Ionic Equilibrium

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1.

SALT

2

2.

IONIC DISSOCIATION

2

3.

OSTWALD’S DILUTION LAW

2

4.

STRENGTHS OF ACIDS AND BASES

4

5.

SELF IONIZATION OF WATER

6

6.

HYDROGEN ION / HYDROXYL ION CONC. OF ACID / BASE

6

7.

pH SCALE

7

8.

COMMON ION EFFECT

8

9.

BUFFER SOLUTIONS

8

10.

BUFFER ACTION

9

11.

SOLUBILITY

12

12.

HYDROLYSIS OF SALT

14

13.

DEGREE OF HYDROLYSIS

14

14.

THEORY OF INDICATOR

18

CONTENTS

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1. SALT

A substance which ionizes in water to produce ions other than H+and OHis called a salt.

Types of Salts:

Neutral Salts :Those salts whose aqueous solutions neither turn blue litmus red nor red litmus blue are called neutral salts. These are prepared by the neutralization of strong acid and strong base. e.g. NaCl, K2SO4, KNO3 etc.

Basic Salts :Those salts whose aqueous solution turn red litmus blue are called basic salts. These are formed by the neutralization of strong bases with weak acids. e.g. Na2CO3, CH3COONa

Mixed salts :Salts formed by the neutralization of one acid by two bases or one base by two acids are called mixed salts. e.g. CaOCl2

Double Salts :Acompound of two salts whose aqueous solution shows the tests for all constituent ions is called double salt e.g.

Mohr Salt FeSO4. (NH4)2SO4. 6H2O

PotashAlum K2SO4.Al2(SO4)3. 24H2O.

Complex Salts :Acompound whose solutions does not give test for the constituent ions is called a complex salt. e.g.

K4[Fe(CN)6] Li (AlH4)

2. IONIC DISSOCIATION

The process in which molecules (acids, bases, and salts) when dissolved in water or when melted break into ions is called ionic dissociation.

Electrolytes. Hence substances which dissociate into ions in aqueous solutions are called electrolytes. e.g. NaCl, NaNO3, HCl, K2SO4etc.

Those electrolytes which dissociate almost completely into ions are known as strong electrolytes. e.g. HCl, HBr, HI, HClO4, NaCl, Na2SO4, KNO3etc.

Those electrolytes which dissociate partially are called weak electrolytes. e.g. H3PO4, HF, H2CO3,HCN, CH3COOH, NH4OH, etc.

.

3. OSTWALD’S DILUTION LAW

Ostwald’s pointed out that like chemicalequilibriumin ionic equilibriumwe can applylaw of mass action. An equilibriumbetweenionized and unionized molecules. Consider a binary electrolyte having conc. C and degree of dissociation is.

 

[A ][B ]C C C 2

K , for a weak electrolyte 1  aq C(1 ) [AB] 1  1  A 0 B 0 C AB C At time = 0 At time = t C(1 –) C 

IONIC EQUILIBRIUM

Strong Electrolyte And Weak Electrolyte and Non Electrolytes:

Those electrolytes which do not dissociate into ions in aqueous solutions are called non electrolytes. Molecules of the substances which do not dissociate into ions in aqueous solutions are called non- electrolytes. e.g. sugar, urea, etc.

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Ionic Equilibrium Keq K C2 ,  eq C

If 1 mole ofAB is present in ‘V’litre of solution. 1 C V   KeqV Keq C C  K C Conc. of A+= Conc. B– eq C Limitation :

This is only for weak electrolytes not for strong electrolytes.

This law is not applicable for strong electrolyte because strong electrolytes are almost completely ion-C

ized at all dilution and hence does not give accurate results. (i)

(ii)



Dissociation of Weak Acids and Bases :

Dissociation of Weak acids : Consider the dissociation of a weak acid HAin water, represented by the equation  – HA + H2O HA H3O + A H+ + A– Initial conc. moles/litre

in moles/lit. Final concentration

 Ka= 1α

This equation is referred to as Ostwald dilution law. In case of weak acids the degree of dissociation is very small, therefore (1 –) may be taken to be equal to unity

Hence

Since

Ka

Also as=  VKa C

This shows that decree of dissociation is inversely proportional to square root of concentration and directly proportional to square root of dilution of the solution.

Dissociation of Weak Base in Water :

The dissociation of a weak base can be represented in the same manner as a weak acid. e.g. C BOH C – C O O (initial)

Final concentration in moles/lit.

B+ OH– C C. C 0 0 (C – C) C C C α2 in moles/lit.

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Where C = Initial conc. in moles/litre

 = Degree of dissociation

C C C2α2 C α2

B OH

 As the base is weak

Kb= C (1α) 1α CBOH   1  1 –= 1. 2   Hence, Kb= C  = Kb/ c Kb [OH–] = C= C × C [OH–] = KbC

4. STRENGTHS OF ACIDS AND BASES

The strength of an acid/base depend upon the number H+/OHpresent in solution.

(i)

(ii) Since=  

VKa or= VKb

increases when V increase

Ionization increases with dilution, hence number of H+/OH–increases. (iii) At infinite dilution the ionization of all acids and bases tends to become almost equal and all acids

and bases behave equally strong at infinite dilution.

All mineral acids or bases which ionize fully at all dilutions are considered as strong acids. While acids/bases like CH3COOH/NH4OH respectively which ionize to a less extent are called weak acids/bases. The relative strenths is generally compared in terms of their dissociation constants. We known Ka1= C1111 (iv) 2 2 Ka2= C22 21 Ka1 Ka1C2 α1 C1  = K when C C a 2C1 α2 Ka2 1 2 C2 Ka1 α1 when C 1 2C α2 α1 Ka2 Kb1C2  when C C α2 Kb 2 C1 1 2 Kb1 α1 when C C 1 2 α2 Kb2

(v) Strength of all strong acids/bases in water is same. This is called levelling effect

Ex.1 Calculate the degree of ionization of 0.01 M solution of HCN, K of HCN is 4.8 × 10–10. Also a

calculate hydronium ion concentration.

The ionization of HCN may be represented as, HCN(aq)H2O() CN -(aq)H3O (aq)

Sol.

– 

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Ionic Equilibrium

[HCN]C(1 –) [CN–]C [H3O ]C

where C = concentration of HCN. Applying law of chemical equilibrium

[CN–][H O] (C)(C) C2 3

K a   

[HCN] C(1 –) (1 –)

Since is very small as compared with unity therefore, 1 –  in the denominator may be taken as 1.

2 Ka C –10 Ka 4.810  2.210–4.    C 0.01  –4 –6 –1 [H3O ]C 0.012.210 2.210 mol L . –4 –5

Ex.2 The Ka for formic acid and acetic acid are 2 × 10 and 2 × 10 respectively. Calculate the relative strength of acids with same molar concentration–

Relative strength of weak acid Sol.

Relative strength for HCOOH to CH3COOH = 10 : 1

The degree of ionization of this weak acid can be calculated by the approximate relation :

–5

Ka 1.810 

0.03310–2.

  

C 0.02

Now let us calculate the degree of ionization when the solution also contains 0.01 M sodium acetate. Sodium acetate being a strong electrolyte would be completely ionized in solution. Let x mol L–1 of acetic acid be ionized.

CH COOH(aq) CH COO–(aq)  H(aq)

x M

3 3

(0.02– x )M x M

CH COONa(aq)CH COO–(aq)  Na(aq)

0.01 M 3 3 K a1 K Relative strength = a2 ( C1= C2) 2 𝑥 10−4 2 𝑥10−5 𝑐2 𝑐1∗ 𝑘𝑎 1 𝑘𝑎2 = 𝑎1 𝑎2

Calculate the degree of ionization of 0.02 M acetic acid if its K 1.810–5. What would be the

Ex.3 a

degree of ionization if the solution also contains 0.01 M sodium acetate ?

CH COOH(aq) CH COO–(aq)H(aq)

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[H]x mol L–1

– –1 –1

[CH3COO ](x0.01) mol L 0.01 mol L

[ x is very small as compared to 0.01]

–1 –1

[CH3COOH](0.02 – x) mol L 0.02 mol L

[H][CH COO–] K a  3 [CH3COOH] 1.810–5 (x)(0.01) (0.02) –5 x1.810 0.023.610–5M 0.01 –5 x 3.610 1.810–3 Degree of ionization,  0.02 0.02

Thus, it may noted that the degree of ionization of acetic acid has decreased from 310–2 due to the presence of sodium acetate.

5. SELF IONIZATION OF WATER

to 1.810–3

Water can behave both as an acid as well as a base. This behaviour of water is due to ionization ofwater to form protons and hydroxyl ions.

+ – H2O Hence K = H + OH C – CH OH CH2O  K CH2O= CH+ COH– K CH2O= Kw= CH+COH–

Kw is the Ionic Product of water and may be defined as the product of concentration of ions [(H+) and (OH–)] ions. Its value depends only on temperature and is found to be 1 × 10–14at 25ºC

At –14 2 2 OºC 10ºC 25ºC Kw= 0.11 × 10 mol /lit –14 2 2 Kw= 0.30 × 10 mol /lit –14 2 2 Kw= 1 × 10 mol /lit –14 2 2 100ºC Kw= 7.5 × 10 mol /lit –7

For pure water CH+= COH– = 10 mol/lit at 25ºC

Degree of dissociation of pure water at 25ºC

For pure water CH+= COH–.

–14 2 2

Also at room temperature CH+COH–= 10 mol /lit

–7  CH+= 10 mol/lit  C= 10–7mol/lit 107 107 = 1.8 × 10–9 =  C 55.6

Hence degree of dissociation = 1.8 × 10–9  % degree of dissociation = 1.8 × 10–7

6. HYDROGEN ION / HYDROXYL ION CONC. OF ACID/ BASE

(i) In case of strong acids (or bases) concentration in water solution is taken as equal to normality of the acid/base since they ionize completely.

[H+] = Normality of acid = Molarity × Basicity

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Ionic Equilibrium

(ii) In case of weak acids/Bases, the H+/OHconcentration is less than normal and may be calculated

by using Oswald's dilution law.

[H+] = [OH–] = = C= N × KaC = C= N × KbC

7. pH SCALE :

It may be defined in number of ways.

pH calculation of solution of a mixture of two weak Acids in water :

Let two weak acids be HA and HB and their conc. are C and C ,1 2 1 is the degree of dissociation of HA in presence of HB (due to common ion effect) and2be degree of dissociation of HB in presence of HA. In aqueous solution of HA and HB following equilibrium exists.

HA + H O(l) H O+ + A– C  2 conc. at eq. C (1 –) 3 C + C 1 1 1 1 2 2 1 1 H O+ B– C  HB + H O(l) + + C  ) 2 C (1 –) 3 (C  2 2 1 1 2 2 2 2   [H3O ][A ][C11C12][C12] Ka[ HA] C1(1 1) [HA]   [H3O ][B ][C11C22][C22] Ka[ HB] [HB] [C2(1 2)]  pH– log[H ]– log[C11C22]

pH of a DibasicAcid and Polyprotic Acid :

Let’s take the eg. of a dibasic acid H A.Assuming both dissociation is weak. Let the initial conc. of H A

2

is C and and be degree of dissociation for first and second dissocation.

2 1 H A 2 HA– C (1 –) + H+ C + C  2 C(1 –1) 1 2 1 A– – C . 1 2 HA– H+ C + C  + C (1 –) 1 2 1 1 2 1 2   [HA ][H ] Ka1 [H2A] [C1(1 2)][C1C12] Ka1 C(1 1)

(i) The pH value of a solution is equal to the negative power to which 10 must be raised in order to express [H+] concentration [H+] = 10–pH.

It can also be defined as the negative logarithm of its [H+] ion concentrations

pH = –log[H+]

pH values do not give instantaneous idea above the relative strengths of the solution (ii)

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  

[H ][A ][C1C12][C12]

Ka2

[HA ] [C1(1 2)]

After solving for and. We can calculate the H+conc.

1 2

[H+] = C+ C  1 1 2

pH = – log [C1+ C 1 2]

pOH SCALE :

It is defined as the negative logarithms of hydroxyl ions concentration. pOH = –log (OH–).

Also it is known that [H+] [OH–] = 10–14= Kw –log[H+] (+) –log[OH] = 14 = pK

w

 pH + pOH = pKw= 14. Ex.4

Sol.

The pH of a 0.05 M solution of H2SO4in water is nearly?

+

pH= – log10 H

The concentration of H+ions is expressed in gm equivalent

Molarity of H2SO4= 0.05   or Normality = 0.05 × 2 = 0.1 pH = –log 0.1 pH = 1

Calculate the pH of solution having H+ion concentration of 5 × 10–4mole/litre

[H+] ion concentration = 5 × 10–4mole/litre

pH = – log [5 × 10–4] = – (log 5 + log 10–4) = – 0.7 + 4 = 3.3 Ex.5 Sol.

8. COMMON ION EFFECT

When a solution of weak electrolyte is mixed with another electrolyte which provides one or more ion common with the weak electrolyte the dissociation of weak electrolyte is suppressed. This suppression of dissociation of weak electrolyte on addition of a common ion is called common ion effect.

For example if we consider dissociation of CH3COOH in the presence of CH3COONa we get following situation :

CH3COOH

CH3COO + H– +

– +

CH3COONa CH3COO + Na

In this case the CH3COO ion contributed by CH3COONa suppresses the dissociation of CH3COOH. This suppression of dissociation of CH3COOH is called common ion effect.

BUFFER SOLUTIONS

A buffer is a solution which resists any change in its pH value on either (a) dilution or (b) addition of acid/base. The process by which the added H+/OH–are removed to maintain the pH of solution, is known as buffer action.

Types of Buffer:

9.

1. Simple buffers :Asalt of weak acid and weak base in water e.g. (a)

(b) (c)

NH4CN or CH3COONH4. Proteins and amino acids.

A mixture of an acid salt and normal salt of a poly basic acid e.g. NaHCO3 + Na2CO3

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Ionic Equilibrium

2. Mixed Buffer : These are of two types

(a) Acid Buffer Solution : An acidic buffer solution of a weak acid and its salt with strong base.

e.g. CH3COOH + CH3COONa

 Initial conc. (mol/lit.) C CH3COOH (C – C) 0 0 – + CH3COO + C H C Final conc. (mol/lit.)

Initial conc.(mol/lit.) C1 CH3COONa 0 0 0 – + CH3COO + Na Final conc.(mol/lit.)

Hence in the solution

C1 C1

CH3COOH concentration = (C – C)

CH3COO concentration = C+ C1 H+concentration = C = CH+ .

But sinceis very small [CH3COOH] = (C – C) C. [CH3COO-] = C+ C1C1 CCH COO–CH 3 C Ka = C => CH+= Ka× C 1 CH3COOH C1  pH = – log CH+= PKa+ log C [Salt]  pH = pKa+ log [Acid]

This equation is called Henderson’s Equation.

10. BUFFER ACTION :

When a few drops of acid is added then the H+ions from the acid combines with excess of CH COO– 3 +

ions to form CH3COOH. Hence there is no rise in [H ] ion concentrations but due to the consumptions

– –

of CH3COO the concentration of CH3COO decreases and the concentration of CH3COOH increases hence the ratio of[Salt]

[Acid] decreases slightly. Thus the pH change is minimal, meaning the solution has

to form acetate ion and water. OH–+ CH COOHCH COO–+ H O

resisted the change in pH. In other wards one can say that the pH change which is very minimal is not due to change in the concentration of [H+] but due to change in buffer capacity of solution.

On the other hand when NaOH is added, the [OH–] ions of the base reacts with the unionized CH COOH3

3 3 2

Ex.6 Calculate the ratio of pH of a solution containing 1 mole of CH COONa + 1 mole of HCl per litre and3 of other solution containing 1 mole CH COONa + 1 mole of acetic acid per litre.3

Case I. pH when 1 mole CH3COONa and 1 mole HCl are present. Sol.

The sodium acetate, being a salt ionizes completely to form CH3COO and Na ion. But acetic acid being a weak acid is ionized to a less extent.Also its ionizations is further suppressed by the acetate ion from sodium acetate.

Let C moles/litre be the concentration of CH3COOH taken and C1be the concentration of CH3COONa.

The degree of dissociation of CH3COOH is𝛼 in the presence of sodium acetate

– +

CH3COOH CH3COO + H (Weakly ionized)

– +

CH3COONa CH3COO + Na (Highly ionized)

– +

CH COONa + HCl3  CH COOH + NaCl3

Before reaction 1 1 0 0

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 [CH COOH] = 1 M3 K a   (K .C) (K )  [H+] = C.= C C = 1 a a C   1 pH = – log K  1 2 a

Case II : pH when 1 mole CH COONa and 1 mole of CH COOH; a acidic buffer solution forms3 [Salt] = 1 M, [Acid] = 1 M [Salt] 3   pH = – log K + log [Acid] 2 a pH = –log K2 pH1 a 1  2  pH 2 Basic Buffers :

A basic buffer solution consists of a mixture of a weak base and its salt with a strong acid. e.g. mixture of NH4OH and NH4Cl.

+ – NH4OH NH4 + OH Weakly ionized Highly ionized + – NH4Cl NH4 + Cl +

The NH4 ions of NH4Cl suppress the ionization of NH4OH due to common ion effect.

Let the concentrations of NH4OH taken be C moles/lit. and-be the degree of ionization after adding NH4Cl. Let C1be the concentration of NH4Cl taken.

Initial conc. (mol/lit.) C

NH4OH

0 0 + –

NH4 OH Final conc. (mol/lit.)

Initial conc.(mol/lit.) (C – C) C1 NH4Cl 0 C C 0 0 + – NH4 + Cl Final conc.(mol/lit.) The concentration of C1 C1 + [NH4 ] = C1+ C  C1[  1] [NH4OH] = C – C  C [  1] [OH–] = C. C C – NH OH CNH4OH Kb= C4  COH–= Kb C  NH4OH NH4 C  C 1 NH4

 pOH = pKb+ log C = pKb+ log

C

NH4OH

[Salt] 

BufferAction :

pOH = pKb+ log [Base]

When few drops of base say NaOH is added then the OH–ions added react with NH4+to form

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Ionic Equilibrium

[NH4 ] –

NH4OH and thus the concentration of [OH ] remains unchanged. But the ratio of[NH

4OH]changes.

Thus the change in pH is very small and that too due to change in buffer capacity.

On the other hand when a few drops of acid (Say HCl) is added, then the [H+] ions of acid combine with

+ + +

excess of NH4OH to form H2O and NH4 ions. i.e. NH4OH + H NH4 + H2O.

Thus the addition of acid does not increase the H+ions but since the concentration of NH OH decreases 4

[NH4 ] +

and [NH4 ] ion conc. increases, the ratio[NH increases and thus pH changes infinitively.

4OH]

Ex.7 Abuffer of pH 9.26 is made by dissolving x moles of ammonium sulphate and 0.1 mole of ammonia into 100 mL solution. If pK of ammonia is 4.74, calculate value of x.

b

Sol. (NH ) SO 4 2 4 2 NH4 + SO42

Thus, every one mole of (NH ) SO gives two moles of NH4 2 4 4.

 millimoles of NH , (NH OH) = 1003 4 0.1 = 10 millimol millimoles of (NH ) SO = 1004 2 4 x = 100 x 1000 millimol  millimoles of NH4 = 200x = 200 x millimol pH = 9.26 pOH = 14 – 9.26 = 4.74 [NH4]    pOHpK blog [ NH4OH] 200x 4.74 = 4.74 + log log 20x = 0 20 x = 1 1 x = = 0.05. 10 20 Buffer Capacity :

It is defined as the number of moles of the acid/base added to the buffer solution to produce a change in pH by one unit.

number of moles of acid/base added Buffer Capacity = change in pH

In general Buffer Capacity would be maximum when both components are present in equimolar proportions.

Agiven acid must have a pH ranging between pKa+ 1 to pKa– 1 if it is to be used in buffer solution. The best buffer will have the acid with pH = pKa.

What amount of sodium propanoate should be added to one litre of an aqueous solution containing Ex.8

0.02 mole of propanoic acid (Ka= 1.0 × Using the expression

[Salt]

10–5 at 25ºC) to obtain a buffer solution of pH 6. Sol.

pH = pKa + log

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[Salt] log[0.02 M] we get, 6 = –log(1.0 ×10–5)+

[Salt] Which gives 6 = 5 + log

[0.02 M] [Salt]

= 10 or [Salt] = 0.2 M [0.02 M]

Ex.9 What will be the pH of the solution, if 0.01 mole of HCl is dissolved in a buffer solution containing 0.03 mole of propanoic acid (Ka = 1.0 × 10–5) and 0.02 moles of salt, at 25ºC.

[Salt] Sol. pH = pKa + log [Acid] (0.020.01) = –log(1.0 × 10–5) + log(0.030.01) 1 = 5 + log  4 = 5 – 0.6 = 4.4

Ex.10 What amount of HCl will be required to prepare one litre of a buffer solution (containing NaCN and HCN) of pH 10.4 using 0.01 mole of NaCN. Given Kion (HCN) = 4.1 × 10–10

Sol. The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have. [NaCN] = (0.01 – x)

[HCN] = x

Substituting these values along with pH and Kain the expression. [Salt]

pH = – log Ka + log

[Acid]

0.01x We get 10.4 = –log[4 × 10–10] + log

x 0.01x or 10.4 = 9.4 + log x 0.01x or log = 1 x 0.01x or or = 10  11x = 10–2 x 10–4 x = 9.9 × M

11. SOLUBILITY

It is the amount of the solute in gram that can be dissolved in 100 gm of a solvent to obtain a saturated solutions at a particular temperature.

However solubility can also be expressed in moles/litre.

Solubility in gm/lit.

Solubility of solution in moles/litre = The factors affecting solubility are.

Molecular weight of solute

(a) (b) (c) (d) Nature of solvent Nature of solute Temperature Pressure

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Ionic Equilibrium

Solubility Product :

If a sparingly soluble salt is added in water, very little amount of it dissolves in water, and thus solution becomes saturated, but a highly soluble salt saturation is reached on dissolving more of salt.

For all salts at saturation an equilibrium is achieved between the undissolved salt and the ions insolutions.

+ ++ A

BA(s) Applying law of mass action we get

[A–] [B]

B (aq) (aq)

+ –

K = [BA]  K[BA] = [B ] [A ]

Since only little of salt dissolves so the concentration of salt remains constant

+ –

For any salt

K[AB] = K × Constant = Ksp= [B ] [A ]

y+ x–

BxAy xB (aq)+ yA (aq)

y+ x x– y

Ksp= [B ] [A ]

Ksp= KcThis expression shows that in a saturated solution of a sparingly soluble salt the ionic product is equal to the solubility product.

Ksp= Solubility Product

It is defined as the product of the molar concentrations of its ion raised to the power equal to its number of ions present at equilibrium representing the ionization of one molecule of salt at a given temperature.

Points to Remember:

(1) (2) (3)

When Ksp= Kisolution saturated When Ki< Kspunsaturated solution When Ki> Kspsuper saturated solution. Hence precipitation occurs to keep Ki= Ksp

Relationships Between Solubility and Solubility Product:

(1) Let the solubility of a salt

BxAyin water be s moles/lit. Thus at equilibrium

y+ x–

BxAy xB (aq)+ yA (aq) At equilibrium xs mol/lit ys moles/lit.

y+ x x– y  Ksp = [B ] [A ] = x y x + y (xs)x(ys)y Ksp = x y s

Hence for salts of type MA (AgCl, BaSO4, etc.)

2

Ksp= s For M2Atype of salts

 s = Ksp 3 Ksp= 4 s  s = 3 Ksp/4 For MA3 4 Ksp= 27 s  s = 4 Ksp/27 Ex.11 Sol.

A salt M2X3dissolves in water such that its solubility is x g. mole/litre. What is Solubility of M2X3= x gm mole/litre KSPof salt ? +3 –2 M2 X3 [M+3] = 2x [X–2] = 3x 2M + 3X  2 3 5 Solubility product KSP= (2x) .(3x) =108 x

Ex.12 The solubility of AgCl in water, in 0.02 M CaCl2, in 0.01M NaCl and in 0.05 M AgNO3are S0, S1,S2,S3respectively. What is the relationships between these quantities ?

(20)

Solubility Product Sol. (B) Solubility =

Concentration of Common ion KSP S1 = S2 = S3 = = 50 KSP  0.02 KSP = 100 KSP 0.01 Ksp = 20 K SP 0.05

So, S2> S1> S3Again solubility will be greatest in water. So, S0 > S2 > S1 > S3

12. HYDROLYSIS OF SALT :

It is the process involving action of water on a salt to form a mixture of acid and alkali. The hydrolysis of a salt is reverse of neutrilization.

Let there be a salt BA. The hydrolysis of such a salt can be represented as

[HA] [BOH] [HA] [BOH]

K = [AB][H O]  K[H2O] = [AB]

2

[HA] [BOH]

 Kh= [AB] where Kh= hydrolysis constant.

The hydrolysis constant is dependent on nature of acid or base which is formed as a result of hydrolysis.

Degree of Hydrolysis :It is defined as the fraction of the total salt, which is hydrolysed at equilibrium.

Anionic Hydrolysis :It is the Hydrolysis of salts of Weak acids and Strong Base e.g. CH3COONa, Na2CO3, K2CO3, KCN, Na2S etc.

Let the salt hydrolysis be represented as

BA + H2O BOH + HA

B++ A–

(1) BA

SinceA–is a strong base is under hydrolysis according to the equation

A–+ H O HA + OH–

2

This is called anionic hydrolysis [HA] [OH–] Kh= [A]

Multiplying and dividing RHS by H+we get

   

[H ][A ]

– 

[HA] [OH ][H ] KwK[H][OH], K

Kh= w a

 [HA] 

[A–][H] Ka

Kw

Hence hydrolysis constant = Kh=K a

1

Kh i.e. hydrolysis constant varies inversely with dissociation constant of acid.

K

a

13. DEGREE OF HYDROLYSIS :

Let C moles/Lof salt be taken then C moles/Lof cation and anion will be formed respectively. H2O + AB

salt

HA + BOH

water acid base

+ B+ A

BA

C mole/L C mole/L. C

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Ionic Equilibrium

Let ‘h’be the degree of hydrolysis of A–

 Ch moles/lit. of A–will take part in hydrolysis and Ch moles/lit. of HAand Ch moles/L of [OH–] will be formed. C A– + HO2O HA + OHO O – Ch Ch At equilibrium conc. in moles/L. (C – Ch) [HA][OH] Ch2  Kh= = [A–]  1– h Kh 2 When h <<<< 1  Ch Kh  h = = KhV . C

This shows degree of hydrolysis increases with dilution.

Degree of hydrolysis is directly proportional to the square root of volume of solution in litres containing one mole of salt.

1

Also h C

Degree of hydrolysis in inversely proportional to the square root of concentration of salt in moles/lit.

pH of solution : Kh [OH–] = Ch = C × C (OH–) = KhC KwC = K a 1 1 1 1

pOH = 2pKw2logC – 2log K a 1 1 1 1 = 2pKw2logC – 2log K a 1 1 1 pOH = 2pKw2logC – 2pKa 1 1 1 = × 14 – logC – 2pKa 2 2 1 1 pOH = 7 – logC – 2pKa 2 1 1  pH = 14 – pOH = 7 + pK2 a+ 2logC Cationic Hydrolysis :

It involves the hydrolysis of a salt of strong acid with a weak base e.g. NH4Cl, (NH4)2SO4, NH4NO3, BaSO4, FeCl3, etc.

These solutions are acidic solution

BA + H2O

salt

HA + BOH

stong acid weak base

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Here the salt dissociates to form the cation and the anion. Let C moles of salt be taken in one litre solution and ‘h’ be the degree of hydrolysis.

B++ A

BA

Initial conc. (mole/lit.) C 0 0

+

B+ + H2O Final conc. (mole/lit.) (C – Ch)

BOH + H Ch Ch (Ch)2 [H][BOH] Kh= = [B] Kw 1– h  – [H][OH–] [H ][OH ] Kh= K = = [B ] [B ][OH ] b [BOH] 1 KhK b Since h <<< 1 (Ch)2= K h Kh/ C KhV  h = = 1 h C and h V pH of solution : Kw [H+] = Ch = K C hC = K b 1 1 1 1 1 1 1 pH = –2logKw2 logK b – 2logC = 2pKw– 2pKb– 2logC 1 pH = 7 – 2[pKb+ logC]

Cationic andAnionic Hydrolysis : Salts of Weak Acids and Weak Bases :

Here both cation and anion undergo hydrolysis. BA + H2O HA + BOH

weak acid weak base

Let C moles per litre of salt solution be taken. Let the degree of hydrolysis be ‘h’

B+ A– (1) BA + C moles/lit. C moles/lit [H][OH–][HA][BOH] [BOH][HA] Kh= =

[B][A–] [OH–][B][A–][H]

B++ A+ H O BOH + HA

2

weak base weak acid

Cationic Hydrolysis B+ + H O BOH + H conc. of 2 equilibrium (C – Ch) Ch + Ch B+ A– BA Anionic Hydrolysis A– + H O HA + OH– conc. of 2 equilibrium (C – Ch) Ch Ch

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Ionic Equilibrium Kw [BOH][HA] Kh = = [B][A–] K Ka b h2 Ch Ch Kh = = (1h)2 C(1–h) C(1–h)  When h <<< 1 h2= K or h = K h h

So degree of hydrolysis is independent of the concentration of salt taken. Also, pH = 21(pKw+ pKa– pKb)

When pKa> pKbthen solution's alkaline and anionic hydrolysis takes place more than cationic hydrolysis when pKa< pKbthen solution acidic and cationic hydrolysis takes place more than anionic hydrolysis. When pKa= pKbsolution neutral and extent of anionic and cationic hydrolysis are equal.

Kafor cyanoacetic acid is 4 × 10 . What is the value of degree of hydrolysis of 0.4 M sodium cyano acetate solution ?

14 –3 Ex.13 Kw 10 4103 Sol. Kh = Ka 10–11 = 0.25 × 11 Kh 0.2510 10–6 h =  = 2.5 × c 0.4

The dissociation constants for aniline, acetic acid and water at 25ºC are 4 × 10–10, 2 × 10–5and 10–14respectively. Calculate degree of hydrolysis of aniline acetate in a deci normal solution ? Ex.14

Aniline++Acetate+H O

Sol. 2  Aniline +Acetic acid

Before hydrolysis

1 1 0 0

After hydrolysis

1–h 1–h h h

Let conc. salt be C mole litre–1 Ch . Ch h2 Kh = = (1 - h)2 C(1 h) . C(1 h) Kw h = K a. Kb 1 - h 1014 = 210641010 % hydrolysis = 54.95 % h = 0.035

Calculate the pH of aqueous solution of 1.0 M HCOONH4assuming complete dissociation (pKa of COOH = 3.8 and pKb of NH3 = 4.8) ?

Ammonium formate undergoes hydrolysis as Ex.15 Sol. NH4+ + HCOO+H 2O

NH4OH + HCOOH KW K h K . Ka b

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Moreover in the solution we have [NH4OH] = [HCOOH] [HCOOH]2 Hence Kh = K [HCOO]2 [H]2 W or K . K = [K ]2 a b a K . K [H+]2 = W a ∴ K b 2pH = pKW + pKa – pKb pH = 1[pK + pK – pK ] or W a b 2 1 = [14 + 3.8 – 4.8] = 6.5 2

A certain weak acid has Ka=1.0×10–4. Calculate the equilibrium constant for its reaction with a strong base. Ex.16 Sol. HA weak or  + BOH strong  BA + H2O HA + + OHB+ – B+ + A+ H 2O or HA– + A– + H2O [A ] K = [HA] [OH ] Also for weak acid HA

H+ [H] [A] A– HA + Ka 104 K 10 Ka = a K or K = = = 10 K [HA] W 14 W 10 K

14. THEORY OF INDICATOR

It was first given by Ostwald. According to him, acid-base indicators are either weak organic acids (HIn) or bases (InOH) having different colours for the ionised and unionised forms. For example, litmus contains azolitmic acid in which the unionised acid molecules are red while the anionic form is blue. HIn H++ In

red blue

When litmus is dissolved in water, it appears violet due to presence of unionised acid molecules aswell as the anionic form. When some acid is added to the solution, equilibrium shift left and the solution turns red. Similarly, when base is added to the solution, equilibrium shift right and the solution turns blue.

pH Range of Indicator :

The ability of the human eye to preceive colours is limited. Nomally the eye fails to detect the presence of one of the coloured forms of an indicator together with the other if the concentration of the former is one-tenth of the concentration of the latter. Now for the process :

HIn H++ In

[H][In] Dissociation or indicator constant, K =

[HIn] In [HIn] or, [H+] = K . [In ] In [In] or, pH = pK + log [HIn] In or

(25)

E

Ionic Equilibrium

[In]

10, or pH(pK + 1) For the solution to For the solution to appear only in the colour of In–,

[HIn] In

[In] 1

[HIn] 10

[HIn]

appear in the colour of HIn, [In] 10 or or pH(pK – 1)In

The pH values of the solution below and above which the solution appears in a single colour is called pH range of indicator. Normally, pH range of an indicator is (pK – 1) to (pK + 1).In In

pH range of some indicators

solution

Selection of Indicator in a particular Acid-Base Titration :

The pH range of the resulting solution just before and after adding one drop of titrate is called pH range of titration. All the indicators for which pH range lie within the pH range of titration, are correct indicator for that titration. The selection of correct indicator for the titration can be understood from the following examples :

Titration of strong acid (say 25 mL of 1 M-HCl solution) against strong base (say 1 M-NaOH solution) :By simple calculation, it may be determined that the complete neutralisation will occur on adding 25 mL of the NaOH solution. Now, see the pH values of resulting solution on adding different volumes of NaOH solution, assuming that the volumes are additive and one drop of solution occupy 0.05 mL.

Volume

(in mL) 0 10 20 24 24.9 24.95 25.05 25.1 26 30 40

pH 0 0.37 0.95 1.69 2.69 3.0 11.0 11.3 12.3 12.9 13.4

Indicator Nature Colour in acidic

solution pH range

Colour in baisc

Phenolphthalein acidic Colourless 8.0 – 10.0 Red

Methyl red Basic Red 4.2 – 6.2 Yellow

Methyl orange Basic Red 3.1 – 4.4 Yellow

Litmus acidic Red 5.0 – 8.0 Blue

(26)

pH range of this titration is 3.0 – 11.0. pH range of almost all indicators lie in this range and hence almost all indicators are correct for this titration. However, a maximum error of one drop may occur. For example, pH range of methyl orange is 3.1 – 4.4. pH of solution on adding 24.95 mL NaOH solution (just one drop before the equivalent point) is 3.0 and hence, the solution will appear red (pink). When one more drop of solution is added, pH increases to 7.0 and the colour of solution suddenly become yellowish. This sudden colour change helps us in deciding end point of titration. However, if the titration is performed in presence of phenolphthalein indicator, the sudden colour change will occur on adding one more drop of solution after equivalent point.

If NaOH solution will be titrated against HCl solution, the titration curve will be just opposite. In this case, phenolphthalein will be the perfect indicator but methyl orange will give error of one drop.

pH range of this titration is 7.44 – 11.0 pH range of phenolphthalein lies in this range and hence, it is suitable indicator for the titration. If methyl orange is used, it will show gradual colour change much before the equivalent point.

With the help of titration curves, it may be determined that methyl orange, not phenolphthalein, is correct indicator for the titration of strong acid and weak base.

The titration curve of weak acid and weak base do not have any large jump in pH of the solution and hence such titration can not be performed by using any of such indicators.

When polyprotic acid or base are titrated, the different stage of titration can be identified by different indicators. For example, when H CO solution is titrated against NaOH solution, the

2 3

first stage of titration can be identified by methyl Volume

(in mL) 0 10 20 24 24.9 24.95 25.05 25.1 26 30 40

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Ionic Equilibrium

Ex.1 The self ionization constant for pure formic acid, K = [HCOOH2][HCOO ] has been

estimated as 10–6at room temperature. What percentage of formic acid molecules in pure formic acid are converted to formate ion? The density of formic acid is 1.22 g/cm3.

Given density of formic acid = 1.22 g/cm3

The value of K is much large than K and K .1 2 3 Also dissociation of II and III steps occurs in presence of H furnished in I step and thus, dissociation of II and III steps is further sup-pressed due to common ion effect.

+  2 For II step H P2 O4 0.024 H ++ HPO 4 ; Sol. 0.024 (0.024 + y) 0 y  Weight of formic acid in 1 litre solution

= 1.22103

(0.024 – y)

The dissociation of H P2 O4 occurs in presence 1.22103 of [H+] furnished in step I. [H][HPO2 ] 26.5 M Thus, [HCOOH] = 46 4 Thus, K =2

Since in case of auto ionization [HCOO–] = 10–6

[HCOOH2] [H PO2 4]

(0.024y) y [HCOO–] = [HCOOH] = 10–3.

 2 or 6.210–8= (0.024 – y)

Now % disso ciat io n of HCOOH =

y is small0.024 – y 0.024 and  [HCOO]100 103 neglecting y2.  100 = 0.004% [HCOOH] 26.5 0.024y   or 6.210–8= 0.024

Ex.2 Calculate the concentration of all species of significant concentrations present in 0.1 M H PO solution. K = 7.5 10–3, y = 6.210–8 2] = K = 6.210–8. [HPO4 3 4 1 2 K = 6.22 10–8, K = 3.63 10–3. HPO2 H++ PO3

For III step : 4 4

I step H PO H++ H PO; 4 (6.210–8–x) (0.024 + x) x 3 4 2 K = 7.5 10–3 1 3 [H ][PO4 ] (0.024x).x  2 II step H PO4 H + HPO4 ; + K    2 3 [HPO2] (6.2108x) 4 K = 6.210–8 2

III step HPO2 Again neglecting x and assuming,6.210–8–x = 6.210–8

0.024x 2 H++ PO3; 4 4 K = 3.6 10–13 3 for I step : H PO H+ + H PO  3.610–13= 8 6.210 4 3 4 0.1 0.1 – C 2 0 C 0 C C.C 13 8 x3.610 6.210 9.31019 .  0.024  

[H ][H2PO4] Ex.3 If CH COOH (Ka = 10–5) reacts with NaOH

K1 3

at 298 K, then find out the value of the maxi-mum rate constant of the reverse reaction at 298 K at the end point of the reaction. Given that the rate constant of the forward reaction is 10–11mol–1L sec–1at 298 K. Also calculate

Arrhenius parameter for backward reaction if [H3PO4] (0.1C) C2 7.510–3= (0.1C)  C = 0.024 [H+] = 0.024 M [H2PO4] = 0.024 M

H298 = 44 kcal and E = 94 kcal.a(f)

[H PO ] = 0.1 – 0.024 = 0.076 M3 4

SOLVED EXAMPLES

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Sol. CH COOH+NaOH CH COONa+H O; [H O][CN–] xx 3 3 2 3 K = 10–11mol–1L sec–1 K a = = 0.1x f

The backward reaction is of hydrolysis of so-dium acetate [HCN] 6.31066.3106 = 6 1 0.16.310 )1 KW Kfor Ka (K  K = 39.6 × 10–11

An aqueous solution contains 10% ammonia by mass and has a density of 0.99 g cm–3. Calculate hydroxyl and hydrogen ion concentration in this solution.

(K for NH+= 5.0 × 10–10M) K   C h Kb a K ac W Ex.6 10111014 K 20  KW  for  10 Kbac105 Ka

Given,H298= 44 kcal and = 94 kcalEaf a 4

[OH–] in aqueous solution of NH4OH can be

determined from the following relation. Sol. H = E af Eab E 44 = 94 – ab [OH] = K bc Eabac/ RT

 Ea b Abe (a) Determination of the value of Kb

14 Kw 110  50103 = 2 × 10–5 K =b = 10–20= A e   2298  Ka 5.01010 b

(b) Determination of concentration (molarity) C of NH4OH solution Wt . o f NH3 in 100

ml of NH4O H solution=10gm

100

 A = 2.71 10+16. b

Ex.4 The pH of pure water at 25°C and 35°C are 7 and 6 respectively. Calculate the heat of for-mation of water from H+and OH.

Vol. of 100 g of NH4OH solution = Sol. At 25°C;  [H +] = 10–7 0.99 K = 10–14 W w1000 1010000.99 = 99 At 35°C; 10–12 [H+] = 10–6 Molarity = mV 17100 = 17  K =W (c) Determination of [OH–] [OH–] = Kb C Kw2  HT2T1   2.303 log K R T T 210599 19.8 10  1 2–2 w1 = = 10 17 17 1012 2.303 log10 1014 H308298 2 298308 –2 = 10  1.16 = 10–2× 1.077 mol l–1 = 1.077 × 10–2mol l–1 14   [OH]  H = 84551.4 cal/mol = 84.551 kcal/mol 10 [H+] = = 9.28 × 10–13mol l–1 2 2 1.07710 H = 84.551 kcal/mol  H++ OH H O;

Ex.7 Calculate the pH of solution obtained by mix-ing 10 mL of 0.1 M HCl and 40 ml of 0.2 M H SO . 2 4 Milli-equivalent of H+from HCl = 100.1 =1 Milli-equvalent of H+from H SO 2 H = – 84.551 kcal/mol

The pH of a 0.10 M hydrocyanic acid solution is 5.2. What is the value of Kafor hydrocyanic

acid ? Ex.5 Sol. 2 4 H3O++ CN– = 400.22 = 16 Sol. HCN + H2O

(0.1 – x) Total meq. of H+in solution = 1 + 16 = 17

17 x x –5.2 = log[H3O+] 3.4101 [H3O+] = 6.3 × 10–6 or x = [H3O+]  [H+] = = 6.3 × 10–6 M/l 50 Thus H O H++ OH

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Ionic Equilibrium

Ex.10 A 0.1 M solution of weak acid HAis 1% dis-sociated at 25°C. What is its K ? If this

solu-

 [H ] Meq. 

a

tion is with respect to NaA 0.2 M, what will be the new degree of dissociation of HA and pH ?

For weak acid HA :

1  Vin mL   pH = – log [H+] = – log 0.34 pH = 0.4685 Sol. Ex.8 Calculate the pH of a solution which contains

100 mL of 0.1 M HCl and 9.9 mL of 1.0 M NaOH. HA  = 0.01, [HA] = 0.1 M 100 2 2 5  Ka C 0.1(0.01) 10 Sol. t = 0 t = t HCl + 1000.1 10–9.9 NaOH  NaCl + H O2 9.91 9.9–9.9 = 0 9.9

Now 0.2 M NaA, a salt of HA,is added to it resulting a buffer solution of [HA] = 0.1 M and [NaA] = 0. 2 M. 0.2 9.9 0.1 9.099104 M  [H+] left from HCl = 109.9 pH = – log 10–5+ log pH = 5.3010 0.1  pH = – log [H+] = – log 9.09910–4 pH = 3.0409 Also HA H+ + A– 0 Ex.9 Calculate [H+] in a solution containing 0.1 M

HCOOH and 0.1 M HOCN. K for HCOOHa 1 0

and HOCN are 1.810–4and 3.310–4.

In this problem both the acids contribute for [H+] due to appreciable dissociation. Thus,

(1 –)

[A–] is provided by NaA since

disso-

Sol.

ciation of HAin presence ofNaAis suppressed due to a common ion effect.

HCOON H+ + HCOO– x + OCN– y 0.1 –x HOCN 0.1–y x + y H+ x + y [H][A] (C)0.2 5 Ka   10  [HA] C(1 )   5105

Because [H+] will remain common in solution.

Thus, Ex.11 Calculate the amount of (NH ) SO in g which

4 2 4

must be added to 500 mL of 0.2 M NH to

 

[H ][HCOO ]

1.8104 3

K = … (1) yield a solution of pH = 9.35. K for NH =

[HCOOH] HCOOH b 3 1.78 10–5.   [H ][OCN ] 3.3104 [Salt]

K = … (2) Sol. pOH = – log K + log

[HOCN] (xy)x HOCN [Base] b  [NH4] 1.8104

or K = …(3) or pOH = – log K + log

b

HCOOH 0.1

[NH OH]4

(xy)y

3.3104

[NH]is obtained from salt (NH ) SO .

K = …(4)     4 pH = 9.35 pOH = 14 – 9.35 = 4.65 Millimole of NH OH in solution 4 = 0.2500 = 100 4 2 4 HOCN 0.1 Thus, by (3) and (4) x 1.8 y 3.3 y = 1.83 x

or … (5) Let millimole ofNH4 added in solution = a

From (3) (x + 1.83x) . x = 1.810–5 x = 2.5210–3 Therefore y = 4.6110–3  Thus, [H+] = x + y = 2.5210–3 + 4.6110–3= 7.1310–3M  

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Nic– + H+ Sol. HNic (1 –)C [NH] a ;[NH OH]100  4 4 C C 500 500 a / 500

whereis the dissociation of the acid and c is the concentration of nicotinic acid.

4.65 = log 1.7810–5+ log 100 / 500 a [H][Nic–] c  c c Ka= = (1–)c or [HNic] 1– 4.65 = 4.7496 + log a = 79.51 100 0.1 mol c = = 0.05 mol/litre 

 Millimole of (NH ) SO added4 2 4 2.0 litre

 79.5139.755 0.05 –5 1.4 × 10 = a 1– = 2 2 = 0.016594 Percent dissociation = 0.016594 × 100 ~ 1.66%.

How many moles of calcium hydroxide must be dissolved to produce 250 ml of an aqueous solution of pH 10.65 ? Assume complete dissociation. pH + pOH = 14 pOH = 14 – 10.65 = 3.35 [OH–] = 4.47 × 10–4mol/litre No. of OH–moles in 250 ml w 100039.755  132 Ex.14 w 5.248 g  ( NH4)2SO4

Ex.12 A solution contains 0.10 M H2S and 0.3 M

HCl. Calculate the concentration of S2–and HS–ions in the solution. For H2S, Sol.

Ka1= 1.0 × 10–7 Ka2= 1.3 × 10–13

Sol. Dissociation of H2S can be represented as

H2S H++ HS–

(a)

(b) HS– H++ S2– 4

4.4710 = 1.12 × 10–4

From the first ionisaiton of

[H][HS]

=

4

1 H2S = [H S] = Ka1 No. of moles of Ca(OH)2dissolved = ×

2

2

–4 –4

1.12 × 10 = 0.56 × 10

Ex.15 Saccharin (Ka = 2 × 10–12) is a weak acid

represented by formula HSac. A 4 × 10–4mole amount of saccharin is dissolved in 200 cm3

water of pH 3. Assuming no change in volume, calculate the concentration of Sac ions in the resulting solution at equilibrium.

Calculation of [H+] and [HSac] at start

We know that [H ] = 10 = 10 = 0.001M

410 1000

Since H2S is weakly ionised and its ionisation

is further decreased in presence of highly ionised HCl, concentration of H+in solution will be mainly due to HCl. Thus [H+] = 0.3

Substituting the values in the above equation

0.3[HS] –7 = 1 × 10 0.1 Sol. [HS–] = 3.33 × 10–8M

From the second ionisation of H2S,

[H][S2–] + –pH –3 4 [Hsac] = = 0.002 M = Ka2 or 200 [HS–] 0.3 [S2]

The dissociation of HSac is as below H+ 0.001 0.001 + x Sac– 0 x HSac. 0.002 0.002 – x + –13 = 1.3 × 10 At start At equb 3.33108 [S2–] = 1.44 × 10–20M  – [H ][Sac ] (0.01x) x = = 2 × 10–12 Nicotinic acid (Ka= 1.4 × 10–5) is represented

by the formula HNic. Calculate its percent dissociation in a solution which contains 0.10 mol. of nicotinic acid per 2.0 litre of solution.

Ex.13 K =

a [H Sac-] 0.002 – x

x = 4 × 10–12M

[Sac–]equb= 4 × 10–12M

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Ionic Equilibrium

Ex.16 Calculate the pH value ofthe mixture containing 50 c.c. of N HCl and 30 c.c. of N NaOH solution both to be completely dissociated in normal solution.

Since we know 30 c.c. of N NaOH will neutralise 30 c.c. of N HCl. = 1.8 mole/litre 1000 ml contains 1.8 moles 500 ml contains = 0.9 moles1.8 2 Sol. Weight of gms.

= Numbers of moles × Mol. wt. of NH4Cl

= 0.90 × 53.5 = 48.150 gm

Calculate the pH at the equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M sodium hydroxide. Kafor

acetic acid is 1.9 × 10–5.

We know that pH of a mixture of solution containing weak acid and strong base can be calculated as below.

 Amount of N HCl left un-neutralised = 50 – 30 = 20 c.c.

Total volume after mixing = 50 + 30 = 80 c.c. In other words 20 c.c of N HCl has been diluted to 80 c.c.

Now since 1000 c.c. of N HCl contain = 1 g eq of HCl Ex.18 Sol. 20 1 20 c.c. of N HCl contain = 1 × gram eq. = 1000 50 1 1 pH = 7 + 2pKa+ 2logc  1g eq.1 1

This50 is the amount of HCl present in = 7 + × 4.7212 + log 0.052 2 = 7 + 2.3606 – 0.65 = 8.71

Calcium lactate is a salt of a weak organic acid and represented as Ca (Lac)2. A saturated

solution of Ca (Lac)2contains 0.13 mol of this

salt in 0.50 litre solution. The pOH of this solution is 5.60. Assuming a complete dissociation of the salt, calculate Kaof lactic

acid. 80 c.c. of the mixture.

Ex.19

1

Thus 80 c.c. of the mixture contain =50 g eq. of HCl

1 1000

1000 c.c of the mixture contain =50 × eq. = 0.25 g eq.

[H+] = 0.25 g eq. per litre

We know that g 80 Ca2+ + 2 Lac– 2 × 0.26 M Sol. Ca (Lac)2 0.13 × 2 M pH = –log [H+] = – log [0.25] = 0.6021

How many grams of NH4Cl should 

Concentration of the lactate ion C = 0.52

Ex.17 be

dissolved in 500 ml of water to have solution of pH 4.5 ? Kbfor ammonium hydroxide is

1.8 × 10–5. pH = –log[H+]

Also we are given pOH K = 5.6  = 10–14or pH = 14 – 5.6 = 8.4 log K = –14 w w

Sol. Substituting the values in the following equation

for the salt of a weak acid and strong alkali. log[H+] = log 1 – 4.5 =

5.5

[H+] = Antilog [5.5] = 3.162 × 10–5

= 1(log Kw + log Ka– log c)

pH

mole per litre 2

K wc 1 8.4 Ka What = – (–14 + log Ka– log 0.52) [H+] = hc = 2 = 8.26 × 10–4 is the pH of 0.10 M CH COONa Kb K wc [H+]2= Ex.20 or K b 3

solution. Hydrolysis constant of sodium

ac-[H]2K 10–10? etate is 5.6 × bK w 3.1621053.1621051.8105 c = 11014

(32)

Sol. Hydrolysis of the salt may be represented as CH3COO– + Na+ + H 2O or CH3COO– + H2O At eq = 0 C CH3COOH+Na++OH– CH3COOH + OH– 0 C 0 C At C(1 – ) equilibrium C2 2 Kh = C(1 -) = C2 when  < 1 5.6 X 10 Kh 10 C = 0.10  = 10–5 = 7.5 × [OH–] pOH pH 10–5 = C = 0.10 × 7.5 × – log (7.5 × 10–6) = 5.126 = = 14 – pOH = 14 – 5.126 = 8.874 

(33)

Ionic Equilibrium

Q.1 The degree of dissociation of acetic acid in a

0.1 N solut ion is 1.32 × 10–2. At what concentration of nitrous acid, its degree of dissociation will be same as that of acetic acid? K (HNO ) = 4×10–4

Q.9 What is the solubility product ofAg CrO if2 4 0.0166 g ofthe salt dissolves in 500 ml of water at 18ºC?

When a 40 mL of a 0.1 M weak base is ti-trated with 0.16 M HCl, the pH of the solution at the end point is 5.23. What will be the pH if 15 mL of 0.12 M NaOH is added to the resulting solution.

A chemist needs a buffer with pH 4.35. How many milliliters of pure acetic acid (density = 1.049 g/mL) must be added to 465 mL of 0.0941 M NaOH solution to obtain such buffer? (K = 1.8 × 10–5)

Q.10

a 2

Q.2 How many times is the H+concentration in the

blood (pH = 7.36) greater than in the spinal fluid (pH = 7.53) ?

A 0.400 M formic acid solution freezes at –0.758°C. Calculate the K of the acid at

Q.3 Q.11

a

that temperature. (Assume molarity equal to molality). K (H O) is 1.86° mol–1kg.

f 2

A sample of AgCl was treated with 5 mL of 1.5 M Na CO solution to give Ag CO . the

a

It is found that 0.1 M solution of three sodium salts NaX, NaY and NaZ have pHs 7.0, 9.0 and 11.0, respectively. Arrange the acids HX, HY and HZ in order of increasing strength. Where possible, calculate the ionisation constants of the acids.

Given a solution that is 0.5 M CH COOH. To Q.4

Q.12

2 3 2 3

remaining solutioncontained 0.00266625 g/litre Cl–ion. Calculate the solubility ofAgCl in

(a) 0.1 M AgNO3 (b) 0.1 M BaCl2 (c) 0.2 M NaNO .3

Given that solubility ofAg CO2 3

Q.13

in500 mL H O = 13.8 mg. 3

what volume at 25°C must one dm3of this

solution be diluted in order to (a) double the pH; (b) double the hyd roxide-ion concentration.

Given that K = 1.8 × 10 M.a

The solubility of Mg(OH) is increased by

2

100 mL of solution S contains 0.17 mg of

Q.5 1

AgNO .Another 200 mLsolution S contains3 2 0.117 mg of NaCl. On mixing these two solutions predict whether the precipitate of AgCl will appear or not K AgCl = 10–10M2

–5

Q.14

sp

An indicator is a weak acid and the pH range of its colour is 3.1 to 4.5. If the neutral point of the indicator lies in the centre of the hydrogen ion concentrations corresponding to given pH range, calculate the ionization constant of the indicator.

Calculate the hydrolysis constant of NH Cl ;

2

addition of NH4+ion. Calculate –

Q.6

(a) K for the reaction,

C

Mg(OH) +2NH2 4+ 2NH + 2H O+Mg 3 2

2+

(b) Find solubility of Mg(OH) in a solution

2

containing 0.5 M NH Cl before addition of4 Mg(OH) (K of Mg(OH) = 1 × 102 sp –11, K

H

2 b

for NH O = 1.8 × 10–5) 4

Q.7 Q.15 A buffer solution was prepared by dissolving

0.05 mol formic acid and 0.06 mol sodium for-mate in enough water to make 1.0 L of solu-tion. K for formic acid is 1.80 × 10 .a

(a) Calculate the pH of the solution.

(b) If this solution were diluted to 10 times its volume, what would be the pH?

(c) If the solution in (b) were diluted to 10 times its volume, what would be the pH?

4

determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution.

K (NH OH) = 1.8 × 10–5 -4

b 4

Calculate the pH of 0.1 M acetic acid solution if its dissociation constant is 1.8 × 10–5. If 1

litre of this solution is mixed with 0.05 mole of HCl, what will be the pH of the mixture ? Q.8

(34)

Q.22 The ionization constant of acetic acid is

Q.16 An unknown volume and unkno wn

concentration of weak acid HX is titrated with NaOH of unknown concentration. After addition of 10.0 cm3of NaOH solution, pH of

solution is 5.8 and after the addition of 20.0 cm3of NaOH solution, the pH is 6.4. Calculate

the pH of aqueous solution of 0.1 M NaX. A solution containing zinc and manganese ions each at a concentration of 0.01 mol dm–3is

saturated with H S. Calculate (i) pH at which

1.74 × 10–5. Calculate the degree of dissocia-tion of acetic acid in its 0.05 M soludissocia-tion. Cal-culate the concentration of acetate ion in the solution and its pH.

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

Assuming complete dissociation, calculate the pH ofthe following solutions:

Q.23 Q.17

Q.24

2

the MnS will form a precipitate (ii) conc. of Zn+2ions remaining. Given : [H S] = 0.1 mol/lit, K (ZnS) = 1 × (a) 0.003 M (c) 0.002 M HBr (b) 0.005 M NaOH (d) 0.002 M KOH 2 sp

10–22mol2lit–2, K (MnS) = 5.6 × 10–16mol2

Q.25 Calculate the pH of the following solutions: (a) 2 g of TlOH dissolved in water to give 500 ml of solution.

(b) 0.3 g of Ca (OH)2 dissolved in water to

give 2 litre of solution

(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

(d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution

If the solubility product of silver oxalate is 1 × 10–11, what will be the weight ofAg2C2O4

in 2.5 litres of a saturated solution ?

Calculate the hydrolysis constant of KF. Determine the degree of hydrolysis of salt in a 0.01 M solution and the pH of the solution.

Ka(HF) = 6.6 × 10–4.

Assuming that t he buffer in blood is CO3––HCO3 , calculate the ratio of conjugate

base to acid necessary to maintain blood at its proper pH, 7.4 K1(H2CO3) =4.5 × 10–7.

How does the solubility of CaC2O4in a 0.1 M

solution of ( NH4) 2C 2O4 decrease in

comparison with its solubility in water ?Assume that the ionisation of (NH4)2C2O4is complete. [Ksp (CaC2O4) = 2 × 10–9]

How many moles of sodium hydroxide can be added to 1.0 L of a solution 0.10 M in NH3and

0.10 M in NH4Cl without changing the

pOH by more than 1 unit. Assume no change in

volume.

Kb= 1.8 × 10–5.

sp

lit–2.

K and K for H S are 1×10–7and 1.1×10–14.

1 2 2

Q.18 For the indicator thymol blue, the value of pH is 2.0, when half of the indicator is present in an unionized form. Calculate the percentage ofthe indicator in the unionized formina solution of 4.0 × 10–3 mol/dm– 3hydr ogen ion concentration.

One of the substances sometimes responsible for the hardness of water is CaSO .Aparticular

Q.26 Q.19

4

water sample has 131 ppm CaSO (131 g4

CaSO per 106g of water). If this water boiled Q.27 4

in a tea kettle, approximately what fraction of water must be evaporated before CaSO (s)4 begins to deposit ?Assume that solubility of

CaSO does not change with temperature in4 Q.28

the range 0 to 100°C, K (CaSO ) = 9.1 ×sp 4 – 10–6.

Q.20 The indicator phenol red is half in the ionic form when pH is 7.2. If the ratio of the undissociated form to the ionic form is 1 : 5, find the pH of the solution. With the same pH for solution, if indicator is altered such that the ratio of undissociated form to dissociated formbecomes 1 : 4, find the pH when 50% of the new indicator is in ionic form.

The first ionization constant of H S is

Q.29

Q.30

Q.21 2

9.1 × 10–8. Calculate the concentration of

HS–ion in its 0.1 M solution. How will this

concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H S is 1.2 × 10–13. Calculate the

2

(35)

Ionic Equilibrium

Calculate the number of H+present in one ml

of solution whose pH is 13.

Calculate change in concentration of H+ion in

one litre of water, when temperature changes from 298 K to 310 K.

Given Kw(298)=10–14Kw(310)=2.56 ×10–14. (i) Kwfor H2O is 9.62 × 10–14at 60°C. What is pH of water at 60°C.

(ii) What is the nature of solutionat 60°C whose

Q.1 Q.11 An acid indicator has a K of 3 × 105. The

a

acid form of the indicator is red & the basic form is blue. By how much must the pH change in order to change the indicator form 75% red to 75 % blue?

What is the OHconcentration of a 0.08 M solution of CH3COONa.

[Ka(CH3COOH)=1.8 × 105]

Calculate the pH of a 2.0 M solution of NH4Cl. [Kb(NH3) = 1.8 × 105]

Calculate OH–concentration at the equivalent point when a solution of 0.1 M acetic acid is titrated with a solution of 0.1 M NaOH. Ka for the acid = 1.9 × 10–5.

Calculate the hydronium ion concentration and pH at the equivalence point in the reaction of 22.0 mL of 0.10M acetic acid, CH3COOH, with 22.0 mL of 0.10 M NaOH. [Ka = 1.8 × 10–5] Q.2 Q.12 Q.3 Q.13 (a) pH = 6.7 (b) pH = 6.35

Q.4 The value of Kw at the physio logical temperature (37°C) is 2.56 × 10–14. What is the pH at the neutral point of water at this temperature?

Calculate pH of following solutions :

(a) 0.1 M H2SO4(50 ml) + 0.4 M HCl 50 (ml)

(b) 0.1 M HA + 0.1 M HB [ Ka(HA) = 2 × 10–5; Ka(HB) = 4 × 10–5]

What are the concentration of H+, H 2C2O4, HC2O4 and C2O4 in a 0.1 M solution of oxalic acid ? [K1= 10–2M and K2= 10–5M ] Q.14 Q.5 Q.15

The values of K for the slightly soluble salts

Q.6 Q.16

sp

MX and QX are each equal to 4.0×10–18.

 2

2

i

Which salt s more soluble? Explain your answer fully.

Q.17 Calculate the Simultaneous solubility ofAgSCN of H+,

Q.7 What ar e the concent r at ions and AgBr. K (AgSCN) = 1.1 × 10–12, sp

 2 K –13

sp(AgBr) = 5 × 10 .

A recent investigation of the complexation of SCN–with Fe3+led of 130, 16, and 1.0 for

K1, K2, and K3, respectively. What is the overall formation constant of Fe(SCN)3from its component ions, and what is the dissociation constant of Fe(SCN) into its simplest ions on HSO4, SO4 and H2SO4in a 0.20 M solution

of sulphuric acid ? Q.18  H2SO4H++HSO 4 Given : ; strong  l H++ SO2 4 ; K2= 10–2M HSO4

Q.8 Calculate the pH of a solution which results from the mixing of 50.0 ml of 0.3 M HCl with 50.0 ml of 0.4 M NH3.

[Kb(NH3) = 1.8 × 105]

Calculate the pH of a solution made by mixing 50.0 ml of 0.2M NH4Cl & 75.0 ml of 0.1 M NaOH. [ Kb(NH3) = 1.8 × 105]

What indicator should be used for the titration of 0.10 M KH2BO3with 0.10 M HCl ? Ka(H3BO3) = 7.2 × 1010.

3

the basis of these data ?

Calculate the percent error in the [H O+]

Q.19 3

concentration made by neglecting the ionization of water in a 10–6M NaOH solution.

A solution was made up to be 0.01 M in chloroacetic acid, ClCH2COOH and also 0. 002 M in so d ium chlo r o a cet a t e ClCH COONa . What is [H+] in the solution?

Q.9 Q.20 Q.10 2 K = 1.5 × 103. a

EXERCISE - II

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