Electrochemistry & Ionic equilibrium

16 Pages·2012·3.86 MB·English

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Knowledge of thermodynamics, reaction quotient and § Rule (3)
m
redox reaction are very important for solving the problem
If both half cell reactions are spontaneous then we
of electrochemistry o
c convert less spontaneous reaction into non-
RULE TO SOLVE PROBLEM OF ELECTROCHEMISTRY . spontaneous reaction.
s
s A AA+2 + 2e D G = –0.5 kcal (more spontaneous)
§ Rule (1 ) A AA+2 + 2e D G = –0.2kcal ....(i) re B B+2 + 2e D G12 = –0.2 kcal (less spontaneous)
B B+2 + 2e D G 1= +0.4 kcal ....(ii) p Same : A AA+2 + 2e D G = –0.5 kcal
2 d 1
According to the rule of spontaneity, reaction (i) is r Change : B+2 + 2e B D G2 = +0.2 kcal
o
spontaneous but reaction (ii) is non-spontaneous. To w A + B+2 B + AA+2 D Grxn
gain full cell reaction, we convert non-spontaneous
y. D G rxn = – 0.5 kcal + 0.2 kcal = – 0.3 kcal
reaction into spontaneous reaction.
r
t
SCahmaneg :e : BA+ 2 + 2e AA+2 + 2 Be;; DD GG1 == ––00..24k kccaall mis § Rule (4 A) AA+2 + 2e D Gº1 = –0.2 kcal ...(i)
2
A + B+2 AA+2 + B D Grxn he B B+2 + 2e D Gº2 = +0.5 kcal...(ii)
D Grxn = D G + D G = (–0.2kcal) + (–0.4 kcal) c If D Gº given in the half cells then spontaneity of the
1 2
= – 0.6 kcal r reaction not determined by D Gº. We convert D Gº into
k
To Calculate Emf of reaction a D G by following equation.
r
D G = – nFE (cid:240) E = –D G/nF d D G = D Gº + RT ln Q
n If Q = 1 , then D G = D Gº
§ Rule (2) e
l D G = D Gº + RT ln [A+2] .......(i)
If both half cell reactions are non-spontaneous, then ai 1 1
we convert more non-spontaneous reaction into h D G = D Gº + RT ln [B+2] .......(ii)
spontaneous reaction to gain full cell reaction. s 2 2
AB AAB++22 ++ 22ee DD GG12 == ++00..25 kkccaall ........((ii)i) ww. §Itfh eRem uspfl eoo fn( 5ttha)en ehiatylf ocef lrle gaivcetino nin b thy ee pmrof.blem then we predict
Reaction (i) is less non-spontaneous and reaction (ii) is w D G = – nFE
more non-spontaneous. : * Spontaneity from E
t
si
Change : B+2 + 2e B D G = –0.5 kcal vi 4
2
,
Same : A AA+2 + 2e D G = +0.2 kcal os 3 Spontaneoity
1 de 2 increases
B+2 + A B + AA+2 D Grxn vi 1

d
D Grxn = D G + D G = (–0.5) + (0.2) kcal = –0.3 kcal n 0 E = 0 (Rxn is at equilibrium)
1 2 a
* Spontaneity from D G s –1
k –2 Non - spontaneoity
o
increases
o –3
4 b
- –4
3 e
Non-spontaneoity
, (cid:240) Greater the reduction potential greater will be
2 increases s
n ease of reduction. Greater the oxidation potential greater
1 o
0 D G = 0 (Rxn is at equilibrium) sti will be ease of oxidation.
e
–1 u § Rule (6)
q
–2 Spontaneoity y, A AA+2 + 2e Eo.p= –0.2 Volt (non-spontaneous)
–3 increases or B B+2 + 2e Eo.p = +0.5 Volt (Spontaneous)
e
–4 h
t Same : B B+2 + 2e EO.P.. = +0.5 Volt
(cid:240) Greater the positive value (+ve) of free energy
e
greater will be non-spontaneity. e Change: A+2 + 2e A ER.P..= +0.2 Volt
r
(cid:240) Greater the negative value (–ve) greater will be f
spontaneity or B + A+2 B+2 +A Ecell
f
Ecell = EO.P + ER.P = 0.5 + 0.2 = 0.7 Volt
2
§ Rule (7) To calculate Ecell we convert all the half cell reactions
m
are in same type (either oxidation half or reduction half)
A AA+2 + 2e EO.P.= +0.5 Volt (more spontaneous) o
(i) Convert both reaction into oxidation half
B B+2 + 2e EO.P.. = +0.3 Volt (Less spontaneous) c
. A AA+2 + 2e EO.P.= 0.2 Volt
s
Same : A AA+2 + 2e EO.P. = +0.5 Volt s B B+2 + 2e EO.P.. = –0.5 Volt
Change: B+2 + 2e B ER.P.= –0.3 Volt e

r (ii) Convert both reaction into reduction half cell
A + B+2 B + AA+2 Ecell p A+2 + 2e A ER.P..= –0.2 Volt
d
Ecell = EO.P. + ER.P. = +0.5 – 0.3 Volt = +0.2 Volt r B+2 + 2e B ER.P.. = 0.5 Volt
o
§ Rule (8) w Same : B+2 + 2e B ER.P.= 0.5 Volt

. Change : A AA+2 + 2e EO.P.= +0.2 Volt
A AA+2 + 2e EO.P.= – 0.5 Volt (more spontaneous) y
B B+2 + 2e EO.P.. = – 0.2 Volt (Less spontaneous) tr B+2 + A AA+2 + B Ecell = 0.7 Volt
s
Same : B B+2 + 2e EO.P.= –0.2 Volt mi § Rule (11) :
Change: A+2 + 2e AA ER.P. = +0.5 Volt e Fquoor tifeonlltosw (mingix ttuyrpee o fh Qalpf acnedll Qreca) cutsioend ihny pblaricde roefa Qctcio onr
h
A + B+2 B + AA+2 Ecell Qp
c
Ecell = EO.P. + ER.P. = –0.2 + 0.5 Volt = +0.3 Volt kr « H2 2H+ + 2e (no. of electron used = 2)
a [H+]2mol/litre
§ Rule (9) r Q =
d Hybrid (P )atm
A AA+2 + 2e EºO.P.= +0.5 Volt.....(i) n H2
B B+2 + 2e EºO.P.. = –0.2 Volt.....(ii) e « 1/2H2 H+ + 1e (no. of electron used = 1)
l
Spontaneity of reaction not determined by standard emf ai Q = [H+]
it is determined by following equation. h Hybrid (P )1/2
s H2
E =EO - (0.0592n)log[A+ 2] .
O.P. O.P. w « Cl + 2e 2Cl– (no. of electron used = 2)
E =EO - (0.0592n)log[B+ 2] w 2
O.P. O.P.
(cid:240) Greater the O.P. greater will be ease of oxidation. w Q =[Cl- ]2
Hybrid (P )
HIT AND TRIAL METHOD sit: Cl2
vi « Cl– Cl + 1e (no. of electron used = 1)
We determined spontaneity of reaction by Eº value, , 2
s
however it is not correct. These are two possible result o (P )1/2
e Q = Cl2
one is correct and other is incorrect. If Emf of the cell is vid Hybrid [Cl- ]
+ve then our result is correct, but Emf of the cell is –ve,

d § Rule(12) :
then our result is incorrect, to gain correct result we n
convert oxidation half cell into reduction half cell and a Emf is mass or mole independent property (intensive
s
reduction half cell into oxidation half cell. k property), but free energy is mass or mole dependent
o
Rule for converting oxidation half cell into o property (Extensive property)
b
reduction half cell and vice-versa e- H2 2H+ + 2e; EO.P.. = x volt, D G = – 2Fx
* A AA+2 + 2e EO.P. = –0.2 Volt ns, 1/2H2 H+ + 1e; EO.P.. = x volt, D G = – Fx
If you want to gain reduction half cell then o Emf of the reaction not depends on stoichiometry coeff.
A+2 + 2e A ER.P..= +0.2 Volt sti Example : 2X+ 3y 5 AA Ecell = x volt
ue X+ y AA Ecell = x volt
* B B+2 + 2e EO.P.. = +0.5 Volt q § Rule (13) :

If you want to gain reduction half cell then y, If reaction is spontaneous in forward direction then non-
r
o spontaneous in reverse direction and vice-versa.
B+2 + 2e B ER.P.. = –0.5 Volt he Magnitude of Emf and D G are same but sign is reverse.
t
§ Rule (10) : e A AA+2 + 2e; EO.P. = X volt , D G = – 2FX
A AA+2 + 2e EO.P. = 0.2 Volt re A+2 + 2e AA D G = +2 FX, D G = – nFE
B+2 + 2e B ER.P.. = 0.5 Volt or f D G = – nFER.P.
Ecell = ? f ER.P. = D G/–nF=2FX / –2F = –X volt
3
§ Rule (14) : This electrochemical cell represented as follows.
m
If two half cell (no. of electron same or different)
produces full cell reaction then for simplicity we add o A|A+2 || B+2 | B
directly emf of half cell not D G. Addition of D G gives c 142[C431] 1[C42]243
.
same result but process is lengthy. s L.H.S. R.H.S.
s
A AA+2 + 2e E Note :
1 e
B+2 + 2e B E r (a) Oxidation half cell always represented in L.H.S
2
p
and reduction half cell always represented in
B+2 + A AA+2 + B Ecell = E + E d
1 2 R.H.S.
If we proceed this problem by D G then. or
A AA+2 + 2e D G = –2FE w (b) Single vertical line placed in L.H.S denote
B+2 + 2e B D G1 = –2FE1 . anode and R.H.S denote cathode
2 2 y
(c) Cathode is known as +ve terminal but anode is
B+2 + A AA+2 + B D Grxn = D G1 + D G2 str known as –ve terminal in electrolytic cell.
–2FEcell = –2FE1 – 2FE2 mi (d) C denotes concentration of A+2 and C denotes
1 2
Ecell = E + E concentration of B+2.
1 2 e
(cid:240) If no. of electron involved and apparent in h (e) Reaction quotient of this representation is:
reaction then reaction is half cell. But if no. of electron c
involved but not apparent in the reaction then reaction kr Q= [A+2][B]
is full cell reaction. a [A][B+2]
Example: r
d (f) Flow of electron takes place from left to right
« Fe Fe+3 + 3e oxidation half cell, n=3 n
(anode to cathode), but flow of current takes
« Fe+3 + 1e Fe+2 reduction half cell, n= 1 e
place from right to left (cathode to anode)
« 2Fe+3 + 3I– 2Fe+2 +I – full cell rxn, n=2 il
« MnO4– + 4H+ +3e 23H2O+MnO2 ha (g) No. of electron involved = 2
reduction half cell, n = 3 s (h) Ecell for this representation calculated as follows
« Fe+2 Fe+3 + 1e oxidation half cell, n=1
(cid:240) If two half cell reaction having different no. of w. E =E0 - 0.0592logQ
electron provide full cell reaction then we also add Emf w cell cell 2
for simplicity. w CONCEPT OF SPONTANEITY
Example :
2× (A AA+3 +3e) E , D G = –6FE sit: For the complete study of reaction knowledge of
3× ( B+2 + 2e B) E1, D G1 = –6FE1 vi thermodynamics and chemical kinetics are essential.
2 2 2 s, Thermodynamics tell us about spontaneity of
2A + 3B+2 2A+3 +3B, D Grxn = –6FE eo reaction (Reaction is spontaneous or not). Chemical
3 d
kinetics tells us about the rate of reaction (Reaction is
D Grxn = D G + D G = –6FE – 6FE = –6FE vi
1 2 1 2 3 d fast or slow). Spontaneity of the reaction is determined
E = E + E n by thermodynamic parameter like D G, D S (universe) and
§ Rule (15) : 3 1 2 a value of emf (For oxidation, Reduction or Redox
s
If two half cell produces third half cell then we should k Reaction)
o
not added Emf directly. To calculate emf of third half o Entropy Change (D S)
b
cell we add D G, -
e Entropy is thermodynamic state quantity that is measure
BA + 2 + 2e AA +3 + 3e B ; ; EE1, DD GG1 == ––23FFEE1 ns, of randomness or disorder of the molecule of the
2, 2 2
o system.
A + B+2 AA+3 + B + 1e E3 ¹ E1 + E2 sti Explanation of entropy on the basis of probability
D G = D G + D G (cid:240) –1FE = –3FE + (–2FE ) ue
3 1 2 3 1 2 q left right
E3 = 3E1 + 2E2 y, bulb bulb
r
§ Rule (16) : o
e no. of molecule relative prob. of finding all
For writing shorthand notation or cell representation of h
t molecules in left bulb
electrochemical cell following conventions are used.
e
e A (1 molecule) PA = ½ = (½)1
A AA+2 + 2e EO.P = 0.5 volt fr A,B (2 molecules) PAB = PA×PB = (½)2
B+2 + 2e B ER.P = 0.2 volt or A,B,C....(n molecules) PABC...... = (½)n
A + B+2 B + AA+2 Ecell = 0.7 volt f 6.023 × 1023 (½)6.023× 1023 = 2–6.023× 1023
4
PA = Probability of finding molecule A Condition (2) melting of ice at 0ºC (at M.P.)
PB = Probability of finding molecule B m Temp = 273 K; D H = 6.03 × 103
PA ×PB = Probability of finding molecule A and B at o D G = D H – TD S = 6.03 × 103 – 263 × 22.1 = 0
the same time c
. D G =0 (process is at equilibrium)
In probability +× ddeennootteess ‘O‘ANRD’’ ss D Suniverse = D Ssystem + D Ssurro
e = 22.1 – (6.03 × 103 /273) = 0
From above example, it is clear that as no of molecules
r
increases probability (chance) of finding of molecule at p Condition (2) melting of ice at 10ºC (above M.P.)
d
the one place decrease. Temp = 283 K; D H = 6.03 × 103
r
According to 2nd law of thermodynamics entropy of o D G = D H – TD S = 6.03 × 103 – 263 × 22.1
universe always increase in spontaneous process. w
Surrounding : The rest part of the universe other than . = –2.2 × 102 Joule
y
the system is called surrounding. r Process is spontaneous.
t
s
i Q : By which statement we can say confirmly process
m
System is spontaneous.
e
h (a) D S system is +ve (f) D S universe is –ve
c (b) D S system is –ve (g) D G is +ve
r
Example : k (c) D S surro is +ve (h) D G is –ve
a
Melting of ice at 10ºC. D S universe = +ve r (d) D S surro is –ve (i) EMF is +ve
Melting of ice at 0ºC D S universe = 0 d
Melting of ice at –10ºC D S universe = –ve n (e) D S universe is +ve (j) EMF is –ve
e Ans: (e), (h) and (i)
l
i
For spontaneous process : a
(+vDe Sosr,y s–tevme) + D(+Svseu roror u–nvdein)g = D S(aulnwivaerysse +ve) sh ** RReellaattiioonn ooff DD GG wwiitthh DemSufn iivse rDseG i s= D –Gn F=E –TD Suniverse

D G = D H – TD S .................. (1) w.
D G = Free energy, D H = Heat enthalpy w § Conditions for Spontaneity
D S = Change in entropy w condition D G D Suniverse EMF
Eq. (1) is multiplied by –(1/T)

t: Spontaneous
D G D H TD S D H si – ve + ve + ve
- =- - =sy-s +D S vi reaction
T T - T T sys ,
s
o Non-spontaneous
D G (cid:230) D H (cid:246) e +ve –ve – ve
- =D S +D S=D S - (cid:231) Q=D S (cid:247) d reaction
T surro sys universeŁ T surroł vi

d
(cid:222) D =G- D T S V.V.I n Reaction is at 0 0 0
universe a equilibrium
s
k
« Consider The melting of Ice o
o
H O (s) + heat fi H O (l) b
2 2 - THERMODYNAMIC EQUILIBRIA
D H = 6.03 × 103 J/mol e

D S system = 22.1 J/K mol s, (a) On the basis of free energy
Condition (1) melting of ice at –10ºC (Below M.P.) on D G = D Gº + RT lnQ (Q may be Qp and Qc)
Temp = 263 K; D H = 6.03 × 103 ti At equilibrium condition Q = 0, D G = 0
s
D S system = 22.1 J/K ue D Gº = –RT lnK (K may be Kp or Kc)
uD nSive srsuerr o= = D –SD sHys/ T+ =D –S6 s.u0r3ro × = 1 2023 ./1 2–6232 .=9 –= 2–20..98 JJ//KK D S ry, q lnK= -D RTG0 (cid:222) K= antiln-D (cid:230)(cid:231)Ł RTG0(cid:246)(cid:247)ł
o
D S universe is –ve hence process is not spontaneous e
DD GG i=s –+ vD eS huneivnecrsee p× rTo c=e –s s(– i s0 n.8o) n×– 2s6p3o n= t2a.n1e ×o u10s.2 J e th (cid:222) K= e(-D G0/RT)
e
r Equilibrium constant (Kp or Kc) can be
we can calculate D G from eq. D G = D H – TD S also, r f calculated by above equation.
D G = 6.03 × 103 – 263 × 22.1 = 2.1 × 102 J o
f
5
(b) On the basis of entropy If at any time Q < K, the forward reaction must occur to
m
D S reaction = D Sº reaction – RlnQ a greater extent than the reverse reaction for equilibrium
o to be established. This is because numerator of Q is
At equilibrium Q = K, D S reaction = 0 c too small and the denominator is too large.
D Sº reaction = R lnK s. Reducing the denominator and increasing the
s
numerator requires forward reaction until equilibrium is
-D S0 -D (cid:230) S0 (cid:246) e
lnK= reaction (cid:222) K= antiln(cid:231)(cid:231) reaction(cid:247)(cid:247) r established.
R Ł R ł p If Q > K, the reverse reaction must occur to
d
greater extent than the forward reaction for equilibrium
r
(cid:222) K= e(DSr0eaction/R) o to be reached. When Q = K, the system is at equilibrium,
w so no further net reaction occurs.
Equilibrium constant (Kp or Kc) can be
. Conclusion:
y
calculated by above equation. r Q < K (cid:240) Forward reaction predominates
(c) On the basis of electromotive force for st until equilibrium is established.
Oxidation, Reduction and Redox reaction. i Q = K (cid:240) System is at equilibrium.
m
D G = D Gº + RT ln Q ........(1) Q > K (cid:240) Reverse reaction predominates
also, D G = – nFE and, D Gº = – nFEº e until equilibrium is established.
h
eq. (1) can be also written as
c
–nFE = –nFEº + RT ln Q........(2) Prob. 1 For the synthesis of ammonia at 500ºC, the
r
eq. (2) is divided by nF k equilibrium constant is 6.0 × 10–2 L2/mol2. Predict the
a direction in which the system will shift to reach
r
(cid:240) E=E0 - RTlnQ d equilibrium in each of following case. Reaction for
nF n
e synthesis of NH3 is N2(g)+3H2(g)(cid:135)(cid:136)(cid:136)(cid:136)(cid:136)(cid:134)(cid:136)(cid:136)2NH3(g)
At 25ºC this equation changes into l
i (a) [NH ] = 1 × 10–3 M, [N ] = 1 × 10–5 M
a 30 20
0.0592 h [H ] = 2 × 10–3 M
E=E0 - n lnQ s (b) [NH20] = 2 × 10–4 M, [N ] = 1.5 × 10–5 M
30 20
[H ] = 3.54 × 10–1 M
.
At equilibrium, Q = K and E = 0. w 20
w Ans: (a) Q = 1.2 × 107 L2/mol2
REACTION QUOTIENT C
w Q > K Hence back reaction predominates.
C C
When reactants and products of a given t: (b) Q = 6.0 × 10–2 L2/mol2
cthhee mmiixctaul rree iasc atito enq aurielib mriuixmed, a itn ids iuf sneoftu, li nto w khniochw dwirheectthioenr visi QC = KCCReaction is at equilibrium.
,
s
the system will shift to reach equilibrium. o
e
If the concentration of one of the reactants or d Your Problem in chemistry
product is zero, the system will shift in the direction that vi

produces the missing component. However, if all the d
n
initial concentrations are non zero, it is more difficult to a Now a days, mostly

determine the direction of the move toward equilibrium. s students suffer trouble and
k
To determine the shift in such case, we use the reaction o fear in vital topic like Acid
o
quotient (Q). The reaction quotient is obtained by b Base Titration, Indicator,
applying the law of mass action, using initial e- Double Indicator, Redox

concentrations instead of equilibrium concentration. s, Titration, Ionic Equilibria,
Reaction quotient in the term of pressure known n Ele ctro chem istr y,
o
as Qp and reaction quotient in the term of concentration sti Thermodynamics, Chemical
known as Q . e Equilibria and Chemical Kinetics. Our main
C u
N2(g)+3H2(g)(cid:135)(cid:136)(cid:136)(cid:136)(cid:136)(cid:134)(cid:136)(cid:136)2NH3(g) y, q macoctutroa ttoe raenldat eea aslyl tcoopnicce wpitt.h Phriogbhllye ma davsaknecde din,
r
Q = [NH3]2 andQ = (PNH3)2 heo E(Mxaaimns )l,i kAeI ECE.EB a.Snd. EI.I T( Maraei nnso)t, cBha.Cpt.eEr. Cw.iEse.
c [N ][H ]3 P (P )(P )3 t
2 2 N2 H2 e problem but concept based problem.
Qc and Qp is defined at any concentration and at any e
r
pressure respectively. f
r
o -Shailendra Kumar
f
6
New Concepts of m 1. Concept of Milliequivalent.
o 2. Auto ionisation of pure water.
c 3. PH Calculation of strong acid
s. and strong base.
Ionic Equilibria s 4. PH Calculation of weak acid and
e weak base.
r
p 5. PH Calculation of weak acid and
d weak base in the present of its salt.
r 6. Common Ion effect.
o
7. Buffer Solution.
MAIN OBJECTIVES w
8. Salt hydrolysis.
.
y
9. Solubility Product Principle
r
t
s
i
1. CONCEPT OF MILLIEQIVALENT. m HSO 2
2 4
AlCl 3
e 3
No of equivalent = wt h Al(SO) 6
eq.wt 2 4 3
c COOH
wt · 1000 r 2HO 2
No of milliequivalent = k COOH 2
eq.wt
a KSO Al (SO) 24HO 8
wt · 1000 No of meq r 2 4 2 4 3 2
N = eq.wt · V(ml ) = V(ml) d
2 When 4.9 gm of HSO mixed with 4 gm of
n
2 4
No of meq = N · V(ml) e NaOH. What is the
l (a) Wt of NaOH reacted.
No of mol = wt , No of milli mol = wt · 1000 ai (b) Wt of HSO reacted.
mol.wt mol.wt h 2 4
(c) Mol of Na SO formed.
s 2 4
No of milli mol = M · V(ml) w. M e q b e Hfo2rSO4.49 +· 1N0a0O0H4 · 1000 Na2SO4 + H2O
0 0
w reaction 49 40
CONVERSION FACTOR
w (100) (100) 100 100
MNo · o fV m.Fo (lV ·a lVen.Fc e= FNaoct oorf )e =q. N sit: Mreeaqc taifotner 0 0
vi
No of milli mol · V.F = No of meq. , (a) Wt of NaOH completely reacted in this reaction,
s
mol.wt o Hence wt of NaOH reacted is 4 gm.
No of eq = e
V.F d (b) HSO completely reacted in this reaction, hence wt
vi of2 HS4O reacted is 4.9 gm.
2X + 3Y XY (By Meq Method) d 2 4
2 3 fi n (c) No of meq of NaSO formed is 100
2 4
Changed No need of a 100
balanced equation s No of milli mol of Na2SO4 = 2
k No of milli mol of Na SO = 50
o
2 4
Milliequivalent X + Y X2Y3 bo No of mol of Na SO = 50 · 10–3
befor reaction 10 6 0 (Suppose) e- 2 4 = 5 · 10–2 mol

,
Milliequivalent s Dilution
after reaction 4 0 6 on means addition of solvent (generally HO)
Meq of X Reacted = 6 sti During dilution, wt of solute fi remain2s constant
Meq of Y Reacted = 6 ue Mol.wt of solute fi ”
Meq of XY formed = 6 q Mol of solute fi ”
During reaction eq2ua3l meq of reactant reacted and equal ry, Milli mol of solute fi ”
meq of product formed. eo No of millimol = M · V(ml)
Species V.F h During dilution MV = constant
t
K+ 1 e MV = MV
e 1 1 2 2
SO–2 2 r or
Al+43 3 r f NV = NV
o * 1 1 2 2
Ca+2 2 f During dilution millimol of solute remains
CaSO 2 constant.
4 7
2 Calculate wt of AgCl formed when 200 ml of 5 N Kw = [H+] [OH–]
HCl reacted with 1.7 gm AgNO m Kw = (x + 10–2)x
Solution : Write unbalanced reaction 3 o 1· 10–14 = (x + 10–2)x
AgNO + HCl fi AgCl + HNO c After solving quardatic equation value of H+ pro-
Milliequivalent 1.7 · 10030 200 · 5 0 0 3 s. vided by H2O is very less in comparison to H+ provided by
befor reaction 170 s HCl, Hence H+ provided by HO is neglected.
(10) (1000) re Hence [H+] = 10–2 2
Mafitlelire rqeuaicvtailoennt 0 990 10 10 dp * In pure wPatHe r= m 2olarity of [H+] is 10–7 M but in acidic
r
Meq of AgCl formed = 10 o medium value of H+ provided by HO is less than 10–6 M due
2
wt · 1000 w to common ion effect, But in rough calculation we also take
=10
143.5 . molarity of [H+] is 10–6 M in acidic solution.
y
Wt = 1.435 gm
r
t *
s If [H+] produced by acid is less than 10–6 M or equal
2. AUTO IONISATION OF PURE WATER i to 10–6 M then [H+] produced by HO is not neglected.
m
2
e
(a) HO H+ + OH– *
2 h If [H+] produced by acid is greater than 10–6 M then
or
c [H+] produced by HO is neglected.
HO + HO HO¯ + OH r 2
2 2 3 k
[H+] [OH–] a * [H+] [OH–] = Kw = 10–14 (At 25º C)
K ionization =
[HO] r
2 d taking log both side.
n log [H+] [OH–] = Kw
K ionization [HO] = [H+] [OH–]
Kw = 2 [H+] [OH] e PH + POH = PKw = 14
l
1· 10–14 = [H+] [OH–] ai
fi h
Value of Kw = 1· 10–14 at 25º C s Strong acid (like HCl, HNO, HSO etc) and strong
3 2 4
rIoeancistiaotnio. nH oern Dcei sisf otceimatpio ins ionfc Hre2aOs eids Evanlduoet hofe rKmwic w. bdaissseo c(liiaktee dN.aOH, Ca(OH)2 , Mg(OH)2 etc) are completely
w
also increase. Meq of S.A = Meq of H+
w
Value of Kw at 25º C is 1· 10–14 . Meq of S.B = Meq of OH–

Kw is known as ionic product of water. t: 2 Calculate Molarity of H+ and Normality of H+ in
si
(b) PH scale (0–14) valid for Kw (value equal to vi 1 M H2SO4 Neglected
1· 10–14 ) s, ›
(c) Q. At certain temp (temp greater than 25º C) Value deo H2SO4 fi 2 H¯ + SO4–2
of kw is 1· 10–13 What is its neutral point and what vi 100% dissociated
is its PH scale. d HSO H+ + SO–2
Solution : PH scale ( 0–13) an Meq befor D. 1·2 2 ·4 V 0 04 Let volume
Its neutral point is 6.5 s (2V) of solution
Kw = [H+] [OH–] ok Meq after D. 0 2V 2V is V(ml)
1· 10–13 = x.x o
x2 = 1· 10–13 -b Meq of H+ = 2V Normality of H+ = 2VV
e
x = 1· 10–6.5 = [H+] = [OH–] , Millimol of H+ = 2V (V.F = 1) = 2
s
Hence PH at neutral point is 6.5 n [H+] = 2 V = 2
(d) [H+] [OH–] fi solution is Neutral o V
[H+] > [OH–] fi solution is Acidic sti Molarity of H+ = Normality of H+
e
[H+] < [OH–] fi solution is Basic u *
q Molarity of H+ and Normality of H+ is equal, and

y, meq of H+ and millimol of H+ is equal because valence
3. PH CALCULATION OF STRONG ACID AND r
o factor is 1.
STRONG BASE e
*
Actual Method for PH Calculation. th Molarity of OH– = Normality of OH– (V.F =1)
Calculate PH for 1· 10–2 M HCl e Meq of OH– = Millimol of OH– (V.F =1)
HO H+ + OH– re * Meq of strong acid = Meq of H ¯ =Millimol of H¯
2 x r f N· V(ml) N· V(ml) M· V(ml)
o
HCl H+ + Cl– f Normality of strong acid=Normality of H+ =Molarity of H+
Common ion effect 10–2 8 Conclusion : If we want to calculate PH of strong acid then
we convert molarity acid into normality of acid, which is di- 2 Calculate the [Cl– ], [Na+], [H+], [OH–] and PH of
rectly equal to molarity of H+ m resulting solution obtained by mixing 50 ml of 0.6 N HCl
2 Calculate PH of 1· 10–2 M HSO . o and 50 ml of 0.3 NaOH.
2 4
Molarity of HSO = 1 · 10–2 M c HCl + NaOH fi NaCl + HO
Normality of H2 SO4 = 2 · 10–2 N s. Meq befor 30 15 0 02
Normality of H2+ 4 = 2 · 10–2 s reaction
Molarity of H+ = 2 · 10–2 e Meq after 15 0 15 15
r reaction
PH = –log [H+] p Meq of HCl = 15
PH = – log (2 · 10–2) d Meq of H+ = 15
r
15
o [H+] = = .15
2 Calculate PH 100 ml of 1 · 10–2 M H2SO4 . w PH = –log 1[0H0+] = – log 0.15
Solution : PH of the solution independent on volume. It y. PH = 0.8239
depends upon normality of acid. r [OH–] [H+] = 1 · 10–14
t
MNoertmhoadli t(y1 o)f HSO = Normality of H¯ = Molarity of H+ mis [OH–] = 1 ·0 .1150– 1 4 = 6.6 · 10–14 M
2 1 4· 10–2 · 2 = Molarity of [H+] Meq of Cl– = meq of Cl– provided by NaCl + meq of Cl–
e
2 · 10–2 = Molarity of H+ h provided by HCl.
PH = 2 – log2 c = 15 + 15
Method (2) r
k Meq of Cl– = 30
Meq of H2SO4 = Millimol of H+ a Millimol of [Cl–] = 30 = 0.3 M
100 · 10–2 · 2 = 100 · [H+] r 100
[H+] = 2 · 10–2 d Meq of Na+ = meq of Na¯ provided by NaCl
n
PH = 2– log2 e = meq of NaCl
l Millimol Na+ = 15
2 Calculate PH of 0.5 · 10–2 M Ca (OH) . ai [Na+] = 1 5 = 0.15 M
2 h 100
Normality of Ca (OH) = Molarity of (OH–)
2 s
0.5 · 10–2 · 2 = [OH–] . 4. PH CALCULATION OF WEAK BASE AND
1 · 10–2 = [OH–] w WEAK ACID.
POH = 2, PH = 12 w
w BOH B+ + OH–
2 Calculate PH of the solution when 100 ml of 1 · : weak base
10–3 M HCl, 100 ml of 1 · 10–4 M HNO and 100 ml of 1 · sit Before C 0 0
3 vi dissociation
10–2 N HSO mixed. ,
Solution 2: 4 os After C – C a Ca Ca
e dissociation
(cid:229) Meq of H + = b my eHqN oOf H ++ pmroevqi doef dH b+y p HroCvli d+e md ebqy HH+pSrOovided vid Kb = [B[+B] O[OHH] – ] = CC a(1 .– C aa )
3 2 4 d fi
(cid:229) Meq of H+ = meq of HCl + meq of HNO + meq of HSO n
100 · 10–3 + 100 · 10–43 + 100 · 102–2 4 a Dissociation constant of weak base
e millimol of H+ =0.1 + 0.01 + 1 ks Kb = Ca 2 (If a is very less)
300 · M = 1.11 oo (1–a ) » 1
b
PH is deteMrm =ine13d.01 0b1y equation s, e- a = KCb
on [OH–] = Ca = C K b = (cid:214) Kbc
PH = – log 1.11 ti C
300 s
e
u [OH–] = (Kbc) ½
2 Calculate PH of 102 M HCl. q POH = – log[OH–]

PH = –2 But this is not true, practically PH of this y, = – log(Kbc) ½
r
solution is near to zero. o
e
2 Calculate PH of 1M HCl. h (1) POH = ½ (PKb – log c) fi V.V.I
t
PH = 0 this solution is most acidic. Greater the ee (2) [OH–] =(cid:214) Kbc = Ca
[H+] greater will be acidity and lesser will be PH. r fr (3) a =(cid:214) KCb
o
Note : If PH of more concentrated acid is less than 0 (–ve) f fi
then PH of this concentrated acid is taken as zero. For weak base
9
PH = ½ (PKa – log c) NH4Cl fi NH4 ¯ +Cl–
[H+] = Ca = (cid:214) KaC m
fi 100%
For weak acid
a =(cid:214) Ka o B.D C1 O O
C c A.D O C C
1 1
s. NHOH NH+ +OH–
2 Calculate PH of s 4 4
(a) 0.002 N acitic acid having 2.3% dissociation. e B.D C O O
(b) 0.002 N NHOH having 2.3% dissociation. r 2
(a) [H+] = 2 · 10–3 · 2.43 = 4.6 · 10–5 M dp A.D C2 – C2a C2 a C2a
100 fi
r
PH = 5 – log 4.6 = 4.3372 o Dissociation of NHOH is
(b) [OH–] = Ca = 2 · 10–2 · 120.30 w depressed due to N4H4Cl
= 4.6 · 10–5 M y. [NH+][OH–] [C a + C][Ca ]
POH = 4.3372 tr Kb = [NH4 OH] = 2[C – C1 a ]2
PH = 9.6627 s 4 2 2
i
m [C a + C][OH–]
5. PH CALCULATION OF WEAK ACID AND e Kb = 2[C – C1 a ]
WEAK BASE IN THE PRESENCE OF ITS SALT. h 2 2
c [C– Ca ]
Example 1: If we want to calculate PH of CH – C – OH kr [OH–] = Kb [C2 + C2a ]
3 = 1 2
(C) in the presence of CH COONa (C) O a [C+ Ca ]
2 CH3 – COONa fi3 CH3CO1O–+ Na¯ ndr POH = PKb + log [C21 – C22a ]
[C]
B.D C 100% O O e POH = PKb + log [C1]
A.D O1 C C ail 2
CHCOOH CH–1 C–O–+ H1¯ h
B.D 3 C O3 =O O w. s POH = PKb + log [[bsaaslet]]fi nToht i1s0 e0q%ua ctioornre icst
2
A.D C2 – C2a C2a C2a w Method for PH Calculation of Following Reaction
w
fi
Due to common ion : Condition 1.
t
O dissociatidoenp orfe sCsHed3COOH is , visi Meq.B.RN 1a0OH + C H53–=OC–H C H 03–=CO–N a + H 20O
s
[CH–=C–O–][H+] [C+Ca ] [Ca ] eo (Suppose)
Ka = [CH3 –C–OH] = 1C –2 Ca 2 d Meq.A.R 5 0 5 5
3=O 2 2 vi If strong acid or strong base present in solution
d
[C+Ca ] [H+] n after reaction then PH is calculated by strong acid or strong
Ka = [1C –2 Ca ] a base. Concentration of salt not considered.
2 2 s
k
[H+] = Ka. [C– Ca ] o Condition 2.
2 2 o
[C1 + C2a ] -b NaOH + CH3COOH CH3COONa + H2O
[C+ Ca ] e Meq.B.R 10 10 0 0
PH = PKa + log [C1 – C2a ] ns, (Suppose)
2 2 o Meq.A.R 0 0 10 10
Value of Ca is very very less, ti If solution contains only salt [CHCOONa] then
hence [C1+2 C2a ] = C1 and [C2 – C2a ] = C2 ues PH calculate by salt hydrolysis. 3
C q
PH = PKa + log C1 y, Condition 3.
2 r
o NaOH + CH COOH CH–COONa + HO
e 3 3 2
PH = PKa + log[[ascaildt]]fi nToht i1s0 e0q%ua ctoiornre icst th Meq.B.R 5 10 0 0
e
e Meq.A.R 0 5 5 5
r
Example 2 : If we want to calculate PH of NH4OH (C2) in r f If solution contains weak acid and its salt then PH
the presence of NHCl (C) o is calculated by.
4 1 f
10