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booklet of ELECTROCHEMISTRY and IONIC EQUILIBRIA

FREE CRASH COURSE

Meq. Approach

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Knowledge of thermodynamics, reaction quotient andredox reaction are very important for solving the problemof electrochemistry

RULE TO SOLVE PROBLEM OF ELECTROCHEMISTRY

§ Rule (1) A AA+2 + 2e ∆G

1 = –0.2kcal ....(i)

B B+2 + 2e ∆G2 = +0.4 kcal ....(ii)

According to the rule of spontaneity, reaction (i) isspontaneous but reaction (ii) is non-spontaneous. Togain full cell reaction, we convert non-spontaneousreaction into spontaneous reaction.

Same : A AA+2 + 2e; ∆G1 = –0.2kcal

Change : B+2 + 2e B; ∆G2 = –0.4 kcal

A + B+2 AA+2 + B ∆Grxn

∆Grxn = ∆G1 + ∆G

2 = (–0.2kcal) + (–0.4 kcal)

= – 0.6 kcal

To Calculate Emf of reaction

∆G = – nFE ð E = –∆G/nF

§ Rule (2)

If both half cell reactions are non-spontaneous, thenwe convert more non-spontaneous reaction intospontaneous reaction to gain full cell reaction.

A AA+2 + 2e ∆G1 = +0.2 kcal ....(i)

B B+2 + 2e ∆G2 = +0.5 kcal ....(ii)

Reaction (i) is less non-spontaneous and reaction (ii) ismore non-spontaneous.

Change : B+2 + 2e B ∆G2 = –0.5 kcal

Same : A AA+2 + 2e ∆G1 = +0.2 kcal

B+2 + A B + AA+2 ∆Grxn

∆Grxn = ∆G1 + ∆G

2 = (–0.5) + (0.2) kcal = –0.3 kcal

* Spontaneity from ∆G

ð Greater the positive value (+ve) of free energygreater will be non-spontaneity.

ð Greater the negative value (–ve) greater will bespontaneity

Non-spontaneoity increases

Spontaneoityincreases

4

3

2

1

0

–1

–2

–3

–4

∆G = 0 (Rxn is at equilibrium)

§ Rule (3)

If both half cell reactions are spontaneous then weconvert less spontaneous reaction into non-spontaneous reaction.A AA+2 + 2e ∆G

1 = –0.5 kcal (more spontaneous)

B B+2 + 2e ∆G2 = –0.2 kcal (less spontaneous)

Same : A AA+2 + 2e ∆G1 = –0.5 kcal

Change : B+2 + 2e B ∆G2 = +0.2 kcal

A + B+2

B + AA+2 ∆Grxn

∆G rxn = – 0.5 kcal + 0.2 kcal = – 0.3 kcal

§ Rule (4) A AA+2 + 2e ∆Gº

1 = –0.2 kcal ...(i)

B B+2 + 2e ∆Gº2 = +0.5 kcal...(ii)

If ∆Gº given in the half cells then spontaneity of thereaction not determined by ∆Gº. We convert ∆Gº into∆G by following equation.

∆G = ∆Gº + RT ln QIf Q = 1 , then ∆G = ∆Gº

∆G1 = ∆Gº

1 + RT ln [A+2] .......(i)

∆G2 = ∆Gº

2 + RT ln [B+2] .......(ii)

§ Rule (5)If emf of the half cell given in the problem then we predictthe spontaneity of reaction by emf.

∆G = – nFE* Spontaneity from E

ð Greater the reduction potential greater will beease of reduction. Greater the oxidation potential greaterwill be ease of oxidation.

§ Rule (6)

A AA+2 + 2e Eo.p = –0.2 Volt (non-spontaneous)

B 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

Spontaneoity increases

Non - spontaneoityincreases

4

3

2

1

0

–1

–2

–3

–4

E = 0 (Rxn is at equilibrium)

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§ Rule (7)

A AA+2 + 2e EO.P. = +0.5 Volt (more spontaneous)

B B+2 + 2e EO.P.. = +0.3 Volt (Less spontaneous)

Same : A AA+2 + 2e EO.P. = +0.5 Volt

Change: B+2 + 2e B ER.P. = –0.3 Volt

A + B+2 B + AA+2 Ecell

Ecell = EO.P. + ER.P. = +0.5 – 0.3 Volt = +0.2 Volt

§ Rule (8)

A AA+2 + 2e EO.P. = – 0.5 Volt (more spontaneous)

B B+2 + 2e EO.P.. = – 0.2 Volt (Less spontaneous)

Same : B B+2 + 2e EO.P. = –0.2 Volt

Change: A+2 + 2e AA ER.P. = +0.5 Volt

A + B+2 B + AA+2 Ecell

Ecell = EO.P. + ER.P. = –0.2 + 0.5 Volt = +0.3 Volt

§ Rule (9)

A AA+2 + 2e EºO.P. = +0.5 Volt.....(i)

B B+2 + 2e EºO.P.. = –0.2 Volt.....(ii)

Spontaneity of reaction not determined by standard emfit is determined by following equation.

2. . . .

2. . . .

(0.0592 )log[ ]

(0.0592 )log[ ]

O P O P

O P O P

E E n A

E E n B

Ο +

Ο +

= −

= −

ð 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 resultone 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 weconvert oxidation half cell into reduction half cell andreduction half cell into oxidation half cell.

Rule for converting oxidation half cell intoreduction half cell and vice-versa

* A AA+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 AA+2 + 2e EO.P.

= 0.2 Volt

B+2 + 2e B ER.P.. = 0.5 Volt

Ecell = ?

To calculate Ecell we convert all the half cell reactionsare in same type (either oxidation half or reduction half)

(i) Convert both reaction into oxidation halfA AA+2 + 2e EO.P.

= 0.2 Volt

B B+2 + 2e EO.P.. = –0.5 Volt

(ii) Convert both reaction into reduction half cellA+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 AA+2 + 2e EO.P. = +0.2 Volt

B+2 + A AA+2 + B Ecell = 0.7 Volt

§ Rule (11) :For following type half cell reaction hybrid reactionquotients (mixture of Qp and Qc) used in place of Qc orQp« H

2 2H+ + 2e (no. of electron used = 2)

2

2[ ] /

( )Hybrid

H

H mol litreQ

P atm

+

=

« 1/2H2 H+ + 1e (no. of electron used = 1)

2

1/2

[ ]

( )Hybrid

H

HQ

P

+

=

« Cl2 + 2e

2Cl– (no. of electron used = 2)

2

2[ ]

( )Hybrid

Cl

ClQ

P

−

=

« Cl– Cl

2 + 1e (no. of electron used = 1)

2

1/2( )

[ ]

Cl

Hybrid

PQ

Cl −=

§ Rule(12) :

Emf is mass or mole independent property (intensiveproperty), but free energy is mass or mole dependentproperty (Extensive property)

H2 2H+ + 2e; EO.P.. = x volt, ∆G = – 2Fx

1/2H2 H+ + 1e; EO.P.. = x volt, ∆G = – Fx

Emf of the reaction not depends on stoichiometry coeff.Example : 2X+ 3y 5 AA Ecell = x volt

X+ y AA Ecell = x volt§ Rule (13) :If reaction is spontaneous in forward direction then non-spontaneous in reverse direction and vice-versa.Magnitude of Emf and ∆G are same but sign is reverse.

A AA+2 + 2e; EO.P. = X volt , ∆G = – 2FXA+2 + 2e AA ∆G = +2 FX, ∆G = – nFE

∆G = – nFER.P.

ER.P. = ∆G/–nF =2FX / –2F = –X volt

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§ Rule (14) :If two half cell (no. of electron same or different)

produces full cell reaction then for simplicity we adddirectly emf of half cell not ∆G. Addition of ∆G givessame result but process is lengthy.

A AA+2 + 2e E1

B+2 + 2e B E2

B+2 + A AA+2 + B Ecell = E1 + E

2

If we proceed this problem by ∆G then.A AA+2 + 2e ∆G

1 = –2FE

1

B+2 + 2e B ∆G2 = –2FE

2

B+2 + A AA+2 + B ∆Grxn = ∆G1 + ∆G

2

–2FEcell = –2FE1 – 2FE

2

Ecell = E1 + E

2

ð If no. of electron involved and apparent inreaction then reaction is half cell. But if no. of electroninvolved but not apparent in the reaction then reactionis 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

« MnO4– + 4H+ +3e 2H

2O+MnO

2

reduction half cell, n = 3« Fe+2 Fe+3 + 1e oxidation half cell, n=1ð If two half cell reaction having different no. ofelectron provide full cell reaction then we also add Emffor simplicity.Example :

2× (A AA+3 +3e) E1, ∆G

1 = –6FE

1

3× ( B+2 + 2e B) E2, ∆G

2 = –6FE

2

2A + 3B+2 2A+3 +3B, ∆Grxn = –6FE3

∆Grxn = ∆G1 + ∆G

2 = –6FE

1– 6FE

2 = –6FE

3

E3 = E

1 + E

2

§ Rule (15) :If two half cell produces third half cell then we shouldnot added Emf directly. To calculate emf of third halfcell we add ∆G,

A AA+3 + 3e ; E1, ∆G

1 = –3FE

1

B+2 + 2e B ; E2,

∆G2 = –2FE

2

A + B+2 AA+3 + B + 1e E3 ¹ E

1 + E

2

∆G3 = ∆G

1 + ∆G

2 ð –1FE

3 = –3FE

1+ (–2FE

2)

E3 = 3E

1 + 2E

2

§ Rule (16) :For writing shorthand notation or cell representation ofelectrochemical cell following conventions are used.

A AA+2 + 2e EO.P = 0.5 voltB+2 + 2e B ER.P

= 0.2 volt

A + B+2 B + AA+2 Ecell = 0.7 volt

This electrochemical cell represented as follows.

1 2

2 2

[ ] [ ]

. . . . . .

| || |C C

L H S R H S

A A B B+ +

14243 14243

Note :

(a) Oxidation half cell always represented in L.H.Sand reduction half cell always represented inR.H.S.

(b) Single vertical line placed in L.H.S denoteanode and R.H.S denote cathode

(c) Cathode is known as +ve terminal but anode isknown as –ve terminal in electrolytic cell.

(d) C1 denotes concentration of A+2 and C

2 denotes

concentration of B+2.

(e) Reaction quotient of this representation is:

2

2

[ ][ ]

[ ][ ]

A B

A BQ

+

+=

(f) Flow of electron takes place from left to right(anode to cathode), but flow of current takesplace from right to left (cathode to anode)

(g) No. of electron involved = 2

(h) Ecell for this representation calculated as follows

0 0.0592log

2cell cell QE E −=

CONCEPT OF SPONTANEITY

For the complete study of reaction knowledge ofthermodynamics and chemical kinetics are essential.

Thermodynamics tell us about spontaneity ofreaction (Reaction is spontaneous or not). Chemicalkinetics tells us about the rate of reaction (Reaction isfast or slow). Spontaneity of the reaction is determinedby thermodynamic parameter like ∆G, ∆S (universe) andvalue of emf (For oxidation, Reduction or RedoxReaction)

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

no. of molecule relative prob. of finding all

molecules in left bulbA (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 × 1023 (½)6.023× 1023 = 2–6.023× 1023

leftbulb

rightbulb

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PA = Probability of finding molecule APB = Probability of finding molecule BPA ×PB = Probability of finding molecule A and B at

the same timeIn probability × denotes ‘AND’

+ denotes ‘OR’From above example, it is clear that as no of moleculesincreases probability (chance) of finding of molecule atthe one place decrease.According to 2nd law of thermodynamics entropy ofuniverse always increase in spontaneous process.Surrounding : The rest part of the universe other thanthe system is called surrounding.

Example :Melting of ice at 10ºC. ∆S universe = +veMelting of ice at 0ºC ∆S universe = 0Melting 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)

∆∆ ∆ ∆− = − − = − + ∆

−

sys

sys

T SG H HS

T T T T

∆ ∆ − = ∆ + ∆ = ∆ − = ∆

Qsurro sys universe surro

G HS S S S

T T

⇒ ∆ = − ∆ . .universeG T S V V I

« Consider The melting of IceH

2O (s) + heat → H

2O (l)

∆H = 6.03 × 103 J/mol

∆S system = 22.1 J/K molCondition (1) melting of ice at –10ºC (Below M.P.)Temp = 263 K; ∆H = 6.03 × 103

∆S system = 22.1 J/K

∆S surro = –∆H/T =–6.03 × 103 / 263 = – 22.9 J/K ∆ S

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 × 102 J∆G is +ve hence process is non–spontaneous.

we can calculate ∆G from eq. ∆G = ∆H – T∆S also,∆G = 6.03 × 103 – 263 × 22.1 = 2.1 × 102 J

System

Condition (2) melting of ice at 0ºC (at M.P.)Temp = 273 K; ∆H = 6.03 × 103

∆G = ∆H – T∆S = 6.03 × 103 – 263 × 22.1 = 0

∆G =0 (process is at equilibrium)∆Suniverse = ∆Ssystem + ∆Ssurro

= 22.1 – (6.03 × 103 /273) = 0

Condition (2) melting of ice at 10ºC (above M.P.)Temp = 283 K; ∆H = 6.03 × 103

∆G = ∆H – T∆S = 6.03 × 103 – 263 × 22.1

= –2.2 × 102 JouleProcess 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

THERMODYNAMIC EQUILIBRIA

(a) 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

ln lnG G

K K antiRT RT

−∆ −∆= ⇒ =

( )0 /G RTK e

−∆⇒ =

Equil ibrium constant (Kp or Kc) can becalculated by above equation.

Spontaneousreaction – ve

+ve

0

condition ∆G ∆Suniverse EMF

Non-spontaneousreaction

Reaction is atequilibrium

+ ve

–ve

0

+ ve

– ve

0

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(b) On the basis of entropy

∆S reaction = ∆Sº reaction – RlnQ

At equilibrium Q = K, ∆S reaction = 0

∆Sº reaction = R lnK

0 0

ln lnreaction reactionS SK K anti

R R

−∆ −∆= ⇒ =

0( / )reactionS RK e ∆⇒ =

Equil ibrium constant (Kp or Kc) can be

calculated by above equation.

(c) On the basis of electromotive force forOxidation, Reduction and Redox reaction.∆G = ∆Gº + RT ln Q ........(1)also, ∆G = – nFE and, ∆Gº = – nFEºeq. (1) can be also written as–nFE = –nFEº + RT ln Q........(2)eq. (2) is divided by nF

ð 0 ln

RTE E Q

nF= −

At 25ºC this equation changes into

0 0.0592

lnE E Qn

= −

At equilibrium, Q = K and E = 0.

REACTION QUOTIENT

When reactants and products of a givenchemical reaction are mixed it is useful to know whetherthe mixture is at equilibrium, and if not, in which directionthe system will shift to reach equilibrium.

If the concentration of one of the reactants orproduct is zero, the system will shift in the direction thatproduces the missing component. However, if all theinitial concentrations are non zero, it is more difficult todetermine the direction of the move toward equilibrium.To determine the shift in such case, we use the reactionquotient (Q). The reaction quotient is obtained byapplying the law of mass action, using init ialconcentrations instead of equilibrium concentration.

Reaction quotient in the term of pressure knownas Q

p and reaction quotient in the term of concentration

known as QC .

+ àààÜáààà2 2 3( ) 3 ( ) 2 ( )N g H g NH g

= = 3

2 2

223

3 32 2

( )[ ]

[ ][ ] ( )( )

NH

c P

N H

PNHQ and Q

N H P P

Qc and Qp is defined at any concentration and at anypressure respectively.

If at any time Q < K, the forward reaction must occur toa greater extent than the reverse reaction for equilibriumto be established. This is because numerator of Q istoo small and the denominator is too large.

Reducing the denominator and increasing thenumerator requires forward reaction until equilibrium isestablished.

If Q > K, the reverse reaction must occur togreater extent than the forward reaction for equilibriumto be reached. When Q = K, the system is at equilibrium,so no further net reaction occurs.Conclusion:

Q < K ð Forward reaction predominates until equilibrium is established.

Q = K ð System is at equilibrium.Q > K ð Reverse reaction predominates

until equilibrium is established.

Prob. 1 For the synthesis of ammonia at 500ºC, theequilibrium constant is 6.0 × 10–2 L2/mol2. Predict thedirection in which the system will shift to reachequilibrium in each of following case. Reaction for

synthesis of NH3 is + àààÜáààà2 2 3( ) 3 ( ) 2 ( )N g H g NH g

(a) [NH3]0 = 1 × 10–3 M, [N

2]0 = 1 × 10–5 M

[H2]0 = 2 × 10–3 M

(b) [NH3]0 = 2 × 10–4 M, [N

2]0 = 1.5 × 10–5 M

[H2]0 = 3.54 × 10–1 M

Ans: (a) QC = 1.2 × 107 L2/mol2

QC > K

C Hence back reaction predominates.

(b) QC = 6.0 × 10–2 L2/mol2

QC = K

CReaction is at equilibrium.

Your Problem in chemistry

Now a days, most lystudents suffer trouble andfear in vital topic like AcidBase Titration, Indicator,Double Indicator, RedoxTitration, 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 mainmotto to relate all topic with highly advanced,accurate and easy concept. Problem asked inExam like C.B.S.E. (Mains), B.C.E.C.E.(Mains), AIEEE and IIT are not chapter wiseproblem but concept based problem.

-Shailendra Kumar

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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

New Concepts of

1. CONCEPT OF MILLIEQIVALENT.

No of equivalent =

No of milliequivalent =

N = =

No of meq = N × V(ml)

No of mol = , No of milli mol =

No of milli mol = M × V(ml)

CONVERSION FACTOR

M × V.F (Valence Factor) = NNo of mol × V.F = No of eq.No of milli mol × V.F = No of meq.

No of eq =

2X + 3Y X2Y

3 (By Meq Method)

X + Y X2Y

3

10 6 0

4 0 6

Meq of X Reacted = 6Meq of Y Reacted = 6Meq of X

2Y

3 formed = 6

During reaction equal meq of reactant reacted and equalmeq of product formed.

Species V.F K+ 1 SO

4–2 2

Al+3 3 Ca+2 2 CaSO

4 2

wteq.wt

wt × 1000eq.wt

wt × 1000eq.wt × V(ml)

No of meq V(ml)

wtmol.wt

wt × 1000mol.wt

mol.wtV.F

→

Changed No need ofbalanced equation

(Suppose)Milliequivalentbefor reaction

Milliequivalentafter reaction

H2SO

4 2

AlCl3

3 Al

2(SO

4)

3 6

K2SO

4 Al

2(SO

4)

3 24H

2O 8

When 4.9 gm of H2SO

4 mixed with 4 gm of

NaOH. What is the(a) Wt of NaOH reacted.(b) Wt of H

2SO

4 reacted.

(c) Mol of Na2SO

4 formed.

H2SO

4 + NaOH Na

2SO

4 + H

2O

(a) Wt of NaOH completely reacted in this reaction, Hence wt of NaOH reacted is 4 gm.(b) H

2SO

4 completely reacted in this reaction, hence wt

of H2SO

4 reacted is 4.9 gm.

(c) No of meq of Na2SO

4 formed is 100

No of milli mol of Na2SO

4 =

No of milli mol of Na2SO

4 = 50

No of mol of Na2SO

4 = 50 × 10–3

= 5 × 10–2 mol

Dilutionmeans addition of solvent (generally H

2O)

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

M1V

1 = M

2V

2

orN

1V

1 = N

2V

2

∗ During dilution millimol of solute remains

constant.

COOH

COOH2H

2O 2

4.9 × 100049

4 × 100040

0 0

(100) (100) 100 100

Meq beforreaction

0 0Meq afterreaction

1002

2

Ionic Equilibria

MAIN OBJECTIVES

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2 Calculate wt of AgCl formed when 200 ml of 5 N

HCl reacted with 1.7 gm AgNO3

Solution : Write unbalanced reaction AgNO

3 + HCl → AgCl + HNO

3

Meq of AgCl formed = 10

Wt = 1.435 gm

2. AUTO IONISATION OF PURE WATER

(a) H2O H+ + OH–

orH

2O + H

2O H

3O + OH

K ionization =

K ionization [H2O] = [H+] [OH–]

Kw = [H+] [OH]1× 10–14 = [H+] [OH–]

Value of Kw = 1× 10–14 at 25º CIonisation or Dissociation of H

2O is Endothermic

reaction. Hence if temp is increased value of Kwalso increase.Value of Kw at 25º C is 1× 10–14 .Kw is known as ionic product of water.

(b) PH scale (0–14) valid for Kw (value equal to1× 10–14 )

(c) Q. At certain temp (temp greater than 25º C) Valueof kw is 1× 10–13 What is its neutral point and whatis its PH scale.

Solution : PH scale ( 0–13)Its neutral point is 6.5Kw = [H+] [OH–]1× 10–13 = x.xx2 = 1× 10–13

x = 1× 10–6.5 = [H+] = [OH–]Hence PH at neutral point is 6.5

(d) [H+] [OH–] → solution is Neutral[H+] > [OH–] → solution is Acidic[H+] < [OH–] → solution is Basic

3. PH CALCULATION OF STRONG ACID ANDSTRONG BASE

Actual Method for PH Calculation.Calculate PH for 1× 10–2 M HCl

H2O H+ + OH–

xHCl H+ + Cl–

10–2

1.7 × 1000170

200 × 5Milliequivalentbefor reaction

0 0

(10) (1000)

Milliequivalentafter reaction 0 990 10 10

wt × 1000143.5

=10

⊕

[H+] [OH–][H

2O]

→

Common ion effect

Kw = [H+] [OH–]Kw = (x + 10–2)x1× 10–14 = (x + 10–2)xAfter solving quardatic equation value of H+ pro-

vided by H2O is very less in comparison to H+ provided by

HCl, Hence H+ provided by H2O 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 H2O is less than 10–6 M due

to common ion effect, But in rough calculation we also takemolarity 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 H2O is not neglected.

∗ If [H+] produced by acid is greater than 10–6 M then

[H+] produced by H2O 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, HNO3, H

2SO

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 H2SO

4

H2SO

42 H + SO

4–2

H2SO

4H+ + SO

4–2

1× 2 × V 0 0 (2V) 0 2V 2V

Meq of H+ = 2VMillimol of H+ = 2V (V.F = 1) = 2

[H+] = = 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 valencefactor 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

⊕↑Neglected

→

100% dissociated

Meq befor D.

Meq after D.

Let volumeof solution

is V(ml)

Normality of H+ =

⊕ ⊕

2VV

2VV

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we convert molarity acid into normality of acid, which is di-rectly equal to molarity of H+

2 Calculate PH of 1× 10–2 M H2SO

4 .

Molarity of H2SO

4 = 1 × 10–2 M

Normality of H2SO

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 H2SO

4 .

Solution : PH of the solution independent on volume. Itdepends upon normality of acid.Method (1)Normality of H

2SO

4 = Normality of H = Molarity of H+

1 × 10–2 × 2 = Molarity of [H+]2 × 10–2 = Molarity of H+

PH = 2 – log2Method (2)

Meq of H2SO

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 HNO3 and 100 ml of 1 ×

10–2 N H2SO

4 mixed.

Solution :

∑ Meq of H+ = meq of H+ provided by HCl + meq H+provided

by HNO3 + meq of H+ provided by H

2SO

4

∑ Meq of H+ = meq of HCl + meq of HNO3 + meq of H

2SO

4

100 × 10–3 + 100 × 10–4 + 100 × 10–2

ε millimol of H+ =0.1 + 0.01 + 1

300 × M = 1.11

M =

PH is determined by equation

PH = – log

2 Calculate PH of 102 M HCl.

PH = –2 But this is not true, practically PH of thissolution 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.

⊕

1.11300

1.11300

2 Calculate the [Cl– ], [Na+], [H+], [OH–] and PH of

resulting solution obtained by mixing 50 ml of 0.6 N HCland 50 ml of 0.3 NaOH.

HCl + NaOH → NaCl + H2O

30 15 0 0

15 0 15 15Meq of HCl = 15Meq of H+ = 15

[H+] = = .15

PH = –log [H+] = – log 0.15PH = 0.8239

[OH–] [H+] = 1 × 10–14

[OH–] = = 6.6 × 10–14 M

Meq of Cl– = meq of Cl– provided by NaCl + meq of Cl–

provided by HCl.

= 15 + 15Meq of Cl– = 30

Millimol of [Cl–] = = 0.3 M

Meq of Na+ = meq of Na provided by NaCl

= meq of NaClMillimol Na+ = 15

[Na+] = = 0.15 M

4. PH CALCULATION OF WEAK BASE ANDWEAK ACID.

BOH B+ + OH–

weak base C 0 0

C – C α Cα Cα

Kb = =

Kb = Cα2 (If α is very less)

(1–α) ≈ 1

α =

[OH–] = Cα = C = Kbc

[OH–] = (Kbc) ½

POH = – log[OH–]= – log(Kbc) ½

(1) POH = ½ (PKb – log c)

(2) [OH–] = Kbc = Cα

(3) α =

Meq beforreaction

Meq afterreaction

15100

1 × 10–14

0.15

30100

⊕

15100

Beforedissociation

Afterdissociation

[B+] [OH–][BOH]

Cα . CαC (1– α)→

Dissociation constant of weak base

KbC

KbC

√

√KbC√

→V.V.I

→

For weak base

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10

PH = ½ (PKa – log c)[H+] = Cα = KaC

α =

2 Calculate PH of

(a) 0.002 N acitic acid having 2.3% dissociation.(b) 0.002 N NH

4OH having 2.3% dissociation.

(a) [H+] = 2 × 10–3 × = 4.6 × 10–5 M

PH = 5 – log 4.6 = 4.3372

(b) [OH–] = Cα = 2 × 10–2 ×

= 4.6 × 10–5 MPOH = 4.3372PH = 9.6627

5. PH CALCULATION OF WEAK ACID ANDWEAK BASE IN THE PRESENCE OF ITS SALT.

Example 1: If we want to calculate PH of CH3 – C – OH

(C2) in the presence of CH

3 COONa (C

1)

CH3 – COONa CH

3COO–+ Na

C1

O OO C

1 C

1

CH3COOH CH

3– C–O–+ H

C2

O O

C2 – C

2α C

2α C

2α

Ka = =

Ka =

[H+] = Ka.

PH = PKa + log

Value of C2α is very very less,

hence [C1+ C

2α] = C

1 and [C

2 – C

2α] = C

2

Example 2 : If we want to calculate PH of NH4OH (C

2) in

the presence of NH4Cl (C

1)

Ka

C√√

2.3100

2.3100

=

O

=O

→

100%

⊕

⊕

B.DA.D

B.D

A.D

Due to common iondissociation of CH

3COOH is

depressed

→

[CH3–C–O–][H+]

[CH3–C–OH]

[C1+C

2α] [C

2α]

C2 – C

2α=

O

=O

[C1+C

2α] [H+]

[C2 – C

2α]

[C2– C

2α]

[C1 + C

2α]

[C1+ C

2α]

[C2 – C

2α]

C1

C2

[salt][acid]

→This equation isnot 100% correct

For weak acid→

PH = PKa + log

PH = PKa + log

NH4Cl NH

4 +Cl–

C1

O OO C

1 C

1

NH4OH NH

4+ +OH–

C2

O O

C2 – C

2α C

2α C

2α

Kb = =

Kb =

POH = PKb + log

POH = PKb + log

Method for PH Calculation of Following Reaction

Condition 1.NaOH + CH

3–C–H CH

3–C–Na + H

2O

10 5 0 0

5 0 5 5If strong acid or strong base present in solution

after reaction then PH is calculated by strong acid or strongbase. Concentration of salt not considered.

Condition 2.NaOH + CH

3COOH CH

3COONa + H

2O

10 10 0 0

0 0 10 10If solution contains only salt [CH

3COONa] then

PH calculate by salt hydrolysis.

Condition 3.NaOH + CH

3 COOH CH

3–COONa + H

2O

5 10 0 0

0 5 5 5If solution contains weak acid and its salt then PH

is calculated by.

→

100%

⊕

B.DA.D

B.D

A.D

Dissociation of NH4OH is

depressed due to NH4Cl

→

[NH4

+][OH–][NH

4OH]

[C2α + C

1][C

2α]

[C2 – C

2α]

[C2α + C

1][OH–]

[C2 – C

2α]

[C2– C

2α]

[C1 + C

2α]

[C1+ C

2α]

[C2 – C

2α]

[C1]

[C2]

[salt][base]

This equation isnot 100% correct

→

Meq.B.R(Suppose)

Meq.A.R

Meq.B.R(Suppose)

Meq.A.R

=

O

=

O

Meq.B.R

Meq.A.R

[OH–] = Kb

POH = PKb + log

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Calculation of PH in the reaction NH4OH with HCl

Condition 1.NH

4OH + HCl NH

4Cl + H

2O

10 5 0 0 5 0 5 5 Solution contains weak base and its salt.

Condition 2.NH

4OH + HCl NH

4Cl + H

2O

5 10 0 0 0 5 5 5PH is calculated by HCl (strong acid)

Condition 3.NH

4OH + HCl NH

4Cl + H

2O

5 5 0 0 0 0 5 5PH is calculated by salt hydrolysis (NH

4Cl)

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 CH3COOH

(b) 50 ml of 0.05 M NaOH + 50 ml of 0.10 M CH3COOH

(c) 50 ml of 0.10 M NaOH + 50 ml of 0.10 M CH3COOH

(d) 50 ml of 0.10 M NH4OH + 50 ml of 0.05 M HCl

(e) 50 ml of 0.05 M NH4OH + 50 ml of 0.05 M HCl

(f) 50 ml of 0.10 M NH4OH + 50 ml of 0.10 M HCl

Solution : (a)NaOH + CH

3 COOH CH

3–COONa + H

2O

50×0.1 50×0.05 0 0

2.5 0 2.5 2.5Solution after reaction contain strong base (NaOH)

Hence PH determined by NaOHMeq of NaOH = 2.5

N of NaOH = = 2.5 × 10–2 M

[OH–] = 2.5 × 10–2

POH = 1.6021PH = 12.3979

(b) NaOH + CH3 COOH CH

3–COONa + H

2O

2.5 5 0 0

0 2.5 2.5Solution after reaction contains weak acid and its

salt, Hence PH determined by the equation

[salt][acid]

Meq.B.RMeq.A.R

[salt][base]

Meq.B.RMeq.A.R

Meq.B.RMeq.A.R

Millimol.B.R

Millimol.A.R

2.5100

Meq.B.R

Meq.A.R

[salt][acid]

PH =PKa + log

POH = PKb + log

PH = PKa + log

[CH3COONa] =

= 4.74 + log , [CH3COOH] =

PH = 4.74(c) NaOH + CH

3 COOH CH

3–COONa + H

2O

5 5 0 0

0 0 5 5If solution contains salt after reaction then PH de-

termined 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

4Cl)

6 & 7. BUFFER SOLUTION AND COMMON IONEFFECT BUFFER SOLUTION AND COMMON

Buffer solution: solution which possess reserve acidic na-ture or alkaline nature or solution which resist change in PH

due to dilution or addition of small quantity of acid or alkaliare known as buffer solution.Example of acidic Buffer

A weak acid + its salt (CH3COOH + CH

3COONa)

Basic BufferA weak base + its salt (NH

4OH + NH

4Cl)

Equation acidic buffer isPH = PKa + log

Equation for basic buffer is

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

3COOH

(c) 1litre solution 0.10 M CH3COOH and 0.10M CH

3COONa

(a) In pure water, [H+] = [OH–] = 10–7

Hence PH1 = 7

After addition of .01 mol of NaOH

[NaOH] = = 0.01

[OH–] = 1 × 10–2

POH = 2, PH2 = 12

∆ PH = PH2 – PH

1 = 12–7 = 5

(b) NaOH reacts with acetic acid.

NaOH + CH3 COOH CH

3–COONa + H

2O

0.01 0.10 0 0 0 (0.01–0.01) 0.01 0.01

0.09Solution contains weak acid and its salt, Hence

2.5/1002.5/100

Meq.B.R

Meq.A.R

→

→

[salt][acid]

[salt][acid]

0.011

Mol.B.R

Mol.A.R

PH = – log Ka + log

PH = PKa + log

[CH3 COONa]

[CH3 COOH]

2.5100

2.5100

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12

PH2 = 3.78

PH of 0.1 M CH3COOH is

PH1 1/2 (PKa – log C)

PH1 = 1/2 (4.74 + 1) = 2.87,

∆ PH = PH2 – PH

1 = 3.78 – 2.87

∆ PH = 0.91(c)

= 4.74 + log

PH1 = 4.74

PH after addition of NaOHCH

3COOH + NaOH CH

3–COONa + H

2O

0.10 0.01 0.10

(0.10–0.01) 0 (0.10 + 0.01)

∆ PH = PH2 – PH

1 = log = 0.087

Conclusion : After addition of NaOH ∆ PH is very less inbuffer 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 baseCH

3COOH + NaOH CH

3–COONa + H

2O

10 5 0 0 5 0 5 5(c) Exess basic salt + HClCH

3COONa + HCl NaCl + CH

3COOH

10 2 0 0 8 0 8 8

II. Basic Buffer(a) Mixing weak base and its salt

NH4OH + NH

4Cl

(b) Exess NH4Cl + HCl

(c) Exess NHaCl + NaOHPH range of buffer solution

PH = PKa + log

PH range of acidic buffer

[salt][acid]0.010.09

[salt][acid]

0.100.10

Mol.B.R

Mol.A.R[salt][acid]0.110.09

0.110.09

Meq.B.R

Meq.A.R

Meq.B.R

Meq.A.R

[salt][acid]

[salt][acid] 0.1 1 10→

Best Buffer (most sensitive buffer)

[salt][acid]

PH2 = PKa + log

PH2 = 4.74 + log

PH1 = PKa + log

PH2 = PKa + log

PH2 = 4.74 + log

PH = PKa + log

ratio vary 0.1 to 10

PH = PKa ± 1In the same manner POH of range of basic buffer

POH = PKb ± 1

2 A physician or lab technician want to prepare buffer

solution having PH = 5 using CH3COOH and CH

3COONa.

What will be ratio of

(PKa of CH3COOH is 4.74)

PH = PKa + log

log = 0.26

= anti log 0.26 = 1.8

Note: Buffer has maximum capicity when its acid hasits PKa as close as possible to the target PH.

2 A physician want to prepare buffer solution hav-

ing PH = 5 using weak acid and its salt, which of the follow-ing weak acid and its salt would be a good choice.

Acid Salt PKa

(a) H3PO

4H

2PO

4–2 2.12

(b) HCOOH HCO2

– 3.74(c) CH

3COOH CH

3COO– 4.75

(d) H2PO

4–2 HPO

4–2 7.21

Ans = c

2 What mass of sodium acetate (CH3COO

Na.3H2O,M.W = 136) and what volume of concentrated

acetic acid (17.45 M) should be used to prepare 0.50 L of abuffer solution at PH = 5.0 that is 0.150 M overall ?

Mass of sodium acetate (hydrated) = 6.6 gmVolume of Acetic acid = 1.5 ml

8. SALT HYDROLYSIS

Concept Suppose you want to calculate PH of sodium ac-etate (basic salt) of certain concentration (c)

CH3–C–ONa CH

3–C–O– + Na

C 0 0 0 C Canionic hydroysis taken place.CH

3COO– + H

2O CH

3COOH + OH–

C 0 0 C–Cα Cα CαCH

3COO– (Acetate ion) bevaves as weak base

[salt][acid]

[salt][base]

[salt][acid]

=OO=

→

100%B.DA.D

(Any salt is100%

dissociated)

B.RA.R

POH = PKb + log

5.0 = 7.74 + log

[CH3COONa]

[CH3COOH]

[CH3COONa]

[CH3COOH]

[CH3COONa]

[CH3COOH]

[CH3COONa]

[CH3COOH]

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Kb(CH3COO–) =

Kb(CH3COO–) = =

Kb(CH3COO–) = 5.6 × 10–10

Kb of (CH3–C–O–) Acetate ion denotes, it is weak base

PKb (CH3COO–) = 10 –log 5.6

= 9.25POH of weak base is denoted by equationPOH = 1/2 (PKb – log c)POH = 1/2 (PKb (CH

3COO–)–log c)

Generally α is replaced by h (degree of hydration)and Kb of CH

3 C–O– is replaced by KH (Hydrolysis con-

stant)

MISCONCEPT

Conjugate base of weak acid is strong. It is nottrue. Conjugate base of weak acid is also weak.Example : Ka of CH

3COOH is 1.8 × 10–5 (W.A)

Ka of CH3C–O– is 5.6 × 10–10 (W.B)

∗ Conjugate acid of weak base is strong →

It is not correct.Conjugate acid of weak base is also weak.

Example : Kb of NH4OH is 1.8 × 10–5 (W.B)

Kb of NH4

+ is 5.6 × 10–10 (W.A)NH

4+ + H

2O NH

4OH + H+

Ka (NH4+) =

= =

Ka [NH4

+] = 5.6 × 10–10

This concept is useful for the solution of problemof 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)

[CH3COOH][OH–][H+]

[CH3COO–][H+]

KwKa(CH

3COOH)

1 × 10–14

1.8 × 10–5

=

O

=

O

[NH4OH][H+][OH–]

[NH4

+][OH–]

KwKa(NH

4+)

1.0 × 10–14

1.8 × 10–5

=

O

Solution :(a) Kb (CH

3COO–) = KH =

=

= 5.6 × 10–10

(b) h = =

= 2.3 × 10–4

(c) POH = 1/2 (PKb – log c)= 1/2 (PKb(CH

3COO–) – 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 con-

taining NO2

– ions.Given Ka for HNO

2 = 4.5 × 10–10

Solution : KH = = = 2.2 × 10–5

2 What is PH of 0.5 M aqueous NaCN solution PKb

of CN– = 4.70.NaCN Na + CN–

CN– + H2O HCN + OH–

Kb (CN–) =

POH = 1/2 (PKb – log c)POH = 1/2 (4.70 – log 0.5)

POH = 2.5PH = 11.5

2 Calculate the percentage hydrolysis in 0.003 M a

queous solution of NaOCN.Ka for HOCN = 3.33 × 10–4

h = =

= =

h = 10–4, % hydrolysis = 10–2 %

Hydrolysis of Acidic salt (like NH4Cl)

NH4Cl NH

4+ + Cl–

C 0 0 0 C C

NH4

+ + H2O NH

4OH + H+

C 0 0 C–Ch Ch Ch

KwKa(CH

3COOH)

1 × 10–14

1.8 × 10–5

Kb(CH3COO–)

C5.6× 10–10

0.01

KwKa

[HCN] [OH–][CN–]

KHC

Kb (OCN–)C

KwKa (HOCN)C

10–14

3.33 × 10–4× 0.003

B.DA.D

Befor hydrolysisAfter hydroly-sis

10–14

5.6× 10–10

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∗ 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 Ca3(PO

4)

2 a slightly soluble compound.

∗ Tooth decay involves solubility, when food lodges

between the teeth, acids form that dissolve enamel, whichcontain a mineral called hydroxyapatite (Ca

5(PO

4)

3OH).

Tooth decay can be reduced by treating teeth with fluoride.Fluoride replaces the hydroxide to hydroxyapatite to producethe 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 BaSO4 to 1.0

litre of water at 25ºC and stir untill the solution is saturatedvery littile BaSO

4 dissolves only 0.0025 gm of BaSO

4

dissolves in 1.0 liter of water, no matter how much moreBaSO

4 is added. Hence BaSO

4 is known as slightly soluble

compound.

BaSO4(s) + H

2O BaSO

4(aq) Ba+2

(aq) + SO

4–2

(aq)

Minimize

BaSO4(s) + H

2O Ba+2

(aq) + SO

4–2

(aq)

K =

K × BaSO4(s) = [Ba+2]

eq [SO

4–2]

eq

Ksp = [Ba+2]eq

[SO4

–2]eq

In equilibrium that involve slightly soluble

compound in water, the equilibrium constant is called a

solubility product constant (Ksp). The activity of solid BaSO4

is one. Hence the concentration of solid is not included in

the equilibrium constant expression.

Other Example :

CaF2

Ca+2(aq) + 2F– (aq)

Ksp = [Ca+2] [F–]2 = 4S3

Bi2S

3(s) 2Bi+3

(aq) + 3S–2

(aq)

Ksp = [Bi+3]2 [S–2]3 = 108 S5

MyXz(s) yM+2(aq)

+ Zx–y(aq)

Ksp = [M+2]y [ X–y]z = (Sy)y x (Sz)z

MgNH4PO

4(s) Mg+2

(aq) + NH

4+

(aq) + PO

4–3

(aq)

Ksp = [Mg+2] [NH4

+] [PO4

–3] = S3

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 BaSO4”. Which means

0.0025 gm of solid BaSO4 dissolves to give a solution that

→

100 % (generally)

[Ba+2]eq

[SO4

–2]eq

BaSO4(s) eq denotes equilibrium

Ka (NH4+ ) = KH =

=

Ka(NH4+) = = 5.6 × 10–10

NH4+ behaves as weak acid.

PH of the weak acid is calculated by the equation.PH = 1/2 (PKa – log c)

Ka (NH4+) = 5.6 × 10–10

PKa (NH4+) = 10 – log 5.6

PKa(NH4+) = 9.25

2 Calculate the PH of 0.1 M NH4Cl.

KbNH4OH = 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 becauseit 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 verylarge, so, Ka of Na+ is very low, Hence hydrolysis of Na+ nottakes place.

If Hx is weak acid (Ka) then value of conjugateacid (X–) is Kb then.

Kb (x)– =

Kw = Kb (X)– × Ka(Hx)taking log both sidePKw = PKb + PKa

14 = PKb + PKa

2 A certain buffer solution contains equal concen-

tration of X– and Hx. Kb for X– is 10–10. Calculate PH ofbuffer.

Kb for X– = 10–10

Kb for Hx = = = 1 × 10–4

PKa for Hx = 4PH = PKa + log

PH = 4

SOLUBILITY PRODUCT PRINCIPLE

∗ Most compounds dissolve in water to some extent

and many are so slightly soluble that they are called“ Insoluble ”

[NH4OH][H+][OH–]

[NH4

+][OH–]

Kw Kb(NH

4OH)

1 × 10–14

1.8 × 10–5

KwKa (Hx)

KwKb

1 × 10–14

1.0 × 10–10

[salt][acid]

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NaF Na+ + F–

0.1

H2O + 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

S = 6.4 × 10–9 × 102

= 6.4 × 10–7 MHence solubility of MgF

2 is decreased in NaF in

comparision to pure water due to common ion (F–) which isprovided by NaF.

Relative Solubilities

A salt’s Ksp value gives as information about itssolubility. 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

CaSO4

Ksp = 6.1 × 10–5

Ksp = S2

S = Ksp = SolubilityIn this case we compare the solubilities for these

solid by comparing the Ksp values :

CaSO4 (s) > CuI(s) > AgI (s)

(2) The salts being compared produce differentnumbers of ions. For Example,

Consider :

CuS (s) Ksp = 8.5 × 10–45

Ag2S (s) Ksp = 1.6 × 10–49

Bi2S

3(s) Ksp = 1.1 × 10–73

These salt prodeces different numbers of ions whenthey dissolve. The Ksp values connot be compared directlyto determine relative solubility.

Relative solubility measured by above mentionedRule

* Remember that relative solubility can be predictedby comparing Ksp value only for salts that produce the sametotal numbre of ions.

For CuS For Bi2S

3

S2 = Ksp 108 S5 = Ksp

S = Ksp S =

For : Ag2S

4S3 = Ksp

S =

→Due to commonion reactionshifted towardsleft

Neglected

→

100%

√

Least solublesmallest Ksp

Most solublelargest Ksp

√Ksp

108

1/5

Ksp4

1/3

contains equal number of Ba+2 and SO4

–2 ions.

Question : 1.0 liter of saturated BaSO4 solution contains

0.0025 gm of dissolved BaSO4. Calculate the solubility

product constant for BaSO4 .

Solution :

Molar solubility (s) =

= 1.1 × 10–5 M

BaSO4(s) Ba+2(aq) + SO4

–2(aq)

1.1 × 10–5 M 0 0

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

Q K

Qsp also known as ion product.

If Qsp < Ksp Forward process is favored. Noprecipitation occure, more solic can

dissolve.

Qsp = Ksp Solution is just saturated, reaction is atequilibrium

Qsp > Ksp Reverse process is favored, precipitation

occures.Question : If 100 mL of 0.00075 M Na

2SO

4 and 50.0 mL of

0.015 M BaCl2 solution are mixed, will a precipitate form ?

Ksp of BaSO4 is 1.1 × 10–10

[Ba+2] =

[SO4–2] =

Qsp = 2.5 × 10–6

Qsp > Ksp

Solid BaSO4 precipitated untill [Ba+2] [SO

4–2] just

equal to Ksp of BaSO4

Common Ion Effect in solubility Calculation :

Example : The molar solubility of MgF2 is 1.2 × 10–3 M in

pure water at 25ºC. Calculate the molar solubility of MgF2

is 0.10 M NaF.

MgF2(s) + H

2O Mg+2 + 2F–

Ksp = [Mg+2] [F–]2 = S. (2S)2 = 4S3

= 4 × (1.2 × 10–3)3

= 6.4 × 10–9

2.5 × 10–3

233 M

Before dissociation

After dissociationat equilibrium

0.015 × 50 × 2

2 × 150= 5 × 10–3 M

100 × 0.00075 × 2

150 × 2= 5 × 10–4 M

No necessarilyat equilibrium

→→

Qsp

Qc

Qp

used in chemicalequilibrium

→→→

→ Qsp

Qc

Qp

used in ionicequilibrium

used inchemical

equilibrium

→

→

→ Necessarily atequilibrium.

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