Ionic Equilibrium 2e3i3a

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CONTENTS S.NO.

TOPIC

PAGE NO.

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

1

Ionic Equilibrium

IONIC EQUILIBRIUM 1. SALT A substance which ionizes in water to produce ions other than H+ and OH– is 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)2 SO4. 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, K2SO4 etc. Strong Electrolyte And Weak Electrolyte and Non Electrolytes: Those electrolytes which dissociate almost completely into ions are known as strong electrolytes. e.g. HCl, HBr, HI, HClO4, NaCl, Na2SO4, KNO3 etc. Those electrolytes which dissociate partially are called weak electrolytes. e.g. H3PO4, HF, H2CO3,HCN, CH3COOH, NH4OH, etc. .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.

3. OSTWALD’S DILUTION LAW Ostwald’s pointed out that like chemicalequilibriuminionic equilibriumwe can applylawof mass action. An equilibriumbetweenionized and unionized molecules. Considera binary electrolyte having conc. C and degree of dissociation is. A   B 0 0 C C(1 – ) C  C AB At time = 0 At time = t 

K aq





[A ][B ]

1   1

[AB]



C  C

C(1 )



C

2

1 

, for a weak electrolyte

2

Ionic Equilibrium

K eq  C 2 ,  

K eq

C If 1 mole ofAB is present in ‘V’litre of solution. C

1 V

  K eq  V  C  C

Conc. of A+ = Conc. B–

(i) (ii)

K eq C

 K 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 completelyionC ized at all dilution and hence  does not give accurate results.  Dissociation of Weak Acids and Bases : Dissociation of Weak acids : Consider the dissociation of a weak acid HAin water, represented by the equation – H3O + A HA + H2O C

Initial conc. moles/litre

HA

Final concentration

(C – C)

0

0

in moles/lit.

H+ + A–

C C in moles/lit.

C α2 Ka = 1 α



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 Also as  =

Ka  C

VK a

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. (initial)

C BOH

Final concentration in moles/lit. C – C

O

O

B + OH–

C

C.

3

Ionic Equilibrium

Where C = Initial conc. in moles/litre  = Degree of dissociation 

Kb =

As the base is weak   1 Kb = C 2    = Kb / c Hence,

CB  COH  CBOH

C 2α2 Cα2  C(1α) 1α

1 –  = 1.



Kb C

[OH–] = C = C × [OH–] =



Kb C

4. STRENGTHS OF ACIDS AND BASES (i) (ii)

(iii) (iv)

The strength of an acid/base depend upon the number H+/OH– present in solution. Since  = V  K a or  = V  Kb   increases when V increase Ionization increases with dilution, hence number of H+ /OH– increases.  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 of their dissociation constants. We known Ka1 = C1 1 2 1  1 Ka2 = C2 2 2 2  1 K a1 C1 K a2 C2

=

α1  α2

(v) Ex.1 Sol.

K a1 C 2 α  1 α2 K a2 C1

K a1 Ka 2

α1  α2

K b1 C 2  K b2 C1

α1  α2

K b1 K b2

when C1  C2

when C1  C2 when C1  C2 when C1  C2

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

Calculate the degree of ionization of 0.01 M solution of HCN, Ka of HCN is 4.8 × 10–10. Also calculate hydronium ion concentration. The ionization of HCN may be represented as, HCN(aq)  H 2O()



CN– - (aq)  H 3O (aq)

If degree of ionization of HCN is  then equilibrium concentration of given species are

4

Ionic Equilibrium

[HCN]  C(1 – ) [CN – ]  C [H3O  ]  C where C = concentration of HCN. Applying law of chemical equilibrium C2 [CN – ][H 3O ] (C)(C)   C(1– ) (1 – ) [HCN] Since  is very small as compared with unity therefore, 1 –  in the denominator may be taken as 1. Ka 

K a  C 2 

Ka 4.8 10–10  2.2 10–4 .  C 0.01

[H3O ]  C  0.01 2.2 10 –4  2.2 10 –6 mol L–1. Ex.2 Sol.

The Ka for formic acid and acetic acid are 2 × 10–4 and 2 × 10–5 respectively. Calculate the relative strength of acids with same molar concentration– Relative strength of weak acid 𝑐 2 𝑘𝑎1 𝑎1 ∗ = 1 𝑐 𝑘𝑎2 𝑎2 Ka

1

 Relative strength =

K a2

( C1 = C2)

2 𝑥 10−4 2 𝑥10−5

Relative strength for HCOOH to CH3COOH = Ex.3

10 : 1

Calculate the degree of ionization of 0.02 M acetic acid if its K a  1.8  10 –5 . What would be the degree of ionization if the solution also contains 0.01 M sodium acetate ?

Sol.

CH 3COOH(aq)

CH COO3– (aq)  H  (aq)

The degree of ionization of this weak acid can be calculated by the approximate relation : Ka 1.810 –5  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 3COOH(aq) (0.02– x )M

CH3 COO – (aq)  H  (aq) xM

xM

CH 3COONa(aq)   CH3 COO– (aq)



Na  (aq) 0.01 M

5

Ionic Equilibrium

[H ]  x mol L–1 [CH3COO –]  (x  0.01) mol L–1  0.01 mol L–1 [ x is very small as compared to 0.01] [CH3COOH]  (0.02 – x) mol L–1  0.02 mol L–1 Ka 

[H  ][CH3COO – ] [CH3COOH]

1.810–5 

(x)(0.01) (0.02) –5

1.8 10  0.02  3.6 10–5 M 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 to 1.8 10–3 due to the presence of sodium acetate. x

5. SELF IONIZATION OF WATER 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 H  COH – CH 2 O

K CH2O = CH+ COH– CH2O = Kw = CH+ COH– K 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 × 10–14 at 25ºC At Kw = 0.11 × 10–14 mol 2/lit 2 OºC Kw = 0.30 × 10–14 mol2/lit2 10ºC 25ºC Kw = 1 × 10–14 mol2 /lit2 –14 2 2 100ºC Kw = 7.5 × 10 mol /lit For pure water CH+ = COH– = 10 –7 mol/lit at 25ºC Degree of dissociation of pure water at 25ºC For pure water CH+ = COH–. Also at room temperature CH+ COH– = 10–14 mol2 /lit2  CH+ = 10–7 mol/lit  C = 10–7 mol/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 [OH–] = Normality of base = Molarity × Acidity

6

Ionic Equilibrium

(ii)

In case of weak acids/Bases, the H+/OH– concentration is less than normal and may be calculated by using Oswald's dilution law. [H+] =

K aC

[OH–] =

= C = N ×  = C = N × 

K bC

7. pH SCALE : It may be defined in number of ways. (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. (ii) It can also be defined as the negative logarithmof its [H+] ion concentrations pH = –log[H+] (iii) pH values do not give instantaneous idea above the relative strengths of the solution 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 C1 and C2, 1 is the degree of dissociation of HA in presence of HB (due to common ion effect) and 2 be degree of dissociation of HB in presence of HA. In aqueous solution of HAand HB following equilibrium exists. conc. at eq.

+

A– C1  1

+ (C1 1 + C2 2)

B– C2  2

HA+ H 2O(l)

H3O+

C1(1 – 1)

C11 + C22

HB + H2O(l)

H3O+

C2(1 – 2) 







K a[HA] 

[H3O ][A ] [C11  C12 ][C12 ]  C1 (1  1 ) [HA]

K a[HB] 

[H3O ][B ] [C11  C22 ][C22 ]  [C2 (1  2 )] [HB]

pH  – log[H  ]  – log[C11  C2 2 ] pH of a DibasicAcid and PolyproticAcid : Let’s take the eg. of a dibasic acid H2A.Assuming both dissociation is weak. Let the initial conc. of H2A is C and 1 and 2 be degree of dissociation for first and second dissocation. HA– C1 (1 – 2)

H2A

C(1 –1) HA– C1 (1 – 2) 

+

H+ + C1 + C1 2

H+ C1 + C1 2 A– – C1 2.



Ka1 

[HA ][H ] [H 2A]

Ka1 

[C1 (1 2 )][C1 C1 2 ] C(1  1 )

7

Ionic Equilibrium

Ka 2 

Ex.4 Sol.

Ex.5 Sol.





[H ][A ] [C1  C12 ][C12 ]  [C1 (1  2 )] [HA  ]

After solving for 1 and 2. We can calculate the H+ conc. [H+] = C1 + C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.  The pH of a 0.05 M solution of H2SO4 in water is nearly? pH= – log10 H + The concentration of H+ ions is expressed in gm equivalent Molarity of H2SO4 = 0.05 Normality = 0.05 × 2 = 0.1  pH = –log 0.1  pH = 1 or Calculate the pH of solution having H+ ion concentration of 5 × 10–4 mole/litre [H+] ion concentration = 5 × 10–4 mole/litre pH = – log [5 × 10–4] = – (log 5 + log 10–4) = – 0.7 + 4 = 3.3

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.

9. 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 : 1. Simple buffers : Asalt of weak acid and weak base in water e.g. (a) NH4CN or CH3COONH4. (b) Proteins and amino acids. (c) Amixture of an acid salt and normal salt of a poly basic acid e.g. NaHCO3 + Na2CO3 Na2 HPO4 + Na3PO4.

8

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. CH 3COOH + CH3COONa – + CH3COOH CH3COO + H (Weakly ionized) – CH3COO + Na+ (Highly ionized) CH3COONa 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 C1 be the concentration of CH3COONa. The degree of dissociation of CH3COOH is 𝛼 in the presence of sodium acetate Initial conc. (mol/lit.) C 0 0 – CH3COOH H+ CH3COO + ⇌ C Final conc. (mol/lit.) (C – C) C Initial conc.(mol/lit.)

Final conc.(mol/lit.) Hence in the solution

C1 CH3COONa 0

0 0 – + CH3COO + Na C1 C1

CH3COOH concentration = (C – C) CH3COO– concentration = C + C1 H+ concentration = CH+ = C. But since  is very small [CH3COOH] = (C – C)  C. [CH3COO-] = C + C1  C1 Ka =

CCH 3COO– CH  C CH3COOH

=> CH+ = Ka ×



pH = – log CH+ = PKa + log



pH = pKa + log [Salt]

C C1

C1 C

[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 CH3COO– 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

resisted the change in pH. In other wards one can saythat 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 CH3COOH to form acetate ion and water. OH– + CH3COOH  CH3COO– + H 2O Ex.6 Sol.

Calculate the ratio of pH of a solution containing 1 mole of CH COONa + 1 mole of HCl per litre and 3 of other solution containing 1 mole CH3COONa + 1 mole of acetic acid per litre. Case I. pH when 1 mole CH3COONa and 1 mole HCl are present. 0 0 1 1 CH COOH + NaCl CH3COONa + HCl   3 After reaction 1 1 0 0

Before reaction

9

Ionic Equilibrium



[CH3COOH] = 1 M



[H+] = C. = C



pH1 = –



pH2 = – log Ka + log

 K a   (K .C)  (K ) a a  C   

 C=1

1

log Ka 2 Case II : pH when 1 mole CH3COONa and 1 mole of CH3COOH; a acidic buffer solution forms [Salt] = 1 M, [Acid] = 1 M  [Salt] [Acid]

pH2 = –log Ka 

pH1 1  pH 2 2

Basic Buffers : Abasic 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 Highlyionized 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 C1 be the concentration of NH4Cl taken. Initial conc. (mol/lit.) C 0 0 + NH4OH NH4 OH – Final conc. (mol/lit.) (C – C) C C C1 Initial conc.(mol/lit.) 0 0 + NH4 + Cl– NH4Cl 0 Final conc.(mol/lit.) C1 C1 The concentration of [NH4+] = C1 + C  C1 [  1] [NH4OH] = C – C  C [  1] [OH–] = C. Kb =

C NH  C OH – 4

C NH 4 OH

 COH– = Kb C NH



C NH 4 OH C NH 4  C



pOH = pKb + log C = pKb + log C1 NH 4 OH



pOH = pKb + log [Base]

4

[Salt]

BufferAction : When few drops of base say NaOH is added then the OH– ions added react with NH 4+ to form

10

Ionic Equilibrium

[NH 4 ]

–

NH4OH and thus the concentration of [OH ] remains unchanged. But the ratio of [NH OH] changes. 4 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 NH4 OH decreases [NH  ]

4 and [NH4+] ion conc. increases, the ratio [NH OH] increases and thus pH changes infinitively. 4

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 pKb of ammonia is 4.74, calculate value of x.

Sol.

(NH4)2SO4

2 NH4 + SO42 

Thus, every one mole of (NH4)2SO4 gives two moles of NH4 . 

 

millimoles of NH3, (NH4OH) = 100  0.1 = 10 millimol millimoles of (NH4)2SO4 = 100  x = 100 x 1000 millimol millimoles of NH4 = 200  x = 200 x millimol pH = 9.26 pOH = 14 – 9.26 = 4.74 

pOH  pK b  log 4.74 = 4.74 + log

[NH 4 ] [ NH4OH] 200x 10

log 20x = 0 20 x = 1 x=

1 = 0.05. 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. Buffer Capacity =

number of moles of acid/base added 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. Ex.8 Sol.

What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid (Ka = 1.0 × 10–5 at 25ºC) to obtain a buffer solution of pH 6. Using the expression pH = pKa + log

[Salt] [Acid]

11

Ionic Equilibrium [Salt]

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

Which gives 6 = 5 + log [0.02 M] [Salt] [0.02 M]

= 10 or [Salt] = 0.2 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.

Sol.

pH = pKa + log

[Salt] [Acid] (0.02  0.01)

= –log(1.0 × 10–5) + log (0.03  0.01) 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 Ka in the expression. pH = – log Ka + log

[Salt] [Acid]

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

or

10.4 = 9.4 + log

or

log

or

0.01 x x

or

x = 9.9 × 10–4 M

0.01 x x

0.01 x x

=1

= 10  11x = 10–2

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 = Molecular weight of solute The factors affecting solubility are. (a) Nature of solvent (b) Nature of solute (c) Temperature (d) Pressure

12

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. B+(aq) + + A– (aq)

BA(s) Applying law of mass action we get K=

[A – ] [B  ] [BA]

K[BA] = [B+] [A–]



Since only little of salt dissolves so the concentration of salt remains constant K[AB] = K × Constant = Ksp = [B+] [A–]  For any salt BxAy xBy+ (aq) + yAx–(aq) Ksp = [By+ ]x [Ax–] y Ksp = Kc This 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 : When Ksp = Ki solution saturated (1) When Ki < Ksp unsaturated solution (2) (3) When Ki > Ksp super saturated solution. Hence precipitation occurs to keep Ki = Ksp Relationships Between Solubility and Solubility Product: (1) Let the solubility of a salt Bx Ay in water be s moles/lit. Thus at equilibrium BxAy xBy+ (aq) + yAx–(aq) At equilibrium xs mol/lit ys moles/lit. Ksp = [By+ ]x [Ax–] y  (xs)x (ys)y = Ksp = xxyy sx + y Hence for salts of type MA (AgCl, BaSO4, etc.) 

s=

Ksp = 4 s 3



s=

Ksp = 27 s 4



s = 4 K sp /27

Ksp = s2 For M2Atype of salts

K sp

3

K sp /4

For MA3 Ex.11 A salt M2 X3 dissolves in water such that its solubility is x g. mole/litre. What is KSP of salt ? Sol. Solubility of M2X3 = x gm mole/litre 2M+3 + 3X –2 M2 X3 +3 [M ] = 2x  [X–2] = 3x Solubility product KSP = (2x) 2.(3x) 3 =108 x 5 Ex.12 The solubility of AgCl in water, in 0.02 M CaCl2, in 0.01M NaCl and in 0.05 M AgNO3 are S0, S1,S2,S3 respectively. What is the relationships between these quantities ?

13

Ionic Equilibrium

Sol.

(B) Solubility = 

S1 =

Solubility Product Concentration of Common ion

K SP 0.02 K SP

= 50 KSP

S2 = 0.01 = 100 KSP K sp = 20 KSP S = 3

So,

0.05

S2 > S1 > S3 Again 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 H2O + AB water salt

HA+ BOH acid base

[HA] [BOH]



K = [AB][H O] 2 

Kh =

[HA] [BOH] [AB]

K[H2O] =

[HA] [BOH] [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 (1)

BOH + HA B +A– BA SinceA– is a strong base is under hydrolysis according to the equation +

HA + OH–

A– + H 2O

This is called anionic hydrolysis Kh =

[HA] [OH – ] [A – ]

Multiplying and dividing RHS by H+ we get Kh =

     K w  [H  ][OH ], K a  [H ][A ]   [HA]   

[HA] [OH – ][H  ] K w  Ka [A – ][H  ]

Kw

Hence hydrolysis constant = Kh = K 

a

1

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

13. DEGREE OF HYDROLYSIS : Let C moles/Lof salt be taken then C moles/Lof cation and anion will be formed respectively. BA C

B+ C mole/L

+

A– C mole/L.

14

Ionic Equilibrium

Let ‘h’be the degree of hydrolysis ofA–  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

O

O

A– + H2O

(C – Ch)

At equilibrium conc. in moles/L. 

Ch

[HA][OH ]

Kh =

=

–

[A ]

When h <<<< 1



Kh C

=



h=

O

HA + OH–

Ch

Ch 2 1– h

Kh  Ch2 K hV .

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. Also

h

1 C

Degree of hydrolysis in inversely proportional to the square root of concentration of salt in moles/lit. pH of solution : Kh C

[OH–] = Ch = C × (OH–) =

K hC

Kw C Ka

= pOH

1

1

1

1

1

1

1

1

1

1

1

= 2 pKw – 2 logC – 2 log K a = 2 pKw – 2 logC – 2 log K a

pOH

= 2 pKw – 2 logC – 2 pKa =

1 2

× 14 – 1 2

1 2

1

logC – 2 pKa 1

pOH

=7–

logC – 2 pKa



pH = 14 – pOH = 7 + 2 pKa + 2 logC

1

1

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 ⇌ HA + BOH salt

stong acid

weak base

15

Ionic Equilibrium

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 ⇌ BOH + H+ Ch Ch Final conc. (mole/lit.) (C – Ch) [H  ][BOH]

Kh =

[B ]

=

(Ch) 2 1– h

[H  ][OH– ]

Kw

Kh = K =  = – b [B ][OH ]

[H  ][OH ] –

[B ]

[BOH] 1

Kh  K b 2 Since h <<< 1  (Ch) = K h 

h=

Kh /C

h

1 C

=

K hV

and h  V

pH of solution : Kw C Kb

[H+]

= Ch =

pH

= – 2 logKw – 2 log Kb – 2 logC = 2 pKw – 2 pKb – 2 logC

pH

= 7 – 2 [pKb + logC]

1

K hC =

1

1

1

1

1

1

1

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’ (1) BA + B+ A– C moles/lit. C moles/lit

Cationic Hydrolysis conc. of B+

+ H2O equilibrium (C – Ch) BA

H

BOH + Ch Ch B+ + A–

Anionic Hydrolysis A– + H2O equilibrium (C – Ch) conc. of

B+ + A– + H 2O

BOH + weak base

[BOH][HA]

HA + OH– Ch Ch

HA weak acid

[H  ][OH – ][HA][BOH]

Kh = [B  ][A – ] = [OH – ][B ][A – ][H  ]

16

Ionic Equilibrium



Kw

[BOH][HA]

Kh

= K K = [B  ][A – ] a b

Kh

= C(1–h) C(1–h) = (1 h) 2

Ch Ch

h2

When h <<< 1 h2 = Kh or h =

Kh

So degree of hydrolysis is independent of the concentration of salt taken.

Also, pH = 21 (pKw + pKa – pKb) When pKa > pKb then solution's alkaline and anionic hydrolysis takes place more than cationic hydrolysis when pKa < pKb then solution acidic and cationic hydrolysis takes place more than anionic hydrolysis. When pKa = pKb solution neutral and extent of anionic and cationic hydrolysis are equal. Ex.13 Ka for cyanoacetic acid is 4 × 10–3 . What is the value of degree of hydrolysis of 0.4 M sodium cyano acetate solution ? 14

Sol.

Kh =

Kw 10  K a 4 10 3

= 0.25 × 10–11 h=

Kh 0.25  1011  c 0.4

= 2.5 × 10–6

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

Aniline++Acetate–+H 2O  Before hydrolysis 1 1 After hydrolysis 1–h 1–h Let conc. salt be C mole litre–1 Ch . Ch

Aniline +Acetic acid 0

0

h

h

h2

Kh = C(1 h) . C(1 h) = (1 - h)2 h 1-h

=

=

Kw Ka. Kb 1014

2  10 6  4  10 10

% hydrolysis = 54.95 % h = 0.035 Ex.15 Calculate the pH of aqueous solution of 1.0 M HCOONH4 assuming complete dissociation (pKa of COOH = 3.8 and pKb of NH3 = 4.8) ? Sol. Ammonium formate undergoes hydrolysis as NH4+ + HCOO–+H2O ⇌ NH4OH + HCOOH Kh 

KW K a . Kb

17

Ionic Equilibrium

Moreover in the solution we have [NH4OH] = [HCOOH] Hence Kh =

[HCOOH]2 [HCOO  ] 2

or

KW Ka . Kb

∴

[H+]2 =

or

2pH

or

pH =

=

[H ] 2 [K a ]2 KW . Ka Kb

= pKW + pKa – pKb

=

1 [pK W 2 1 2

+ pKa – pKb ]

[14 + 3.8 – 4.8] = 6.5

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

K =

[A ] [HA] [OH ]

Also for weak acid HA H+ + A– HA Ka =

[H ] [A  ] [HA]

Ka  KW K

or

Ka

10 4

K = K = = 1010 10 14 W

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– blue red 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 onetenth of the concentration of the latter. Now for the process : HIn H+ + In– [H  ][In  ] Dissociation or indicator constant, KIn = [HIn] [HIn] [In  ]

or,

[H+] = KIn .

or,

[In  ] pH = pKIn + log [HIn]

18

Ionic Equilibrium

For the solution to appear only in the colour of In–,

[In  ]  10, or pH  (pKIn + 1) For the solution to [HIn]

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

pH range of some indicators Indicator

Colour in acidic solution Colourless Red Red Red Yellow

Nature

E

Phenolphthalein Methyl red Methyl orange Litmus Bromphenol blue

acidic Basic Basic acidic Acidic

pH range 8.0 – 10.0 4.2 – 6.2 3.1 – 4.4 5.0 – 8.0 3.0 – 4.6

Colour in baisc solution Red Yellow Yellow Blue Blue

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) pH

0 0

10

20

24

24.9 24.95

0.37 0.95 1.69 2.69

3.0

25.05 25.1 11.0

26

30

40

11.3 12.3 12.9 13.4

19

Ionic Equilibrium

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. Volume 0 10 20 24 24.9 (in mL) pH 2.37 4.56 5.34 6.12 7.14

24.95

25.05 25.1

7.44

11.0

26

30

40

11.3 12.3 12.9 13.4

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 H2CO3 solution is titrated against NaOH solution, the first stage of titration can be identified by methyl

20

Ionic Equilibrium

SOLVED EXAMPLES Ex.1

Sol.

The self ionizationconstant for pure formic acid, K = [HCOO H 2 ][HCOO –] has been estimated as 10–6 at 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  Weight offormic acid in 1 litre solution = 1.22  103 1.22 103  26.5 M 46

Thus, [HCOOH] =

Calculate the concentration of all species of significant concentrations present in 0.1 M H3PO4 solution. K1 = 7.5  10–3, K2 = 6.2  10–8, K 3 = 3.6  10–3.

K3 = 3.6  10–13



y = 6.2  10–8 [HP O42  ] = K2 = 6.2 10–8.

⇌

H+ + P O34 (6.2  10–8 –x) (0.024 + x) x K3 

[H ][PO43] (0.024  x).x  2 [HPO4 ] (6.2 108  x)

H+ + HP O42 ;



H+ + P O34 ;

Again neglecting x2 and assuming, 6.2  10–8 –x = 6.2  10–8

for I step : H3PO4 0.1 0.1 – C



H+ + H2P O4 0 0 C C

  C.C K1  [H ][H 2 PO4 ]  [H3PO4 ] (0.1 C)

C2 7.5  10 = (0.1 C) –3



6.2  10–8 =

For III step : HP O24

H+ + H 2P O 4;

⇌

0.024y 0.024

 or

K2 = 6.2  10–8 III step HP O 24

The dissociation of H2P O4 occurs in presence of [H+] furnished in step I.

or

[HCOO ]100 103 100 = 0.004%  [HCOOH] 26.5

K1 = 7.5  10–3 II step H2P O4

HP O42 ; H+ + 0.024 0 (0.024 + y) y

(0.024  y)y 6.2  10–8 = (0.024 – y) y is small  0.024 – y  0.024 and  neglecting y2.

[HCOO–] = [HCOO H 2 ] = 10–3.  Now % dissociat ion of HCOOH =

I step H3PO4

For II step H2P O 4 0.024 (0.024 – y)

[H  ][HPO 2 ] 4 Thus, K2 = [H 2PO4 ]

Since in case of auto ionization [HCOO H 2 ] [HCOO–] = 10–6

Ex.2

The value of K1 is much large than K2 and K3. 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 suppressed due to common ion effect.

C = 0.024  [H+] = 0.024 M [H2 PO 4 ] = 0.024 M [H3PO4] = 0.1 – 0.024 = 0.076 M

13

3.6 10

0.024x 6.210 8 8

 6.2 10  9.31019 . 0.024 If CH3COOH (Ka = 10–5) reacts with NaOH at 298 K, then find out the value of the maximum 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 10–11 mol–1 L sec–1 at 298 K. Also calculate Arrhenius parameter for backward reaction if H298 = 44 kcal and Ea(f) = 94 kcal.  x

Ex.3

3.6  10–13 =

21

Ionic Equilibrium

Sol.

CH3COOH+NaOH CH 3COONa+H 2O; –11 –1 –1 Kf = 10 mol L sec The backward reaction is of hydrolysis of sodium acetate

Ka

=

1

Ka KW  K 1  for  K C K  (K h )   K   K W bac  a  K for 10 10  K bac  K  K W  105 a 11

14

 10

Ex.6 20

Sol.

H = E a f  E a b 44 = 94 – E a b

Ex.4

Sol.

Ea b  A b  e

 Eabac / RT

Ab = 2.71  10+16.

The pH of pure water at 25°C and 35°C are 7 and 6 respectively. Calculate the heat of formation of water from H+ and OH–. [H+] = 10–7 At 25°C;  KW = 10–14 [H+] = 10–6 At 35°C; –12 KW = 10  Kw2

2.303 log10 K w1

H  T2  T1   R  T1  T2 

1012 H  308  298  2.303 log10 1014  2  298  308 

H = 84551.4 cal/mol = 84.551 kcal/mol Thus H2O H+ + OH– H = 84.551 kcal/mol H+ + OH– ⇌ H O;2  Ex.7 H = – 84.551 kcal/mol The pH of a 0.10 M hydrocyanic acid solution is 5.2. What is the value of Ka for hydrocyanic Sol. acid ? H3O+ + CN– HCN + H2O x x (0.1 – x) –5.2 = log[H3O+] [H3O+] = 6.3 × 10–6 or x = [H3O+] –6 = 6.3 × 10 M/l 

Ex.5

Sol.

6.3  10 6  6.3  10 6 0.1  6.3  10

6

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. (Ka for NH4+ = 5.0 × 10–10 M) [OH–] in aqueous solution of NH4OH can be determined from the following relation. [OH–] =

 50 103  –20 10 = A b  e  2 298    

[HCN]

xx

= 0.1  x

= 39.6 × 10–11

Given, H298 = 44 kcal and E a f = 94 kcal



=

[H3O  ][CN – ]

Kb  c

(a) Determination of the value of Kb 14 Kw 110 –5 Kb = K = 10 = 2 × 10 5.0  10 a

(b) Determination of concentration (molarity) C of NH4OH solution Wt . o f NH 3 in 100 ml of NH 4 O H solution=10gm

Vol. of 100 g of NH4OH solution = w  1000

10  1000  0.99

Molarity = m V = 17  100 – (c) Determination of [OH ] [OH–] =

100 0.99

=

99 17

Kb C

99 5 = 2  10  17 = 10–2

19.8 17

= 10–2 1.16 [OH–] = 10–2 × 1.077 mol l–1 = 1.077 × 10–2 mol l–1 14

 [H+] =

10 –13 mol l–1 2 = 9.28 × 10 1.077 10

Calculate the pH of solution obtained by mixing 10 mL of 0.1 M HCl and 40 ml of 0.2 M H2SO4. Milli-equivalent of H+ from HCl = 10  0.1 =1 Milli-equvalent of H+ from H2SO4 = 40  0.2  2 = 16  Total meq. of H+ in solution = 1 + 16 = 17 

[H+] =

17  3.4 101 50

22

Ionic Equilibrium



Ex.10 A0.1 M solution of weak acid HAis 1% dissociated at 25°C. What is its Ka ? If this solution is with respect to NaA 0.2 M, what will be the new degree of dissociation of HA and pH ? Sol. For weak acid HA :

  [H ]  Meq.   Vin mL   pH = – log [H+] = – log 0.34 pH = 0.4685

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. Sol. HCl + NaOH   NaCl + H2O t = 0 100  0.1 9.9  1 9.9 t = t 10–9.9 9.9–9.9 = 0 9.9 

Sol.

1 = 0.01, [HA] = 0.1 M 100

K a  C 2  0.1 (0.01) 2  10 5  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.

0.1  9.099 104 M 109.9 pH = – log [H +] = – log 9.099  10–4 pH = 3.0409

[H+] left from HCl = 

Ex.9

 HA 

Calculate [H+] in a solution containing 0.1 M HCOOH and 0.1 M HOCN. K a for HCOOH and HOCN are 1.8  10–4 and 3.3  10–4. In this problem both the acids contribute for [H+] due to appreciable dissociation. Thus, H+ + HCOO– HCOON x 0.1 –x x+y H+ HOCN + OCN– 0.1–y x+y y + Because [H ] will remain common in solution. Thus, 

KHCOOH =



[H ][HCOO ]  1.8 10 4 … (1) [HCOOH]

[H  ][OCN  ]  3.3104 … (2) KHOCN = [HOCN] or KHCOOH =

(x  y)x 0.1

 1.8 104 …(3)

(x  y)y  3.3 10 4 0.1 Thus, by (3) and (4)

KHOCN =

…(4)

x  1.8 y 3.3 y = 1.83 x or … (5) 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 = 2.52  10–3 + 4.61  10–3 = 7.13  10–3M



pH = – log 10–5 + log

0.2 0.1

pH = 5.3010 H+ + A– Also HA 0 0 1 (1 – )   [A–] is provided by NaA since disso ciation ofHAin presence ofNaAis suppressed due to a common ion effect. [H  ][A  ] (C)  0.2  105  C(1 ) [HA]



Ka 



  5 105

Ex.11 Calculate the amount of (NH4)2SO4 in g which must be added to 500 mL of 0.2 M NH 3 to yield a solution of pH = 9.35. Kb for NH 3 = 1.78  10–5. Sol.

pOH = – log Kb + log

or

[Salt] [Base]

[NH 4 ] pOH = – log Kb + log [NH 4OH]

 [NH 4 ] is obtained from salt (NH4) 2SO 4.  pH = 9.35 pOH = 14 – 9.35 = 4.65   Millimole of NH 4OH in solution = 0.2  500 = 100 Let millimole of NH4 added in solution = a

23

Ionic Equilibrium

a 100 [NH ]  ; [NH 4OH]  500 500  4



4.65 = log 1.78  10–5 + log

a / 500 a

100 a = 79.51 Millimole of (NH4)2SO4 added

=

a  79.51  39.755 2 2

w 1000  39.755 132



Nic– + H+ HNic (1 – )C C C where  is the dissociation of the acid and c is the concentration of nicotinic acid.

100 / 500

4.65 = 4.7496 + log  

Sol.

Ka = c=

[H  ][Nic – ] c  c = (1–)c [HNic]

or

c 1–

0.1 mol = 0.05 mol/litre 2.0 litre

1.4 × 10–5 =

   0.05 1–

 = 0.016594  Percent dissociation = 0.016594 × 100 ~ 1.66%.

Ex.14 How many moles of calcium hydroxide must be dissolved to produce 250 ml of an aqueous Ex.12 A solution contains 0.10 M H2S and 0.3 M solution of pH 10.65 ? Assume complete 2– dissociation. HCl. Calculate the concentration of S and – pH + pOH = 14 Sol. HS ions in the solution. For H2S, –7 –13 Ka1 = 1.0 × 10 Ka2 = 1.3 × 10 pOH = 14 – 10.65 = 3.35 Sol. Dissociation of H2S can be represented as [OH–] = 4.47 × 10–4 mol/litre H2S H+ + HS– No. of OH– moles in 250 ml (a) – + 2– HS (b) H +S 4 = 4.47 10 = 1.12 × 10–4 From the first ionisaiton of w ( NH 4 ) 2 SO4  5.248 g



[H  ][HS ] H2S = = Ka1 [H 2S]

4

No. of moles of Ca(OH)2 dissolved = 1 × 2

Since H2S is weakly ionised and its ionisation 1.12 × 10–4 = 0.56 × 10 –4 is further decreased in presence of highly Ex.15 Saccharin (Ka = 2 × 10–12) is a weak acid ionised HCl, concentration of H+ in solution represented by formula HSac. A 4 × 10–4 mole will be mainly due to HCl. Thus [H+] = 0.3 amount of saccharin is dissolved in 200 cm3 Substituting the values in the above equation water of pH 3. Assuming no change in volume,  calculate the concentration of Sac– ions in the 0.3[HS ] –7 = 1 × 10 resulting solutionat equilibrium. 0.1 Sol. Calculation of [H+] and [HSac] at start [HS–] = 3.33 × 10–8 M We know that [H +] = 10 –pH = 10 –3 = 0.001M From the second ionisation of H2S, 4 10 4 1000

[H  ][S2– ] [HS– ] 0.3  [S2 ] 3.33  108

[S2–]

= Ka2 or = 1.3 ×

10–13

= 1.44 × 10–20 M

Ex.13 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.

[Hsac] =

200

= 0.002 M

The dissociation of HSac is as below HSac. + H+ 0.002 0.001 At start At equb 0.002 – x 0.001 + x  Ka = 

Sac– 0 x

[H  ][Sac – ] (0.01  x) x = = 2 × 10–12 [H Sac- ] 0.002 – x

x = 4 × 10–12 M [Sac–]equb = 4 × 10–12 M

24

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. Sol. Since we know 30 c.c. of N NaOH will neutralise 30 c.c. of N HCl. Amount of N HCl left un-neutralised  = 50 – 30 = 20 c.c. Ex.18 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 Sol. = 1 g eq of HCl 20 c.c. of N HCl contain = 1 ×

20 1 = 1000 50

= 1.8 mole/litre 1000 ml contains 1.8 moles 500 ml contains 1.8 = 0.9 moles 2

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. Ka for 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. pH

gram eq.   1 This  50g eq.  is the amount of HCl present in 

1

=7+



1

= 7 + 2 pKa + 2 logc 1 2

× 4.7212 +

1 log 0.05 2

= 7 + 2.3606 – 0.65 = 8.71 Ex.19 Calcium lactate is a salt of a weak organic acid 1 Thus 80 c.c. of the mixture contain = 50 g eq. and represented as Ca (Lac)2. A saturated solution of Ca (Lac)2 contains 0.13 mol of this of HCl salt in 0.50 litre solution. The pOH of this 1 1000 solution is 5.60. Assuming a complete 1000 c.c of the mixture contain = 50 × 80 g dissociation of the salt, calculate Ka of lactic eq. = 0.25 g eq. acid.  [H+] = 0.25 g eq. per litre Ca2+ + 2 Lac– Sol. Ca (Lac)2 We know that 2 × 0.26 M 0.13 × 2 M pH = –log [H+] = – log [0.25] = 0.6021 Concentration of the lactate ion  Ex.17 How many grams of NH 4Cl should be C = 0.52 dissolved in 500 ml of water to have solution Also we are given of pH 4.5 ? Kb for ammonium hydroxide is pOH = 5.6  pH = 14 – 5.6 = 8.4 1.8 × 10–5. –14 K log Kw = –14 = 10 or w Sol. pH = –log[H+] Substituting the values inthe following equation log[H+] = log 1 – 4.5 = 5.5 for the salt of a weak acid and strong alkali. + –5 [H ] = Antilog [5.5] = 3.162 × 10  = 1 (log Kw + log Ka – log c) pH mole per litre 80 c.c. of the mixture.

2

[H+] = hc = or

[H+]2 =



[H  ]2  K b Kw

Kw  c Kb

Kw  c Kb

1

8.4

= – 2 (–14 + log Ka – log 0.52)

Ka

= 8.26 × 10–4

Ex.20 What is the pH of 0.10 M CH 3COONa solution. Hydrolysis constant of sodium acetate is 5.6 × 10–10?

3.162 10 5  3.162  10 5  1.8  105

c=

1  10 14

25

Ionic Equilibrium

Sol.

Hydrolysis of the salt may be represented as CH3COO– + Na+ + H2O CH3COOH+Na++OH– CH3COOH + OH– or CH3COO– + H2O 0 C

At eq = 0 C At C(1 – )

0 C

equilibrium Kh =

 =

C 22 C(1 - )

Kh C

= C2 when  < 1

=

5.6 X 10 0.10

10

= 7.5 × 10–5 [OH–]

= C = 0.10 × 7.5 × 10–5

pOH

= – log (7.5 × 10–6) = 5.126

pH

= 14 – pOH = 14 – 5.126 = 8.874

26

Ionic Equilibrium

EXERCISE - I Q.1

The degree of dissociation of acetic acid in a 0.1 N solution is 1.32 × 10 –2. At what concentration of nitrous acid, its degree of dissociation willbe same as that of acetic acid? Ka (HNO2) = 4×10–4

Q.2

How many times is the H+ concentration in the blood (pH = 7.36) greater than in the spinal fluid (pH = 7.53) ?

Q.3

A 0.400 M formic acid solution freezes at –0.758°C. Calculate the Ka of the acid at that temperature. (Assume molarity equal to molality). Kf (H2O) is 1.86° mol–1 kg.

Q.4

Asample ofAgCl was treated with 5 mL of 1.5 M Na2CO3 solution to give Ag 2CO 3. the remaining solutioncontained 0.00266625 g/litre Cl– ion. Calculate the solubility ofAgCl in (a) 0.1 M AgNO 3 (b) 0.1 M BaCl2 (c) 0.2 M NaNO3. Given that solubility ofAg2CO3 in500 mL H 2O = 13.8 mg.

Q.5

100 mL of solution S 1 contains 0.17 mg of AgNO3 .Another 200 mLsolution S2 contains 0.117 mg of NaCl. On mixing these two solutions predict whether the precipitate of AgCl will appear or not Ksp AgCl = 10–10 M2

Q.6

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.

Q.7

Calculate the hydrolysis constant of NH4Cl ; determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution. Kb (NH4OH) = 1.8 × 10–5

Q.8

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 with0.05 mole of HCl, what will be the pH of the mixture ?

Q.9

What is the solubility product ofAg 2CrO 4 if 0.0166 g ofthe salt dissolves in 500 ml of water at 18ºC?

Q.10

When a 40 mL of a 0.1 M weak base is titrated 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.

Q.11

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? (Ka = 1.8 × 10–5)

Q.12

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.

Q.13

Given a solution that is 0.5 M CH3COOH. To what volume at 25°C must one dm3 of this solution be diluted in order to (a) double the pH; (b) double the hyd roxide-ion concentration. Given that Ka = 1.8 × 10–5 M.

Q.14

The solubility of Mg(OH)2 is increased by addition of NH4+ ion. Calculate – (a) KC for the reaction, 2NH3+ 2H2 O+Mg2+ Mg(OH)2+2NH 4+ (b) Find solubility of Mg(OH)2 in a solution containing 0.5 M NH4Cl before addition of Mg(OH)2 (Ksp of M g(OH)2 = 1 × 10–11, Kb for NH4O H = 1.8 × 10–5 )

Q.15

Abuffer solution was prepared by dissolving 0.05 mol formic acid and 0.06 mol sodium formate in enough water to make 1.0 L of solution. Ka for formic acid is 1.80 × 10-4. (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?

27

Ionic Equilibrium

Q.16

An unknown volume and unknown Q.22 concentration of weak acid HX is titrated with NaOH of unknown concentration. After addition of 10.0 cm3 of NaOH solution, pH of solution is 5.8 and after the addition of 20.0 cm3 of NaOH solution, the pH is 6.4. Calculate Q.23 the pH of aqueous solution of 0.1 M NaX.

The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate 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 Q.17 Asolution containing zinc and manganese ions constant of the acid and its pKa . each at a concentration of 0.01 mol dm–3 is saturated with H S.2 Calculate (i) pH at which Q.24 Assuming complete dissociation, calculate the pH ofthe following solutions: the MnS will form a precipitate (ii) conc. of (a) 0.003 M (b) 0.005 M NaOH Zn+2 ions remaining. (c) 0.002 M HBr (d) 0.002 M KOH Given : [H2S] = 0.1 mol/lit, Ksp(ZnS) = 1 × –22 2 –2 –16 2 10 mol lit , Ksp(MnS) = 5.6 × 10 mol Q.25 Calculate the pH of the following solutions: lit–2. (a) 2 g of TlOH dissolved in water to give 500 ml K1 and K2 for H 2S are 1×10–7 and 1.1×10–14. of solution. (b) 0.3 g of Ca (OH)2 dissolved in water to Q.18 For the indicator thymol blue, the value of pH give 2 litre of solution is 2.0, when half of the indicator is present in (c) 0.3 g of NaOH dissolved in water to give an unionized form. Calculate the percentage 200 mLof solution. ofthe indicator in the unionized formina solution (d) 1 mL of 13.6 M HCl is diluted with water to of 4.0 × 10 –3 mol/dm– 3 hydrogen ion give 1 litre of solution concentration. Q.26 If the solubility product of silver oxalate is Q.19 One of the substances sometimes responsible 1 × 10–11, what will be the weight ofAg2C2O4 for the hardness of water is CaSO4.Aparticular in 2.5 litres of a saturated solution ? water sample has 131 ppm CaSO (131 g 4 CaSO4 per 106 g of water). If this water boiled Q.27 Calculate the hydrolysis constant of KF. Determine the degree of hydrolysis of salt in a in a tea kettle, approximately what fraction of 0.01 M solution and the pH of the solution. water must be evaporated before CaSO4(s) Ka(HF) = 6.6 × 10–4. begins to deposit ?Assume that solubility of CaSO4 does not change with temperature in Q.28 Assuming that the buffer in blood is CO3––HCO3 – , calculate the ratio of conjugate the range 0 to 100°C, Ksp (CaSO4) = 9.1 × base to acid necessary to maintain blood at its 10–6. proper pH, 7.4 K1 (H2CO3) =4.5 × 10–7. Q.20 The indicator phenolred is half inthe ionic form when pH is 7.2. If the ratio of the undissociated Q.29 How does the solubility of CaC2O4 in a 0.1 M solution of ( NH 4 ) 2 C 2 O 4 decrease in form to the ionic form is 1 : 5, find the pH of comparison with its solubility in water ?Assume that the solution. With the same pH for solution, if the ionisation of (NH4)2C2O4 is complete. [Ksp indicator is altered such that the ratio of (CaC2O4) = 2 × 10–9] undissociated formto dissociated formbecomes 1 : 4, find the pH when 50% of the new indicator Q.30 How many moles of sodium hydroxide can be is in ionic form. added to 1.0 L of a solution 0.10 M in NH3 and 0.10 M in NH4Cl without changing the Q.21 The first ionization constant of H 2S is pOH by more than 1 unit. Assume no change in 9.1 × 10–8. Calculate the concentration of volume. – HS ion in its 0.1 M solution. How will this Kb = 1.8 × 10–5. concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13. Calculate the concentration of S2– under both conditions.

28

Ionic Equilibrium

EXERCISE - II Q.1

Calculate the number of H+ present in one ml Q.11 of solution whose pH is 13.

Q.2

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. Q.12

Q.3

(i) Kw for H2O is 9.62 × 10–14 at 60°C. What is pH of water at 60°C. (ii) What is the nature of solutionat 60°C whose (a) pH = 6.7 (b) pH = 6.35

Q.4

The value of K w 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?

Q.5

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 ]

Q.6

What are the concentration of H+, H2C2O4,  2 HC 2O 4 and C 2O 4 in a 0.1 M solution of oxalic acid ? [K1 = 10–2 M and K2 = 10–5 M ]

Q.7

What are the concentrations of H + ,

Q.14

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.

Q.15 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.16 The values of Ksp for the slightly soluble salts MX and QX 2 are each equal to 4.0×10–18. Which salt is more soluble? Explain your answer fully. Q.17

Calculate the SimultaneoussolubilityofAgSCN and AgBr. K sp (AgSCN) = 1.1 × 10–12 , Ksp(AgBr) = 5 × 10–13.

Q.18

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)3 from its component ions, and what is the dissociation constant of Fe(SCN)3 into its simplest ions on the basis of these data ?

Q.19

Calculate the percent error in the [H O+] 3 concentration made by neglecting the ionization of water in a 10–6 M NaOH solution.

Q.20

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 ClCH2COONa . What is [H+] in the solution? Ka = 1.5 × 103.

Given : H2SO4  H+ + HSO 4 ; strong –2 HSO4 l H+ + SO 2 4 ; K2 = 10 M

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] Q.9 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 ] Q.10 What indicator should be used for the titration of 0.10 M KH2BO3 with 0.10 M HCl ? Ka (H3BO3) = 7.2 × 1010 .

What is the OH concentration of a 0.08 M solution of CH3COONa. [Ka(CH3COOH)=1.8 × 105]

Q.13 Calculate the pH of a 2.0 M solution of NH4Cl. [Kb (NH3) = 1.8 × 105]

HSO4 , SO 2 4 and H2SO4 in a 0.20 M solution of sulphuric acid ?

An acid indicator has a K aof 3 × 105. The 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?

Q.8

29

Ionic Equilibrium

A solution contains HCl, Cl2HC COOH & Q.27 How many moles of sodium hydroxide can be CH3 COOH at concentrations 0.09 M in HCl, added to 1.00 L of a solution 0.1 M in NH3 & 0.1 M in NH4Cl without changing the pOH by 0.09 M in Cl2HC COOH & 0.1 M in CH3 COOH. pH for the solution is 1. more than 1.00 unit ? Assume no change in Ionization constant of CH3 COOH = 105. volume. Kb(NH3) = 1.8 × 105. What is the magnitude of K for dichloroacetic Q.28 20 ml of a solution of 0.1 M CH3COOH acid ? solution is being titrated against 0.1 M NaOH Q.22 Determine the [S2] in a saturated (0.1M) H2S solution. The pH values after the addition of 1 ml solution to which enough HCl has been added & 19 ml of NaOH are (pH) &1(pH) , what 2 to produce a [H+] of 2 × 104 . K1 = 107, is pH ? K2 = 1014. Q.29 Aweak base BOH was titrated against a strong Q.23 What is the pH of a 1.0 M solution of acetic acid . The pH at 1/4th equivalence point was acid ? To what volume must 1 litre of the 9.24. Enough strong base was now added solution be diluted so that the pH of the resulting (6m.eq.) to completely convert the salt. The solution will be twice the original value. Given total volume was 50ml. Find the pH at this Ka = 1.8 × 10–5. point. Q.24 It is desired to prepare 100 ml of a buffer of Q.30 An organic monoprotic acid [0.1M] is titrated pH 5.00.Acetic, benzoic and formic acids and against 0.1M NaOH. By how much does the their salts are available for use. Which acid pH change between one fourth and three fourth should be used for maximum effectiveness stages of neutralization? If at one third stage of against increase in pH? What acid-salt ratio neutralization, the pH is 4.45 what is the should be used ? pKa values of these acids are: dissociation constant of the acid? Between acetic 4.74; benzoic 4.18 and formic 3.68. what stages of neutralisation may the pH change Q.25 When a 40 mL of a 0.1 M weak base is titrated by 2 units ? with 0.16M 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. Q.21

Q.26 Abuffer solution was prepared by dissolving 0.05 mol formic acid & 0.06 mol sodium formate in enough water to make 1.0 L of solution. Ka for formic acid is 1.80 × 104. (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?

30

Ionic Equilibrium

EXERCISE- III Q.1

The conjugate acid of NH2 is (A) NH3 (B) NH2OH  (D) N2H4 (C) NH 4

Q.2

Out of the following, amphiprotic species are II I: HPO32– III OH– – IV H2PO4 (A) I, HCO3– III, IV (C) III (B) I and III and IV (D)All

Q.3

Q.9

Which of the following solution will have pH close to 1.0? (A) 100 ml of M/100 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

pH of an aqeous solution of NaCl at 85°C should be (A) 7 (B) > 7 (C) < 7 (D) 0

Q.10

Asolution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of: (A) 3 (B) 4 (C) 3000 (D) 10000

Q.4

1 cc of 0.1 N HCl is added to 99 cc solution of NaCl. The pH of the resulting solution will be (A) 7 (B) 3 (C) 4 (D) 1

Q.11

The first and second dissociation constants of an acid H2A are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall dissociatio n constant of the acid will be : (A) 5.0 ×10–5 (B) 5.0 ×1015 (C) 5.0 ×10–15 (D) 0.2 ×105

Q.5

10 ml of

Q.6

Q.7

M H SO is mixed with 40 ml of 200 2 4

Q.12 An aqueous solution contains 0.01 M RNH2 (Kb = 2 ×10–6) & 10–4 M NaOH. M H2SO4. The pH of the resulting solution The concentration of OH¯ is nearly : 200 (A) 2.414 ×10–4 (B) 10–4 M is (B) 2 (A) 1 (C) 1.414 × 10–4 (D) 2 × 10–4 (D) none of these (C) 2.3 Q.13 The degree of hydrolysis of a salt of weak acid If pKb for fluoride ion at 25° C is 10.83, the and weak base in it’s 0.1 M solution is found to be 50%. If the molarity of the solution is 0.2 ionisation constant of hydrofluoric acid in water M, the percentage hydrolysis of the salt should at this temperature is : (A) 1.74 × 105 (B) 3.52 × 103 be 4 2 (C) 6.75 × 10 (D) 5.38 × 10 (A) 100% (B) 50% (C) 25% (D) none of these The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is (A) 1 (B) 2 (C) 3 (D) 11

Q.8

If K1 & K2 be first and second ionisation constant of H3PO4 and K1 >> K2 which is incorrect. (A) [H+] = [ H 2PO4 ] (B) [H+] = K1[H 3PO 4 ] (C) K2 = [HPO4 ]

Q.14 What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is 1.3 × 10–9 and Kw = 1.0 × 10–14 (A) 2.48 (B) 5.26 (C) 8.2 (D) 9.6 Q.15

The compound whose 0.1 M solution is basic is (A) Ammonium acetate (B) Ammonium chloride (C) Ammonium sulphate (D) Sodium acetate

(D) [H+] = 3[PO3 ] 4

31

Ionic Equilibrium

Q.16

The  pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is (A) 1 (B) 6 (C) 7 (D) 9

Q.24

The solubility of A2X3 is y mol dm–3. Its solubility product is (A) 6 y2 (B) 64 y4 (C) 36 y5 (D) 108 y5

Q.17

If equilibrium constant of CH3COOH + H2O  CH3COO– + H3O+ Is 1.8 × 10–5, equilibrium constant for CH3COOH + OH– l CH3COO– + H2O is (A) 1.8 ×10–9 (B) 1.8 × 109 (C) 5.55 × 10–9 (D) 5.55 × 1010

Q.25

If Ksp for HgSO4 is 6.4 ×10–5, then solubility of this substance in mole per m3 is (A) 8 ×10–3 (B) 6.4 ×10–5 –6 (C) 8 × 10 (D) None of these

Q.26

The solubility of a sparingly soluble saltAB2 in water is 1.0 × 10–5 mol L–1. Its solubility product is: (A) 10–15 (B) 10–10 –15 (C) 4 × 10 (D) 4 × 10–10

Q.18

Q.19

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 solution ofthe corresponding salt, BA, will be : Q.27 (A) 8.58 (B) 4.79 (C) 7.01 (D) 9.22 The range of most suitable indicator which should be used for titration of X– Na+ (0.1 M, 10 ml) wit h 0.1 M HCl sho uld be (Given: k b(X  ) =10–6) (A) 2–3 (C) 6–8

Q.20

(B) 3–5 (D) 8–10

How many gm of solid NaOH must be added to 100 ml of a buffer solution which is 0.1 M each w.r.t. Acid HA and salt Na+ A– to make Q.29 the pH of solution 5.5. Given pka (HA) = 5. (Use antilog (0.5) = 3.16) (A) 2.08 × 10–1 (B) 3.05 × 10–3 –2 (C) 2.01 × 10 (D) None of these

Q.21

If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka = 2×10–4], the pOH of the resulting solution is (A) 3.4 (B) 3.7 (C) 7 (D) 10.3

Q.22

1 M NaCl and 1M HCl are present in an aqueous solution. The solution is (A) not a buffer solution and with pH < 7 (B) not a buffer solution with pH > 7 (C) a buffer solution with pH < 7 (D) a buffer solution with pH > 7

Q.23

Q.28

The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HAin which 50% of the acid is ionized is : (A) 4.5 (B) 2.5 (C) 9.5 (D) 7.0

Which ofthe following is most soluble in water? (A) MnS (K sp= 8×10–37) (B) ZnS (Ksp= 7×10–16) (C) Bi2S3 (Ksp= 1×10–72) (D) Ag3(PO4) (Ksp= 1.8×10–18) When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10–10) will occur only with: (A) 10–4 M (Ag+) and 10–4 M (Cl¯ ) (B) 10–5 M (Ag+) and 10–5 M (Cl¯ ) (C) 10–6 M (Ag+) and 10–6 M (Cl¯ ) (D) 10–10 M (Ag+) and 10–10 M (Cl¯ )

The precipitate of CaF2(Ksp = 1.7 × 10–10) is obtained when equal volumes of the following are mixed (A) 10–4 M Ca3+ + 10–4 M F– (B) 10–2 M Ca2+ + 10–3 M F– (C) 10–5 M Ca2+ + 10–3 M F– (D) 10–3 M Ca2+ + 10–5 M F– Q.30 50 litre of a solution containing 10–5 mole of Ag+ is mixed with 50 litre of a 2 × 10–7 M HBr solution. [Ag+] in resultant solution is : [Given : Ksp (AgBr) = 5 × 10–13] (A) 10–5 M (C) 10–7 M

Q.31

(B) 10–6 M (D) None of these

pH of a saturated solution of silver salt of monobasic acid HAis found to be 9. Find the Ksp of sparingly soluble saltAgA(s). Given : Ka (HA) = 10–10 (A) 1.1 × 10–11 (B) 1.1 × 10–10 –12 (D) None of these (C) 10

32

Ionic Equilibrium

Q.32 The solubility of metal sulphides in saturated solution of H2S {[H2S]= 0.1 M} can be represented by MS + 2H+  M2+ +H2S;Keq=

[M 2 ][H S]  2

2

[H ]

The value of Keq is given for few metal sulphide. If conc. of each metal ion in solution is 0.01 M, which metal sulphides are selectively ppt at Q.37 total [H+]= 1M in saturated H2S solution. Metal sulphides MnS ZnS CoS PbS [M 2 ][H2S] –7 10 –2 Keq = 3 × 10 3 × 10 3 3 × 10 [H  ]2 (A) MnS, ZnS, CoS (C) PbS, ZnS Q.33

Q.34

Select incorrect statement(s). (A) Phenolphthalein is suitable indicator for the titration of HCl (aq) with NH4OH (aq). (B) An acid-base indicator in a buffer solution of pH = pKIn + 1 is ionized to the extent of 90%. (C) In the titration of a monoacidic weak base with a strong acid, the pH at the equivalent point is always calculated by

(B) PbS, ZnS, CoS (D) PbS

Solid Ba(NO3) 2 is gradually dissolved in a 1.0 × 10–4 M Na2CO 3 solution. At what concentration of Ba2+ will a precipitate begin to form ? (Ksp for BaCO3 = 5.1 × 10–9) (A) 4.1 × 10–5 M (B) 5.1 × 10–5 M (C) 8.1 × 10–8 M (D) 8.1 × 10–7 M Ksp of MX4 and solubility of MX4 is S mol/litre is related by : (A) S = [KSP /256]1/5 (B) S = [128 KSP ]1/4 (C) S = [256 KSP]1/5 (D) S = [KSP /128]1/4

(A) Kb of base is less than 10–6 (B) Concentration of salt (C) at equivalent point is 0.25 M (C) Volume of HCl is used at equivalent point is 100 ml (D) Weight percentage of base in given sample is 80%.

1 pH = 2 [pK w – pK b – log C]

(D) When Na3PO 4 (aq) is titrated with HCl (aq), the pH of solution at second equivalent point is calculated by Q.38

Q.35 Which of the following statement(s) is/are correct ? (A) the pH of 1.0×108 M solution of HCl is 8 (B) the conjugate base of H2PO4 is HPO42 Q.39 (C) autoprotolysis constant of water increases with temperature (D) when a solution of a weak monoprotic acid is titrated again a strong base, at half neutralization point pH = (1/2) pKa . Q.36 A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl Q.40 th 1 when   of the base was neutralised the 5 pH was found to be 9 and at equivalent point pH of solution is 4.5 . Given :All data at 25°C & log 2 = 0.3. Select correct statement(s) .

1 [pK a1  pK a 2 ] 2

Which of the following is true for alkaline aqueous solution? pk w (A) pH > (B) pH > pOH 2 pk w (C) pOH < (D) pH < pOH 2 An acid-base indicator has a Ka of 3.0 ×10–5. The acid form of the indicator is red and the basic form is blue. then: (A) pH is 4.05 when indicator is 75% red (B) pH is 5.00 when indicator is 75% blue. (C) pH is 5.00 when indicator is 75% red (D) pH is 4.05 when indicator is 75% blue. The equilibrium constant for the reaction HONO(aq) + CN– (aq) HCN(aq) – +6 + ONO (aq) is 1.1 × 10 . From the magnitude of this Keq one can conclude that (A) CN– is stronger base than ONO– (B) HCN is a stronger acid than HONO (C) The conjugate base of HONO is ONO– (D) The conjugate acid of CN– is HCN

33

Ionic Equilibrium

Q.41

Q.42

All of the following are acid-base conjugate pairs: (A) HONO, NO 2– (B) H3O+, OH– (C) CH3NH3+, CH3NH 2 (D) HS–, S 2– In the following reaction:

(A)Ais an acid and B the base (B) Ais a base and B the acid (C) C is the conjugate acid ofA, and D is the conjugate base of B (D) C is the conjugate base ofA, and D is the conjugate acid of B Q.43 Which does not react with NaOH or which is not acid salt? (A) NaH2PO2 (B) Na2HPO3 (C) Na2HPO4 (D) NaHCO3 Q.44 pH of the following solution is not affected by dilution: (A) 0.01 M CH 3COONa (B) 0.01 M NaHCO3 (C) buffer of 0.01 M CH 3COONa and 0.01 M CH3COOH (D) 0.01 M CH 3COONH 4 Q.45 Which of t he following mixtures constitute a buffer? (A) CH3COOH + CH 3COONa (B) Na2CO3 + NaHCO 3 (C) NaCl + HCl (D) NH4Cl + (NH 4) 2SO 4 Q.46 Which of t he following mixtures constitute a buffer? (A) Na2CO 3 + HCl (B) NaOH + CH 3COOH (C) NH3 + CH 3COONH 4 (D) NaOH + BaCl 2 Q.47 In which of the following pairs of solutions is there no effect on the pH upon dilution? (A) 0.1 M NH 3 and 0.1M(NH4) 2SO 4 (B) 0.1 M NaH 2PO 4 and 0.1 M Na 2HPO 4 (C) 0.1 M HCl and 0.01 M NaOH (D) 0.1 M KCl and 0.1 M HCl

Q.48 Which of the following will suppress the ionization of phthalic acid in an aqueous solution. (A) KCl (B) H 2SO (C) HNO3 (D) NaOH Assertion and Reason : Q.49 Statement-1 : pH of 10–7 M NaOH solution is exist between 7 to 7.3 at 25°C. Statement-2 : Due to common ion effect ionization of water is reduced. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Q.50 Statement-1 : In general phenolphthalein is used as an indicator for the titration of weak acid (HA) against strong base (NaOH). Statement-2 : At equivalent point solution is basic. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Q.51

Statement-1 : Moles of Sr2+ furnished by sparingly soluble substance Sr(OH)2 decreases due to dilution. Statement-2 : Solubility product constant of Sr(OH)2 is not affected by dilution. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

34

Ionic Equilibrium

Paragraph 1 Paragraph 2 1.2 g of a monoprotic acid HA, is titrated with Potash alum is K Al(SO4)2·12H2O. As a strong electrolyte, it is considered to be 100 0.222 M NaOH solution. The pH of the % dissociated into K+, Al3+ and SO42–. The solution is monitored with pH meter. Aportion solution is acidic because of the hydrolysis of of the titration curve is shown in the diagram. Al3+, but not so acidic as might be expected, Expanded titration because the SO42– can sponge up some of the 12.00 curve of H3O+ by forming HSO4¯. Given a solution HA vs NaOH 11.00 ma de by disso lving 11. 85 g m o f K 10.00 Al(SO4)2·12H2O in enough water to make pH 9.00 100 cm3 of solution. What is [H3O+] of the 8.00 solution if 7.00 6.00 5.00 15 16

Q.55 17 18

19

20

Volume of NaOH (mL)

Q.52

Q.53

Q.54

How many mL of NaOH is required to bring about the titration to its equivalence point ? (A) 4.00 (B) 9.00 (C) 19.00 (D) None of these

Q.56

What is the pH of solution at the equivalence point ? (A) 3.50 (B) 7.00 (C) 8.40 (D) None of these

Q.57

What is the molar mass of HA ? (A) 180 (B) 222 (C) 282 (D) None of these

Q.58

Match the column Q.59 Match the following List I (A) (P) CH3COOH (B) (Q) H2SO4 NaOH (C) (R) NH3 (D) (S) Q.60 Column I (pH of resultant solution) 200 ml of H2SO4 solution (A) (specific gravity 1.225 containing 25% H2SO4 by weight) + 800 ml of 0.525 M strong triacidic base X (OH)3 50 ml of 0.1 M HCO3 + (B)

None of the ion is hydrolysing. (A) 10–7 M (B) less than 10–7 M (C) More than 10–7M (D) 0.0 OnlyAl3+ is hydrolysing and its first hydrolysis constant is 1.4 × 10–5 M (A) 1.87 × 10–3 M (B) 6.24 × 10–4 M (C) 0.09 M (D) None of these 2– Only SO4 is hydrolysing and acid dissociation constant of HSO4¯ in water is 1.25× 10–2. (A) 1.26 × 10–3 M (B) 6.32 × 10–7 M –8 (C) 1.58 × 10 M (D) None of these 3+ 2– BothAl and SO4 are hydrolysing. (A) 2.93 × 10–4 M (B) 0.0114 M (D) None of these (C) 5.43 × 10–6 M

List II weak Base Weak acid Strong acid Stronge base Column II (Exist between Colour transition range of an indicator) (P) Phenol Red (6.8 to 8.4) (Q)

Propyl red (4.6 to 6.4)

(R)

Phenolphtalein (8.3 to 10.1)

(S)

Malachite green (11.4 to 13)

2 50 ml of 0.8 M CO3

(C)

(H2CO3 : K a1 = 4 × 10 –7 , K a 2 = 2 × 10 –11) 50 ml of 0.2 M HA (aq) (Ka = 10–5) + 50 ml of 0.1 M HCl (aq) + 100 ml of 0.13 M NaOH (aq)

35

Ionic Equilibrium

ANSWER KEY EXERCISE -I Q.1 Q.4 Q.6 Q.9 Q.12 Q.14 Q.16 Q.18 Q.21 Q.22 Q.24 Q.26

1.44 × 10–4 2.3 mol/lit Q.2. 1.5 times Q.3 –4 –5 (a) 1.76 × 10 mg/L, (b) 8.8 × 10 mg/L, (c) 1.588 mg/L Q.5 No –4 –10 –4 Q.8 2.87, 1.3 4.11 × 10 M 5.6 × 10 , 2.4 × 10 , 5.63 Q.7 Q.10 pH = 9.168 Q.11 8.7 mL Ksp (Ag2CrO4) = 4 × 10–12 HZ < HY < HX , K a(HY) = 10–5 M, K a(HZ) = 10–9 M Q.13 (a) 3.7 × 104 dm3 (b) 4 dm3 Q.15 (a) pH = 3.83 (b) pH = 3.85 (c) 3.99 KC = 3.02 × 10–2, S = 0.123 M Q.17 (i) pH = 4.35 (ii) 1.79 × 10–9 mol/lit 9.55 (i) 28.57% (ii) 4.11 × 10–4mol dm–3 Q.19 67% Q.20 pH = 7.3 [HS–] 9.54 × 10–5, in 0.1M HCl [HS–] = 9.1 × 10–8 M, [S2–] = 1.2 × 10–13 M, in 0.1 M HCl [S2–] = 1.09 × 10–19 M [Ac–] = 0.00093, pH = 3.03 Q.23 [A–] = 7.08 × 10–5 M, K a = 5.08 × 10–7, pK a = 6.29 (a) 2.52 (b) 11.70 (c) 2.7 (d) 11.30 Q.25 (a) 11.65 (B) 12.21 (c) 12.57 (d) 1.87 Q.27 pH = 7.59, kH = 1.5 × 10–11 Q.28 11 0.103 g

Q.29

1 of its solubility in water solubility decreases 2200

Q.30

0.082 mol of NaOH can be added

EXERCISE - II ×107

Q.1 Q.4 Q.7 Q.10 Q.12 Q.16 Q.18 Q.21 Q.24 Q.26

6.022 Q.2 0.6 ×10–7 Q.3 (i) 6.51 ; (ii) (a) Basic , (b) Acidic Q.5 Q.6 0.027 M, 0.073 M, 0.027 M, 10–5 M 6.81 (a) 0.522 , (b) 2.61 Q.8 0.2116 M, 0.1884 M, 0.0116 M, 0 8.7782 Q.9 9.7324 Q.11 pH = 0.954 (methyl red), one with pH = 5.22 as midpoint of colour range Q.14 5.12 ×10–6 M Q.15 8.71 [OH] = 6.664 × 106 Q.13 pH = 4.477 QX2 is more soluble Q.17 4 × 10–7mol/LAgBr, 9 × 10 –7mol/LAgSCN Kd = 1/Kf = 4.8 × 10–4 Q.19 error = 1% Q.20 [H+]=2.5×10-3 2-15 -2 Ka = 1.25 × 10 Q.22 [S ] = 2.5 × 10 Q.23 V = 2.77 × 104 litre Q.25 9.168 acetic acid, salt-acid molar ratio 1.8 :1 Q.27 0.0818 moles Q.28 2.558 (a) pH = 3.83 (b) pH = 3.85 , (c) = 3.99

Q.29

11.22

Q.30

0.9542, pKa = 4.751,

Q.1 Q.8 Q.15 Q.22 Q.30 Q.37 Q.44 Q.51 Q.54 Q.60

A Q.2 C Q.9 D D Q.16 B D A Q.23 C C Q.31 A A,B,C Q.38 A,B,C B Q.45 C A, B Q.52 B,C C Q.54 A (A) S, (B) S, (C) Q

Q.3 Q.10 Q.17 Q.24 Q.32 Q.39 Q.46 Q.53 Q.56

C D B D D A, B B,C B A

1 10 th & th stages of neutralisation 11 11

EXERCISE - III Q.4 Q.11 Q.18 Q.25 Q.33 Q.40 Q.47 Q.40 Q.57

B Q.5 B Q.6 Q.12 D Q.13 C Q.19 B Q.20 C D Q.26 C Q.27 B Q.34 A Q.35 A, C, D Q.41 D Q.42 A, B Q.48 B, C, D Q.49 Q.41 D B Q.42 Q.58 A C Q.59

C Q.7 B Q.14 A B Q.21 D A A Q.29 B B,C Q.36 B,C A, C, DQ.43 A A,B Q.50 A,B C Q.43 C A-Q, B-R, C-S, D-P

36

ROUGH WORK

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