Electrochemistry & Ionic equilibrium - Shailendra Kumar Chemistry

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Knowledge of thermodynamics, reaction quotient and
redox reaction are very important for solving the problem
of electrochemistry
RULE TO SOLVE PROBLEM OF ELECTROCHEMISTRY
§ Rule (1)
A
B
A +2 + 2e ∆G 1
= –0.2kcal ....(i)
B +2 + 2e ∆G 2
= +0.4 kcal ....(ii)
According to the rule of spontaneity, reaction (i) is
spontaneous but reaction (ii) is non-spontaneous. To
gain full cell reaction, we convert non-spontaneous
reaction into spontaneous reaction.
Same : A A +2 + 2e; ∆G 1
= –0.2kcal
Change : B +2 + 2e B; ∆G 2
= –0.4 kcal
A + B +2 A +2 + B ∆Grxn
∆Grxn = ∆G 1
+ ∆G 2
= (–0.2kcal) + (–0.4 kcal)
= – 0.6 kcal
To Calculate Emf of reaction
§ Rule (2)
∆G = – nFE E = –∆G/nF
If both half cell reactions are non-spontaneous, then
we convert more non-spontaneous reaction into
spontaneous reaction to gain full cell reaction.
A A +2 + 2e ∆G 1
= +0.2 kcal ....(i)
B B +2 + 2e ∆G 2
= +0.5 kcal ....(ii)
Reaction (i) is less non-spontaneous and reaction (ii) is
more non-spontaneous.
Change : B +2 + 2e B ∆G 2
= –0.5 kcal
Same : A A +2 + 2e ∆G 1
= +0.2 kcal
B +2 + A B + A +2 ∆Grxn
∆Grxn = ∆G 1
+ ∆G 2
= (–0.5) + (0.2) kcal = –0.3 kcal
* Spontaneity from ∆G
4
3
2
1
0
–1
–2
–3
Non-spontaneoity
increases
∆G = 0 (Rxn is at equilibrium)
Spontaneoity
increases
–4
Greater the positive value (+ve) of free energy
greater will be non-spontaneity.
Greater the negative value (–ve) greater will be
spontaneity
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§ Rule (3)
If both half cell reactions are spontaneous then we
convert less spontaneous reaction into nonspontaneous
reaction.
A A +2 + 2e ∆G 1
= –0.5 kcal (more spontaneous)
B B +2 + 2e ∆G 2
= –0.2 kcal (less spontaneous)
Same : A A +2 + 2e ∆G 1
= –0.5 kcal
Change : B +2 + 2e B ∆G 2
= +0.2 kcal
§ Rule (4)
A
A + B +2 B + A +2 ∆Grxn
∆G rxn = – 0.5 kcal + 0.2 kcal = – 0.3 kcal
B
A +2 + 2e ∆Gº 1
= –0.2 kcal ...(i)
B +2 + 2e ∆Gº 2
= +0.5 kcal...(ii)
If ∆Gº given in the half cells then spontaneity of the
reaction not determined by ∆Gº. We convert ∆Gº into
∆G by following equation.
∆G = ∆Gº + RT ln Q
If Q = 1 , then ∆G = ∆Gº
∆G 1
= ∆Gº 1
+ RT ln [A +2 ] .......(i)
∆G 2
= ∆Gº 2
+ RT ln [B +2 ] .......(ii)
§ Rule (5)
If emf of the half cell given in the problem then we predict
the spontaneity of reaction by emf.
∆G = – nFE
* Spontaneity from E
Greater the reduction potential greater will be
ease of reduction. Greater the oxidation potential greater
will be ease of oxidation.
§ Rule (6)
A
B
4
3
2
1
0
–1
–2
–3
–4
Spontaneoity
increases
E = 0 (Rxn is at equilibrium)
Non - spontaneoity
increases
A +2 + 2e Eo.p = –0.2 Volt (non-spontaneous)
B +2 + 2e Eo.p = +0.5 Volt (Spontaneous)
Same : B B +2 + 2e EO.P. = +0.5 Volt
Change: A +2 + 2e A ER.P. = +0.2 Volt
B + A +2 B +2 +A Ecell
Ecell = EO.P + ER.P = 0.5 + 0.2 = 0.7 Volt

§ Rule (7)
A
B
A +2 + 2e EO.P. = +0.5 Volt (more spontaneous)
B +2 + 2e EO.P. = +0.3 Volt (Less spontaneous)
Same : A A +2 + 2e EO.P. = +0.5 Volt
Change: B +2 + 2e B ER.P. = –0.3 Volt
A + B +2 B + A +2 Ecell
Ecell = EO.P. + ER.P. = +0.5 – 0.3 Volt = +0.2 Volt
§ Rule (8)
A
B
A +2 + 2e EO.P.= – 0.5 Volt (more spontaneous)
B +2 + 2e EO.P. = – 0.2 Volt (Less spontaneous)
Same : B B +2 + 2e EO.P. = –0.2 Volt
Change: A +2 + 2e A ER.P. = +0.5 Volt
A + B +2 B + A +2 Ecell
Ecell = EO.P. + ER.P. = –0.2 + 0.5 Volt = +0.3 Volt
§ Rule (9)
A
B
A +2 + 2e EºO.P. = +0.5 Volt.....(i)
B +2 + 2e EºO.P. = –0.2 Volt.....(ii)
Spontaneity of reaction not determined by standard emf
it is determined by following equation.
E E n A
E E n B
Ο + 2
= − (0.0592 )log[ ]
O . P . O . P .
Ο + 2
O . P .
=
O . P .
− (0.0592 )log[ ]
Greater the O.P. greater will be ease of oxidation.
HIT AND TRIAL METHOD
We determined spontaneity of reaction by Eº value,
however it is not correct. These are two possible result
one is correct and other is incorrect. If Emf of the cell is
+ve then our result is correct, but Emf of the cell is –ve,
then our result is incorrect, to gain correct result we
convert oxidation half cell into reduction half cell and
reduction half cell into oxidation half cell.
Rule for converting oxidation half cell into
reduction half cell and vice-versa
* A A +2 + 2e EO.P. = –0.2 Volt
If you want to gain reduction half cell then
A +2 + 2e A ER.P. = +0.2 Volt
* B B +2 + 2e EO.P. = +0.5 Volt
If you want to gain reduction half cell then
B +2 + 2e B ER.P. = –0.5 Volt
§ Rule (10) :
A A +2 + 2e EO.P. = 0.2 Volt
B +2 + 2e B ER.P. = 0.5 Volt
Ecell = ?
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To calculate Ecell we convert all the half cell reactions
are in same type (either oxidation half or reduction half)
(i) Convert both reaction into oxidation half
A A +2 + 2e EO.P. = 0.2 Volt
B B +2 + 2e EO.P. = –0.5 Volt
(ii) Convert both reaction into reduction half cell
A +2 + 2e A ER.P. = –0.2 Volt
B +2 + 2e B ER.P. = 0.5 Volt
Same : B +2 + 2e B ER.P. = 0.5 Volt
Change : A A +2 + 2e EO.P. = +0.2 Volt
B +2 + A
A +2 + B
Ecell = 0.7 Volt
§ Rule (11) :
For following type half cell reaction hybrid reaction
quotients (mixture of Qp and Qc) used in place of Qc or
Qp
« H 2
2H + + 2e (no. of electron used = 2)
Q
Hybrid
=
+ 2
[ ] /
H mol litre
( P ) atm
H2
« 1/2H 2
H + + 1e (no. of electron used = 1)
Q
Hybrid
+
[ H ]
=
( P )
1/2
H 2
« Cl 2
+ 2e 2Cl – (no. of electron used = 2)
Q
Hybrid
− 2
[ Cl ]
=
( P )
Cl 2
« Cl – Cl 2
+ 1e (no. of electron used = 1)
Q
Hybrid
§ Rule(12) :
( P )
=
[ Cl − ]
1/2
Cl 2
Emf is mass or mole independent property (intensive
property), but free energy is mass or mole dependent
property (Extensive property)
H 2
2H + + 2e; EO.P. = x volt, ∆G = – 2Fx
1/2H 2
H + + 1e; EO.P. = x volt, ∆G = – Fx
Emf of the reaction not depends on stoichiometry coeff.
Example : 2X+ 3y 5 A Ecell = x volt
X+ y A Ecell = x volt
§ Rule (13) :
If reaction is spontaneous in forward direction then nonspontaneous
in reverse direction and vice-versa.
Magnitude of Emf and ∆G are same but sign is reverse.
A A +2 + 2e; EO.P. = X volt , ∆G = – 2FX
A +2 + 2e A ∆G = +2 FX, ∆G = – nFE
∆G = – nFER.P.
ER.P. =
∆G/–nF =2FX / –2F = –X volt

§ Rule (14) :
If two half cell (no. of electron same or different)
produces full cell reaction then for simplicity we add
directly emf of half cell not ∆G. Addition of ∆G gives
same result but process is lengthy.
A A +2 + 2e E 1
B +2 + 2e B E 2
B +2 + A A +2 + B Ecell = E 1
+ E 2
If we proceed this problem by ∆G then.
A A +2 + 2e ∆G 1
= –2FE 1
B +2 + 2e B ∆G 2
= –2FE 2
B +2 + A A +2 + B ∆Grxn = ∆G 1
+ ∆G 2
–2FEcell = –2FE 1
– 2FE 2
Ecell = E 1
+ E 2
If no. of electron involved and apparent in
reaction then reaction is half cell. But if no. of electron
involved but not apparent in the reaction then reaction
is full cell reaction.
Example:
« Fe Fe +3 + 3e oxidation half cell, n=3
« Fe +3 + 1e Fe +2 reduction half cell, n= 1
« 2Fe +3 + 3I – 2Fe +2 –
+I 3
full cell rxn, n=2
«
–
MnO 4
+ 4H + +3e 2H 2
O+MnO 2
reduction half cell, n = 3
« Fe +2 Fe +3 + 1e oxidation half cell, n=1
If two half cell reaction having different no. of
electron provide full cell reaction then we also add Emf
for simplicity.
Example :
2× (A A +3 +3e) E 1
, ∆G 1
= –6FE 1
3× ( B +2 + 2e B) E 2
, ∆G 2
= –6FE 2
2A + 3B +2 2A +3 +3B, ∆Grxn = –6FE 3
∆Grxn = ∆G 1
+ ∆G 2
= –6FE 1
– 6FE 2
= –6FE 3
E 3
= E 1
+ E 2
§ Rule (15) :
If two half cell produces third half cell then we should
not added Emf directly. To calculate emf of third half
cell we add ∆G,
A A +3 + 3e ; E 1
, ∆G 1
= –3FE 1
B +2 + 2e B ; E 2,
∆G 2
= –2FE 2
A + B +2 A +3 + B + 1e E 3
¹ E 1
+ E 2
∆G 3
= ∆G 1
+ ∆G 2
–1FE 3
= –3FE 1
+ (–2FE 2
)
E 3
= 3E 1
+ 2E 2
§ Rule (16) :
For writing shorthand notation or cell representation of
electrochemical cell following conventions are used.
A A +2 + 2e EO.P = 0.5 volt
B +2 + 2e B ER.P = 0.2 volt
A + B +2 B + A +2 Ecell = 0.7 volt
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This electrochemical cell represented as follows.
Note :
(a)
(b)
(c)
(d)
(e)
A | A || B | B
14243 14243
+ 2 + 2
[ C1] [ C
2]
. . . R. H . S .
L H S
Oxidation half cell always represented in L.H.S
and reduction half cell always represented in
R.H.S.
Single vertical line placed in L.H.S denote
anode and R.H.S denote cathode
Cathode is known as +ve terminal but anode is
known as –ve terminal in electrolytic cell.
C 1
denotes concentration of A +2 and C 2
denotes
concentration of B +2 .
Reaction quotient of this representation is:
Q =
+ 2
[ A ][ B ]
+ 2
[ A][ B ]
(f) Flow of electron takes place from left to right
(anode to cathode), but flow of current takes
place from right to left (cathode to anode)
(g) No. of electron involved = 2
(h) Ecell for this representation calculated as follows
E = E −
0 0.0592
cell cell log Q
2
CONCEPT OF SPONTANEITY
For the complete study of reaction knowledge of
thermodynamics and chemical kinetics are essential.
Thermodynamics tell us about spontaneity of
reaction (Reaction is spontaneous or not). Chemical
kinetics tells us about the rate of reaction (Reaction is
fast or slow). Spontaneity of the reaction is determined
by thermodynamic parameter like ∆G, ∆S (universe) and
value of emf (For oxidation, Reduction or Redox
Reaction)
Entropy Change (∆S)
Entropy is thermodynamic state quantity that is measure
of randomness or disorder of the molecule of the
system.
Explanation of entropy on the basis of probability
left
bulb
right
bulb
no. of molecule relative prob. of finding all
molecules in left bulb
A (1 molecule) PA = ½ = (½) 1
A,B (2 molecules) PAB = PA×PB = (½) 2
A,B,C....(n molecules) PABC...... = (½) n
6.023 × 10 23 (½) 6.023× 1023 –6.023× 1023
= 2

PA = Probability of finding molecule A
PB = Probability of finding molecule B
PA ×PB = Probability of finding molecule A and B at
the same time
In probability × denotes ‘AND’
+ denotes ‘OR’
From above example, it is clear that as no of molecules
increases probability (chance) of finding of molecule at
the one place decrease.
According to 2nd law of thermodynamics entropy of
universe always increase in spontaneous process.
Surrounding : The rest part of the universe other than
the system is called surrounding.
System
Example :
Melting of ice at 10ºC. ∆S universe = +ve
Melting of ice at 0ºC ∆S universe = 0
Melting of ice at –10ºC ∆S universe = –ve
For spontaneous process :
∆Ssystem + ∆Ssurrounding = ∆Suniverse
(+ve or, –ve) (+ve or –ve) (always +ve)
∆G = ∆H – T∆S .................. (1)
∆G = Free energy, ∆H = Heat enthalpy
∆S = Change in entropy
Eq. (1) is multiplied by –(1/T)
∆G ∆H T ∆Ssys
∆H
− = − − = − + ∆S
T T −T T
∆G
⎛ ∆H
− = ∆ S + ∆ S = ∆S ⎜Q
− = ∆S
T
⎝ T
sys
surro sys universe surro
⇒ ∆ G = −T ∆ Suniverse
V. V.
I
« Consider The melting of Ice
H 2
O (s) + heat → H 2
O (l)
∆H = 6.03 × 10 3 J/mol
∆S system = 22.1 J/K mol
Condition (1) melting of ice at –10ºC (Below M.P.)
Temp = 263 K; ∆H = 6.03 × 10 3
∆S system = 22.1 J/K
∆S surro = –∆H/T =–6.03 × 10 3 / 263 = – 22.9 J/K ∆
universe = ∆S sys + ∆S surro = 22.1–22.9 = –0.8 J/K
∆S universe is –ve hence process is not spontaneous
∆G = – ∆Suniverse × T = – (– 0.8) × 263 = 2.1 × 10 2 J
∆G is +ve hence process is non–spontaneous.
we can calculate ∆G from eq. ∆G = ∆H – T∆S also,
∆G = 6.03 × 10 3 – 263 × 22.1 = 2.1 × 10 2 J
⎞
⎟
⎠
S
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Condition (2) melting of ice at 0ºC (at M.P.)
Temp = 273 K; ∆H = 6.03 × 10 3
∆G = ∆H – T∆S = 6.03 × 10 3 – 263 × 22.1 = 0
∆G =0 (process is at equilibrium)
∆Suniverse = ∆Ssystem + ∆Ssurro
= 22.1 – (6.03 × 10 3 /273) = 0
Condition (2) melting of ice at 10ºC (above M.P.)
Temp = 283 K; ∆H = 6.03 × 10 3
∆G = ∆H – T∆S = 6.03 × 10 3 – 263 × 22.1
= –2.2 × 10 2 Joule
Process is spontaneous.
Q : By which statement we can say confirmly process
is spontaneous.
(a) ∆S system is +ve (f) ∆S universe is –ve
(b) ∆S system is –ve (g) ∆G is +ve
(c) ∆S surro is +ve (h) ∆G is –ve
(d) ∆S surro is –ve (i) EMF is +ve
(e) ∆S universe is +ve (j) EMF is –ve
Ans: (e), (h) and (i)
* Relation of ∆G with emf is ∆G = –nFE
* Relation of ∆G with ∆Suniverse is ∆G = –T∆Suniverse
§ Conditions for Spontaneity
(a)
condition ∆G ∆Suniverse EMF
Spontaneous
reaction
Non-spontaneous
reaction
Reaction is at
equilibrium
– ve
+ve
0
THERMODYNAMIC EQUILIBRIA
On the basis of free energy
∆G = ∆Gº + RT lnQ (Q may be Qp and Qc)
At equilibrium condition Q = 0, ∆G = 0
∆Gº = –RT lnK (K may be Kp or Kc)
0 0
−∆G
⎛ −∆G
⎞
lnK = ⇒ K = anti ln⎜ ⎟
RT
⎝ RT ⎠
( −∆G 0 / RT )
⇒ K = e
+ ve
–ve
Equilibrium constant (Kp or Kc) can be
calculated by above equation.
0
+ ve
– ve
0

(b)
(c)
On the basis of entropy
∆S reaction = ∆Sº reaction – RlnQ
At equilibrium Q = K, ∆S reaction = 0
∆Sº reaction = R lnK
0 0
−∆S
⎛
reaction
−∆S
⎞
reaction
lnK = ⇒ K = anti ln
R
⎜
R ⎟
⎝ ⎠
0
( reaction / )
⇒ K = e ∆S R
Equilibrium constant (Kp or Kc) can be
calculated by above equation.
On the basis of electromotive force for
Oxidation, Reduction and Redox reaction.
∆G = ∆Gº + RT ln Q ........(1)
also, ∆G = – nFE and, ∆Gº = – nFEº
eq. (1) can be also written as
–nFE = –nFEº + RT ln Q........(2)
eq. (2) is divided by nF
RT
nF
0
E = E − lnQ
At 25ºC this equation changes into
0 0.0592 E = E − lnQ
n
At equilibrium, Q = K and E = 0.
REACTION QUOTIENT
When reactants and products of a given
chemical reaction are mixed it is useful to know whether
the mixture is at equilibrium, and if not, in which direction
the system will shift to reach equilibrium.
If the concentration of one of the reactants or
product is zero, the system will shift in the direction that
produces the missing component. However, if all the
initial concentrations are non zero, it is more difficult to
determine the direction of the move toward equilibrium.
To determine the shift in such case, we use the reaction
quotient (Q). The reaction quotient is obtained by
applying the law of mass action, using initial
concentrations instead of equilibrium concentration.
Reaction quotient in the term of pressure known
as Q p
and reaction quotient in the term of concentration
known as Q C
.
N ( g) + 3 H ( g) 2 NH ( g)
Q
c
2 2 3
2
2
[ NH3
]
( PNH
)
3
= and QP
=
[ N ][ H ] ( P )( P )
3 3
2 2
N2 H2
Qc and Qp is defined at any concentration and at any
pressure respectively.
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If at any time Q < K, the forward reaction must occur to
a greater extent than the reverse reaction for equilibrium
to be established. This is because numerator of Q is
too small and the denominator is too large.
Reducing the denominator and increasing the
numerator requires forward reaction until equilibrium is
established.
If Q > K, the reverse reaction must occur to
greater extent than the forward reaction for equilibrium
to be reached. When Q = K, the system is at equilibrium,
so no further net reaction occurs.
Conclusion:
Q < K Forward reaction predominates
until equilibrium is established.
Q = K System is at equilibrium.
Q > K Reverse reaction predominates
until equilibrium is established.
Prob. 1 For the synthesis of ammonia at 500ºC, the
equilibrium constant is 6.0 × 10 –2 L 2 /mol 2 . Predict the
direction in which the system will shift to reach
equilibrium in each of following case. Reaction for
synthesis of NH 3
is N ( g) + 3 H ( g) 2 NH ( g)
2 2 3
(a) [NH 3
] 0
= 1 × 10 –3 M, [N 2
] 0
= 1 × 10 –5 M
[H 2
] 0
= 2 × 10 –3 M
(b) [NH 3
] 0
= 2 × 10 –4 M, [N 2
] 0
= 1.5 × 10 –5 M
[H 2
] 0
= 3.54 × 10 –1 M
Ans: (a) Q C
= 1.2 × 10 7 L 2 /mol 2
Q C
> K C
Hence back reaction predominates.
(b) Q C
= 6.0 × 10 –2 L 2 /mol 2
Q C
= K C
Reaction is at equilibrium.
Your Problem in chemistry
Now a days, mostly
students suffer trouble and
fear in vital topic like Acid
Base Titration, Indicator,
Double Indicator, Redox
Titration, Ionic Equilibria,
E l e c t r o c h e m i s t r y ,
Thermodynamics, Chemical
Equilibria and Chemical Kinetics. Our main
motto to relate all topic with highly advanced,
accurate and easy concept. Problem asked in
Exam like C.B.S.E. (Mains), B.C.E.C.E.
(Mains), AIEEE and IIT are not chapter wise
problem but concept based problem.
-Shailendra Kumar

→
New Concepts of
Ionic Equilibria
1. CONCEPT OF MILLIEQIVALENT.
No of equivalent = wt
eq.wt
wt × 1000
No of milliequivalent =
eq.wt
wt × 1000 No of meq
N = =
eq.wt × V(ml) V(ml)
No of meq = N × V(ml)
No of mol =
wt
mol.wt
, No of milli mol =
No of milli mol = M × V(ml)
CONVERSION FACTOR
M × V.F (Valence Factor) = N
No of mol × V.F = No of eq.
No of milli mol × V.F = No of meq.
No of eq =
mol.wt
V.F
2X + 3Y X 2
Y 3
(By Meq Method)
Milliequivalent
befor reaction
Milliequivalent
after reaction
MAIN OBJECTIVES
Changed
X + Y X 2
Y 3
10 6 0
4 0 6
wt × 1000
mol.wt
No need of
balanced equation
(Suppose)
Meq of X Reacted = 6
Meq of Y Reacted = 6
Meq of X 2
Y 3
formed = 6
During reaction equal meq of reactant reacted and equal
meq of product formed.
Species
V.F
K + 1
–2
SO 4
2
Al +3 3
Ca +2 2
CaSO 4
2
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7
1. Concept of Milliequivalent.
2. Auto ionisation of pure water.
3. PH Calculation of strong acid
and strong base.
4. PH Calculation of weak acid and
weak base.
5. PH Calculation of weak acid and
weak base in the present of its salt.
6. Common Ion effect.
7. Buffer Solution.
8. Salt hydrolysis.
9. Solubility Product Principle
H 2
SO 4
2
AlCl 3
3
Al 2
(SO 4
) 3
6
COOH
COOH 2H O 2
2
K 2
SO 4
Al 2
(SO 4
) 3
24H 2
O 8
2 When 4.9 gm of H 2
SO 4
mixed with 4 gm of
NaOH. What is the
(a) Wt of NaOH reacted.
(b) Wt of H 2
SO 4
reacted.
(c) Mol of Na 2
SO 4
formed.
H 2
SO 4
+ NaOH Na 2
SO 4
+ H 2
O
Meq befor 4.9 × 1000 4 × 1000
reaction
0 0
49 40
(100) (100) 100 100
Meq after
reaction 0 0
(a) Wt of NaOH completely reacted in this reaction,
Hence wt of NaOH reacted is 4 gm.
(b) H 2
SO 4
completely reacted in this reaction, hence wt
of H 2
SO 4
reacted is 4.9 gm.
(c) No of meq of Na 2
SO 4
formed is 100
No of milli mol of Na 2
SO 4
=
100
2
No of milli mol of Na 2
SO 4
= 50
No of mol of Na 2
SO 4
= 50 × 10 –3
∗
= 5 × 10 –2 mol
Dilution
means addition of solvent (generally H 2
O)
During dilution, wt of solute → remains constant
Mol.wt of solute → ”
Mol of solute → ”
Milli mol of solute → ”
No of millimol = M × V(ml)
During dilution MV = constant
M 1
V 1
= M 2
V 2
or
N 1
V 1
= N 2
V 2
During dilution millimol of solute remains
constant.

→
→
2 Calculate wt of AgCl formed when 200 ml of 5 N
HCl reacted with 1.7 gm AgNO 3
Solution : Write unbalanced reaction
AgNO 3
+ HCl → AgCl + HNO 3
Milliequivalent 1.7 × 1000
befor reaction 170
200 × 5 0 0
(10) (1000)
Milliequivalent
after reaction 0 990 10 10
Meq of AgCl formed = 10
wt × 1000
143.5
=10
Wt = 1.435 gm
2. AUTO IONISATION OF PURE WATER
(a) H 2
O H + + OH –
or
H 2
O + H 2
O H 3
O ⊕ + OH
(b)
(c) Q.
K ionization =
K ionization [H 2
O] = [H + ] [OH – ]
Kw = [H + ] [OH]
1× 10 –14 = [H + ] [OH – ]
Value of Kw = 1× 10 –14 at 25º C
Ionisation or Dissociation of H 2
O is Endothermic
reaction. Hence if temp is increased value of Kw
also increase.
Value of Kw at 25º C is 1× 10 –14 .
Kw is known as ionic product of water.
PH scale (0–14) valid for Kw (value equal to
1× 10 –14 )
At certain temp (temp greater than 25º C) Value
of kw is 1× 10 –13 What is its neutral point and what
is its PH scale.
Solution : PH scale ( 0–13)
Its neutral point is 6.5
Kw = [H + ] [OH – ]
1× 10 –13 = x.x
x 2 = 1× 10 –13
x = 1× 10 –6.5 = [H + ] = [OH – ]
Hence PH at neutral point is 6.5
(d)
[H + ] [OH – ]
[H 2
O]
[H + ] [OH – ] → solution is Neutral
[H + ] > [OH – ] → solution is Acidic
[H + ] < [OH – ] → solution is Basic
3. PH CALCULATION OF STRONG ACID AND
STRONG BASE
Actual Method for PH Calculation.
Calculate PH for 1× 10 –2 M HCl
H 2
O H + + OH –
x
HCl H + + Cl –
Common ion effect 10 –2
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8
Kw = [H + ] [OH – ]
Kw = (x + 10 –2 )x
1× 10 –14 = (x + 10 –2 )x
After solving quardatic equation value of H + provided
by H 2
O is very less in comparison to H + provided by
HCl, Hence H + provided by H 2
O is neglected.
Hence [H + ] = 10 –2
PH = 2
∗ In pure water molarity of [H + ] is 10 –7 M but in acidic
medium value of H + provided by H 2
O is less than 10 –6 M due
to common ion effect, But in rough calculation we also take
molarity of [H + ] is 10 –6 M in acidic solution.
∗ If [H + ] produced by acid is less than 10 –6 M or equal
to 10 –6 M then [H + ] produced by H 2
O is not neglected.
∗ If [H + ] produced by acid is greater than 10 –6 M then
[H + ] produced by H 2
O is neglected.
∗ [H + ] [OH – ] = Kw = 10 –14 (At 25º C)
taking log both side.
log [H + ] [OH – ] = Kw
PH + POH = PKw = 14
Strong acid (like HCl, HNO 3
, H 2
SO 4
etc) and strong
base (like NaOH, Ca(OH) 2
, Mg(OH) 2
etc) are completely
dissociated.
Meq of S.A = Meq of H +
Meq of S.B = Meq of OH –
2 Calculate Molarity of H + and Normality of H + in
1 M H 2
SO 4
H 2
SO 4
Neglected
↑
⊕ –2
2 H + SO 4
100% dissociated
H 2
SO 4
H + –2
+ SO 4
Meq befor D. 1× 2 × V 0 0 Let volume
(2V)
of solution
Meq after D.
is V(ml)
0 2V 2V
Meq of H + = 2V
Normality of H + =
2V
V
Millimol of H + = 2V (V.F = 1) = 2
2V
[H + ] = V
= 2
Molarity of H + = Normality of H +
∗ Molarity of H + and Normality of H + is equal, and
meq of H + and millimol of H + is equal because valence
factor is 1.
∗ Molarity of OH – = Normality of OH – (V.F =1)
Meq of OH – = Millimol of OH – (V.F =1)
∗ Meq of strong acid = Meq of H ⊕=Millimol of H⊕
N× V(ml) N× V(ml) M× V(ml)
Normality of strong acid=Normality of H + =Molarity of H +
Conclusion : If we want to calculate PH of strong acid then

→
→
we convert molarity acid into normality of acid, which is directly
equal to molarity of H +
2 Calculate PH of 1× 10 –2 M H 2
SO 4
.
Molarity of H 2
SO 4
= 1 × 10 –2 M
Normality of H 2
SO 4
= 2 × 10 –2 N
Normality of H + = 2 × 10 –2
Molarity of H + = 2 × 10 –2
PH = –log [H + ]
PH = – log (2 × 10 –2 )
2 Calculate PH 100 ml of 1 × 10 –2 M H 2
SO 4
.
Solution : PH of the solution independent on volume. It
depends upon normality of acid.
Method (1)
⊕
Normality of H 2
SO 4
= Normality of H = Molarity of H +
1 × 10 –2 × 2 = Molarity of [H + ]
2 × 10 –2 = Molarity of H +
PH = 2 – log2
Method (2)
Meq of H 2
SO 4
= Millimol of H +
100 × 10 –2 × 2 = 100 × [H + ]
[H + ] = 2 × 10 –2
PH = 2– log2
2 Calculate PH of 0.5 × 10 –2 M Ca (OH) 2
.
Normality of Ca (OH) 2
= Molarity of (OH – )
0.5 × 10 –2 × 2 = [OH – ]
1 × 10 –2 = [OH – ]
POH = 2, PH = 12
2 Calculate PH of the solution when 100 ml of 1 ×
10 –3 M HCl, 100 ml of 1 × 10 –4 M HNO 3
and 100 ml of 1 ×
10 –2 N H 2
SO 4
mixed.
Solution :
∑ Meq of H + = meq of H + provided by HCl + meq H + provided
by HNO 3
+ meq of H + provided by H 2
SO 4
∑ Meq of H + = meq of HCl + meq of HNO 3
+ meq of H 2
SO 4
100 × 10 –3 + 100 × 10 –4 + 100 × 10 –2
ε millimol of H + =0.1 + 0.01 + 1
300 × M = 1.11
M = 1.11
300
PH is determined by equation
PH = – log
1.11
300
2 Calculate PH of 10 2 M HCl.
PH = –2 But this is not true, practically PH of this
solution is near to zero.
2 Calculate PH of 1M HCl.
PH = 0 this solution is most acidic. Greater the
[H + ] greater will be acidity and lesser will be PH.
Note : If PH of more concentrated acid is less than 0 (–ve)
then PH of this concentrated acid is taken as zero.
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9
2 Calculate the [Cl – ], [Na + ], [H + ], [OH – ] and PH of
resulting solution obtained by mixing 50 ml of 0.6 N HCl
and 50 ml of 0.3 NaOH.
Meq befor
reaction
Meq after
reaction
HCl + NaOH → NaCl + H 2
O
30 15 0 0
15 0 15 15
Meq of HCl = 15
Meq of H + = 15
15
[H + ] =
100
= .15
PH = –log [H + ] = – log 0.15
PH = 0.8239
[OH – ] [H + ] = 1 × 10 –14
[OH – ] =
1 × 10 –14
= 6.6 × 10 –14 M
0.15
Meq of Cl – = meq of Cl – provided by NaCl + meq of Cl –
provided by HCl.
= 15 + 15
Meq of Cl – = 30
Millimol of [Cl – ] =
30
100
= 0.3 M
Meq of Na + = meq of Na
⊕
provided by NaCl
= meq of NaCl
Millimol Na + = 15
[Na + ] =
15
100
= 0.15 M
4. PH CALCULATION OF WEAK BASE AND
WEAK ACID.
Before
dissociation
BOH B + + OH –
weak base
C 0 0
After
dissociation
C – C α Cα Cα
Kb =
[B + ] [OH – ] Cα . Cα
[BOH]
=
C (1– α)
Dissociation constant of weak base
Kb = Cα 2 (If α is very less)
(1–α) ≈ 1
α =
Kb
C
[OH – ] = Cα = C
Kb
C
= √ Kbc
[OH – ] = (Kbc) ½
POH = – log[OH – ]
= – log(Kbc) ½
(1) POH = ½ (PKb – log c) →V.V.I
(2) [OH – ] = √ Kbc = Cα
(3) α =
√
Kb
C
For weak base

→
→
=
→
=
→
=
=
=
=
PH = ½ (PKa – log c)
[H + ] = Cα = √KaC
α =
√
Ka
C
2 Calculate PH of
(a) 0.002 N acitic acid having 2.3% dissociation.
(b) 0.002 N NH 4
OH having 2.3% dissociation.
(a) [H + ] = 2 × 10 –3 ×
2.3
= 4.6 × 10 –5 M
100
PH = 5 – log 4.6 = 4.3372
2.3
(b) [OH – ] = Cα = 2 × 10 –2 ×
100
= 4.6 × 10 –5 M
POH = 4.3372
PH = 9.6627
5. PH CALCULATION OF WEAK ACID AND
WEAK BASE IN THE PRESENCE OF ITS SALT.
Example 1: If we want to calculate PH of CH 3
– C – OH
(C 2
) in the presence of CH 3
COONa (C 1
) O
CH 3
– COONa CH 3
COO – + Na⊕
B.D C 1
100%
O O
A.D O C 1
C 1
CH 3
COOH CH 3
– C–O – + H⊕
O
B.D C 2
O O
A.D C 2
– C 2
α C 2
α C 2
α
O
Ka =
[CH 3
–C–O – ][H + ]
[CH 3
–C–OH] =
O
Ka =
[C 1
+C 2
α] [H + ]
[C 2
– C 2
α]
Due to common ion
dissociation of CH 3
COOH is
depressed
[C 1
+C 2
α] [C 2
α]
C 2
– C 2
α
[H + ] = Ka . [C 2
– C 2
α]
[C 1
+ C 2
α]
[C
PH = PKa + log 1
+ C 2
α]
[C 2
– C 2
α]
Value of C 2
α is very very less,
hence [C 1
+ C 2
α] = C 1
and [C 2
– C 2
α] = C 2
C
PH = PKa + log 1
C 2
[salt]
PH = PKa + log
[acid] →
→For weak acid
This equation is
not 100% correct
Example 2 : If we want to calculate PH of NH 4
OH (C 2
) in
the presence of NH 4
Cl (C 1
)
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B.D
A.D
B.D
A.D
⊕
NH 4
Cl NH 4
+Cl –
C 1
100%
O O
O C 1
C 1
NH 4
OH
+
NH 4
+OH –
C 2
O O
C 2
– C 2
α C 2
α C 2
α
[NH
Kb = 4+
][OH – ]
[NH
=
4
OH]
Kb =
[OH – ] = Kb
[C 2
α + C 1
][OH – ]
[C 2
– C 2
α]
POH = PKb + log
POH = PKb + log
[C 2
– C 2
α]
[C 1
+ C 2
α]
Dissociation of NH 4
OH is
depressed due to NH 4
Cl
[C 1
+ C 2
α]
[C 2
– C 2
α]
[C 1
]
[C 2
]
[salt]
POH = PKb + log →
[base]
[C 2
α + C 1
][C 2
α]
[C 2
– C 2
α]
This equation is
not 100% correct
Method for PH Calculation of Following Reaction
Condition 1.
NaOH + CH 3
–C–H CH 3
–C–Na + H 2
O
Meq.B.R 10 5 O 0 O 0
(Suppose)
Meq.A.R 5 0 5 5
If strong acid or strong base present in solution
after reaction then PH is calculated by strong acid or strong
base. Concentration of salt not considered.
Condition 2.
NaOH + CH 3
COOH CH 3
COONa + H 2
O
Meq.B.R 10 10 0 0
(Suppose)
Meq.A.R 0 0 10 10
If solution contains only salt [CH 3
COONa] then
PH calculate by salt hydrolysis.
Condition 3.
NaOH + CH 3
COOH CH 3
–COONa + H 2
O
Meq.B.R 5 10 0 0
Meq.A.R 0 5 5 5
If solution contains weak acid and its salt then PH
is calculated by.
10

PH =PKa + log
[salt]
[acid]
Calculation of PH in the reaction NH 4
OH with HCl
Condition 1.
NH 4
OH + HCl NH 4
Cl + H 2
O
Meq.B.R 10 5 0 0
Meq.A.R 5 0 5 5
Solution contains weak base and its salt.
[salt]
POH = PKb + log
[base]
Condition 2.
NH 4
OH + HCl NH 4
Cl + H 2
O
Meq.B.R 5 10 0 0
Meq.A.R 0 5 5 5
PH is calculated by HCl (strong acid)
Condition 3.
NH 4
OH + HCl NH 4
Cl + H 2
O
Meq.B.R 5 5 0 0
Meq.A.R 0 0 5 5
PH is calculated by salt hydrolysis (NH 4
Cl)
2 Calculate PH of the following mixtures, given that
Ka = 1.8 × 10 –5 and Kb = 1.8 × 10 –5
(a) 50 ml of 0.10 M NaOH + 50 ml of 0.05 M CH 3
COOH
(b) 50 ml of 0.05 M NaOH + 50 ml of 0.10 M CH 3
COOH
(c) 50 ml of 0.10 M NaOH + 50 ml of 0.10 M CH 3
COOH
(d) 50 ml of 0.10 M NH 4
OH + 50 ml of 0.05 M HCl
(e) 50 ml of 0.05 M NH 4
OH + 50 ml of 0.05 M HCl
(f) 50 ml of 0.10 M NH 4
OH + 50 ml of 0.10 M HCl
Solution : (a)
NaOH + CH 3
COOH
Millimol.
B.R
50×0.1 50×0.05
CH 3
–COONa + H 2
O
0 0
Millimol.
A.R 2.5 0 2.5 2.5
Solution after reaction contain strong base (NaOH)
Hence PH determined by NaOH
Meq of NaOH = 2.5
N of NaOH =
2.5
100
= 2.5 × 10 –2 M
[OH – ] = 2.5 × 10 –2
POH = 1.6021
PH = 12.3979
(b) NaOH + CH 3
COOH CH 3
–COONa + H 2
O
Meq.B.R 2.5 5 0 0
Meq.A.R 0 2.5 2.5
Solution after reaction contains weak acid and its
salt, Hence PH determined by the equation
[salt]
PH = PKa + log
[acid]
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PH = – log Ka + log
[CH 3
COONa]
[CH 3
COOH]
2.5
[CH 3
COONa] =
100
= 4.74 + log 2.5/100
2.5
, [CH 3
COOH] =
2.5/100
100
PH = 4.74
(c) NaOH + CH 3
COOH CH 3
–COONa + H 2
O
Meq.B.R 5 5 0 0
Meq.A.R 0 0 5 5
If solution contains salt after reaction then PH determined
by salt hydrolysis. Salt hydrolysis discused latter.
Problem (d), (e) and (f) do yourself.
Hints :
(d) PH determined by buffer equation.
(e) PH determined by strong acid (HCl)
(f) PH determined by salt hydrolysis (NH 4
Cl)
6 & 7. BUFFER SOLUTION AND COMMON ION
EFFECT BUFFER SOLUTION AND COMMON
Buffer solution: solution which possess reserve acidic nature
or alkaline nature or solution which resist change in PH
due to dilution or addition of small quantity of acid or alkali
are known as buffer solution.
Example of acidic Buffer →
A weak acid + its salt (CH 3
COOH + CH 3
COONa)
Basic Buffer →
A weak base + its salt (NH 4
OH + NH 4
Cl)
Equation acidic buffer is
[salt]
PH = PKa + log
[acid]
Equation for basic buffer is
[salt]
PH = PKa + log
[acid]
2 Calculate change in PH (∆ PH ) which 0.01 mol
NaOH added into.
(a) 1.0 litre pure water
(b) 1 litre of 0.10M CH 3
COOH
(c) 1litre solution 0.10 M CH 3
COOH and 0.10M CH 3
COONa
(a) In pure water, [H + ] = [OH – ] = 10 –7
Hence PH 1
= 7
After addition of .01 mol of NaOH
[NaOH] =
0.01
= 0.01
1
[OH – ] = 1 × 10 –2
POH = 2, PH 2
= 12
∆ PH = PH 2
– PH 1
= 12–7 = 5
(b) NaOH reacts with acetic acid.
NaOH + CH 3
COOH CH 3
–COONa + H 2
O
Mol.B.R 0.01 0.10 0 0
Mol.A.R 0 (0.01–0.01) 0.01 0.01
0.09
Solution contains weak acid and its salt, Hence
11

=
=
→
→
[salt]
PH 2
= PKa + log
[acid]
0.01
PH 2
= 4.74 + log
0.09
PH 2
= 3.78
PH of 0.1 M CH 3
COOH is
PH 1
1/2 (PKa – log C)
PH 1
= 1/2 (4.74 + 1) = 2.87,
∆ PH = PH 2
– PH 1
= 3.78 – 2.87
∆ PH = 0.91
(c)
[salt]
PH 1
= PKa + log
[acid]
0.10
= 4.74 + log
0.10
PH 1
= 4.74
PH after addition of NaOH
CH 3
COOH + NaOH CH 3
–COONa + H 2
O
Mol.B.R 0.10 0.01 0.10
Mol.A.R (0.10–0.01) 0 (0.10 + 0.01)
[salt]
PH 2
= PKa + log
[acid]
0.11
PH 2
= 4.74 + log
0.09
0.11
∆ PH = PH 2
– PH 1
= log = 0.087
0.09
Conclusion : After addition of NaOH ∆ PH is very less in
buffer solution (mixture of weak acid and its salt).
Preparation of Buffer
I. Acidic Buffer
(a) Mixing weak acid and its salt.
(b) Exess weak acid acid and strong base
CH 3
COOH + NaOH CH 3
–COONa + H 2
O
Meq.B.R 10 5 0 0
Meq.A.R 5 0 5 5
(c) Exess basic salt + HCl
CH 3
COONa + HCl NaCl + CH 3
COOH
Meq.B.R 10 2 0 0
Meq.A.R 8 0 8 8
II.
Basic Buffer
(a) Mixing weak base and its salt
NH 4
OH + NH 4
Cl
(b) Exess NH 4
Cl + HCl
(c) Exess NHaCl + NaOH
PH range of buffer solution
PH = PKa + log
[salt]
[acid]
[salt]
[acid] 0.1 1 10
Best Buffer (most sensitive buffer)
PH range of acidic buffer
[salt]
PH = PKa + log
[acid]
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12
[salt]
[acid]
ratio vary 0.1 to 10
PH = PKa ± 1
In the same manner POH of range of basic buffer
[salt]
POH = PKb + log
[base]
POH = PKb ± 1
2 A physician or lab technician want to prepare buffer
solution having PH = 5 using CH 3
COOH and CH 3
COONa.
What will be ratio of [CH 3
COONa]
[CH 3
COOH]
(PKa of CH 3
COOH is 4.74)
PH = PKa + log [salt]
[acid]
[CH
5.0 = 7.74 + log 3
COONa]
[CH 3
COOH]
log
[CH 3
COONa]
[CH 3
COOH]
= 0.26
[CH 3
COONa]
[CH 3
COOH]
= anti log 0.26 = 1.8
Note: Buffer has maximum capicity when its acid has
its PKa as close as possible to the target PH.
2 A physician want to prepare buffer solution having
PH = 5 using weak acid and its salt, which of the following
weak acid and its salt would be a good choice.
Acid Salt PKa
(a) H 3
PO 4
–2
H 2
PO 4
2.12
(b) HCOOH
–
HCO 2
3.74
(c) CH 3
COOH CH 3
COO – 4.75
(d)
–2
H 2
PO 4
–2
HPO 4
7.21
Ans = c
2 What mass of sodium acetate (CH 3
COO
Na.3H 2
O,M.W = 136) and what volume of concentrated
acetic acid (17.45 M) should be used to prepare 0.50 L of a
buffer solution at PH = 5.0 that is 0.150 M overall ?
Mass of sodium acetate (hydrated) = 6.6 gm
Volume of Acetic acid = 1.5 ml
8. SALT HYDROLYSIS
Concept Suppose you want to calculate PH of sodium acetate
(basic salt) of certain concentration (c)
O
O
CH 3
–C–ONa CH 3
–C–O – + Na (Any salt is
100%
dissociated)
B.D
A.D
B.R
A.R
C
100%
0 0
0 C C
anionic hydroysis taken place.
CH 3
COO – + H 2
O CH 3
COOH + OH –
C 0 0
C–Cα Cα Cα
CH 3
COO – (Acetate ion) bevaves as weak base

=
=
=
[CH
Kb(CH 3
COO – 3
COOH][OH – ][H + ]
) = [CH 3
COO – ][H + ]
Kw 1 × 10
Kb(CH 3
COO – ) = =
–14
Ka(CH 3
COOH) 1.8 × 10 –5
Kb(CH 3
COO – ) = 5.6 × 10 –10
Kb of (CH 3
–C–O – ) Acetate ion denotes, it is weak base
O
PKb (CH 3
COO – ) = 10 –log 5.6
= 9.25
POH of weak base is denoted by equation
POH = 1/2 (PKb – log c)
POH = 1/2 (PKb (CH 3
COO – )–log c)
Generally α is replaced by h (degree of hydration)
and Kb of CH 3
C–O – is replaced by KH (Hydrolysis constant)
O
MISCONCEPT
Conjugate base of weak acid is strong. It is not
true. Conjugate base of weak acid is also weak.
Example : Ka of CH 3
COOH is 1.8 × 10 –5 (W.A)
Ka of CH 3
C–O – is 5.6 × 10 –10 (W.B)
O
∗ Conjugate acid of weak base is strong →
It is not correct.
Conjugate acid of weak base is also weak.
Example : Kb of NH 4
OH is 1.8 × 10 –5 (W.B)
+
Kb of NH 4
is 5.6 × 10 –10 (W.A)
+
NH 4
+ H 2
O NH 4
OH + H +
[NH 4
OH][H + ][OH – ]
Ka (NH 4+
) = [NH 4+
][OH – ]
Kw 1.0 × 10 –14
= =
Ka(NH 4+
) 1.8 × 10 –5
Ka [NH 4+
] = 5.6 × 10 –10
This concept is useful for the solution of problem
of salt hydrolysis.
∗ Weaker the acid stronger its conjugate base .
Weak Acid Ka Cunjugate Kb
Base
HX 1 × 10 –3 X– 1 × 10 –11
HY 1 × 10 –4 Y– 1 × 10 –10
HZ 1 × 10 –5 Z– 1 × 10 –9
Weaker the acid stronger its conjugate base
∗ Weaker the base stronger its conjugate acid.
2 Calculate for 0.01 N solution of sodium acetate.
(a) Hydrolysis constant
(b) Degree of Hydrolysis
(c) PH (Ka = 1.8 × 10 –5 )
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13
Solution :
(a) Kb (CH 3
COO – ) = KH =
1 × 10
=
–14
1.8 × 10 –5
= 5.6 × 10 –10
Kb(CH
(b) h = 3
COO – )
=
C
= 2.3 × 10 –4
(c) POH = 1/2 (PKb – log c)
= 1/2 (PKb(CH 3
COO – ) – log c)
= 1/2 (9.25 – log 0.01)
= 1/2 (9.25 + 2)
= 5.625
PH = 8.375
2 Calculate the hydrolysis constant of the salt containing
NO 2
–
ions.
Given Ka for HNO 2
= 4.5 × 10 –10
Solution :
Kw 10
KH = = –14
= 2.2 × 10
Ka 5.6× 10 –10
–5
2 What is PH of 0.5 M aqueous NaCN solution PKb
of CN – = 4.70.
NaCN Na + CN –
CN – + H 2
O HCN + OH –
[HCN] [OH – ]
Kb (CN – ) =
[CN – ]
POH = 1/2 (PKb – log c)
POH = 1/2 (4.70 – log 0.5)
POH = 2.5
PH = 11.5
2 Calculate the percentage hydrolysis in 0.003 M a
queous solution of NaOCN.
Ka for HOCN = 3.33 × 10 –4
KH
h = =
C
Kb (OCN – )
C
Kw
=
Ka (HOCN)C
=
h = 10 –4 , % hydrolysis = 10 –2 %
Hydrolysis of Acidic salt (like NH 4
Cl)
B.D
A.D
Befor hydrolysis
After hydrolysis
NH 4
Cl
+
NH 4
+ Cl –
C 0 0
0 C C
Kw
Ka(CH 3
COOH)
5.6× 10 –10
0.01
10 –14
3.33 × 10 –4 × 0.003
NH 4
+
+ H 2
O NH 4
OH + H +
C 0 0
C–Ch Ch Ch

→
Ka (NH 4
+
) = KH =
Kw
=
Kb(NH 4
OH)
1 × 10
Ka(NH 4+
) = –14
= 5.6 × 10
1.8 × 10 –5 –10
+
NH 4
behaves as weak acid.
PH of the weak acid is calculated by the equation.
PH = 1/2 (PKa – log c)
Ka (NH 4+
) = 5.6 × 10 –10
PKa (NH 4+
) = 10 – log 5.6
PKa(NH 4+
) = 9.25
2 Calculate the PH of 0.1 M NH 4
Cl.
KbNH 4
OH = 1.8 × 10 –5
PH = 1/2 (PKa – log c)
= 1/2 (9.25 – log 0.1) = 1/2 (10.25)
PH = 5.12
Note : Hydrolysis of Na and Cl – not taken place because
it comes from strong base and strong acid.
Ka for strong acid is very large.
So Kb of Cl – is very-very small, Hence hydrolysis
of Cl – not takes place. Value of Kb for strong base is very
large, so, Ka of Na + is very low, Hence hydrolysis of Na + not
takes place.
If Hx is weak acid (Ka) then value of conjugate
acid (X – ) is Kb then.
Kw
Kb (x) – =
Ka (Hx)
Kw = Kb (X) – × Ka(Hx)
taking log both side
PKw = PKb + PKa
14 = PKb + PKa
2 A certain buffer solution contains equal concentration
of X – and Hx. Kb for X – is 10 –10 . Calculate PH of
buffer.
Kb for X – = 10 –10
Kw 1 × 10 –14
Kb for Hx =
Kb
= = 1 × 10
1.0 × 10 –10
–4
PKa for Hx = 4
PH = PKa + log
PH = 4
[salt]
[acid]
[NH 4
OH][H + ][OH – ]
[NH 4+
][OH – ]
SOLUBILITY PRODUCT PRINCIPLE
∗ Most compounds dissolve in water to some extent
and many are so slightly soluble that they are called
“ Insoluble ”
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14
∗ Compounds that dissolve in water to the extent of
0.02 mole per litre or more are classified as soluble.
∗ Slightly soluble compounds are important in many
natural phenomenon. Our bones and teeth are mostly calcium
phosphate Ca 3
(PO 4
) 2
a slightly soluble compound.
∗ Tooth decay involves solubility, when food lodges
between the teeth, acids form that dissolve enamel, which
contain a mineral called hydroxyapatite (Ca 5
(PO 4
) 3
OH).
Tooth decay can be reduced by treating teeth with fluoride.
Fluoride replaces the hydroxide to hydroxyapatite to produce
the corresponding fluorapatite Ca 5
(PO 4
) 3
F and CaF 2
, both
of which are less soluble in acid than original enamel.
SOLUBILITY PRODUCT CONSTANT
Suppose we add one gram of solid BaSO 4
to 1.0
litre of water at 25ºC and stir untill the solution is saturated
very littile BaSO 4
dissolves only 0.0025 gm of BaSO 4
dissolves in 1.0 liter of water, no matter how much more
BaSO 4
is added. Hence BaSO 4
is known as slightly soluble
compound.
BaSO 4
(s) + H 2
O BaSO 4
(aq) Ba +2 + SO –2
(aq) 4 (aq)
Minimize
BaSO 4
(s) + H 2
O Ba +2 + SO –2
(aq) 4 (aq)
K =
[Ba +2 ] eq
[SO 4
–2
] eq
BaSO 4
(s)
100 % (generally)
K × BaSO 4
(s) = [Ba +2 ] eq
[SO 4
–2
] eq
eq denotes equilibrium
Ksp = [Ba +2 –2
] eq
[SO 4
] eq
In equilibrium that involve slightly soluble
compound in water, the equilibrium constant is called a
solubility product constant (Ksp). The activity of solid BaSO 4
is one. Hence the concentration of solid is not included in
the equilibrium constant expression.
Other Example :
CaF 2
Ca +2 (aq) + 2F – (aq)
Ksp = [Ca +2 ] [F – ] 2 = 4S 3
Bi 2
S 3
(s) 2Bi +3 + (aq) 3S–2 (aq)
Ksp = [Bi +3 ] 2 [S –2 ] 3 = 108 S 5
MyXz(s) yM +2 + (aq) Zx–y (aq)
Ksp = [M +2 ] y [ X –y ] z = (Sy) y x (Sz) z
MgNH 4
PO 4
(s) Mg +2 + NH +
+ PO –3
(aq) 4 (aq) 4 (aq)
Ksp = [Mg +2 –3
] [NH 4+
] [PO 4
] = S 3
Molar Solubility :- Molar solubility of a compound is the
number of mole that dessolve to give one litre of saturated
solution. It is generally respresented by ‘S’.
∗ We frequently use statements such as “ The solution
contains 0.0025 gm of dissolved BaSO 4
”. Which means
0.0025 gm of solid BaSO 4
dissolves to give a solution that

→
–3
contains equal number of Ba +2 –2
and SO 4
ions.
Question : 1.0 liter of saturated BaSO 4
solution contains
0.0025 gm of dissolved BaSO 4
. Calculate the solubility
product constant for BaSO 4
.
Solution :
2.5 × 10
Molar solubility (s) =
233
M
= 6.4 × 10 –9 = 1.1 × 10 –5 M
BaSO4(s) Ba +2 –2
(aq) + SO 4
(aq)
Before
dissociation
1.1 × 10 –5 M 0 0
After dissociation
at equilibrium
0 1.1 × 10 –5 M 1.1 × 10 –5 M
Ksp = (1.1 × 10 –5 ) (1.1 × 10 –5 )
Ksp = 1.2 × 10 –10
Reaction Quotient in precipitation Reation
used in ionic
Qsp
Qsp →
No necessarily
equilibrium
Q →Qc
K Qc
at equilibrium
used in
Qp
Qp
→
chemical
used in chemical equilibrium
equilibrium
Necessarily at
→
equilibrium.
Qsp also known as ion product.
If Qsp < Ksp Forward process is favored. No
precipitation occure, more solic can
dissolve.
Qsp = Ksp Solution is just saturated, reaction is at
equilibrium
Qsp > Ksp Reverse process is favored, precipitation
occures.
Question : If 100 mL of 0.00075 M Na 2
SO 4
and 50.0 mL of
0.015 M BaCl 2
solution are mixed, will a precipitate form ?
Ksp of BaSO 4
is 1.1 × 10 –10
0.015 × 50 × 2
[Ba +2 ] =
= 5 × 10
2 × 150
M
–2
[SO 4
] =
100 × 0.00075 × 2
150 × 2
= 5 × 10 –4 M
Qsp = 2.5 × 10 –6
Qsp > Ksp
Solid BaSO 4
precipitated untill [Ba +2 –2
] [SO 4
] just
equal to Ksp of BaSO 4
Common Ion Effect in solubility Calculation :
Example : The molar solubility of MgF 2
is 1.2 × 10 –3 M in
pure water at 25ºC. Calculate the molar solubility of MgF 2
is 0.10 M NaF.
MgF 2
(s) + H 2
O Mg +2 + 2F –
Ksp = [Mg +2 ] [F – ] 2 = S. (2S) 2 = 4S 3
= 4 × (1.2 × 10 –3 ) 3
→
→
→ →→
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15
NaF Na + + F –
100% 0.1
H 2
O + MgF 2
(s) Mg +2 + (aq) 2F– (aq)
–S +S +2S
Ksp = (S) (2S + 0.1) 2 , (2S + 0.1) = 0.1
6.4 × 10 –9 = (0.1) 2 × S
Neglected
S = 6.4 × 10 –9 × 10 2
= 6.4 × 10 –7 M
Hence solubility of MgF 2
is decreased in NaF in
comparision to pure water due to common ion (F – ) which is
provided by NaF.
Relative Solubilities
A salt’s Ksp value gives as information about its
solubility. They are two possible cases :
(1) The salts being compared produce the same number
of ions. For Example,
Consider :
AgI (s) Ksp = 1.5 × 10 –16
CuI (s) Ksp = 5.0 × 10 –12
CaSO 4
Ksp = 6.1 × 10 –5
Ksp = S 2
S = √ Ksp = Solubility
In this case we compare the solubilities for these
solid by comparing the Ksp values :
CaSO 4
(s) > CuI(s) > AgI (s)
Most soluble
largest Ksp
(2) The salts being compared produce different
numbers of ions. For Example,
Consider :
CuS (s) Ksp = 8.5 × 10 –45
Ag 2
S (s) Ksp = 1.6 × 10 –49
Bi 2
S 3
(s) Ksp = 1.1 × 10 –73
These salt prodeces different numbers of ions when
they dissolve. The Ksp values connot be compared directly
to determine relative solubility.
Relative solubility measured by above mentioned
Rule
* Remember that relative solubility can be predicted
by comparing Ksp value only for salts that produce the same
total numbre of ions.
For CuS For Bi 2
S 3
S 2 = Ksp
108 S 5 = Ksp
S = √ Ksp S =
For : Ag 2
S
4S 3 = Ksp
S =
Ksp
4
1
/ 3
→
Least soluble
smallest Ksp
Ksp
108
Due to common
ion reaction
shifted towards
left
1
/ 5

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