Contents - arride learning Equilibrium.pdf · Weak acid and Strong acid both will contribute H+...

Topic Page No.

Theory 01 - 11

Exercise - 1 12 - 21




Answer Key 32 - 33

Contents

Ionic Equilibrium

SyllabusIonic Equilibrium

Solubility product, common ion effect, pH and buffer solutions;Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.

Name : ____________________________ Contact No. __________________

IONIC EQUILIBRIUM_ADVANCED # 1A-479 Indra vihar, kota

Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)Physical & Inorganic By

NV SirB.Tech. IIT Delhi

Organic Chemistry By

VKP SirM.Sc. IT-BHU

IONIC EQUILIBRIUMStrong electrolytes :(i) Those substance which are almost completely ionize into ions in their aqueous solution are

called strong electrolytes.(ii) Degree of ionization for this type of electrolyte is one i.e. 1. eg. HCl, H2SO4, NaCl.HNO3,

KOH, NaOH, HNO3, AgNO3, CuSO4, etc. Means all strong acids and bases and all types ofsalts.

Weak electrolytes :(i) Those substance which are ionize to a small extent in their aqueous solution are known weak

electrolytes.eg. H2O, CH3COOH, NH4OH, HCN, HCOOH, Liq. SO2 etc. Means all weak acidsand bases.

(ii) Degree of ionization for this types of electrolytes in <<< 1.

DEGREE OF DISSOCIATION When an electrolyte is dissolved in a solvent (H2O), it spontaneously dissociates into ions. It may dissociate partially (<< < 1 ) or sometimes completely ( 1 )

Eg. NaCl + aq Na+ (aq) + Cl– (aq)CH3COOH + aq CH3COO– (aq) + H+ (aq)

The degree of dissociation of an electrolyte () is the fraction of one mole of the electrolytethat has dissociated under the given conditions.

= initially taken moles of No.ddissociate moles of No.

For a weak electrolyte A+B– dissolved is water, if is the degree of dissociation thenAB (aq) AA+ (aq) + B– (aq)

initial conc C 0 0conc-at eq. C(1 – ) C CThen according to law of mass action,

Keq = )1(C

)1(CC.C

]AB[]B][A[ 2–

= dissociation constant of the weak electrolyte

[C = V1

, then V = 1/C(volume of solution in which 1 mole is present) is called dilution,

so Keq = V)1(

2





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If is negligible in comparison to unity then, 1 – ~– 1. So Keq = 2 C = c

Keq= V.Keq

ionconcentrat1

{ Thumb rule }

ACIDS BASES AND SALTS : Arrhenius concept :Arrhenius Acid : Substance which gives H+ ion on dissolving in water (H+ donor)eg. HNO3, HClO4, HCl, HI, HBr, H2SO4, H3PO4 etc.

Types of acids

H3BO3 is not Arrhenius acid. H+ ion in water is extremely hydrated (in form of H3O+, H5O2

+, H7O3+) and high charge density.

The structure of solid HClO4 is studied by X-ray, It is found to be consisting of H3O+ and ClO4¯.HClO4 + H2O H3O+ + ClO4¯ (better representation)

Arrhenius base : Any substance which releases OH¯ (hydroxyl) ion in water (OH¯ ion donor)

Types of base

OH¯ ion also in hydrated form of H3O2¯, H7O4¯, H5O3¯

First group elements (except Li.) form strong bases.

Bronsted - Lowery concept : (Conjugate acid - base concept) (Protonic concept)Acid : Substances which donate H+ are Bronsted Lowery acids (H+ donor)Base : Substances which accept H+ are Bronsted Lowery bases (H+ acceptor)

Conjugate acid - base pairsIn a typical acid base reaction

HX + B X– + HB+





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Forward reaction –Here HX, being a proton donor, is an acidB, being a proton acceptor, is a base.

Backward reaction – Here HB+, being a proton donor, is an acidX–, being a proton acceptor, is a base.

Acid Base Conjugate Conjugate Acid Base

HCl + H2O H3O+ + Cl¯

HSO4– + NH3 NH4

+ + SO42–

[Fe(H2O)6]3+ +H2O H3O+ + [Fe(H2O)5 (OH)]2+

Conjugate acid - base pair differ by only one proton Strong acid will have weak conjugate base and vise versa Reaction will always proceed from strong acid to weak acid or from strong base to weak base.

eg. Acid Conjugate base Base Conjugate acidHCl Cl¯ NH3 NH4

+

H2SO4 HSO4¯ H2O H3O+

HSO4¯ SO42– RNH2 RNH3

+

H2O OH¯

Amphoteric (amphiprotic) : substances which can act as acid as well as base are known as amphotericHCl + H2O H3O+ + Cl¯

base

NH3 + H2O NH4+ + OH¯

acid

LEWIS CONCEPT (ELECTRONIC CONCEPT) : An acid is a molecule/ion which can accept an electron pair with the formation of a coordinatebond.

Acid e– pair acceptore.g. Electron deficient molecules : BF3, AlCl3

Cations : H+, Fe2+, Na+

Molecules with vacant orbitals : SF4, PF3

A base is any molecule/ion which has a lone pair of electrons which can be donated.Base (electron pair donor)

e.g. Molecules with lone pairs (ligands) : NH3, PH3, H2O, CH3OH

PROPERTIES OF WATER : Amphoteric (amphiprotic) Acid/base nature:

Water acts as an acid as well as base according to Arhenius and Bronsted-Lowry theory butaccording to Lewis concept it can be generally taken as base.In pure water [H+] = [OH–] so it is Neutral.

Molar concentration / Molarity of water :

Molarity No. of moles/litre = mole/gm18litre/gm1000

= 55.55 mole /litre = 55.55 M (density

= 1 gm/cc) Ionic product of water : According to Arrhenius concept

H2O H+ + OH– so, ionic product of water, Kw = [H+][OH–] = 10–14 at 25° (exp.)





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dissociation of water, is endothermic, so on increasing temperature Keq. increasesKw increases with increase in temperature.

Now pH = –log[H+] = 7 and pOH log[OH–] = 7 for water at 25° (experimental)

pH = 7 = pOH neutralpH < 7 or pOH > 7 acidic at 25°CpH > 7 or pOH < 7 Basic

Ionic product of water is always a constant whatever has been dissolved in watersince its an equilibrium constant so will be dependent only on temperature.

Degree of dissociation of water :

H2O H+ + OH– takeninitiallymolesof.nototalddissociatemolesof.no

= %10x8.1or10x1855.55

10 7107

Absolute dissociation constant of water :

H2O H+ +OH– Ka = Kb = ]OH[]OH][H[

2

= 1677

108.155.551010

So, pKa of H2O = pKb of H2O = – log (1.8 × 10-16) = 16 – log 1.8 = 15.74

pH CALCULATIONS OF DIFFERENT TYPES OF SOLUTIONS :Strong acid Solution :(i) If concentration is greater than 10–6 M

In this case H+ ions coming from water can be neglected,so [H+] = normality of strong acid solution.

(ii) If concentration is less than 10–6 MIn this case H+ ions coming from water cannot be neglected,So [H+] = normality of strong acid + H+ ions coming from water in presence of this strong acid

Strong base Solution :Calculate the [OH–] which will be equal to normality of the strong base solution and then useKW = [H+] × [OH–] = 10–14 , to calculate [H+].

pH of mixture of two strong acids :If V1 volume of a strong acid solution of normality N1 is mixed with V2 volume of another strongacid solution of normality N2 , thenm. equi. of H+ ions from -solution = N1V1m. equi. of H+ ions from -solution = N2 V2

If final normality is N and final volume is V, thenNV = N1 V1 + N2 V2

[dissociation equilibrium of none of these acids will be disturbed as both are strong acid]

[H+] = N =21

2211

VVVNVN

pH of mixture of two strong bases :similar to above calculation

[OH–] = N =21

2211

VVVNVN

& [H+] = ]OH[10 14





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pH of mixture of a strong acid and a strong base :

Acid Base neutralisation reaction will take place. The solution will be acidic or basic depending on which component has been taken in excess. If V1 volume of a strong acid solution of normality N1 is mixed with V2 volume of a strong

base solution of normality N2 , thenNumber of m. eq. H+ ions from -solution = N1V1Number of m. eq. OH– ions from -solution = N2 V2

pH of a weak acid(monoprotic) Solution : Weak acid does not dissociated 100% therefore we have to calculate the percentage

dissociation using Ka dissociation constant of the acid. We have to use Ostwald’s Dilution law (as we have been derived earlier)

HA H+ + OH–

t = 0 C 0 0

t = teq C(1–) C C Ka = ]HA[]OH[]H[

=

1C 2

If <<1 1 – 1 Ka C2 = CKa ( is valid if < 0.1 or 10%)

[H+] = C = C CKa = CKa So pH = ClogpK

21

a

on increasing dilution C and [H+] pH

pH of a mixture of weak acid(monoprotic) and a strong acid Solution : Weak acid and Strong acid both will contribute H+ ion.

For the first approximation we can neglect the H+ ions coming from the weak acid solutionand calculate the pH of the solution from the concentration of the strong acid only.

To calculate exact pH, we have to take the effect of presence of strong acid on the dissociationequilibrium of the weak acid.If [SA] = C1 and [WA] =C2 , then [H+] from SA = C1the weak acid will dissociate as follows.HA H+ + A –

C2 0 0

C2(1– C2C1 C2 Ka = )1(CC)CC(

2

212

( )

(The weak acids dissociation will be further suppressed because of presence of strong acid,common ion effect)

Ka = (C2C1)Total H+ ion concentration = C1 + C2

If the total [H+] from the acid is greater than 10–6 M, then contribution from the water can beneglected at 25ºC temp., if not then we have to take [H+] from the water also.





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pH of a mixture of two weak acid (both monoprotic) solutions : Both acids will dissociate partially. Let the acid are HA1 & HA2 and their concentrations in the mixture are C1 & C2 respectively,then

HA1 H+ + A1– HA2 H + + A 2

–

t= 0 C1 0 0 C2 0 0At eq. C1(1– 1) C11+C22 C11 C2(1– 2) C22+C11 C22

Ka1 = )1(C)CC(C

11

22111

Ka2 = )1(CC)CC(

22

221122

(Since 1, 2 both are small in comparision to unity)

Ka1 = (C1 C22) Ka2 = (c1 c2 2) 2a

1aKK

=2

1

[H+] = C11 + C2= 2a21a1

1a1

KCKCKC +

2a21a1

2a2

KCKCKC [H+]= 2a21a1 KCKC

If the dissociation constant of one of the acid is very much greater than that of the secondacid then contribution from the second acid can be neglected.

So, [H+] = C11 + C2 C11

pH of a Solution of a polyprotic weak acid :

Diprotic acid is the one, which is capable of giving 2 protons per molecule in water.Let us take a weak diprotic acid (H2A) in water whose concentration is C molar.In an aqueous solution, following equilbria exist.1 = degree of ionization of H2A in presence of HA¯ Ka1 = first ionisation constant of H2A.2 = degree of ionisation of HA¯ in presence of H2A. Ka2 = second ionisation constant ofH2A.I step II step

H2A HA¯ + H+ HA¯ A2– + H+

c(1 –1) c(1 – 2) (c1 + c1 ) c(1 –2) c2 (c1 + c1 )

Ka1 = ]AH[]HA[]H[

2

Ka2 = ]HA[

]A[]H[ 2

Ka1 = )1(c)]1(c)[cc(

1

21211

Ka2 = )1(c)cc)(cc(

21

21211

= 1

2121

1)]1()][1(c[

........(i) = 2

221

1)]1(c[

..........(ii)

Knowing the values of Ka1 , Ka2 and c, the values of 1 and 2 can be calculated using equations(i) and (ii).After getting the values of 1 and [H3O+] can be calculated as.

[H3O+]T = c1 + c12

Finally, for calculating pH

If the total [H3O+] < 10–6 M, the contribution of H3O+ from water should be added

If the total [H3O+] > 10–6 M, then [H3O+] contribution from water can be ignored.

Using this [H3O+], pH of the solution can be calculated.





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ApproximationFor diprotic acids, Ka2

< < Ka1 and 2 would be even smaller than 1 .

1 – 2 1 and 1 + 2 1

Thus, equation (i) can be reduced to Ka1 = 1

11

1C

This is an expression similar to the expression for a weak monoprotic acid.

Hence, for a diprotic acid (or a polyprotic acid) the [H3O+] can be calculated from its firstequilibrium constant expression alone provided Ka2 << Ka1

.

Neutralisation :A reaction between acid and base to form salt and water molecule is known as neutralisation. In thistype of reaction acid gives H+ ion and base gives OH– ion.

HydrolysisIt is defined as a process involving the reaction of water on a salt to form mixture of acid and base.(a) It is the just reverse process of neutralization

Salt + water Acid + Base(b) In this reaction the solution is always neutral when both acid and base are strong.(c) If acid is stronger than base, the solution is acidic and if base is stronger than acid, the

solution is basic.(d) Depending upon the nature of an acid or a base, there can be four types of salt -

(i) Salt of strong acid and strong base.(ii) Salt of strong acid and weak base.(iii) Salt of weak acid and weak base.(iv) Salt of weak acid and strong base.

Strong Weak Strong Weakacids acids bases basesHCl CH3COOH NaOH NH4OHH2SO4 HCN KOH LiOHHNO3 H2CO3 RbOH Ca(OH)2HClO4 H3PO4 CsOH Be(OH)2HI H3PO3 Ba(OH)2 Zn(OH)2H2SO3 HOCl Al(OH)3

COOH

COOH|

32

3

NHRNH

)OH(Fe

ANIONIC HYDROLYSISAnions can function as a base on reaction with water and hydrolyse as follows :

A–(aq) + H2O (l) HA (aq) + OH– (aq)The extent of hydrolysis of a given anion depends on its basic strength





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CATIONIC HYDROLYSISCations can function as acid on reaction with water and hydrolyze as follows.B+(aq) + 2H2O(l) BOH(aq) H3O+(aq)The extent of hydrolysis of a given cation depends on its acidic strength.

(i) Hydrolysis of Salt of Strong acid and Strong base :Salt of strong acid and strong base does not hydrolysed.

(ii) Salt of strong acid and weak baseB+ + H2O BOH + H+

Kh = [ ] [ ]

[ ]BOH H

B

b

Wh K

KK

h = KC

h

pH = 7 – 12 pKb –

12 log C

Here, on the study of above equation, we can say that the pH of the strong acid weak base is less than7.

(iii) Salt of weak acid and strong baseA– + H2O OH– + HA

Kh = [ ] [ ]

[ ]HA OH

A

a

Wh K

KK

h = KC

h

pH = 7 + 12

pKa + 12

log C

(iv) Salt of weak acid and weak baseB+ + A– + H2O HA + BOH

Kh = [ ] [ ][ ] [ ]HA BOHB A

Kh = K

K KW

a b

h = K

K KW

a b

pH = 7 + 12 pKa –

12 pKb





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Hydrolysis of Amphiprotic Anion.(Cation is not Hydrolysed)NaHCO3, NaHS, etc., can undergo ionisation to from H+ ion and can undergo hydrolysis to form OH¯(Na+ ion is not hydrolysed)

(a) (i) HCO3¯ + H2O 23CO +

(ii) HCO3¯ + H2O H2CO3 +

pH(HCO3–) =

2

pKpK 2a1a

(b) Similarly for H2PO4¯ and HPO42– amphiprotic anions.

pH(H2PO4–) =

2

pKpK 2a1a and pH(HPO42–) =

2

pKpK 3a2a

Buffer Solution(if the acids and bases are mixed in different amounts (equivalents))(i) In certain applications of chemistry and biochemistry we require solutions of constantpH. Such solution are called buffer solution.(ii) A solution whose pH is not altered to any great extent by the addition of smallquantities of either an acid (H+ ions) or a base (OH– ions) is called buffer solution.(iii) Buffer solutions are also called solutions of reverse acidity or alkalinity.(iv) Following are the characteristics of buffer solutions(a) It must have constant pH.(b) Its pH should not be changed on long standing(c) Its pH should not be changed on dilution.(d) It pH should not be changed to any great extent on addition of small quantity of acid

or base.(v) Buffer solutions can be classified as follows.

(A) Simple buffer (B) Mixed buffer

Acidic bufferThese are the mixture of a weak acid and its salt with strong base.e.g.(a) CH3COOH + CH3COONa(b) Boric acid (H3BO3) + Borax (Na2 B4O7)

pH = pKa + log[ ][ ]SaltAcid





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Basic BufferThese are the mixture of a weak base and its salt with strong acid.e.g.(a) NH4OH + NH4Cl(b) Glycine (NH2CH2COOH) + Glycine hydrochloride (Cl

+NH3CH2COOH)

or pOH = pKb + log[ ][Base]

Salt

Buffer Capacity(i) The property of a buffer solution to resist alteration in its pH value is known as buffer capacity.(ii) Buffer capacity is number moles of acid or base added in one litre of solution as to changethe pH by unity, i.e.

Buffer capacity () = Number of moles acid or baseadded to sol

ChangeinpH1l .

or =

b

pH( )

where b is number of moles of acid or base added and )pH( is change in pH.

INDICATOR :

pH = pKn + log ]formUnionised[]formIonised[

SOLUBILITY(S) AND SOLUBILITY PRODUCT (KSP)At a constant temperature, the mass of a solute or electrolyte dissolved in the 100 gm of solvent inits saturated solution is called as solubility. Or number of gm mole of a solute dissolved in one litreof water at constant temperature is called as solubility of that solute.Solubility of a solute in moles / litre

= Solubility of solute in gm / litremolecular weight of the solute

Relationship between Solubility and Solubility Product :The equilibrium for a saturated solution of a salt Ax By may be expressed as,

Ax By xAy+ + yBx–

Thus, solubility product KSP = [Ay+]x [Bx–]y

Let the solubility of the salt Ax By in water at a particular temperature be ‘s’ moles per litre then

Ax By xAy+ + yBx–

xs ysSo, KSP = [xs]x [ys]y

KSP = xx . yy(s)x + y

(a) 1 : 1 types salts or AB type of salts :eg. AgCl, AgI, BaSO4, PbSO4, etc.

AB AA+ + B–

let the solubility of AB is s moles per litre.So, KSP = [A+] [B–] = s × s = s2

s = KSP





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(b) 1 : 2 or 2 : 1 type of salts or AB2 or A2B type of salts :eg. Ag2CrO4, PbI2, Ag2CO3, CaF2, CaCl2 etc.

(i) AB2 A2+ + 2B–

s 2slet the solubility of AB2 is ‘s’ moles per litreSo, KSP = [A2+][B–]2 = s × (2s)2 = 4s3

s = 34

KSP

(ii) A2B 2A+ + B2–

let s the solubility of A2B

So, A2B 2A+ + B2–

2s sKSP = [A+]2 [B–2] = (2s)2 (s) = 4s3

s = 34

KSP

(c) 1 : 3 type of salts or salts of AB3 or A3B type of salt -AB3 = Valency of A = 3 × Valency of Beg. FeCl3, AlCl3, PCl3, Al(OH)3, Fe(OH)3 etc.A3B = 3 × Valency of A = Valency of Beg. Na3BO3, Na3PO4, H3PO4 etc.(i) AB3 AA3+ + 3B–

let the solubility of A3B is ‘s’ mole / litre.AB3 AA3+ + 3B–

s 3sKSP = [A3+] [B–]3 = s × (3s)3 = 27s4

s = 427

KSP

(ii) A3B 3A+ + B–3

let the solubility of A3B is ‘s’ moles/litre.A3B 3A+ + B–3

3s sKSP = [A+]3 [B–3] = (3s)3 × s = 27s4

s = 427

KSP

(d) 2 : 3 or A2B3 type of salts :eg. Al2(SO4)3

A2B2 2A+ 3 + 3B– 2

let the solubility of salt A2B3 is ‘s’ -So, A2B3 2A+3 + 3B–2

2s 3sKSP = [A+3]2 [B–2]3

= (2s)2 × (3s)3

= 4s2 × 27 s3

KSP = 108 s5

s = 5108KSP





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PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.Section (A) : Basics and acid - base concept

A-1. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid.HCl (aq) + CH3COOH (aq) Cl– (aq) + CH3 COOH2

+(aq).The set that characterises the conjugate acid-base pairs is :(A) (HCl, CH3COOH) and (CH3COOH2

+, Cl–) (B) (HCl, CH3COOH2+) and (CH3COOH, Cl–)

(C) (CH2COOH2+, HCl) and (Cl–, CH3COOH) (D) (HCl, Cl–) and (CH3COOH2

+, CH3COOH).

A-2.* Identify acid and base in the following reaction according to Bronsted - Lowry concept.(a) [Cu(H2O)3(OH)]+ (aq.) + [Al(H2O)6]3+ (aq.) [Cu(H2O)4]

2+ (aq.) + [Al(H2O)5OH]2+ (aq.)(b) [Fe(H2O)5(OH)]2+ (aq.) + [Al(H2O)6]

3+ (aq.) [Fe(H2O)6]3+ (aq.) + [Al(H2O)5(OH)]2+ (aq.)

(c) O2– (aq.) + H2O (aq.) 2OH¯ (aq.)(d) CH3OH (aq.) + H¯ (aq.) CH3O¯ (aq.) + H2 (g)

Acid Base(A) [Al(H2O)6]3+ [Cu(H2O)3(OH)]+(B) [Al(H2O)6]

3+ [Fe(H2O)5(OH)]2+

(C) H2O O2–

(D) CH3OH H¯

A-3. In the following reaction HC2O4– (aq) + PO4

3–(aq) HPO42–(aq) + C2O4

2–(aq), which are the two Bronstedbases ?(A) HC2O4

– and PO43– (B) HPO4

2– and C2O42– (C) HC2O4

– and HPO42– (D) PO4

3– and C2O42–

A-4. If the acid-base reaction HA(aq) + B¯(aq) HB(aq) + A¯(aq) has a Keq. = 10–4, which of the followingstatements are true ?(i) HB is stronger acid than HA. (ii) HA is stronger acid than HB.(iii) HA and HB have the same acidity. (iv) B¯ is stronger base than A¯.(v) A¯ is stronger base than B¯. (vi) B¯ and HB are conjugate acid-base pair.(vii) the acid and base strengths connot be compared.(A) i, iii, v (B) i, v, vi (C) ii, iv, vi (D) iv, v, vi

Section (B) : Ostwald dilution concept, property of water, pH definitionB-1. Which of the following expression is not true ?

(A) [H+] = [OH] = KW for a neutral solution at all temperatures.

(B) [H+] > KW & [OH] < KW for an acidic solution

(C) [H+] < KW & [OH] > KW for an alkaline solution(D) [H+] = [OH] = 107 M for a neutral solution at all temperatures .

B-2.* 0.1 mole of which of the reagents listed below [(i) to (vi)] could be added to one litre of water to make 0.10 Msolutions of each of the following ions (a) NH4

+ ; (b) CH3COO¯ ; (c) Cl¯ separatly, respectively ?(i) NH3 ; (ii) NH4Cl ; (iii) CH3COOH ; (iv) CH3COONa ;(v) HCl ; (vi) NaCl(A) ii, iv, ii (B) ii, iv, v (C) ii, iv, vi (D) ii, iv, iii

B-3. Which of the following relations is correct ?(A) Gº = RT ln Keq (B) [H3O

+] = 10pH

(C) log 12

KwKw

= R303.2ºH

21 T1

T1

(D) [OH–] = 10–7, for pure water at all temperatures.





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B-4. pOH of H2O is 7.0 at 298 K . If water is heated at 350 K, which of the following statement should be true ?(A) pOH will decrease.(B) pOH will increase.(C) pOH will remain 7.0.(D) concentration of H+ ions will increase but that of OH will decrease.

B-5. Percentage ionisation of water at certain temperature is 3.6 × 10–7 %, Calculate Kw and pH of water.(A) 10–14, pH = 6.7 (B) 4 ×10–14, pH = 6.7 (C) 2 ×10–14, pH = 7 (D) 10–14, pH = 7

B-6. Which of the following has the highest degree of ionisation ?(A) 1 M NH3 (B) 0.001 M NH3 (C) 0.1 M NH3 (D) 0.0001 M NH3.

B-7. A 50 ml solution of strong acid of pH = 1 is mixed with a 50 ml solution of strong acid of pH = 2. The pH of themixture will be nearly : (log 5.5 = 0.74)(A) 0.74 (B) 1.26 (C) 1.76 (D) 1.5

B-8.* Kw of H2O at 373 K is 1 × 10–12. Identify which of the following is/are correct ?(A) pKw of H2O is 12 (B) pH of H2O is 6 (C) H2O is neutral (D) H2O is acidic.

Section (C) : SA,. SB and their mixture

C-1. A 1 litre solution of pH = 4 (solution of a strong acid) is added to the 7/3 litre of water. What is the pH ofresulting solution ? (log 3 = 0.48)(A) 4 (B) 4.48 (C) 4.52 (D) 5

C-2. Which of the following solution will have a pH exactly equal to 8 ?(A) 10–8 M HCl solution at 25ºC (B) 10–8 M H+ solution at 25ºC(C) 2 × 10–6 M Ba(OH)2 solution at 25ºC (D) 10–5 M NaOH solution at 25ºC

C-3. The [OH–] in 100.0 ml of 0.016 M-HCl (aq) is :(A) 5 × 1012 M (B) 3 × 10–10 M (C) 6.25 × 10–13 M (D) 2.0 × 10–9 M.

C-4. How many moles of NaOH must be removed from one litre of aqueous solution to change its pH from12 to 11 ?(A) 0.009 (B) 0.01 (C) 0.02 (D) 0.1

C-5. Which of the following solution will have pH close to 1.0 ?(A) 100 ml of M/10 HCl + 100 ml of M/10 NaOH (B) 55 ml of M/10 HCl + 45 ml of M/10 NaOH(C) 10 ml of M/10 HCl + 90 ml of M/10 NaOH (D) 75 ml of M/5 HCl + 25 ml of M/5 NaOH.

C-6. 0.1mol HCl is dissolved in distilled water of volume V then at lim V (pH)solution is equal to :(A) zero (B) 1 (C) 7 (D) 14

C-7. The pH of a solution obtained by mixing 50 ml of 0.4 N HCl and 50 ml of 0.2 N NaOH is :(A) 13 (B) 12 (C) 1.0 (D) 2.0

Section (D) : pH of weak acid and weak base, polyprotic acid and base

D-1. Which statement/relationship is correct ?

(A) pH of 0.1 M HNO3, 0.1M HCl, 0.1M H are not equal. (B) pH = – log]H[

1

(C) At 25°C the pH of pure water is 7. (D) The value of pKw at 25 °C is 7.

D-2. Ka for a monobasic acid whose 0.10 M solution has pH of 4.52. (log 3 = 0.48)(A) 9 × 10–6 (B) 9 × 10–9 (C) 3 × 10–9 (D) 10–12





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D-3. Kb for a monoacidic base whose 0.10 M solution has a pH of 10.48.(A) 9 × 10–6 (B) 9 × 10–9 (C) 9 × 10–7 (D) 3 ×10–7

D-4. One litre of solution contains 10–5 moles of H+ ions at 25°C. Percentage ionisation of water in solution is :(A) 1.8 × 10–7 % (B) 1.8 × 10–9 % (C) 3.6 × 10–9 % (D) 1.8 × 10–11 %.

D-5. Calculate the pH of 500 ml of 1M hydrazoic acid (HN3). (Ka = 2.5 × 10–5, log 5 = 0.7)(A) 2.7 (B) 2.3 (C) 2.4 (D) 1.7

D-6. For ortho-phosphoric acid,

H3PO4 (aq) + H2O (aq) H3O+ (aq) + H2

4PO (aq) ; Ka1

H2PO4 (aq) + H2O (aq) H3O+ (aq) + 2

4HPO (aq) ; Ka2

24HPO (aq) + H2O (aq) H3O

+ (aq) + 34PO (aq) ; Ka3

The correct order of Ka values is :

(A) 1aK >

2aK < 3aK (B) 1aK <

2aK < 3aK (C) 1aK >

2aK > 3aK (D) 1aK <

2aK > 3aK

D-7. What is the pH of 0.01 M H2S solution ? 1aK = 9 × 108 , 2aK = 1.2 × 1013 .(A) 4.52 (B) 3.52 (C) 4.48 (D) 3.48

Section (E) : Mixtures of acids and basesE-1. Consider an aqueous solution, 0.1 M each in HOCN, HCOOH, (COOH)2 and H3PO4, for HOCN, we can write

Ka(HOCN) = ]HOCN[

]OCN[]H[ . [H+] in this expression refers to :

(A) H+ ions released by HOCN.(B) Sum of H+ ions released by all monoprotic acids.(C) Sum of H+ ions released only the first dissociation of all the acids.(D) Overall H+ ion concentration in the solution.

E-2. What concentration of Ac– ions will reduce H3O+ ion to 2 × 10–4 M in 0.40 M solution of HAc ?Ka (HAc) = 1.8 × 10– 5 ?(A) 0.018 M (B) 0.0036 M (C) 00018 M (D) 0.036 M

E-3. The dissociation constant of acetic acid at a given temperature is 1.69 × 10–5. The degree of dissociation of0.01 M acetic acid in the presence of 0.01 M HCl is equal to :(A) 0.41 (B) 0.13 (C) 1.69 × 10–3 (D) 0.013.

E-4. What are [H+], [A¯] and [B¯] in a solution that contains 0.03 M HA and 0.1 M HB ? Ka for HA and HB are3.0 × 10–4 and 1.0 × 10–10 respectively.(A) [H+] = [A¯] = 3 × 10–3 M, [B¯] = 3.3 × 10–9 M (B) [H+] = [A¯] = 4 × 10–3 M, [B¯] = 4 × 10–9 M(C) [H+] = [A¯] = 2 × 10–3 M, [B¯] = 4 × 10–9 M (D) [H+] = [A¯] = 5 × 10–3 M, [B¯] = 6 × 10–9 M

Section (F) : Salt hydrolysis

F-1. Which of the following salts undergoes anionic hydrolysis ?(A) CuSO4 (B) NH4Cl (C) AlCl3 (D) K2CO3.

F-2. An aqueous solution of aluminium sulphate would show :(A) An acidic reaction. (B) A neutral reaction.(C) A basic reaction. (D) Both acidic and basic reactions.

F-3. Which of the following is an acidic salt ?(A) Na2SO4 (B) Ca(OH)Cl (C) Pb(OH)Cl (D) Na2HPO4





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F-4. Expression pKh = pKw – pKa – pKb is not applicable to :(A) Ammonium acetate (B) Ammonium cyanide(C) Aniline acetate (D) Ammonium chloride

F-5. What is the pH of an aqueous solution of ammonium acetate ? (Ka = Kb = 1.8 × 10–5)(A) > 7 (B) 7.0 (C) < 7.0 (D) Zero

F-6. If pKb > pKa then the solution of the salt of weak acid and weak base will be :(A) Neutral (B) Acidic (C) Basic (D) Amphoteric

F-7. pOH = 7– 0.5 pKa + 0.5 pKb is true for which pair of cation and anion ?(A) C6H5NH3

+, CH3 COO¯ (B) Na+, CN–

(C) Al+3, Cl– (D) NH4+, NO3

–

F-8. For a salt of weak acid and weak base [pKa – pKb ] would be equal to :(A) 2 pH + pKw (B) 2 pH – log 10–14 (C) 2 pH – pKw (D) None of these

F-9. Which of the following compound forms an aqueous solution which is acidic when compared with water?(A) NaOH (B) K2CO3 (C) BaCl2 (D) Al2(SO4)3

F-10. The salt of which of the following four weak acids will be most hydrolysed ?(A) HA ; Ka = 1 × 10–8 (B) HB ; Ka = 2 × 10–6

(C) HC ; Ka = 3 × 10–8 (D) HD ; Ka = 4 × 10–10

F-11. Formula for degree of hydrolysis ‘h’; h = [ 10–7 (KaKb)–½] is applicable to the salt :

(A) NH4CN (B) (NH4)2SO4 (C) NH4Cl (D) NH4NO3

F-12. [H+] = K K

Cw a is suitable for :

(A) NaCl, NH4Cl (B) CH3COONa, NaCN(C) CH3COONa, (NH4)2SO4 (D) CH3COONH4 , ( NH4)2CO3

F-13. The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1Msolution at 250C. Given that the ionic product of water is 1014 at this temperature, what is the dissociationconstant of the acid ?(A) 1 x 1010 (B) 1 x 109 (C) 3.33 x 109 (D) 3.33 x 1010

F-14. The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be :(pKa for CH3COOH = 4.74)(A) 4.74 (B) 8.87 (C) 9.10 (D) 8.57

F-15. pH of 0.1M Na2HPO4 and 0.2M NaH2PO4 respectively are : (pKa for H3PO4 are 2.2, 7.2 and 12.0).(A) 4.7, 9.6 (B) 9.6, 4.7 (C) 4.7, 5.6 (D) 5.6, 4.7

F-16. The pH of 0.1 M solution of the following salts increases in the order : [JEE-1999, 2/80](A) NaCl < NH4Cl < NaCN < HCl (B) HCl < NH4Cl < NaCl < NaCN(C) NaCN < NH4Cl < NaCl < HCl (D) HCl < NaCl < NaCN < NH4Cl





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Section (G) : Buffer Solution

G-1. The pH of buffer of NH4OH + NH4Cl - type is given by :(A) pH = pKb (B) pH = 1/2pKb – 1/2 log [salt ] / [base](C) pH = 14 – pKb – log [salt] / [base] (D) pH = pOH – pKb + [salt] / [base]

G-2. A basic buffer solution can be prepared by mixing the solution of :(A) ammonium chloride and ammonium acetate(B) ammonium acetate and acetic acid(C) ammonium chloride and ammonium hydroxide(D) ammonium cyanide and ammonium hydroxide

G-3. In the neutralization process of H3PO4 and NaOH, the number of buffers formed will be :(A) 3 (B) 1 (C) 2 (D) 4

G-4. Addition of sodium acetate solution to acetic acid causes the following change :(A) pH increases. (B) pH decreases.(C) pH remains unchanged. (D) pH becomes 7.

G-5. H+ ion concentration of water does not change by adding :(A) CH3 COONa (B) NaNO3 (C) NaCN (D) Na2CO3

G-6. H2CO3 + NaHCO3 found in blood helps in maintaining pH of the blood close to 7.4 . An excess of acidentering the blood stream is removed by :

(A) HC O3 (B) H2CO3 (C) H+ ion (D) CO3

2 ion

G-7. A solution is 0.1 M CH3COOH and 0.1 M CH3COONa. Which of the following solution will change its pH significantly?(A) Addition of water(B) Addition of small amount of CH3COONa with out change in volume(C) Addition of small amount of CH3COOH with out change in volume(D) None will change the pH significantly.

G-8. Ka for HCN is 5 × 10–10 at 25°C. For maintaining a constant pH of 9, the volume of 5 M KCN solution requiredto be added to 10 ml of 2M HCN solution is : (log 2 = 0.3)(A) 4 ml (B) 8 ml (C) 2 ml (D) 10 ml

G-9. Which of the following solutions would have same pH ?(A) 100 ml of 0.2 M HCl + 100 ml of 0.4 M NH3 (B) 50 ml of 0.1 M HCl + 50 ml of 0.2 M NH3(C) 100 ml of 0.3 M HCl + 100 ml of 0.6 M NH3 (D) All will have same pH.

G-10.* Which of the following is/are correct regarding buffer solution ?(A) It contains weak acid and its conjugate base.(B) It contains weak base and its conjugate acid.(C) It shows large change in pH on adding small amount of acid or base.(D) All of the above.

G-11.* A buffer solution can be prepared from a mixture of : [JEE-1999, 3/80](A) sodium acetate and acetic acid in water (B) sodium acetate and hydrochloric acid in water(C) ammonia and ammonium chloride in water (D) ammonia and soldium hydroxide in water

G-12. 50 ml of 0.20 M solution of the acid HA (Ka = 1.0 x 105) & 50 ml of an NaA solution are given. What shouldbe the concentration of the NaA solution to make a buffer solution with pH = 4.00 ?(A) 0.02 M (B) 0.01M (C) 0.1 M (D) 0.2 M

G-13. Calculate the pH of a solution resulting from the addition of 12.5 ml of 0.1 M HCl to 50 ml of a solutioncontaining 0.15 M CH3COOH & 0.2 M CH3COONa. (pKa = 4.74)(A) 4.26 (B) 3.74 (C) 4.74 (D) 5.74





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Section (H) : Acid-base Titration

H-1.* Which one is the correct graph (figure) for the corresponding acid-base titration ?

(A)

pH

volume of strong base added to a monobasic strong acid

(B)

pH

volume of strong acid added to a monoacidic weak base

(C)

pH

volume of strong base added to a weak dibasic acid

(D)

pH

volume of strong acid added to a weak diacidic base

H-2. 100 ml of 0.02 M benzoic acid (pKa = 4.2) is titrated using 0.02 M NaOH. pH after 50 ml and100 ml of NaOH have been added are(A) 3.50, 7 (B) 4.2, 7 (C) 4.2, 8.1 (D) 4.2, 8.25

H-3.* A weak acid (or base) is titrated against a strong base (or acid), volume v of strong base (or acid) is plottedagainst pH of the solution (as shown in figure). The weak electrolyte (i.e. acid or base) could be :

(A) Na2CO3 (B) Na2C2O4 (C) H2C2O4 (D) CH2(COOH)2

H-4. If the solubility of lithium sodium hexafluorido aluminate, Li3Na3 (AlF6)2 is ‘s’ mol lt–1, its solubility product isequal to :(A) 729 s8 (B) 12 s8 (C) 3900 s8 (D) 2916 s8

H-5. The solubility product Mg(OH)2 in water at 25ºC is 2.56 × 1013(mol/lt)3 while that of Al(OH)3 is 4.32 × 1034

(mol/lt)4. If s1 and s2 are the solubilities of Mg(OH)2 and Al(OH)3 in water in mol/lt at 250C, what is the ratio,s1/s2 ?(A) 2 × 105 (B) 2 × 104 (C) 3 × 106 (D) 3 × 103

H-6. The solubility of CaF2 (Ksp = 3.4 × 10–11) in 0.1 M solution of NaF would be :(A) 3.4 × 10–12 M (B) 3.4 × 10–10 M (C) 3.4 × 10–9 M (D) 3.4 × 10–13 M.

H-7. In a saturated solution of Ag2CO3, silver ion concentration is 2 × 10–4 M. Its solubility product is :(A) 4 × 10–12 (B) 3.2 × 10–11 (C) 8 × 10–12 (D) 10–12

H-8. The solubility of Ag2CO3 in water at 250C is 1 × 104 mole/litre. What is its solubility in 0.01 M Na2CO3solution? Assume no hydrolysis of CO3

2 ion.(A) 6 × 106 mole/litre (B) 4 × 105 mole/litre (C) 105 mole /litre (D) 2 × 105 mole/litre

H-9. Let the solubilities of AgCl in pure water, 0.01 M CaCl2, 0.01 M NaCl & 0.05 M AgNO3 be s1, s2, s3 & s4respectively what is the correct order of these quantities . Neglect any complexation.(A) s1 > s2 > s3 > s4 (B) s1 > s2 = s3 > s4 (C) s1 > s3 > s2 > s4 (D) s4 > s2 > s3 > s1





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Section (I) : Solubility in buffer, Complex formation, Selective precipitation

I-1. The solubility product of AgCl is 1.8 × 10–10. Precipitation of AgCl will occur only when equal volumes ofsolutions of(A) 10–4M Ag+ and 10–4 M Cl– are mixed. (B) 10–7 M Ag+ and 10–7 M Cl– are mixed.(C) 10–5 M Ag+ and 10–5 M Cl– are mixed. (D) 2 × 10–5 M Ag+ and 2 × 10–5 M Cl– are mixed.

I-2. The solubility product of BaCrO4 is 2.4 × 10–10 M2. The maximum concentration of Ba(NO3)2 possible withoutprecipitation in a 6 × 10–4 M K2CrO4 solution is :(A) 4 × 10–7 M (B) 1.2 × 1010 M (C) 6 × 10–4 M (D) 3 × 10–4 M.

I-3. What is the solubility of Al(OH)3, (Ksp = 1 × 10–33) in a buffer solution pH = 4 ?(A) 10–3 M (B) 10–6 M (C) 10–4 M (D) 10–10 M.

I-4. The solubility of Fe(OH)3 would be maximum in :(A) 0.1 M NaOH (B) 0.1 M HCl (C) 0.1 M KOH (D) 0.1 M H2SO4.

I-5. Which of the following statements is correct for a solution saturated with AgCl and AgBr if their solubilities inmoles per litre in separate solutions are x and y respectively ?(A) [Ag+] = x + y (B) [Ag+] = [Br–] + [Cl–] (C) [Br–] = y (D) [Cl–] > x.

I-6. If s is the molar solubility of Ag2SO4, then :

(A) 3 [Ag+] = s (B) [Ag+] = s (C) [2Ag+] = s (D) [ SO42– ] = s

I-7. Which of the following would increase the solubility of Pb (OH)2 ?(A) Add hydrochloric acid(B) Add a solution of Pb(NO3)2(C) Add a solution of NaOH(D) None of the above–the solubility of a compound is constant at constant temperature

I-8. The aqueous solution of which of the following sulphides would contain maximum concentration of S2– ions?(A) MnS (Ksp = 1.1 × 10–21) (B) ZnS (Ksp = 1.1 × 10–23)(C) PbS (Ksp = 1.1 × 10–35) (D) CuS (Ksp = 1.1 × 10–30)

I-9. Which of the following salts has maximum solubility ?(A) HgS, Ksp = 1.6 × 10–54 (B) PbSO4 , Ksp = 1.3 × 10–8

(C) ZnS, Ksp = 7.0 × 10–26 (D) AgCl, Ksp = 1.7 × 10–10

I-10. The necessary condition for saturated solution is :(A) Product of ionic concentrations = Solubility product(B) Product of ionic concentrations < solubility product(C) Product of ionic concentrations > solubility product(D) None of the above

I-11. At 30ºC, the solubility of Ag2CO3 (Ksp = 8 × 10–2) will be maximum in :(A) 0.05 M Na2CO3 (B) 0.05 M AgNO3 (C) Pure water (D) 0.05 NH3

I-12. Which of the following expressions shows the saturated solution of PbSO4 ?

(A) Ksp (PbSO4) = [Pb2+] [ SO42– ] (B) Ksp (PbSO4) > [Pb2+] [ SO4

2– ]

(C) Ksp (PbSO4) = [Pb+] [ SO4– ] (D) Ksp (PbSO4) < [Pb2+] [ SO4

2– ]

I-13. The correct relation between Ksp and solubility for the salt KAl(SO4)2 is :(A) 4s3 (B) 4s4 (C) 27s4 (D) none

I-14. The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is1.1 × 10–11. The possibilities are :(A) x = 1, y = 2 (B) x = 2, y = 1 (C) x = 1, y = 3 (D) x = 3, y = 1





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PART - II : MISCELLANEOUS QUESTIONS

COMPREHENSIONS TYPEComprehension # 1

Read the following passage carefully and answer the questions.Pure water is a weak electrolyte and neutral in nature, i.e. H+ ion concentration is exactly equal to OH– ionconcentration [H+] = [OH–]. When this condition is disturbed by decreasing the concentration of either of thetwo ions, the neutral nature changes into acidic or basic. When [H+] > [OH–], the water becomes acidic andwhen [H+] < [OH–], the water acquires basic nature. This is exactly the change which occurs during thephenomenon known as salt hydrolysis. The pH of salt solution can be calculated using the following relations:

pH = 21

[pKw + pKa + log C]

for salt of weak acid and strong base.

pH = 21

[pKw – pKb – log C]

for salt of weak base and strong acid.

pH = 21

[pKw + pKa – pKb]

for salt of weak acid and weak base.where ‘C’ represents the concentration of salt.When a weak acid or a weak base is not completely neutralized by strong base or strong acid respectively,then formation of buffer takes place. The pH of buffer solution can be calculated using the following relation :

pH = pKa + log ]Acid[]Salt[

; pOH = pKb + log ]Base[]Salt[

Answer the following questions using the following data :pKa = 4.7447, pKb = 4.7447, pKw = 14

1. When 50 ml of 0.1 M NH4OH is added to 50 ml of 0.05 M HCl solution, the pH is :(A) 1.6021 (B) 12.3979 (C) 4.7447 (D) 9.2553

2. 50 ml 0.1 M NaOH is added to 50 ml of 0.1 M CH3COOH solution, the pH will be :(A) 4.7447 (B) 9.2553 (C) 8.7218 (D) 1.6021

3. When 50 ml of 0.1 M NaOH is added to 50 ml of 0.05 M CH3COOH solution. The pH of the solution is :(A) 1.6021 (B) 12.3979 (C) 4.7447 (D) 8.7218

COMPREHENSION # 2

Strontium fluoride (SrF2) is a sparingly soluble salt. Let s1 be its solubility (in mol/lt.) in pure water at 25°C,assuming no hydrolysis of F– ions. Also, let s2 be its solubility (in mol/lt.) in 0.1 M NaF solution at 25°C,assuming no hydrolysis of F– ions and no complex formation.However, it is known that s1 : s2 = 106 : 256. Now, answer the following questions.

4. The Ksp value of SrF2 at 25°C is :(A) 2.048 10–9 (B) 1.372 10–9 (C) 1.864 10–9 (D) 2.916 10–9

5. The mass of NaF to be added to 100 ml solution of 0.0011 M Sr+2 ions to reduce its concentration to2 10–4 M is : [Assume no hydrolysis of F– ions](A) 0.42 g (B) 0.063 g (C) 0.021 g (D) 0.084 g

6. The solubility of SrF2(in mol/L) in a buffer solution of pH =5 at 25°C is : [Given : Ka for HF = 71

10–5]

(A) 1.6 10–3 (B) 3.2 10–3 (C) 4.8 10–3 (D) 6.4 10–3





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MATCH THE COLUMN

7. (Use log 1.8 = 0.26, Ka of formic acid = 1.8 × 10–4, Ka of acetic acid = 1.8 × 10–5, Kb of ammonia = 1.8 × 10–

5, 1aK of H2S = 10–7 and 2aK of H2S = 10–14, for the following matchings)

Match the entries of column II for which the equality or inequality given in the column I are satisfied.

Column I Column II

(A) 10–5 M HCl solution > 0.1 M H2S solution (p) water ( degree of dissociation of water)

(B) CH3COOH solution at pH equal to 4.74 (q) [OH–] = NH4OH solution at pH equal to 9.26

(C) 0.1 M CH3COOH solution (r) (degree of dissociation of electrolytes) = 1.0 M HCOOH solution

(D) 0.1 M of a weak acid HA1(Ka = 10–5) solution (s) pH < 0.01 M of a weak acid HA2(Ka = 10–6) solution

8. Column I Column II

(A) AgBr (p) Solubility in water is more than expectation.

(B) AgCN (q) Solubility in acidic solution is more than that in pure water.

(C) Fe(OH)3 (r) Solubility in strongly basic solution is more than that in pure water.

(D) Zn(OH)2 (s) Solubility decreases in presence of common anion.

ASSERTION / REASONING

DIRECTIONS :Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.

9. Statement-1 : The H3O+ has additional water molecules closely associated with it.

Statement-2 : In solid state the species H5O2+ and H9O4

+ have been found to exist.

10. Statement-1 : The proton transfer reaction between NH3 and H2O proceeds only to a slight extent.Statement-2 : Proton transfer reaction is virtually complete in the case of HCl in dilute solution.

11. Statement-1 : Acids that have more than one proton that can be donated to base are called polyprotic acids.Statement-2 : For all diprotic acids, the equilibrium constant 2aK , for the second stage of ionisation is

smaller than the equilibrium constant, 1aK , for the first stage of ionisation.

12. Statement-1 : 0.20 M solution of NaCN is more basic than 0.20 M solution of NaF.Statement-2 : Ka of HCN is very much less than that of HF.

13. Statement-1 : A substance that can either act as an acid as well as a base is called ampholyte.Statement-2 : Bisulphide ion (HS–) and biscarbonate ion (HCO3

–) are ampholytes.





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14. Statement-1 : Addition of HCl(aq) to HCOOH(aq), decrease the ionization of HCOOH(aq)Statement-2 : Due to common ion effect of H+, ionization of HCOOH decreased.

15. Statement-1 : pH of 10–7 M HCl is less than 7 at 25ºC.Statement-2 : At very low concentration of HCl, contribution of H+ from water is considerable.

16. Statement-1 : In a titration of weak monoprotic acid with strong base, the pH at the half equivalent point ispKa.Statement-2 : At half equivalence point, it will form acidic buffer at it's maximum capacity where[acid] = [salt].

17. Statement-1 : Solubility of AgCl in NH3(aq) is greater than in pure water.Statement-2 : When AgCl dissolve in NH3(aq), complex ion [Ag(NH3)2

+] formation takes place and solubilityequilibrium of AgCl shifted in forward direction.

18. Statement-1 : In the titration of Na2CO3 with HCl using methyl orange indicator, the volume required at theequivalence point is twice that of the acid required using phenolphthalein indicator.Statement-2 : Two mole of HCl are required for the complete neutralization of one mole of Na2CO3.

TRUE / FALSE

19. The ionic product of water changes if a few drops of acid or base are added to it .

20. The reaction, HCN + OH CN + H2O is displaced to the right indicating that the acid strength of HCNis greater than water & the base strength of CN is greater than that of OH.

21. All aqueous solutions whether neutral, acidic or basic contain both H+ & OH ions .

22. The ionic product of a saturated solution is equal to solubility product constant of its solute.

23. A salt of strong acid with a strong base does not undergo hydrolysis .

24. To a solution of 20 ml of 0.1 M acetic acid, a solution of 0.1M NaOH is added from burette. If ‘r’ is the ratio of

]acid[]salt[

, the pH changing with respect to r when 5 ml of the alkali have been added is 303.23

.

FILL IN THE BLANKS25. The dissociation constant of NH4OH is 1.8 × 105 . The hydrolysis constant of NH4

+ ions at 25º C would be______ .

26. The colour of unionized form of phenolphthalein is ______ whereas that of ionized form is ______

27. The smaller the value of Ka of a weak acid, ______ is the hydrolysis constant of its conjugate base .

28. At 90°C, pure water has [H+] = 10–6 M, then the value of Kw at this temperature would be _________.

29. The pH of pure water ____________ with increase of temperature.

30. In a mixture of weak acid (HA) and its salt (NaA), the ratio of concentration of salt to acid is increased tenfold. The pH of the solution would ___________ by _____________unit.

31. An indicator (Hn) will exhibit the colour of its ionized form (n–) when the solution contains more than___________ % of the indicator in n– form.





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PART - I : MIXED OBJECTIVE

Single Correct Answer Type1. If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 M HCOOH. the pH of the resulting solution is :

(Ka = 1.8 × 10–4, log 18 = 1.26)(A) 3.74 (B) 5.64 (C) 7.57 (D) 3.42

2. When 100 ml of 0.4 M CH3COOH are mixed with 100 ml of 0.2 M NaOH, the [H3O+] in the solution is

approximately : [Ka(CH3COOH) = 1.8 × 10–5](A) 1.8 × 10–6 M (B) 1.8 × 10–5 M (C) 9 × 10–6 M (D) 9 × 10–5 M.

3. What % of the carbon in the H2CO3 – HCO3– buffer should be in the form of HCO3

– so as to have a neutralsolution? (Ka = 4 × 10–7)(A) 20 % (B) 40 % (C) 60 % (D) 80%

4. Buffer capacity of a buffer solution is x, the volume of 1 M NaOH added to 100 ml of this solution if change thepH by 1 is :(A) 0.1 x ml (B) 10 x ml (C) 100 x ml (D) x ml

5. What amount of HCl will be required to prepare one litre of a buffer solution of NaCN and HCN of pH 8.5 using0.01 mole of NaCN ? pKb of CN¯ = 4.6 (log 2 = 0.3)(A) 2 × 10–3 mole (B) 8.9 × 10–3 mole (C) 8.9 × 10–2 mole (D) 3 × 10–3 mole

6. To a 200 ml of 0.1 M weak acid HA solution 90 ml of 0.1 M solution of NaOH be added. Now, what volume of0.1 M NaOH be added into above solution so that pH of resulting solution be 5 ? [(Ka(HA) = 10–5](A) 2 ml (B) 20 ml (C) 10 ml (D) 15 ml

7. 50 ml of 0.1 M NaOH is added to 60 ml of 0.15 M H3PO4 solution (K1, K2 and K3 for H3PO4 are 10–3, 10–8

and 10–13 respectively). The pH of the mixture would be about : (log 2 = 0.3)(A) 3.1 (B) 5.5 (C) 4.1 (D) 6.5

8. The correct relationship between the pH of isomolar solutions of Na2O (pH1), Na2S (pH2) Na2Se(pH3) andNa2Te(pH4) is :(A) pH1 > pH2 > pH3 > pH4 (B) pH1 < pH2 < pH3 < pH4(C) pH1 < pH2 < pH3 = pH4 (D) pH1 > pH2 = pH3 > pH4

9. The pH of which salt is independent of its concentration ?1. (CH3COO)C5H5NH 2. NaH2PO4 3. Na2HPO4 4.NH4CN(A) 1, 2, 3, 4 (B) 1, 4 (C) 2, 3 (D) 1, 2, 3

10. What will be the pH at the equivalence point during the titration of a 100 ml 0.2 M solution of CH3COONa with0.2 M solution of HCl ? Ka = 2 × 10–5.(A) 3 – log 2 (B) 3 + log 2 (C) 3 – log 2 (D) 3 + log 2

11. 1 M benzoic acid (pKa = 4.20) and 1M C6H5 COONa solutions are given separately. What is the volume ofbenzoic acid required to prepare a 300 ml buffer solution of pH = 4.5 ? [log 2 = 0.3](A) 200 ml (B) 150 ml (C) 100 ml (D) 50 ml

12. What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH.(A) 2 log 3 (B) 2 log (1/4) (C) 2 log (2/3) (D) 2 log 2





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13. A weak acid (HA) after treatment with 12 ml of 0.1 M strong base (BOH) has a pH of 5. At the end point, thevolume of same base required is 27 ml. Ka of acid is : (log2 = 0.3)(A) 1.8 × 10–5 (B) 8 × 10–6 (C) 1.8 × 10–6 (D) 8 × 10–5

14. To prepare a buffer of pH 8.26 amount of (NH4)2 SO4 to be added to 500 ml of 0.01 M NH4OH solution [pKa(NH4

+) = 9.26] is :(A) 0.05 mole (B) 0.025 mole (C) 0.10 mole (D) 0.005 mole

15. What fraction of an indicator Hln is in basic form at a pH of 6 if the pKa of the indicator is 5 ?

(A) 21

(B) 111

(C) 1110

(D) 101

16. An acid-base indicator which is a weak acid has a pKIn value = 5.45. At what cocentration ratio of sodiumacetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugatebase forms? [pKa of acetic acid = 4.75, log 2 = 0.3](A) 4 : 1 (B) 6 : 1 (C) 5 : 1 (D) 3 : 1

17. Pure water is added into the following solutions causing a 10% increase in volume of each. The greatest %change in pH would be observed in which case (A), (B), (C) or (D) ?(A) 0.1 M NaHCO3 (B) 0.2 M NaOH(C) 0.3 M NH3 – 0.2 M NH4

+ system (D) 0.4 M CH3COONH4

18. A well is dug in a bed of rock containing fluorspar (CaF2). If the well contains 20000 L of water, what is theamount of F– in it ? Ksp = 4 × 10–11 (101/3 = 2.15)(A) 4.3 mol (B) 6.8 mol (C) 8.6 mol (D) 13.6 mol

19. What is the minimum pH when Fe(OH)3 starts precipitating from a solution containing 0.1M FeCl3? Ksp ofFe(OH)3 = 8 × 10–13 M3

(A) 3.7 (B) 5.7 (C) 10.3 (D) 8.3

20. The solubility product of AgCl is 10–10. The minimum volume (in L) of water required to dissolve 1.722 mg ofAgCl is : (molecular weight of AgCl = 143.5).(A) 10 lt. (B) 2.2 lt. (C) 1.2 lt. (D) 20 lt.

21. A solution prepared by dissolving 2.8 g of lime (CaO) in enough water to make 1.00 L of lime water(Ca(OH)2(aq.)). If solubility of Ca(OH)2 in water is 1.48 gm/lt. The pH of the solution obtained will be :[log 2 = 0.3, Ca = 40 , O = 16, H = 1](A) 12.3 (B) 12.6 (C) 1.3 (D) 13

22. At what molar concentration of HCl will its aqueous solution have an [H+] to which equal contributions comefrom HCl and H2O ?

(A) 60 × 10–7 M (B) 50 × 10–8 M (C) 40 × 10–9 M (D) 30 × 10–8

23. The best explanation for the solubility of MnS in dilute HCl is that :(A) Solubility product of MnCl2 is less than that of MnS.(B) Concentration of Mn2+ is lowered by the formation of complex ions with chloride ions.(C) Concentration of sulphide ions is lowered by oxidation to free sulphur.(D) Concentration of sulphide ions is lowered by formation of weak acid H2S.

24. At 25ºC, the solubility product values of AgCl and AgCNS are 1.8 x 1010 and 1.6 x 1011 respectively. Whena solution is saturated with both solids, calculate the ratio [Cl] / [CNS] and also [Ag+] in the solution.(A) 1.125, 4 × 106 M (B) 11.25, 1.4 × 105 M (C) 1.25, 4 ×105 M (D) 1.25, 4 × 106 M

25. Arrange in increasing order of solubility of AgBr in the given solutions.(i) 0.1 M NH3 (ii) 0.1 M AgNO3 (iii) 0.2 M NaBr (iv) pure water(A) (iii) < (ii) < (iv) < (i) (B) (iii) < (ii) < (i) < (iv) (C) (iii) < (ii) = (i) < (iv) (D) (ii) < (iii) < (iv) < (i)





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One or More Than One Correct Type

26. Which is/are correct statements :

(a) when 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with

volume of HCl added will be (as shown in figure):

(b) variation of degree of dissociation with concentration for a weak electrolyte at a particular temperature

is best represented by (as shown in figure) :

(c) 0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. The difference in pH between 1/4 and3/4 stages of neutralization of acid will be 2 log 3.

(A) a & c (B) b & c (C) a, b & c (D) b only

27. Choose the correct statement :(A) pH of acidic buffer solution decrease if more salt is added.(B) pH of acidic buffer solution increases if more salt is added.(C) pH of basic buffer solution decreases if more salt is added.(D) pH of basic buffer solution increases if more salt is added.

28. Which of following can act as buffer ?(A) NaCl + NaOH (B) Borax + Boric acid(C) NaH2PO4 + Na2HPO4 (D) NH4Cl + NH4OH.

29. Which of the following will show common ion effect and form a buffer solution ?(A) CH3COONH4 and CH3COOH (B) NH4Cl + NH4OH(C) H2SO4 + NaHSO4 (D) NaCl + NaOH.

30. Which of the following solutions when added to 1L of a 0.01 M CH3COOH solution will cause no change in thedegree of dissociation of CH3COOH and pH of the solution ? Ka = 1.6 × 10–5 for CH3COOH.(A) 0.6 mm HCOOH (Ka = 8 × 10–4) (B) 0.1 M CH3COONa(C) 0.4 mm HCl (D) 0.01 M CH3COOH

31. Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be averagevalue of pH of two solutions ?(A) aqueous HCl of pH = 2, aqueous NaOH of pH = 12.(B) aqueous HCl of pH = 2, aqueous HCl of pH = 4.(C) aqueous NaOH of pH = 12, aqueous NaOH of pH = 10.(D) aqueous CH3COOH of pH = 5, aqueous NH3 of pH = 9. [Ka (CH3COOH) = Kb (NH3)]





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32. Ka values for HA, HB and HD are 10-5, 10-7 and 10-9 respectively. Which of the following will be correct fordecimolar aqueous solutions of NaA, NaB and NaD at 250C?(A) (pH)NaA < (pH)NaB (B) (pH)NaD < (pH)NaB (C) (pH)NaA < (pH)NaD (D) (pH)NaB = 7

33. Which of the following are true for an acidbase titration ?(A) Indicators catalyse the acid base reactions by releasing or accepting H+ ions.(B) Indicators do not significantly affect the pH of the solution to which they are added.(C) Acidbase reactions do not occur in absence of indicators.(D) Indicators have different colours in dissociated and undissociated forms.

34. Let the colour of the indicator Hn (colourless) will be visible only when its ionised form (pink) is 25% or morein a solution. Suppose Hn (pKa = 9.0) is added to a solution of pH = 9.6 predict what will happen ?(Take log 2 = 0.3)(A) Pink colour will be visible. (B) Pink colour will not be visible.(C) % of ionised form will be less than 25%. (D) % of ionised form will be more than 25%.

35. Which of the following mixtures will act as buffer ?(A) H2CO3 + NaOH (1.5 : 1 molar ratio) (B) H2CO3 + NaOH (1.5 : 2 molar ratio)(C) NH4OH + HCl (5 : 4 molar ratio) (D) NH4OH + HCl (4 : 5 molar ratio)

36. 0.1 M CH3COOH is diluted at 25°C (Ka = 1.8 × 10–5), then which of the following will be found correct ?(A) [H+] will increase. (B) pH will increase.(C) number of H+ will increase. (D) all the above are correct.

37. Which of the following statements are correct at 25°C ?(A) pKa for H3O+ is 15.74 (B) pKb for OH– is – 1.74(C) pKa + pKb = pKw for HCl & ClOH (D) degree of dissociation of water is 1.8 × 10–7 %

PART - II : SUBJECTIVE QUESTIONS

1. Classify the following into acid, base and amphiprotic in terms of protonic concept.(i) H2PO2¯ (ii) H2PO3¯ (iii) H2PO4¯ (iv) HPO3

2–

(v) HPO42– (vi) NH4

+ (vii) CH3COOH2+

2. Calculate [H+], [HCOO¯] and [OCN¯] in a solution that contains 0.1M HCOOH (Ka = 2.4 x 10-4) and 0.1 MHOCN (Ka = 4 × 104).

3. A 0.25 M solution of pyridinium chloride, C5H5NH+Cl was found to have a pH of 2.75. What is Kb for pyridine,C5H5N? (log 2 = 0.3)

4. Calculate the percentage hydrolysis & the pH of 0.02 M CH3COONH4 : Kb(NH3) = 1.6 × 10–5, Ka(CH3COOH)= 1.6 × 10–5.

5. Calculate the degree of hydrolysis of 0.005 M K2CrO4. K2 = 5.0 × 10–7 for H2CrO4. (It is essentially strong forfirst ionization).

6. What is the pH of 0.1M NaHCO3? K1 = 4.5 × 10-7, K2 = 4.5 × 10-11 for carbonic acids. (Given : log 2 = 0.3, log3 = 0.48)

7. Calculate pH for :(i) 0.001 N NaOH, (ii) 0.01 N Ca(OH)2 , (iii) 0.01 M Ca(OH)2(iv) 10–8 M NaOH, (v) 0.0008 M Mg(OH)2 .

8. Calculate the pH of the following solutions :(i) 2.21 g of TlOH dissolved in water to give 2 litre of solution.(ii) 0.37 g of Ca(OH)2 dissolved in water to give 500 ml of solution.(iii) 0.32 g of NaOH dissolved in water to give 200 ml of solution.(iv) 1 ml of 12 M HCl is diluted with water to give 1 litre of solution.





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9. Calculate the pH of the resulting solution formed by mixing the following solutions :(a) 20 ml of 0.2 M Ca(OH)2 + 25 ml of 0.1 M HCl(b) 10 ml of 0.01 M H2SO4 + 10 ml of 0.01 M Ca(OH)2(c) 10 ml of 0.1 M H2SO4 + 10 ml of 0.1 M KOH.

10. Calculate pH of the following mixtures. Given the Ka of CH3COOH = 2 × 10–5 and Kb of NH4OH = 2 × 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 HCl.(d) 50 ml of 0.10 M NH4OH + 50 ml of 0.05 M HCl.(e) 50 ml of 0.10 M NH4OH + 50 ml of 0.10 M HCl.(f) 50 ml of 0.05 M NH4OH + 50 ml of 0.05 M CH3COOH.

11. 15 ml sample of 0.15M NH3(aq) is titrated against 0.1M HCl(aq). What is the pH at the end point? Kb ofNH3(aq) = 1.8 x 105.

12. pH at the mid point of titration of 20 ml of 1M formic acid with 1M NaOH is 3.7. Find the pH at the end pointof titration.

13. Calculate the pH during the titration of 40.00 ml of 0.1 M propanoic acid (HPr; Ka = 1 × 10–5) after adding thefollowing volumes of 0.1 M NaOH : (log 2 = 0.3, log 3 = 0.48)(a) 0.00 ml (b) 30.00 ml (c) 40.00 ml (d) 50.00 ml.

14. How many milligrams of gold(III) ion are there per litre of solution in a saturated aqueous solution of gold (III)iodide ? AuI3(s) Au3+(aq) + 3 I (aq) . Ksp = 2.7 × 10–47 [Au = 197]

15. Calculate the solubility of silver phosphate (Ag3PO4) in mg/ml (a) in pure water and (b) in a solution that is0.02 M in AgNO3 . [Ksp(Ag3PO4) = 2.7 × 10–23]

16. A mixture of solid SrSO4 and solid BaSO4 is shaken up with water until equilibrium is established. Given thatKsp (SrSO4) = 7.5 × 10–7, Ksp (BaSO4) = 6 × 10–8, calculate the concentrations of Sr2+, Ba2+, SO4

2– in thesolution at equilibrium.

17. Equal volumes of 0.002 M solution of sodium iodate and cupric chlorate are mixed together. Will it lead toprecipitation of copper iodate? (For cupric iodate KSP = 7.4 × 10–8).

18. What is the maximum concentration of equimolar solution of ferrous sulphate and sodium sulphide so thatwhen mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide KSp = 6.25 × 10–18)

19. Although it is a violent poison if swallowed, mercury (II) cyanide, Hg(CN)2 has been used as a typical (skin)antiseptic. (KSP = 1.35 × 10–23, Hg = 200)(a) What is the molar solubility of this salt in pure water ?(b) How many mili grams of Hg(CN)2 dissolve per litre of pure water ?(c) How many mili litres of water are required to dissolve 1.134 g of the salt ?

20. A solution which is 0.1 M in NaI and also 0.1 M in Na2SO4 is treated with solid Pb(NO3)2. Which compound,PbI2 or PbSO4, will precipitate first ? What is the concentration of anions of the least soluble compound whenthe more soluble one starts precipitating ? Ksp(PbI2) = 9 × 10–9 , Ksp(PbSO4) = 1.8 × 10–8

21. Calculate the change in pH when 4 gm of solid NaOH & 10 mmol of H2SO4 are added to a solution of volume5 litre, which was prepared by mixing 20 mmol of HCl, 40 mmol of H2SO4 and 2 gm of NaOH and sufficientwater. Ignore the volume change. (log 2 = 0.3, log 3 = 0.48)





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PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions are having more than one correct option.

1. For a sparingly soluble salt ApBq, the relationship of its solubility product (Ls) with its solubility (S) is :[JEE-2001, 1/35]

(A) LS = Sp+q . pp . qq (B) LS = Sp+q . pq . qp (C) LS = Spq . pp . qq (D) LS = Spq . (pq)p+q

2. 500 ml of 0.2 M aqueous solution of acetic acid is mixed with 500 ml of 0.2 M HCl at 25º C.(a) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.(b) If 6 g of NaOH is added to the above solution, determine final pH. Assume there is no change in volumeon mixing. Ka of acetic acid is 1.75 × 10-5 M. [JEE-2002, 5/60]

3. Will the pH of water be same at 4ºC and 25º C? Explain. [JEE-2003, 2/60]

4. A solution which is 10–3 M each in Mn2+, Fe2+, Zn2+ and Hg2+ is treated with 10–16 M sulphide ion. If Ksp valuesof MnS, FeS, ZnS and HgS are 10–15 , 10–23 , 10–20 and 10–54 respectively, which one will precipitate first ?

[JEE-2003, 3/84](A) FeS (B) MgS (C) HgS (D) ZnS

5. A weak acid HX has the dissociation constant 1 × 10–5 M. It forms a salt NaX on reaction with alkali. Thepercentage hydrolysis of 0.1 M solution of NaX is : [JEE-2004, 3/84](A) 0.0001% (B) 0.01 % (C) 0.1 % (D) 0.15 %

6. 0.1 M NaOH is titrated with 0.1 M HA till the end point; Ka for HA is 5.6 × 10–6 and degree of hydrolysis is lesscompared to 1. Calculate pH of the resulting solution at the end point. [JEE-2004, 2/60]

7. 0.1 mole of CH3NH2 (Kb = 5 × 10–4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+

concentration in the solution? What will be the H+ concentration in the solution? [JEE-2005, 3/84](A) 8 × 10–2 M (B) 8 × 10–11 M (C) 1.6 × 10–11 M (D) 8 × 10–5 M

8. Ag+ + NH3 [Ag(NH3)]+ ; K1 = 3.5 × 10–3

[Ag(NH3)]+ + NH3 [Ag(NH3)2]+ ; K2 = 1.8 × 10–3

then the overall formation constant of [Ag(NH3)2]+ is : [JEE-2006, 3/184](A) 6.3 × 10–6 (B) 6.3 × 106 (C) 6.3 × 10–9 (D) none of these

9. 2.5 ml of 52

M weak monoacidic base (Kb = 1 x 10–12 at 25° C) is titrated with 152

M HCl in water at 25°C.

The concentration of H+ at equivalence point is (Kw = 1 x 10–14 at 25°C) [JEE-2008, 3/163](A) 3.7 x 10–14 M (B) 3.2 x 10–7 M (C) 3.2 x 10–2 M (D) 2.7 x 10–2 M

10. Solubility product constant (Ksp) of salts of types MX, MX2 and M3X at temperature T are 4.0 × 10–8,3.2 × 10–14 and 2.7 × 10–15, respectively. Solubilities (mol dm–3) of the salts at temperature T are in the order:

[JEE-2008, 3/163](A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2

11. The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10–4 . The pH of 0.01 M solution ofits sodium salt is : [JEE-2009, 4/160]

12.* Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa of identical concentrations are provided. Thepair (s) of solutions which form a buffer upon mixing is (are) : [JEE-2010, 3/163](A) HNO3 and CH3COOH (B) KOH and CH3COONa(C) HNO3 and CH3COONa (D) CH3COOH and CH3COONa





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13. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blueis : [JEE-2010, 3/163]KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2FeCl3 K2CO3 NH4NO3 LiCN

14. In 1L saturated solution of AgCl[Ksp(AgCl) = 1.6 × 10–10] 0.1 mol of CuCl[Ksp(CuCI) = 1.0 × 10–6] is added. The resultant concentration of Ag+ in the solution is 1.6 × 10–x. The value of“x” is : [JEE-2011, 4/160]

15. The unbalanced chemical reaction given in List I show missing reagent or condition (?) which are provided inList II. Match List I and II and select the correct answer using the code given below the lists : [JEE-2013]

List I List II

P. PbO2 + H2SO4 ? PbSO4 + O2 + other product 1. NO

Q. Na2S2O3 + H2O ? NaHSO4 + other product 2. I2

R. N2H4 ? N2 + other product 3. Warm

S. XeF2 ? Xe + other product 4. Cl2

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. 1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is : [AIEEE-2002](1) not a buffer solution with pH < 7 (2) not a buffer solution with pH > 7(3) a buffer solution with pH < 7 (4) a buffer solution with pH > 7

2. Species acting as both bronsted acid and base is : [AIEEE-2002](1) HSO4

– (2) Na2CO3 (3) NH3 (4) OH–

3. The solubility of Mg(OH)2 is s moles/litre. The solubility product under the same condition is[AIEEE-2002](1) 4S3 (2) 3S4 (3) 4S2 (4) S3.

4. The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product will be :[AIEEE-2003]

(1) 4 × 10–15 (2) 4 × 10–10 (3) 1 × 10–15 (4) 1 × 10–10.

5. Which one of the following statements is not true ? [AIEEE-2003](1) The conjugate base of H2PO4

– is HPO42–.

(2) pH + pOH = 14 for all aqueous solutions at 25ºC.(3) The pH of 1 × 10–8 M HCl is 8.(4) 96, 500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent ofcopper at the cathode.

6. When rain is accompanied by a thunderstorm, the collected rain water will have a pH value :[AIEEE-2003](1) slightly lower than that of rain water without thunderstorm.(2) slightly higher than that when the thunderstorm is not there.(3) uninfluenced by occurrence of thunderstorm.(4) which depends on the amount of dust in air.

7. The conjugate base of H2PO4– is : [AIEEE-2004]

(1) PO43– (2) P2O5 (3) H3PO4 (4) HPO4

2–

8. The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is s. The corresponding solubility product isKsp.. s is given in terms of Ksp by the relation : [AIEEE-2004](1) s = (Ksp/128)1/4 (2) s = (128Ksp)

1/4 (3) s = (256Ksp)1/5 (4) s = (Ksp/256)1/5

9. What is the conjugate base of OH– ? [AIEEE-2005](1) O2 (2) H2O (3) O– (4) O2–





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10. The solubility product of a salt having general formula MX2, in water is : 4 × 10–12. The concentration of M2+

ions in the saturated aqueous solution of the salt is : [AIEEE-2005](1) 2.0 × 10–6 M (2) 1.0 × 10–4 M (3) 1.6 × 10–4 M (4) 4.0 × 10–10 M

11. Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be [AIEEE-2005](1) 3.98 × 108 (2) 3.88 × 106 (3) 3.68 × 10–6 (4) 3.98 × 10–6

12. The first and second dissociation constants of an acid H2A are 1.0 × 10–5 and 5.0 × 10–10 respectively. Theoverall dissociation constant of the acid will be : [AIEEE-2007, 3/120](1) 5.0 × 10–15 (2) 0.2 × 105 (3) 5.0 × 10–5 (4) 5.0 × 1015

13. The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA, in which 50% of the acidis ionized, is : [AIEEE-2007, 3/120](1) 9.5 (2) 7.0 (3) 4.5 (4) 2.5

14. In a saturated solution of the sparingly soluble strong electrolyte AgO3 (Molecular mass = 283) the equilibriumwhich sets in is :

AgO3(s) Ag+(aq) + O–3(aq)

If the solubility product constant Ksp of AgO3 at a given temperature is 1.0 × 10–8, what is the mass of AgO3

contained in 100 ml of its saturated solution ? [AIEEE-2007, 3/120](1) 1.0 × 10–7 g (2) 1.0 × 10–4 g (3) 28.3 × 10–2 g (4) 2.83 × 10–3 g

15. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solutionof the corresponding salt, BA, will be : [AIEEE-2008, 3/105](1) 4.79 (2) 7.01 (3) 9.22 (4) 9.58

16. Solid Ba(NO3)2 is gradually dissolved in 1.0 × 10–4 M Na2CO3 solution. At what concentration of Ba2+ will aprecipitate begin to form ? (Ksp for BaCO3 = 5.1 × 10–9) : [AIEEE-2009, 4/144](1) 5.1 × 10–5 M (2) 8.1 × 10–8 M (3) 8.1 × 10–7 M (4) 4.1 × 10–5 M

17. Three reactions involving H2PO4– are given below : [AIEEE-2010, 4/144]

(i) H3PO4 + H2O H3O+ + H2PO4

–

(ii) H2PO4– + H2O HPO4

2– + H3O+

(iii) H2PO4– + OH– H3PO4

+ O2–

In which of the above does H2PO4– act as an acid ?

(1) (ii) only (2) (i) and (ii) (3) (iii) only (4) (i) only

18. In aqueous solution the ionization constants for carbonic acid are : [AIEEE-2010, 4/144]K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11

Select the correct statement for a saturated 0.034 M solution of the carbonic acid.(1) The concentration of CO3

2– is 0.034 M.(2) The concentration of CO3

2– is greater than that of HCO3–.

(3) The concentration of H+ and HCO3– are approximately equal.

(4) The concentration of H+ is double that of CO32–.

19. Solubility product of silver bromide is 5.0 × 10–13 . This quantity of potassium bromide (molar mass taken as120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :

[AIEEE-2010, 4/144](1) 1.2 × 10–10 g (2) 1.2 × 10–9 g (3) 6.2 × 10–5 g (4) 5.0 × 10–8 g

20. At 25°C, the solubility product of Mg(OH)2 is 1.0 10–11. At Which pH, will Mg2+ ions start precipitating in theform of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ? [AIEEE-2010, 4/144](1) 9 (2) 10 (3) 11 (4) 8

21. The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :(1) 3 × 10–1 (2) 1 × 10–3 (3) 1 × 10–5 (4) 1 × 10–7 [AIEEE-2012]

54. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create anaqueous solution with pH of 2 ? [AIEEE-2013](1) 0.1 L (2) 0.9 L (3) 2.0 L (4) 9.0 L





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NCERT QUESTIONS

1. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species :HNO2, CN–, HClO4, F

–, OH–, CO32–, and S2–

2. Which of the followings are Lewis acids?H2O, BF3, H

+, and NH4+

3. What will be the conjugate bases for the Bronsted acids:HF, H2SO4 and HCO–

3?

4. Write the conjugate acids for the following Brönsted bases : NH2

–, NH3 and HCOO–.

5. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what is its pH?

6. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

7. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively.Calculate the ionization constants of the corresponding conjugate base.

8. The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solutionof phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

9. The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1M solution.How will this concentration be affected if the solution is 0.1M in HCl also ? If the second dissociationconstant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

10. The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

11. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of theanion, the ionization constant of the acid and its pKa.

12. Assuming complete dissociation, calculate the pH of the following solutions :(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

13. Calculate the pH of the following solutions :a) 2 g of TlOH dissolved in water to give 2 litre of solution.b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

14. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution andthe pKa of bromoacetic acid.

15. The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

16. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociationaffected when its solution also contains(a) 0.01M (b) 0.1M in HCl ?

17. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution.What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?





VKP SirM.Sc. IT-BHU

18. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below :(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2(c) Human blood, 7.38 (d) Human saliva, 6.4.

19. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively.Calculate corresponding hydrogen ion concentration in each.

20. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations ofpotassium, hydrogen and hydroxyl ions. What is its pH?

21. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium andhydroxyl ions and the pH of the solution.

22. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

23. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and itsdegree of ionization in the solution.

24. The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution andalso its degree of hydrolysis.

25. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

26. Predict if the solutions of the following salts are neutral, acidic or basic :

27. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1Msodium salt solution?

28. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

29. Calculate the pH of the resultant mixtures :a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HClb) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

30. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate theratio of the molarities of their saturated solutions.

31. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead toprecipitation of copper iodate? (For cupric iodate Ksp = 7.4 ×10–8).

32. The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 ×10–13. How manytimes is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

33. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so thatwhen mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp =6.3 × 10–18).

34. What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calciumsulphate, Ksp is 9.1 × 10–6).

35. The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in whichof these solutions precipitation will take place?





VKP SirM.Sc. IT-BHU

EXERCISE # 1PART # I

A-1. (D) A-2.* (ABCD) A-3. (D) A-4. (B) B-1. (D) B-2.* (ABC) B-3. (C)B-4. (A) B-5. (B) B-6. (D) B-7. (B) B-8.* (ABC) C-1. (C) C-2. (B)C-3. (C) C-4. (A) C-5. (D) C-6. (C) C-7. (C) D-1. (C) D-2. (B)D-3. (C) D-4. (B) D-5. (B) D-6. (C) D-7. (A) E-1. (D) E-2. (D)E-3. (C) E-4. (A) F-1. (D) F-2. (A) F-3. (D) F-4. (D) F-5. (B)F-6. (B) F-7. (A) F-8. (C) F-9. (D) F-10. (D) F-11. (A) F-12. (B)F-13. (A) F-14. (B) F-15. (B) F-16. (B) G-1. (C) G-2. (C) G-3. (A)G-4. (A) G-5. (B) G-6. (A) G-7. (D) G-8. (C) G-9. (D) G-10.* (AB)G-11.* (AC) G-12. (A) G-13. (C) H-1.* (ABCD) H-2. (C) H-3.* (C) H-4. (D)H-5. (B) H-6. (C) H-7. (A) H-8. (C) H-9. (C) I-1. (A) I-2. (A)I-3. (A) I-4. (D) I-5. (B) I-6. (D) I-7. (A) I-8. (A) I-9. (B)I-10. (A) I-11. (D) I-12. (A) I-13. (B) I-14. (A)

PART # II1. (C) 2. (C) 3. (B) 4. (A) 5. (C) 6. (B)7. (A) p, q, r, s; (B) p, r ; (C) r ; (D) p, q, s 8. (A) s; (B) p, q; (C – q, s) ; (D – q, r)9. (B) 10. (B) 11. (B) 12. (A) 13. (B) 14. (A) 15. (A)16. (A) 17. (A) 18. (B) 19. F 20. F 21. T 22. T

23. T 24. T 25. 5.56 x 10 -10 26. colourless, pink 27. greater

28. 10–12 M2 29. decreases 30. Increase, one 31. 91


1. (A) 2. (B) 3. (D) 4. (C) 5. (B) 6. (C) 7. (A)8. (A) 9. (A) 10. (A) 11. (C) 12. (D) 13. (B) 14. (B)15. (C) 16. (C) 17. (B) 18. (C) 19. (C) 20. (C) 21. (B)22. (B) 23. (D) 24. (B) 25. (A) 26. (A) 27. (BC) 28. (BCD)29. (AB) 30. (ACD) 31. (AD) 32. (AC) 33. (BD) 34. (AD) 35. (ABC)36. (BC) 37. (BD)

PART # II1. Acidic – (vi), (vii) ; Basic – (i), (iv) ; Amphiprotic – (ii), (iii), (v)

2. [H+] = 8 × 10-3 M, [HCOO¯] = 3 × 10–3 M, [OCN¯] = 5 × 10–3 M 3. Kb = 8 × 10–10

4. 0.625%, pH = 7 5. 2 × 10–3 6. pH = 8.34

7. (i) 11, (ii) 12, (iii) 12.3, (iv) 7.02, (v) 11.2 8. (i) 11.7, (ii) 12.3, (iii) 12.6, (iv) 1.92

9. (a) 12.82 ; (b) 7 ; (c) 1.3. 10. (a) 12.4, (b) 4.7, (c) 7, (d) 9.3, (e) 1.6, (f) 7

11. 5.24 12. pH = 8.7

13. (a) pH = 3 (b) pH = 5.48, (c) pH = 8.85, (d) pH = 12.05

14. 1.97 × 10-7 mg/L 15. (a) 4.19 × 10-4 mg/ml (b) 1.414 × 10-15 mg/ml

16. 8.33 × 10–4 M, 6.67 × 10–5 M, 9 × 10–4 M

17. No precipitate 18. 5 × 10–9 M

19. (a) 1.5 × 10–8 M (b) 3.78 × 10–3 mg (c) 3 × 108 ml 20. PbSO4, 0.02 M 21. 9.78





VKP SirM.Sc. IT-BHU


1. (A) 2. (a) 0.0175%, 1 (b) 4.7573. It will not be same at two different temperatures.4. Lesser the Ksp, more is the precipitation.5. (B) 6. 8.98 9 7. (B) 8. (A) 9. (D) 10. (D)11. 8 12.* (CD) 13. 3 14. (7) 15. (D)

PART # II1. (1) 2. (1) 3. (1) 4. (1) 5. (3) 6. (1) 7. (4)8. (4) 9. (4) 10. (2) 11. (4) 12. (1) 13. (1) 14. (4)15. (2) 16. (1) 17. (1) 18. (3) 19. (2) 20. (2) 21. (3)54. (4)

EXERCISE # 41. NO2

–, HCN, ClO–4, HF, H2O, HCO3

–, HS–

2. BF3, H+, NH4

+

3. NH3, NH4+, CO3

2–

4. NH3, NH4+, HCOOH

5. 2.426. 1.7 × 10–4 M7. F– = 1.5 × 10–11, HCOO– = 5.6 × 10–11, CN = 2.08 × 10–6

8. [phenolate ion] = 2.2 × 10–6, = 4.47 × 10–5, in sodium phenolate = 10–8

9. [HS–] = 9.54 × 10–5, in 0.1 M HCl [HS–] = 9.1 × 10–8 M, [S2–] = 1.2 × 10–13 M, in 0.1 MHCl [S2–] = 1.09 × 10–19 M

10. [Ac–] = 0.00093, pH = 3.0311. [A–] = 7.08 × 10–5 M, Ka = 5.08 × 10–7, pKa = 6.2912. (a) 2.52 (b) 11.70 (c) 2.70 (d) 11.3013. (a) 11.65 (b) 12.21 (c) 12.57 (d) 1.8714. pH = 1.88, pKa = 2.7015. Kb = 1.6 × 10–6, pKb = 5.816. (a) 0.0018 (b) 0.0001817. = 0.005418. (a) 1.48 × 10–7 M (b) 0.063 (c) 4.17 × 10–8 M (d) 3.98 × 10–7

19. (a) 1.5 × 10–7 M (b) 10–5 M (c) 6.31 × 10–5 M (d) 6.31 × 10–3 M20. [K+] = [OH–] = 0.05 M, [H+] = 2.0 × 10–13 M21. [Sr2+] = 0.158 M, [OH–] = 0.3162 M, pH = 13.5022. = 1.63 × 10–2, pH = 3.09. In presence of 0.01 M HCl, = 1.32 × 10–3

23. Ka = 2.09 × 10–4 and degree of ionization = 0.045724. pH = 7.97. Degree of hydrolysis = 2.36 × 10–5

25. Kb = 1.5 × 10–9

26. NaCl, KBr solutions are neutral, NaCN, NaNO2 and KF solutions are basic and NH4NO3 solution is acidic.27. (a) pH of acid = solution = 1.9 (b) pH of its salt solution = 7.928. pH = 6.7829. (a) 12.6 (b) 7.00 (c) 1.330. Silver chromate is more soluble and the ratio of their molarities = 91.931. No precipitate32. Silver benzoate is 3.317 times more soluble at lower pH33. The highest molarity for the solution is 2.5 × 10–9 M34. 2.43 litre of water35. Precipitation will take place in cadmium chloride solution.

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