqlm. position can be shifted by disturbing the equilibrium
system – Le Chaterlier's Principle.
note: solubility product gives the concentrations of associated
ions existing in equilibrium/their co-existing concentrations.
Reaction quotient, Q = ‘conc. product ratio’ of non. eq. system.
For solubility problems note;
Q<K  dissolution; Q=K @ equilibrium; Q>K  pptn.
Example of a solubility equilibrium.
dissolution
Note: Reverse of solubilization is precipitation
+2
Ba
-2
(aq) + SO4 (aq)
BaSO4(s)
Ba+2(aq) + SO4-2 (aq)
K = Ksp
BaSO4(s) K=1/Ksp
In equilibrium calculations identify the possible equilibria,
and pick the concentrations of species relevant to solve the
problem.
How much Cl- can be added to a Ag+ (0.10 M) solution
without creating a ppt.? Ksp,AgCl=1.810-10. i.e. to reach
saturation solubility of AgCl in water.
AgCl(s) = Ag+(aq) + Cl-(aq)
Calculate [OH-] in a. 0.200M NaOH?
b. 0.100M HCl?
Solubility of Hg2Cl2 in a. water?
b. 0.030M Cl-?
Ksp,Hg2Cl2 = 1.2  10-18
2



Hg 2Cl2 ( s) 
 Hg 2 (aq)  2Cl ( aq)
Salts are less soluble in solvents carrying common ions
- common ion effect!
How much Cl- can be added to a Ag+ solution (0.10 M)
solution without creating a precipitate?
Ksp,AgCl= 1.8  10-10
Q  [ Ag  ][Cl  ]
non-eqlm.
general case
What would be the [Cl-] so the above system is at
equilibrium? That is when would Q=K.
Q<K ; Q>K ; non equilibrium: so that Q=K i.e. @eqlm.
Q  [ Ag  ][Cl  ]  K  1.8  1010
0.1 [Cl  ]  1.8  1010

[Cl ]  1.8 10
Add Cl- until [Cl-] = 1.8  10-9M
[Cl-] > 1.8  10-9M  pptn;  Q>K
9
Alternate approach – next page. Ag+(aq) + Cl-(aq) = AgCl(s)
Ag+(aq) + Cl-(aq) = AgCl(s)
K
1
 5.5 109
K sp , AgCl
Q
1
[ Ag  ][Cl  ]
[Ag+]=0.10M
1
Before adding Cl ; Q 
   K
[0.1]0
1
 1012  K
After adding a trace Cl  ; Q 
[0.1][1011 ]

Note: Precipitation starts when the concentration just
exceeds the saturation point (solubility wise).
Ag+(aq) + Cl-(aq) = AgCl(s)
[Ag+]=0.10M
K sp , AgCl  1.8 1010
K = 1/Ksp,AgCl
Q=?
Q<K ; Q>K ; non equilibrium: so that Q=K i.e. @eqlm.
adding more Cl  ; Q 
1
 1010  K
[0.1][109 ]
adding more Cl  ; Q 
1
 5.5 109  K
[0.1][1.81109 ]
For [Cl-] >1.8110-9, Q < K and precipitation occurs.
6
Can you separate Pb(II) ‘completely’ from Hg(I) from a mixture
using iodide. (PbI2(s), Hg2I2(s))?
Initial conc. of both ions Pb(II) and Hg(I) is 0.1M.
Ksp,lead iodide= 7.910-9
Ksp,mercury(I) iodide = 1.110-28
At the start of 'complete' precipitation [Hg  ] in solution.
 0.1
2
0.01
 0.1
 105
100
100
The ions in solution (saturation) satisfies;
[Hg  ]  0.1
[Hg  ][I  ]2  K spHgI2  [I  ] 
Less soluble compound precipitates first.
K spHgI2
[Hg  ]
2
For the system; PbI 2 (s) = Pb (aq) + 2I  (aq), at the
By inspection, Hg2I2 is more insoluble.
point of 'complete' precipitation of mercurous ion;
In general one must solve for solubility to make that decision.
Q PbI2  [Pb 2 ][I  ]2  [Pb 2 ]
Criterion: "complete pptn." = 99.99%
Q PbI2  1.11024  K PbI2 (K PbI2  7.9  109 )
K spHgI2
[Hg  ]
 0.1
1.11028
105
no precipitation because no trend to left in the above system.
Would Pb(II) remain in solution when 99.99% of Hg(I) is
removed from the solution by precipitation?
If so; at the point of Hg(I) ‘removal’, the Q for the equilibrium;
PbI2(s) = Pb+2(aq) + 2l-(aq)
Ksp,Pbl2
must be less than or equal to it’s K value - (Ksp,Pbl2) so the
equilibrium do not tend to left (i.e. precipitation)
Complex Ions:
Formation constant of complex ions n;
In general more than one ligand combines to form metal
complexes; occurs when [L]>>0.
M+x + nL
MLn+x
overall formation constant
n 
To find out we need to calculate QPbI2 at the point of
‘complete precipitation’ of mercurous ion.
[MLn ]
[M x ][L]n
Q<K ; Q>K ; non equilibrium: so that Q=K i.e. @eqlm.
All aquated species.
Stepwise Formation constants, K1, K2,.. Kn.
Pb2  I  PbI
K1 
[PbI ]
[Pb2 ][I ]
[1]
PbI  I  PbI2
K2 
[PbI2 ]
[PbI ][I ]
[2]
Complexes are formed in a stepwise manner.
Stepwise equilibria exist.
 molecular species
in dissolved state (ion pairs)
PbI2  I  PbI3
K3 
[PbI3 ]
[PbI2 ][I ]
[3]
PbI3  I  PbI42
K4 
[PbI42 ]
[PbI3 ][I ]
[4]
7
 - cumulative equilibrium constants
2

[PbI  ]
1 
[Pb 2 ][I  ]

[1]
Pb  I  PbI
[1]+[2]
Pb 2  2I   PbI 2
2 
[PbI 2 ]
[Pb 2 ][I  ]2
[1]+[2]+[3]
Pb 2  3I   PbI3
3 
[PbI3 ]
[Pb 2 ][I  ]3
Sum all
Pb 2  4I   PbI 42
4 
[PbI 42 ]
[Pb 2 ][I  ]4
Polyprotic Acids:
O
O
P
HO
OH
HO
P
OH
+
Ka1
O
P
O
HO
P
OH
+
O
H
Ka2
O
O
O
P
+
O
H3PO4
H2PO4-1
+
H+
K1a
H2PO4-1
HPO4-2
+
H+
K2a
HPO4-2
PO4-3
+
H+
K3a
H
+
+ H PO  HPO 2  2H 
3
4
4
3
4
H 3PO 4  PO  3H
O
P
Equilibrium constants relate the concentrations of species
that can coexist at equilibrium (i.e. if the system is not
disturbed).
+
O
O
HO
H
OH
O
HO
+
O
Overall formation constant, n = K1K2.. Kn, is the overall
eq. const. of a series of stepwise formation constants
(1 = K1).
+
K1a K 2a

K1a K 2a K 3a
Ka3
O
Polybasic species:
O
O
O
P
+
O
H2O
HO
O
P
O
+ OH-
O
O
HO
Kb1
PO4-2
+
H2O
HPO4-1
+
OH-
K1b
HPO4-2
+
H 2O
H2PO4-1
+
OH-
K2b
H3PO4
+
OH-
K3b
O
P
O
+
H2O
HO
P
O
O
Kb2
H2PO4-1+ H2O
OH
+ OH-
O
HO
O
P
O
OH
+
H2O
HO
P
OH
OH
Kb3
+ OH-
8
Learn to recognize the acidic and basic entities in large molecules.
For triprotic acids:
Ka1 Kb3 = Kw
HO
Ka2 Kb2 = Kw
HO
Ka3 Kb1 = Kw
Note the general fact, that the product of K’s of conjugated
acid-base pairs in aqueous solution (solvent) is equal
to the ion product of water (solvent).
HO
NH2+
HO
NH
O
HO
Equivalence point pH - acid/base rxn or pH of salt solutions:
Strong (acids + bases)  salt (e.g. Na+ + Cl-) + water
At least one component of such salts reacts with water,
hydrolysis;
Weak acid + strong base
CH3COO   H2O  CH3COOH + OH
CH3COOH  Na  OH  CH3COO  Na   H2O
good base
acid
At equivalence point; the resultant solution of a weak
acid/base titration is not neutral, pH≠7.
base
conj.
base
conj.
acid
weak acid
Weaker component  (apart from water) a salt of the weak
component.
Overall; the product side is weaker than reactant side, if
the reaction takes place as written.
i.e. salt of a weak acid or weak base.
Conjugate base of weaker acid is a stronger base.
Conjugate base of stronger acid is a weaker base.
In the above solution;
CH3COO , H2O, CH3COOH, OH
and
H from
H2O  H  OH
Identify all possible equilibria among species
9