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Physical(I) and Chemical or Ionic Equilibrium(II)

Properties of equilibrium:

1) Chemical equilibrium is dynamic in nature 2) It attain in reversible reaction 3) It can approaches from both sides 4) Catalyst have no effect on equilibrium

At equilibrium the rate of forward reaction is equal to the rate of backward reaction ie

(dx/dt)f =(dx/dt)b [ f=forward reaction ,b=backward reaction]

Equilibrium are of two types; a)Homogeneous equilibrium (all reactants and products are in same states or phases. b) Heterogeneous equilibrium (the reactants or products are not same state or phase.

Active Mass Law: At a particular temperature ,at any instant the rate of a reaction is directly proportional to the active mass of the reactants of that reaction. Active mass means the mass present per unit volume.

A+B �Pdt rate �[A][B] .that means, rate=k[A][B] [k=rate constant]

Let a reaction

xA+yB���mC+nD

(dx/dt)f =kf[A]x[B]y , ( dx/dt)b=kb[C]m[D]n At equilibrium kf[A]x[B]y =kb[C]m[D]n

Kf/kb=[C]m[D]n/[A]x[B]y,

ie Keq=[C]m[D]n/[A]x[B]y

keq=kf/kb [Kf and Kb are the rate constant for forward and backward reaction respectively and Keq is the equilibrium constant. Keq

Kc [equilibrium constant in terms of concentration]

Kp [ quilibrium constant in terms of pressure]

Kx [equilibrium constant in terms of mole fraction]

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The relation between these constant are

Kp=Kc(RT)�n =KxP�n [�n=total number of gaseous moles of products –total number of moles of gaseous reactants, and P is the total pressure]

If 1)�n=0, means Kp=Kc [no unit]

2) �n=+ve , Kp>Kc [Unit of Kc=(mol/lit)�n and Kp=(atm)�n ]

3) �n=-ve Kp<Kc [Unit of Kc=(mol/lit)�n and Kp=(atm)�n ]

Note:

1) If the reaction is reverse, the equilibrium constant is inverse of initial 2) If the reaction is multiplied by m, the new equilibrium constant K1= Km ,K =initial

equilibrium constant 3) If the reaction is divided by m ,then new equilibrium constant K2= K1/m , K= initial

equilibrium constant 4) If two equilibrium constants(K1 and K2) are added ,then the new equilibrium constant

ie K3=K1×K2 5) The value of equilibrium constant does not change in the presence of a catalyst.

The Degree of Ionisation /Dissociation (�)=(D-d)/(y-1)d [D=V.D, d=V.D after dissociation, y=no of moles of product formed from 1 moles of reactant].

Concentration quotient (Q)

Qc = [C]m[D]n/[A]x[B]y

If Q=Keq ; the reaction is in equilibrium

If Q < K ; the reaction has not attain equilibrium and has a tendency to proceed in forward direction in a greater extant

If Q > K ; the reaction has attain equilibrium and it has a tendency to proceed in backward direction in a greater extant

La-chisellers principal:By changing any conditions of equilibrium ,The equilibrium itself changes in such a direction that minimize the effect or system attain re-equilibrium again.

Factors:

� Concentration: If concentration of reactant increases, the equilibrium shift in forward direction. If concentration of products increases, the equilibrium shift in backward direction.

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� Temperature: Exothermic reaction prefer at low temp and endothermic prefer at high temperature.

� Pressure effect:

Reaction with Position of equilibrium

�ng=Positive/+ve Shift towards Right at low pressure

�ng=Negative/-ve Shift towards reight at high pressure

�ng=0 No effect on equilibrium

� Catalyst: Generally catalyst have no effect on equilibrium

Addition of Inert gas: Addition of inert gases at constant volume, does not change the equilibrium. But addition of inert gases at constant pressure changes the equilibrium such a way:

Reaction with Position of equilibrium

�ng=Positive/+ve Shift towards Right

�ng=Negative/-ve Shift towards Left

�ng=0 Remain uneffected

Note: � If any one of the product escape from the system, there is no equilibrium or in a reaction

if gaseous products are formed and the reaction performed in a open container; then there is no equilibrium. eg; CaCO3(s)�CaO(s) + CO2(g)

� If in a chemical reaction, one of the product exixt as a precipitate (ppt) formed, there is no questions of equilibrium.

�G= — 2.303logK or �G=-RTlnK; K=e—�G/RT

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IONIC EQUILIBRIUM

The equilibrium which are existing in between the ions and compound that is called ionic equilibrium. This equilibrium is possible only for weak electrolytes (weak acids and weak bases).

CH3COOH�� CH3COO-- +H+

Dissociation of water: At 250C the value of dissociation constant (Kw) is 10— 14 and pH + pOH=14.

PKa+ PKb=14; Ka and Kb are the ionization constant of acid and base

pH= — log[H+] ; [H+] = total H+ ion concentration.

pOH== — log[OH--]; [OH- ] = total [OH-] ion concentration

The value of Kw increases with increasing of temperature, as a result the pH scale schussed but water remains neutral at any temperature.

Dissociation of weak acid and weak bases or weak electrolytes (Ostwald’s dilution law):

For acid Ka= C�2/1- � , for base Kb= C�2/1- � [�=degree of dissociation, C=concentration and Ka or Kb are dissociation constant of acid or base]. For very very weak acid; 1- � � 1. So Ka= C�2 , �=(Ka/C)1/2 ie � � 1/�c. On dilution of weak acids or bsaes, concentration decreases, degree of dissociation increases and conc of H+ or OH- increases hence acid or base become stronger or Degree of dissociation of weak electrolytes is inversely proportional to square root of its dilution.

pH of weak acid

pH= ½ [logKa+log C] where Ka= dissociation constant of weak acid and C =concentration of acid .

For weak base

pOH= 1/2[ logKb +logC] ;Kb= dissociation constant of weak base and C= concentration of that base.

Note:

� All organic acid are weak acid and all non-metal hydroxide are weak base. � All Inorganic acid are strong acid and metal hydroxide are strong base � Among all the inorganic acids, perchloric acid (HClO4) is a strongest acid � In organic acids, phenol and picric acid have no –COOH acid group

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Concept of acids and bases:

Concept Acids Bases

Arrenhius Ionise to give H+ in H2O Ionise to give HO- in H2O

Bronsted-Lowry A proton donor A proton acceptor

Lewis An electron pair acceptor An electron pair donor

Calculation of pH; in a mixture of two acids or two bases: (milimole concept):

n1V1S1+ n2V2S2 =(V1+V2)S

n1 and n2 are the ‘n’ factor of acids or bases,V1,V2 are the volume of acids or bases and S1,S2 are the strengths of acids or bases and S is the final strength of the mixture.

If the mixture is one acid and one base; V1S1n1 – V2S2n2 = (V1+V2)S

If the strength(S) is in normality unit; then n factor is not consider.

OR

Mixture of two weak acids (HA and HB):

[H+] = �(Ka1C1+Ka2C2): �1 and �2 of both acids are negligible w. e .t unity; Ka1 and Ka2 are the ionization constant of acids and C1, C2 are the concentration of two acids

Mixture of two bases (AOH and BOH)

[OH-] = (Kb1C1+Kb2C2)1/2 ; Kb1 and Kb2 are the ionization constant of bases,C1 C2 are the concentration of bases and �1 and �2 are the degree of dissociations of two bases are not consider.

Conjugate acid base pair:

Acid H+ = Conjugate base.

Base+H+ =Conjugate Acid

Note: The difference between the acid base conjugate pair is only one H+ ion

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Conjugate Pairs Conjugate Pairs Acid Conjugate base Acid Conjugate base

HCl Cl CH3COOH CH3COO HNO3 NO3- H2S HS HClO4 ClO4

HCO3 CO3

2

HSO4- SO42 H2CO3 HCO3

NH4+ NH3 HS S2

H2PO4- HPO4 H2O OH

H3O+ H2O [Fe(H2O)6]3+ [Fe(H2O)5OH] H3PO4 H2PO4

[Al(H2O)6]3+ [Al(H2O)5OH] H2SO4 HSO4

� H2PO2� No

All bi-anions (HSO4--/HCO3

- etc) are act as a conjugate acid as well as conjugate base.

Salt Hydrolysis:

� Hydrolysis of salt means the interactions between the ions of salt with water. � Kh=Kw/Kb = Kw/Ka ; Kh=hydrolysis constant, Kw= ionization constant of water, Kb=

ionization constant of base and Ka= ionization constant of acid � Salts of strong acid (SA) and strong base(SB) is not hydrolised and the solution is

neutral,pH=7 � Salt of strong acid (SA) and weak base(WB); solution is acedic and pH= 7- ½ [PKb+log C] at

25oC. � Salt of weak acid(WA) and strong base(SB); solution is basic in nature and pH=

7+1/2[PKa+logC]; at 25oC. � Salt of weak acid(WA) and weak base(WB), solution in neutral in nature and pH= 7+1/2[Pka

–Pkb] ; at 25oC S.No

Type of salt

Example Hydrolysis constant(Kh)

Degree of hydrolysis(h)

pH of solution Nature of solu

1. Strong acid and weak base

NH4Cl, (NH4)2SO4,CuSO4, NH4NO3

Kh=Kw/Kb h= (Kw/Kb.C)1/2

pH= 7- ½ [PKb+log C]

Acedic

2. Weak acid and strong

CH3COONa,C6H5COO

Kh=Kw/Ka h= (Kw/Ka.C)1/2

pH= 7+1/2[PKa+logC]

Basic

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base Na 3. Weak acid

and weak base

CH3COONH4,NH4CN,(CH3COO)2Cu

Kh= Kw/Ka.Kb h= (Kw/Ka.Kb)1/2 pH= 7+1/2[Pka –Pkb]

Neutral, alkaline or acedic

4. Strong acid and strong base

NaCl,Na2

SO4,KCl etc

No hydrolysis No No No

� For amphiprotic salt; like HS , HCO3

: pH = (PKa1 +Pka2)/2 and [H+] =(K1.K2)1/2, K1 and K2 are the first and second ionization constant of acid.

Buffer Solution: Buffer solutions are those solutions whose pH does not change significantly by adding little amount of acids or base. It is prepared by weak acid and its strong base salt or weak base and its strong acid salt.

Buffers are two types:

� Acid buffer: Weak acid and its conjugate base/ Weak acid and its salt with strong base; eg CH3COOH+CH3COONa; H2CO3 +NaHCO3; Na2B4O7 + H3BO3; H3PO4 + NaH2PO4 ; tartaric acid +sodium tartarate; sodium tartarate +disodium tartarate

� Basic buffer: Weak base and its conjugate acid/ Weak base and its salt with strong acid. eg: NH4OH + NH4Cl

� pH of Buffer calculation( Hendersons Equation); For acid buffer pH=PKa + log[salt/acid] and for basic buffer POH=PKb + log[salt/base]

� Buffer Capacity: buffer capacity = Number of moles of acid or base added to one litre/Change in pH

� For good buffer capacity; salt/acid or salt/base ratio should be closed to one as possible. In such case pH=pKa, ie that is also a midpoint of titration.

Indicator: Indicator is a organic substance which indicate the point of equivalent in titration by change its colour.

The theory of indicator is that the ionized and unionized forms of an indicator have different colour. If 90% of a particular form( ionized or unionized) is present then the colour distinctly seen. In general an indicator is a weak acid, HIn � H+ +In; the ratio of ionized and unionized form can be determined by

pH= pKa + log([In-]/[HIn]) and for detectable colour change, pH=pKa ±1.

Some acid base indicator and their colour in different pH

Indicator pH range for Colour in acedic Colour change pKln

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colour change medium in basic medium Methyl orange 3.1 to 4.4 Red Orange 3.7 Methyl Red 4.2 to 6.3 Red Yellow 5.0 Phenopthelene 8.3 to 10.0 Colourless Red 9.6 Litmus 5.0 to 8.0 Red Blue 7.0 Bromocresol 3.8 to 4.6 Yellow Blue 4.6

Name of Indicator which are use in acid base neutralization reaction:

Sl.No Titration pH change at the end point

Indicator which can be used

1. Strong acid with strong base

3-7 Any Indicator may be used ,eg Methyl Orange, Phenopthalene etc

2. Weak acid with strong base

6-11 Phenolpthalene

3. Strong acid with weak base

3-8 Methyl orange, Methyl red

4. Weak acid with weak base

No sharp change in pH

No suitable indicator is used

Solubility(S) and Solubility Product( Ksp):

It is applicable only for sparingly soluble salts

The solubility product of a sparingly soluble salt (AxBy) having solubility(S) is

Ksp = xx × yy× S(x+y) where x=no of cations and y= no of anions

The ionic product (Q): It is define as the product of concentration of all the ions of a sparingly soluble salt at a time t.

Case 1: If Q < Ksp ; the solution is unsaturated, no ppt take place

Case 2: If Q=Ksp: the solution is saturated

Case 3: If Q> Ksp ; solution is supersaturated, ppt take place

Sl No Nature of salt Ksp Example of salt 1. x=y=1 S2 AgCl,AgBr,SrSO4,BaSO4 2. x=2,y=1 or x=1,y=2 4S3 CaF2, PbCl2,BaF2 3. x=1,y=3 or x=3,y=1 27S4 Al(OH)3

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4. x=2,y=3 or x=3,y=2 108S5 Ca3(PO4)2

Common ion effect: The suppression of dissociation (decreases the degree of dissociation) of a weak electrolytes on addition of a strong electrolyte containing a common ion is called common ion effect.

Common ion effect leads to increase the possibility of ppt formation. Common ion is the ion (cation or anion) of solvent,which is similar to the cation or anion of salt. eg AgCl in HCl, here Cl- is the common ion.

Problem Based On This Chapter:

1. Consider the following reactions, In which case the product formation favoured by decreased pressure? i) CO2(g)+ C(s) � 2CO(g) ii) N2(g) +3H2(g) �2NH3(g) ii) N2(g) + O2(g) � 2NO(g) iv) 2H2O(g) �2H2(g) + O2(g)

a) ii, iii b) iii, iv c) ii, iv d) i, iv 2. The relation between the equilibrium constant Kp and Kc is

a) Kc=Kp(RT)�n b) Kp =Kc (RT) �n c) Kp = (Kc/RT) �n d) Kp –Kc = (RT) �n

3. The values of Kp in the reaction MgCO3(s) � MgO(s) + CO2(g) is :

a) Kp=PCO2 b) Kp = PCO2×( PCO2×PMgO/PMgCO3)

c) Kp = PCO2×PMgO/PMgCO3 d) Kp = PMgCO3/ PCO2×PMgO

���� 4. An equilibrium mixture that has formed from dissociation of PCl5(g) at 250oC contains 2 moles of PCl5, 2 moles of PCl3 and 2 moles of Cl2. The mixture has a total pressure of 2 atm. The values of Kp for the dissociation of PCl5 is

a) 2 b) 3 c) 4 d) 1

5. The pH of the solution that is obtained by mixing of 50mL of 0.4(N) HCl solution with 50mL of 0.2(N) Na(OH) solution is

a) –log2 b) –log0.2 c) 1 d) 2.0

6. When a solution of a weak base is titrated with a solution of a strong acid, the pH of the soliution near the end point

a) rises very sharply b) decreses very sharply c) increases slowly d) does not show any changes

7.

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