b/9w øãç 9ø Ù {½ i - studymate · equilibrium e z øãç ø Ù ò 222 ans.for the given...

Equilibrium

NCERT Textual Exercise (Solved)

219

1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

(a) What is the initial effect of the change on vapour pressure ? (b) How do rates of evaporation and condensation change initially ? (c) What happens when equilibrium is restored finally and what will be

the final vapour pressure ? Ans. (a) Initially, the vapour pressure will decrease. (b) The rate of evaporation remains constant at constant temperature

in a closed vessel. However, the rate of condensation will be low initially because there are fewer molecules per unit volume in the vapour phase and hence the number of collisions per unit time with the liquid surface decreases.

(c) When equilibrium is restored, rate of evaporation = rate of condensation. The final vapour pressure will be same as it was originally.

2. What is Kc for the following equilibrium when the equilibrium concentration of each substance is:

[SO2] = 0.60M, [O2] = 0.82M and [SO3] = 1.90M 2SO2(g) + O2(g) 2SO3(g)

Ans. KSO

SO OM

M Mc = =

2[ ][ ] [ ]

( . )( . ) ( . )

3

22

2

2

21 90

0 60 0 82= − −12 229 12 2291 1. .M or Lmol

3. At a certain temperature and a total pressure of 105 Pa, iodine vapour contain 40% by volume of iodine atoms [I2 (g) 2I (g)]. Calculate Kp for the equilibrium.

Ans. Partial pressure of I atoms (PI) = (40 / 100) × l05 Pa = 0.4 × 105 Pa

Partial pressure of I2 P P PI 2= × = ×

60100

10 0 6 105 5a a.

∴ = =×

×= ×K P

PI2

I 2p

( . ).

.0 4 100 60 10

2 67 105 2

54 Pa

4. Write the expressions for the equilibrium constant, Kc for each of the following reactions :

(i) 2NOCl (g) 2NO (g) + Cl2 (g) (ii) 2Cu (NO3)2 (s) 2 CuO (s) + 4 NO2 (g) + O2 (g) (iii) CH3COOC2H5(aq) + H2O (l) CH3COOH (aq) + C2H5OH (aq)

Equilibrium


220

(iv) Fe3+ (aq) + 3 OH– (aq) Fe (OH)3 (s) (v) I2(s) + 5F2 (g) 2 IF5(l)

Ans. (i) K NO CNOC

c =2[ ( )] [ ( )]

[ l (g)]g l g2

2

(ii) K CuO s NO g O gCu NO s

c =[ ( )] [ ( )] [ ( )]

[ ( ) ( )]

22

42

3 22 = [NO2 (g)]4 [O2 (g)]

(iii) KCH COOH aq C H OH aqCH COOC H aq H l

3 2 5

3c =

[ ( )] [ ( )][ ( )][ ]( )]2 5 2O

=[ ( )] [ ( )]

[ ( )]CH COOH aq C H OH aq

CH COOC H aq3 2 5

3 2 5

(iv) KFe OH s

Fe aq OH aq Fe aq OH aqc = =

+ − + −[ ( ) ( )]]

[ ( )][ ( )] [ ( )] [ ( )]3

3 3 3 31

(v) KIF

I s F gIFF g

c = =[ ( )]

[ ( )][ ( )][ ( )][ ( )]

52

2 25

52

25

l l

5. Find out the value of Kc for each of the following equilibria from the value of Kp

(a) 2NOCl (g) 2NO (g) + Cl2 (g), Kp = 1.8 × 10–2 at 500 K (b) CaCO3 (s) CaO (s) + CO2 (g), Kp = 167 at l073 K. Ans. (a) ∆ng = 3 – 2 = 1, Kp = Kc (RT) or Kc = Kp / RT = (1.8 × 10–2)/(0.0831 ×500) (R = 0.0831 bar litre mol–1

K–1)= 4.33×10–4

(b) ∆ng = 1– 0 = 1, Kc = Kp / RT = 167

0831 10731 87

..

×=

6. For the following equilibrium, Kc = 6.3 × 1014 at 1000 K NO(g) + O3(g) NO2(g) + O2(g) Both the forward and reverse reactions in the equilibrium are elementary

bimolecular reactions. What is Kc for the reverse reaction ?

Ans. For the reverse reaction K'c = 1/Kc = ×

16 3 1014.

= 1.59 × 10–15

7. Explain why pure liquids and solids are ignored while writing the equilibrium constant expression.

Ans. [Pure liquid] or [Pure solid] = =Number of moles

Volumein litreMass molar mass

Volume /

Equilibrium


221

= × =Mass

Volume Molar massDensity

Molar mass1

As density of a pure liquid or pure solid is constant at constant temperature and molecular mass is also constant, therefore, their molar concentrations are constant and included into the equilibrium constant.

8. Reaction between nitrogen and oxygen takes place as follows: 2 N2 (g) + O2 (g) 2 N2O (g) lf a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction

vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0×10–37. Determine the composition of the equilibrium mixture.

Ans. 2N2 (g) + O2 (g) 2N2O (g) Initial 0.482 mol 0.933 mol O mol At eqilibrium 0.482 – x mol 0.933 –x/2 mol x mol

Molar conc. 0 48210

. − x 0 93310

. − x/2 x10

As K = 2.0×10–37 is very small, this means that the amount of N2 and O2 reacted (x) is very very small. Hence, at equilibrium, we have

[N2] = 0.0482 moI L–1, [O2] = 0.0933 mol L–1, [N2O] = 0.1 x

Kc

x=

( . )( . ) ( . )

0 10 0482 0 0933

2

2 = 2.0 × 10–37 (Given) On solving, this gives x = 6.6 ×10 –20

[N2O] = 0.1, x = 6.6 × l0–21 mol L–1

9. Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction given below: 2 NO (g) + Br2 (g) 2 NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

Ans. 0.0581 mol of NOBr is formed from 0. 0581 mol of NO and 0. 0581/2 = 0.0259 mol of Br2.

At equilibrium, Amount of NO = 0. 087– 0. 0518 = 0. 0352 mol Amount of Br2 = 0. 0437 – 0. 0259 = 0. 0178 mol. 10. At 450 K, Kp = 2.0 × 1010 / bar for the given reaction at equilibrium 2SO2 (g) + O2 (g) 2SO3 (g) What is Kc at this temperature ?

Equilibrium


222

Ans. For the given reaction, ∆ng = 2 – 3= – 1 Kp = Kc (RT)∆n or Kc= Kp(RT)–∆n = Kp(RT) = (2. 0×1010 bar –1) (0. 0831 L bar K–1mol–1) (450 K) = 74. 8 × 1010 L mol–1 = 7. 48 × 1011 L mol–1

11. A sample of HI (g) is placed in a flask at a pressure of 0·2 atm. At equilibrium, the partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium ?

Ans. 2HI (g) → H2(g) + I2 (g) Initial pressure 0. 2 atm 0 0 At eqilibrium 0. 04 atm 0.16 /2 atm 0. 16 / 2 atm = 0. 08 atm = 0. 08 atm (Decrease in the pressure of HI = 0. 2 – 0. 04 = 0. 16 atm. )

∴ Kp atm atm

atmH

HIp

lp

p=

×=

×=2 2

2 20 08 0 08

0 044 0. .

( . ).

12. A mixture of 1·57 mol of N2, 1·92 mol of H2 and 8·13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction

N2 (g) + 3H2 (g) 2NH3 (g) is l.7 × l02. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction ?

Ans. The reaction is: N2 (g) + 3 H2 (g) ––––→ 2 NH3 (g)

QNH

N HmolL

molLc = =

−

−[ ]

[ ][ ]( )

( ) (3

2

2 23

1 2

18.13/20

1.57/20 1.92/20mmol−1 3)= 23.79 × 102

As Qc ≠ Kc, the reaction mixture is not in equilibrium. As Qc > Kc, the net reaction will be in the backward direction. 13. The equilibrium constant expression for a gas reaction is Kc,Write the

balanced chemical equation corresponding to this expression.

[ ] [ ][ ] [ ]NH ONO H O

34

25

42

6

Ans. 4 NO (g) + 6 H2O –––→ 4 NH3 (g) + 5 O2 (g) 14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated

at 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation : H2O (g) + CO (g) H2 (g) + CO2 (g)

Calculate the equilibrium constant for the reaction. Ans. At equilibrium, [H2O] = (1 – 0. 40) / 10 mol L–1=0. 06 mol L–1

Equilibrium


223

[CO] = 0. 06 mol L–1, [H2] = 0. 4 mol L–1 = 0. 04 mol L–1. [CO2] = 0. 04 mol L–1

K H CO

H O CO= =

××

=[ ][ ][ ][ ]

. .

. ..2 2

2

0 04 0 040 06 0 06

0 444

15. At 700 K, equilibrium constant for the reaction H2 (g) + I2 (g) 2 HI(g) is 54.8. If 0.5 mo1 L–1 of HI (g) is present at

equilibrium at 700 K, what are the concentrations of H2 (g) and I2 (g) assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700 K.

Ans. 2HI (g) H2(g) + I2 (g), K= 1

54 8. At equilibrium, [HI] = 0. 5 mol L–1, [H2 ] = [I2] = x mol L–1

∴ K = (x)(x)/(0. 5)2 = 1÷54.8 (Given). This gives x = 0. 068, i.e., [H2] = [I2] = 0. 068 mol L–1. 16. What is the equilibrium concentration of each of the substances in the

equilibrium when the initial concentration of ICl was 0·78 M? 2 I Cl (g) → I2 (g) + Cl2 (g), Kc = 0.14 Ans. Suppose at equilibrium, [I2] = [Cl2] = x mol L–1. Then 2ICl I2(g) + Cl2(g) Initial concentration 0.78 M 0 0 At eqilibrium 0.78 – 2x x x

K [I ][Cl ][ICl ]2 2

22c = ∴ 0 14

0 78 2 2. ( )( )( . )

=−

x xx

or, x2 = 0.14 (0.78 –2x)2 or, x/(0. 78 –2x) = 0 14. = 0·374 or, x = 0.292 – 0.748 x or, 1.748x = 0.292 or, x = 0.167 Hence, at equilibrium, [I2] = [Cl2] = 0. 167 M, [I Cl] = 0.78 – (2 × 0.167 M)

= 0.446 M 17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the

equilibrium concentration of C2H6 when it is placed in a flask at 4·0 atm. pressure and allowed to come in equilibrium?

C2H6 (g) C2H4 (g) + H2 (g) Ans. C2H6 (g) ––––→ C2H4 (g) + H2 (g) Initial pressure 4. 0 atm 0 0

Equilibrium


224

At eqilibrium 4 – p p p

K

P ×PP

C H H

C H

2 4 2

2 6p =

∴ 0. 04 = p2 / (4 – p) or p2 = 0.16 – 0. 04 p or, p2 + 0. 04p – 0. 16 = 0

∴ p =− ± − −0 04 0 0016 4 0 16

2. . ( . )

=− ±0 04 0 80

2. .

Taking positive value, p = 0.76 / 2 = 0. 38 ∴ [C2H6]eq = 4 – 0.38 atm = + 3.62 atm 18. Ethyl acetate is formed by the reaction between ethanol and acetic acid. The

equilibrium is represented as CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction

(Note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1·00 mol of acetic acid and 0.18 mol of

ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

Ans. (i) QCH COOC H H OCH COOH C H OHc =[ ][ ][ ][ ]

3 2 5 2

3 2 5

(ii) CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial 1. 00 mol 0. 180 mol At eqilibrium 1 – 0.171 0. 180 – 0.171 0. 171 mol 0. 171 mol = 0. 829 mol = 0. 009 mol Molar 0. 829/V 0. 009/V 0. 171/V 0. 171/V concentration

K

[CH COOC H ][H O][CH COOH][C H OH]

(0.171/V)(0.171/V)(0

3 2 5 2

3 2 5c = =

..829/V)(0.009/V)= 3 92.

(iii) CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial 1. 000 mol 0. 500 mol After time ‘t’ 1 – 0.214 0.500 – 0.214 0.214 mol 0.214 mol = 0.786 mol = 0.286 mol

Equilibrium


225

Reaction quotient ( ) .Q (0.214/V)(0.214/V)(0.786/V)(0.286/V)c = = 0 204

As Qc ≠ Κc, equilibrium has not been attained. 19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K.

After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium ?

Ans. At eqilibrium PCl PCl g Cl gmol L mol L5

0 5 103 2

1 1 1.( ) ( )

× − − −+� ⇀�↽ ��

x x mmol L−1

K Givenc

x=

×= ×

−−

2

13

0 5 108 3 10

( . ). ( )

or, x2 = (8.3 × 10–3) (0.5 × 10–1) = 4.15 × 10–4

or, x = 4 15 10 4. × − = 2.04 × 10–2 M = 0.02 M 20. One of the reactions that takes place in producing steel from iron ore is the

reduction of iron (II) oxide by carbon monoxide to given iron metal and CO2.

FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 at 1050 K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the

initial pressures are : pCO = 1.4 atm and pCO2 = 0.80 atm? Ans. FeO (s) + CO (g) ––––→ Fe (s) + CO2 (g) Initial pressures 1. 4 atm 0. 80 atm.

Q p

co

co= = =pp

2 0 801 4

0 571..

.

As Qp > Kp, reaction will move in the backward direction, i.e., pressure of CO2 will decrease and that of CO will increase to attain equilibrium. Hence, if pressure of CO2 decrease, increase in pressure of CO = p

At equilibrium, pCO2 = (0.80 – p) atm , pCO = (1.4 + p) atm

K CO

COp

pp

= ∴ =−+

2 0 265 0 801 4

. ..

pp

or, 0.265 (1.4 + p) = 0.80 – p or 0.371 + 0.265 P = 0.80 – p

or, 1.265 p = 0.429 or p = 0.339 at eqm. ∴ pCO

= 1.4 + 0.339 = 1.739 atm, pCO2 = 0.80 – 0.339 = 0.461 atm

Equilibrium


226

21. Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061. At a particular time, the analysis shows that composition of the reaction

mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Ans. Q[NH ]

[N ][H ]3

2

2 23c = = =

( . )( . )( . )

.0 53 0 2 0

0 01042

3

As Qc ≠ Kc reaction is not in equilibrium. As Qc < Kc reaction will proceed in the forward direction. 22. Bromine monochloride (BrCl) decomposes into bromine and chlorine and

attains the equilibrium 2BrCl (g) Br2 (g) + Cl2 (g) for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.30 × 10–3 mol L–1. What is its molar concentration in the mixture at equilibrium?

Ans. 2BrCl ––––→ Br2 + Cl2

Initial 3.30 × 10–3 mol L–1 0 0 At eqm. (3.30 × 10–3 – x) x /2 x / 2

Kcx x

x=

× −−( / )( / )

( . )2 2

3 30 10 3 2 = 32 (Given)

∴ xx

/( . )

23 30 10

323× −=

−

or, x/2 = 5.66 × (3.30 × 10–3 – x) x = 11.32(3.30 × 10–3 – x) or, 12.32x = 11.32(3.30 × 10–3) or x = 3.0 × 10–3

at eqm. [BrCl] = ( 3.30 × 10–3 – 3.0 × 10–3) = 0.30 × 10–3 = 3.0 × 10–4 mol L–1

23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass in the reaction

C (s) + CO2 (g) 2 CO (g). Calculate Kc for this reaction at the above temperature. Ans. If total mass of mixture of CO and CO2 is 100 g, then CO = 90.55 g and CO2

= 100 – 90.55 = 9.45 g. Number of moles of CO = 90·55/28 = 3.234 Number of moles of CO2 = 9. 45/44 = 0. 215 pCO = 3. 234/(3. 234 + 0. 215) × 1atm = 0.938 atm

Equilibrium


227

pCO2 = 0. 215/(3. 234 + 0. 215) × 1 atm = 0. 062 atm

K CO

COp = =

pp

2 2

2

0 9380 062

( . ). = 14.19

∆ng = 2 –1 = 1

∴ Kp = Kc (RT) or KRTp =

×14 19

0 0821 1127.

.= 0.153.

24. Calculate (a) ∆G° and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

NO (g) + ½ O2 (g) NO2 (g) where ∆fG° (NO2) = 52 .0 kJ/mol, ∆fG° (NO) = 87.0 kJ/mol, ∆f.G°(O2)

= 0 kJ/mol. Ans. (a) ∆rG° = Σf G° (Products) — Σf G° (Reactants) = ∆fG° (NO2) – [{∆f G°(NO) + ½ ∆fG°(O2)}] = 52.0 – (87.0 + ½ × 0) = –35. 0 kJ mol–1 (b) –∆G° = 2. 303 RT log K. Hence, – (– 35000) = 2.303 × 8. 314 × 298 × log K or log K = 6. 1341 or K = 1. 361×106 25. Does the number of moles of reaction products increase, decrease or remain

same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?

(a) PCl5 (g) PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) CaCO3 (s) (c) 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g) Ans. Applying Le Chatelier’s principle, on decreasing the pressure, moles of

reaction products will (a) increase (b) decrease (c) remain same ( ∆ng = 4 – 4 = 0) 26. Which of the following reactions will get affected by increasing the pressure?

Also mention whether change will cause the reaction to go into forward or backward direction?

(i) COCl2 (g) CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) 2 CO (g) (iv) 2 H2 (g) + CO (g) CH3OH (g)

Equilibrium


228

(v) CaCO3 (s) CaO (s) + CO2 (g) (vi) 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g) Ans. Reactions affected will be those in which(np ≠ nr) (gaseous). Hence, reactions

(i), (iii), (iv), (v) and (vi) will be affected. By applying Le Chatelier’s principle, we can predict the direction.

(i) np = 2, nr = 1, i.e., np > nr reaction will go in the backward direction. (ii) np = 3, nr = 3, i.e., np = nr reaction will not be affected by pressure. (iii) np = 2, nr = 1, i.e., np > nr reaction will go in the backward direction. (iv) np = 1, nr = 3, i.e., np < nr reaction will go in the forward direction. (v) np = 1, nr = 0, i.e., np > nr reaction will go in the backward direction. (vi) np = 10, nr = 9, i.e., np > nr reaction will go in the backward direction. 27. The equilibrium constant for the reaction H2 (g) + Br2 (g) 2 HBr (g) at

1024 K is 1·6 × 105. Find the equilibrium pressure of all gases if 10·0 bar of HBr is introduced into a sealed container at 1024 K.

Ans. 2HBr (g) ––––→ H2 (g) + Br2(g), K = 1/(1.6×105) Initial 10 bar 0 0 At eqm. 10 – p p/2 p/2

K /2 /2

pp p

p=

−=

×

( )( )( ) .10

11 6 102 5

or, pp

2

2 54 101

1 6 10( ) .−=

×

Taking square root of both sides, we get p

p2 101

4 0 102( ) ( . )−=

×

or, 4×102 p = 2(10 – p) or 402p = 20 or p = 20/402 = 4. 98×10–2 bar Hence, at equilibrium, pH2 = pBr2 = p/2 = 2. 5 ×10–2 bar, pHBr =10–p ≈10 bar 28. Hydrogen gas is obtained from natural gas by partial oxidation with steam

according to the following endothermic reaction: CH4 (g) + H2O (g) CO (g) + 3H2 (g) (a) Write an expression for Kp for the above reaction. (b) How will the value of Kp and composition of the equilibrium mixture

be affected by (i) increasing the pressure? (ii) increasing the temperature? (iii) using a catalyst?

Ans. (a) K CO H

CH H O

2

2

pp pp p

=×

×

( )3

4

Equilibrium


229

(b) (i) By Le Chatelier’s principle, equilibrium will shift in the backward direction,

(ii) By Le Chatelier’s principle, equilibrium will shift in the forward direction,

(iii) Equilibrium composition will not be disturbed but equilibrium will be attained quickly.

29. Describe the effect of: (a) addition of H2 (b) addition of CH3OH (c) removal of CO (d) removal of CH3OH, on the equilibrium of the reaction: 2H2 (g) + CO (g) CH3OH (g) Ans. (a) Equilibrium will shift in the forward direction (b) Equilibrium will shift in the backward direction (c) Equilibrium will shift in the backward direction (d) Equilibrium will shift in the forward direction 30. At 473 K, equilibrium constant, Kc, for the decomposition of phosphorus

pentachloride, PCl5 is 8.3 × 10–3. If decomposition is depicted as : PCl5 (g) PCl3 (g) + Cl2 (g), ∆rH° = 124.0 KJ mol – 1

(a) Write an expression for Kc for the reaction ? (b) What is the value of Kc for the reverse reaction at the same

temperature ? (c) What would be the effect on Kc if (i) more PCl5 is added (ii) pressure

is increased (iii) temperature is increased?

Ans. (a) K[PC C

PCc =l g l g

l g3 2

5

( )[ ( )][ ( )]

(b) K K8. 3×10 3' / .= = =

−1 1 120 48c

(c) (i) No effect as Kc is constant at constant temperature. (ii) Moves backward (iii) As given reaction is endothermic, on increasing the temperature

kf will increase as Kc = kf /kb, hence Kc will increase with increase of temperature.

31. Dihydrogen gas used in Haber’s process is obtained by reacting methane from natural gas with high temperature steam. The first stage of two stage

Equilibrium


230

reaction involves the formation of CO and H2, In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H2O (g) CO2 (g) + H2 (g). If a reaction vessel at 400°C is charged with an equi-molar mixture of CO and steam such that PCO = PH2 = 4.0 bar, what will be the partial pressure of H2 at equilibrium ? K = 0.1 at 400°C.

Ans. Suppose the partial pressure of H2 at equilibrium = p bar CO (g) + H2O (g) ––––→ CO2 (g) + H2 (g) Initial pressure 4.0 bar 4.0 bar 0 0 At eqm. (4 – p) (4 – p) p p Kp = p2

/(4 – p)2 = 0. 1 (Given) ∴ p/(4 – p) = 0 1. = 0. 316 p = l. 264 – 0. 316p or l. 316 p = 1. 264 or, p = 0. 96 bar 32. Predict which of the following reaction will have appreciable concentration

of reactants and products: (a) Cl2 (g) 2 Cl (g), Kc = 5 × 10–39 (b) Cl2 (g) + 2 NO (g) 2 NO2Cl (g), Kc = 3.7×108 (c) Cl2 (g) + 2 NO2 (g) 2 NO2Cl (g), Kc = 1.8 Ans. For reaction (c), as Kc is neither high nor very low, reactants and products

will be present in comparable amounts. 33. The value of Kc for the reaction 3 O2 (g) 2 O3(g) is 2.0 × 10–50 at 25°C. If

the equilibrium concentration of O2 in air at 25°C is 1.6 × 10–2, what is the concentration of O3?

Ans. KOO

c =[ ][ ]

32

23

∴ 2.0 × 10–50 =× −

[ ]( . )

O32

2 31 6 10 or, [O3 ]

2 = (2.0 ×10 –50) (1.6 × 10 –2)3 = 8.192 ×10–56 or, [O3] = 2.86 ×10 –28 M 34. The reaction, CO (g) + 3 H2 (g) CH4 (g) + H2O (g), is at equilibrium at

1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc, for the reaction at the given temperature is 3·90.

Equilibrium


231

Ans. K [CH ][H O][CO][H ]

4 2

23c =

∴ 3.90 = [CH4 ][ . ]( . ) ( . )

0 020 30 0 10 3

(Molar concentration = Number of moles because volume of flask = 1L) or, [CH4] = 0.0585 M = 5.85 × 10–2 M 35. What is meant by the conjugate acid–base pair? Find the conjugate acid/

base for the following species: HNO2, CN–, HCIO4, F–, OH–, CO3

2– and S2–

Ans. An acid–base pair which differ by a proton is called conjugate acid base pair. NO2

–, HCN, ClO4–, HF, H2O (acid) or O2– (base), HCO3

– and HS–. 36. Which of the following are Lewis acids ? H2O, BF3, H

+, NH4+.

Ans. BF3, H+ and NH4

+ (Remember that all cations are Lewis acids). 37. What will be the conjugate bases for the Bronsted acids: HF, H2SO4 and

HCO3– ?.

Ans. Conjugate acid = Conjugate base + H +

or Conjugate base = Conjugate acid – H+. ∴ Conjugate bases of the given acids will be F–, HSO4

–, CO3–2.

38. Write the conjugate acids for the following Bronsted bases: NH2

–, NH3 and HCOO–

Ans. NH3, NH4+, HCOOH

39. The species: H2O, HCO–3, HSO–

4 and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Ans. Conjugate acids: H3O+, H2CO3, H2SO4, NH4

+

Conjugate bases: OH–, CO32–, SO4

–2, NH2–

40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

(a) OH– (b) F– (c) H+ (d) BCl3

Ans. (a) OH– can donate electron pair. Hence, it is a Lewis base (b) F– can also donate electron pair. Hence, it is a Lewis base (c) H+ can accept electron pair. Hence, it is a Lewis acid (d) BCl3 is deficient in electrons. Hence, it can accept electron pair and

is, therefore, a Lewis acid.

Equilibrium


232

41. The concentration of hydrogen ion in a sample of soft drink is 3. 8 × 10–3

M. What is its pH? Ans. pH = –log[H+] = – log(3.8 × 10 –3) = –log 3.8 + 3 = 3 – 0. 5798 =2.4202 = 2. 42 42. ThepH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen

ion in it. Ans. pH = – log [H+] or log [H+] = – pH = – 3. 76 = 4.24 ∴ [H+] = Antilog 4.24 = 1. 738 ×10– 4M = 1·74 × 10–4 M 43. The ionisation constants of HF, HCOOH and HCN at 298 K are is 6.8 × 10–4,

1.8 × 10–4 and 4. 8 × 10–9 respectively. Calculate the ionisation constant of the corresponding conjugate base.

Ans. For F–, Kb = Kw / Ka= 10–14/(6.8 × 10–4) =1. 47 × 10–11 ≈ 1. 5 × 10–11. For HCOO–, Kb = 10–14/(1.8 × 10–4) = 5. 6× 10–11

For CN–, Kb = 10–14/(4.8×10–9) = 2. 08 × 10–6

44. The ionisation constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionisation if the solution is also 0. 01 M in sodium phenolate?

Ans. C6H5OH ––––→ C6H5O– + H+

Initial 0. 05 M 0 0 After dissociation 0. 05 – x x x ∴ Ka = (x)(x)/(0. 05 – x) = 1·0 × 10–10 (Given) or, x2/0. 05 = 1. 0 × 10–10

or, x2 = 5 × 10–12 or x = 2.2 ×10– 6 M In presence of 0. 01 C6H5ONa, suppose y is the amount of phenol dissociated,

then at equilibrium [C6H5OH] = 0. 05 – y ≈ 0. 05, y << 0.05 [C6H5O

– ] = 0. 01 + y ≈ 0. 01 M, [H+] = y M Ka = (0. 01)(y)/0. 05 = 1.0 ×10–10 (Given)

or, y = 5 × 10–10 ⇒ = =×

=−

−αyc

5 100 05

1010

8

. 45. The first ionisation constant of H2S is 9.1×10–8. Calculate the concentration

of HS– ions 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 H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

Ans. To calculate [HS–]

Equilibrium


233

H2S H+ + HS–

Initial 0. 1M 0 0 After dissociation 0. 1– x x x ≈ 0.1 Ka= (x)(x)/0. 1 = 9.1 × 10–8 or x2 = 9. l × 10–9 or x = 9.54 × 10–5. In presence of 0.1 M HCl, suppose H2S dissociated is y. Then at equilibrium

[H2S] = 0. 1– y ≈ 0.1, [H+] = 0.1+ y ≈ 0.1, [HS–] = y M Ka = (0. 1× y)/0. 1 = 9. 1×10–8 (Given) or y = 9.1 × 10–8 M To calculate [S2–]

H S H HS HS H S2K Ka a1 21 1 2� ⇀��↽ �� ⇀��↽ ��+ − − + −+ +;

For the overall reaction, H2S 2H+ + S2–

Ka = K Kla a×

2 = 9.1×10–8 × 1.2 × 10 –13

= 1. 092 × 10 –20

K H S

H Sa =+ −[ ][ ][ ]

2

2 In the absence of 0.1 M HCl,

H S H S222

0 1 0 00 1 2

� ⇀�↽ �� + −+

−.

. x x x [H+] = 2 [S2–] Hence, if [S2–] = x, [H+] = 2x (2x)2 x / 0.1 = 1.092 × 10–20

or, 4x3 = 1.092 × 10–21 or, x3 = 273 × 10–24

3 log x = log 273 – 24 = 2.4362 – 24 log x = 0.8127– 8 = 8.8127 or x = Antilog – (7.1879) = 6.487 × 10–8 = 6.5 × 10–8 M. In presence of 0·1 M HCl,

Equilibrium


234

H S H S222

0 1 0 00 1 2

� ⇀�↽ �� + −+

−.

. y y y suppose [S2–] = y, then [H2S] = 0.1– y ≈ 0.1 M, y << x [H+] = 0.1 + 2y ≈ 0.1 M Ka = (0.1)2 × y/0.1 = l.09 × 10–20 or y = 1.09 × 10–19 M. 46. The ionisation 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.

Ans. CH3COOH CH3COO– + H+

K

[CH COO ][H ][CH COOH]

[H ][CH COOH]

3+

3

+ 2

3a = =

−

or, [H+] = K CH COOHa[ ] ( . )( )35 21 74 10 5 10= × ×− − = 9.33 × 10–4 M

[CH3COO–] = [H+] = 9.33×10–4 M pH = – log (9.33 × 10–4) = 4 – 0.9699 = 4 – 0.97 = 3.03 47. 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 ionisation constant of the acid and its pKa.

Ans. HA H+ + A–1

pH= –log[H+] or log[H+] = – pH = – 4.15 Taking antilog [H+] = 7.08×10–5 M [A–1] = [H+] = 7.08 × 10–5 M

K H A

HAa = =× ×

= ×+ − − −

−−[ ][ ]

[ ]( . )( . )1 5 5

277 08 10 7 08 10

105 10

pKa = –log Ka = –log (5.0×10–7) = 7– 0.699 = 6.301 48. Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH Ans. (a) HCl + aq ––––→ H+ + Cl–1,

Equilibrium


235

∴ [H+] = [HCl] = 3 ×10–3 M, pH = –log (3×10–3) = 2.52 (b) NaOH + (aq) ––––→Na+ + OH–

∴ [OH–1] = 5 × 10 –3 M, [H+] = 10–14/(5 × 10–3) = 2 × 10–12 M pH = –log (2× 10–12) = 11.70 (c) HBr + aq ––––→ H+ + Br–1, ∴ [H+] = 2 × 10–3 M, pH = –log (2×10–3) = 2.70 (d) KOH + aq ––––→ K+ + OH–1, ∴ [OH–1] = 2 × 10–3 M, [H+] = 10–14/(2× 10–3) = 5 × 10–12

pH = –log (5 × 10–12) = 11.30 49. Calculate the pH of the following solutions: (a) 2 g of TlOH dissolved in water to give 2 litre of the solution (b) 0. 3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution (c) 0. 3 g of NaOH dissolved in water to give 200 mL of the solution (d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of the

solution. Ans. (a) Molar conc. of TlOH = 2 g/(204 + 16 + 1) g mol–1 × 1/2 L = 4. 52

× 10–3 M [OH–] = [TlOH] = 4. 52 × 10–3 M [H+] = 10–14/(4·52 × 10–3) = 2. 21×10–12 M pH = – log(2. 21 × 10–12) = 12 – (0. 3424) = 11.66 (b) Molar conc. of Ca(OH)2 = 0.3 / [(40 + 34) g mol–1] × 1 / 0. 5 L =

8.11×10–3 M Ca(OH)2 ––––→ Ca2+ + 2 OH–

[OH–] = 2[Ca(OH)2] = 2 × (8. 11 × 10–3) M = 16.22 × 10–3 M pOH = – log (16.22 × 10–3) = 3 – 1.2101 = 1. 79 pH = 14 – 1.79 = 12. 21 (c) Molar conc. of NaOH = 0.3 g/40 g mol–1 × (1/0.2) L = 3.75 × 10–2

M pOH = – log(3.75 × 10–2) = 2 – 0.0574 = 1.43 pH = 14 –1. 43 = 12. 57 (d) M1V1 = M2V2 ∴ 13.6 M × 1 mL = M2 × 1000 mL ∴ M2 = 1. 36 ×10–2 M [H+] = [HCl] = 1. 36 ×10–2 M,

Equilibrium


236

pH = –log(1.36×10–2) = 2 – 0.1335 = 1. 87 50. The degree of ionisation of a 0.1 M bromoacetic acid solution is 0.132.

Calculate the pH of the solution and the pKa bromoacetic acid. Ans. CH2(Br)COOH CH2(Br)COO–1 + H+

Initial concentration C 0 0 Concentration at C – Cα C α C α equilibrium

K C CC

Ca =

×−

=−

α αα

αα( )1 1

2= 0. 1×(0·132)2 = 1.74 × 10–3

pKa = – log (1. 74 ×10–3) = 3 – 0.2405 = 2.76 [H+] = Cα = 0.1× 0.132 = 1. 32 ×10–2 M pH = – log(1.32×10–2) = 2 – 0.1206 = 1. 88 Ka = Cα2

= (0.1) (0.0132)2 = 0.0017 pKa = – log [ka] = – log 0.0017 = 2.75. 51. The pH of 0.005 M codeine (C18H21NO3) solution is 9.95. Calculate the

ionisation constant and pKb. Ans. Cod + H2O Cod H+ + OH–

pH = 9. 95 ∴ pOH = 14 – 9. 95 = 4. 05, i. e. – log [OH–] = 4.05

or log[OH–] = – 4.05 = 5 95. or [OH–] = 8.913 ×10–5

K [CodH ][OH ][Cod]

[OHCod

+

b = = =×

×

− − −

−]

[ ]( . )2 5 2

38 91 10

5 10 = 1.588 × 10–6

pKb = – log (1.588 ×10–6) = 5.79 52. What is the pH of 0.001 M aniline solution ? The ionisation constant of aniline is

4.27×10–10. Calculate degree of ionisation of aniline in the solution. Also calculate the ionisation constant of the conjugate acid of aniline.

Ans. (i) C6H5NH2 + H2O C6H5NH3+ + OH–

K

[C H NH OHC H NH

[OH][C H NH ]

6 5

6 5

2

6 5 2a = =

+ −3

2

][ ][ ]

∴ at equilibrium [ ] [ ]C H NH OH6 5 3+ −=

Equilibrium


237

[OH–] = K [C H NH ]6 5 2a

= × − −( . )( )4 27 10 1010 3

= 6.534 × 10–7 M pOH = – log (6.534×10–7) = 7 – 0.8152 = 6.18 pH = 14 – 6.18 = 7.82 (ii) Also C6H5NH2 + H2O C6H5NH3

+ + OH–

Initial C At. equilibrium C – C α C α C α Kb= (C α)(Cα ) / C(1–α ) = (C α2 ) / (1 – α) ∴ α2 = Kb/C = 4. 27 × 10–10 / 10–3 α << 1 α = 6.53 × 10–4 (iii) pKb + pKa = 14 (for a pair of conjugate acid and base) pKb = – log (4. 27 × 10–10) = 10 – 0.62 = 9. 38 pKa = 14 – 9. 38 = 4. 62 – log Ka = 4.62 or log Ka = – 4.62 or Ka = Antilog (– 4.62) = 2.399 × 10–5 ≈ 2·4×10–5 53. Calculate the degree of ionisation of 0.05 M acetic acid if its pKa value

is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0. 01 M (b) 0. 1 M in HCl.

Ans. pKa = 4. 74, i.e. – log Ka = 4. 74 or log Ka = – 4.74 Taking Antilog Ka = Antilog ( – 4.74) ∴ Ka = 1. 82 × 10–5

α = K Ca / = ( . ) / ( )1 82 10 5 105 2× ×− −

= 1. 908×10 –2

In presence of HCl, due to high concentration of H+ ion, dissociation equilibrium will shift backward, i.e., dissociation of acetic acid will decrease.

(a) In presence of 0. 01 M HCl, if x is the amount dissociated, then CH3COOH CH3COO– + H+

Initial 0.05 M 0 0 After dissociation 0.05 – x x 0.01 + x » 0.05 » 0.01 M (0. 01 M H+ ions are obtained from 0. 01 M HCl)

Equilibrium


238

∴ Ka = x (0.01) / 0.05 or x/0.05 = Ka/0.01 ⇒ Ka = 1.8 × 10–5

∴ x = × ×= ×

−−1 8 10 0 05

0 019 0 10

55. .

..

∴ α = (1. 82 × 10–5) / 10–2 = 1.82 × 10–3 since Amount dissociatedamount taken

α =

(b) In the presence of 0.1 M HCl, if y is the amount of acetic acid dissociated, then at equilibrium

[CH3COOH] = 0.05 – y ≈ 0.05 M [CH3COO–] = y, [H+] = 0.1M + y = 0.1M Ka =(0.1) / 0.05 or y/0. 05 = Ka/0.1 ∴ α = (1.82 × 10–5 )/10–1 = 1. 82 × 10–4, i.e. α = 1. 82 ×10–4

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

Ans. α = = × × =− −K Cb / ( . ) / ( ) .5 4 10 2 10 0 1644 2

In presence of 0. 1 M NaOH, if x is the amount of dimethyl amine dissociated, (CH3)2NH + H2O (CH3)2NH2

+ + OH–

Initial 0. 02 M After dissociation 0.02 – x x 0.1+ x ≈ 0.02 ≈ 0.1

Kb

x xx

x=

+−

=×( . )

..

.0 1

0 020 1

0 02

⇒ x Kb=×

=× × −0 02

0 10 02 5 4 10

0 1

4..

. ..

⇒ x = 1.08 × 10–4

α = =

×= ×

−−x

0 021 08 10

0 020 54 10

42

..

..

α% = 0.54% 55. Calculate the hydrogen ion concentration in the following biological fluids

whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4 Ans. (a) log [H+] = pH = – 6.83 ∴ [H+] = Antilog 7 17. = l.479×10–7 M

Equilibrium


239

(b) log [H+] = –pH = –1. 2

∴ [H+] = Antilog 2 8. = 6. 31×10–2 M (c) log [H+] = –pH = –7.38 ∴ [H+] = Antilog 8 62. = 4.169 ×10–8 M (d) log [H+] = –pH = – 6.4 ∴ [H+] = Antilog 7 60. = 3.981× 10 –7 M 56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are

6.8, 5·0, 4.2, 2.2 and 7·8, respectively. Calculate corresponding hydrogen ion concentration in each.

Ans. (i) Milk → – log[H+] = pH = – 6.8 ⇒

= −

+log H 6 87 2. .

∴ [H+] = Antilog ( 7 2. ) = 1.5 × 10–7 M (ii) Black coffee → – log[H+] = pH ⇒ log [H–] = – 5 ∴ [H+]= 10–5 M (iii) Tomato juice → – log[H+] = pH = 4.2 ⇒ = − =+log[ ] . .H 4 2 5 8 ∴ [H+] = Antilog (– 4.2) = Antilog 5 8 6 31 10 5. .= × − M (iv) Lemon juice → –log[H+] = pH = 2.2 ⇒ log [H+] = – 2.2 ∴ [H+] = Antilog (–2.2) = Antilog ( . ) .3 8 6 31 10 3= × − M (v) Egg white → – log [H+] = pH = 7.8 ⇒ log[H+]= – 7.8 ∴ [H+] = Antilog (–7.8) = Antilog ( . ) .8 2 1 5 10 8= × − M 57. If 0. 561 g KOH is dissolved in water to give 200 mL of solution at 298 K,

calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH ?

Ans. [KOH] = (0.561/56) × 1000/200 = 0.05 M As KOH –––→ K+ + OH–

[K+] = [OH–] = 0. 05 M [H+] = Kw/[OH–] = 10–14 / 0. 05 = 10–14/(5 × 10–2) = 2. 0 × 10–13 M. pH = (–log 2.0 × 10–13) = 13 – log 2 = 12.699 58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the

concentrations of strontium and hydroxyl ions and the pH of the solution. (Atomic mass of Sr = 87. 6)

Ans. Molar mass of Sr(OH)2 = 87.6 + 34 =121.6 g mol– 1

Solubility of Sr(OH)2 in moles L–1= 19.23 gL–1/121.6 g mol–1 = 0.1581 M

Equilibrium


240

Assuming complete dissociation, Sr(OH)2 –––→ Sr2+ + 2 OH–

∴ [Sr2+] = 0.1581 M, [OH–] = 2×0.1581= 0.3162 M, pOH = –log 0.3162 = 0.5 ∴ pH = 14 – 0. 5 = 13.5 59. The ionisation constant of propanoic acid is 1.32×10–5. Calculate the degree

of ionisation of the acid in its 0.05 M solution and its pH . What will be its degree of ionisation if the solution is 0.01M in HCl?

Ans. Assuming α to be very small, applying formula directly, we have α = = × −Ka c/ ( . ) / .1 32 10 0 055 = 1.62 ×10–2

CH3CH2COOH � ⇀�↽ �� CH3CH2COO– + H+

In presence of HCl, equilibrium will shift in the backward direction, i.e., concentration of CH3CH2COOH will increase dissociation will be less. If ‘c’ is the initial concentration and ‘x’ is the amount now dissociated, then at equilibrium

[CH3CH2COOH] = c – x [ CH3CH2COO–] = x , [H+] = 0.01 + x Ka = x (0.01 + x)/(c – x) ≈ x(0.01)/c Or α = x / c = Ka/0.01 = 1.32× 10–5/10–2

= 1.32 ×10–3. 60. The pH of 0. 1 M solution of cyanic acid (HCNO) is 2.34. Calculate the

ionisation constant of the acid and its degree of ionisation in the solution. Ans. HCNO � ⇀�↽ �� H+ + CNO–

pH = 2.34 means –log [H+] = 2.34 or log [H+] = –2.34 = 3 66. or [H+] = Antilog 3 66. = 4.57 × 10–3M [CNO–] = [H+] = 4.57×10–3 M

Ka =× −( . ).

4 57 100 1

3 2= 2.09 × 10–4

α = Ka c/ ( . / . )= × −2 09 10 0 14 = 0.0457 61. The ionisation constant of nitrous acid is 4.5× 10–4. Calculate the pH of 0.04

M sodium nitrite solution and also its degree of hydrolysis. Ans. Sodium nitrite is a salt of weak acid, strong base. Hence,

Equilibrium


241

Kh = Kw/Ka = 10–14/(4·5 × 10–4) = 2.22 × 10–11

h = Kh c/ ( . / . ) .= × = ×− −2 22 10 0 04 5 5 1011 10 = 2. 36 × 10–5

NO2– + H2O HNO2 + OH–

Initial c After hydrolysis c–ch ch ch [OH–] = ch = 0. 04 × 2.36 ×10–5 = 9. 44 ×10–7

pOH = –log(9. 44 ×10–7) = 7 – 0.9750 = 6. 03 pH = 14 – pOH = 14 – 6.03 = 7.97. 62. A 0.02 M solution of pyridinium hydrochloride has pH =3·44. Calculate the

ionisation constant of pyridine. Ans. pH = 3.44, i. e., log [H+] = – 3.44 = 4 56. Taking Antilog ∴ [H+] = 3.63 × 10–4 M

C H N.HC aq C H N C H5 5 5 5+

l l+ +− +� ⇀�↽ ��

K[C N N Cl ][H ]

[C H N HCl]5 5

+ +

5 5a = =

× ×

×

− − −

−.( . )( . )3 63 10 3 63 10

2 10

4 4

2 = 6.588×10–6

pKa = –log(6.588 × 10–6) = 6 – 0. 8187 = 5. 18 pKb+ pKa= 14 ∴ pKb= 14 – 5.18 = 8.82 –log Kb = 8.82 or log Kb = – 8.82 ∴ Kb, = l. 514 × 10–9 63. Predict the acidic, basic or neutral nature of the solutions of the following

salts: NaCl, KBr, NaCN, NH4NO3, NaNO2, KF Ans. NaCN, NaNO2, KF solutions are basic, as they are salts of strong base and

weak acid. NaCl and KBr solutions are neutral, as they are salts of strong acid, strong base. NH4NO3 solution is acidic, as it is a salt of strong acid, weak base.

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

Ans. ClCH2COOH ClCH2COO– + H+

Equilibrium


242

K [ClCH COO ][H ]

[ClCH COOH]H2

+

2a c= =

− +[ ]2

[H+] = × = × ×−Ka c 1 35 10 0 13. . = 1.16× 10–2 M pH = – log(1.16 × 10–2) = 2 – 0.06 = 1.94 Sodium salt of chloroacetic acid is a salt of strong base and weak acid. Hence, pH = –1/2 [log Kw + log Ka – log c] ∴ pH = – ½ [log 10–14 + log 1.35×10–3 –log 0.1] = – ½ [–14 + (– 3 + 0.1303) – (–1)] = 7 + 1.44 – 0.5 = 7.94 65. The ionic product of water at 310 K is 2.7×10 –14. What is the pH of neutral

water at this temperature? Ans. As [H+] [OH–] = Kw

For neutral water [H+] = [OH–]

∴ [H+]2 = 2.7 × 10–14 or [ ] .H+ −= ×2 142 7 10 ∴ [H+] = 1.643 × 10–7

pH = – log [H+] = – log (1.643 × 10–7) = 7– 0.2156 = 6.78 66. Calculate the pH of the resultant mixtures (a) 10 mL of 0.2 M Ca(OH)2 + 25 mL of 0.1 M HCl (b) 10 mL of 0. 01 M H2SO4 + 10 mL of 0. 01 M Ca(OH)2

(c) 10 mL of 0.1 M H2SO4 + 10 mL of 0.1 M KOH Ans. (a) 10 mL of 0.2 M Ca(OH)2 = 10 × 0.2 millimoles = 2 millimoles of Ca(OH)2

25 mL of 0.1 M HCl = 25 × 0.1 millimoles = 2.5 millimoles of HCl Ca(OH)2 + 2 HCl ––––→ CaCl2 + 2 H2 O 1 millimole of Ca(OH)2 reacts with 2 millimoles of HCl ∴ 2.5 millimoles of HCl will react with 1.25 millimoles of Ca(OH)2 ∴ Ca(OH)2 left = 2 – 1.25 = 0.75 millimoles (HCl is the limiting

reactant) Total volume of the solution = 10 + 25 mL = 35 mL ∴ Molarity of Ca(OH)2 in the mixture solution = 0. 75/35 M = 0. 0214 M [OH–] = 2 × 0. 0214 M = 0. 0428 M = 4.28 × 10–2 pOH = – log (4.28 × 10–2) = 2 – 0.6314 = 1.3686 ≈ 1.37

Equilibrium


243

pH = 14 – 1.37 = 12.63 (b) 10 mL of 0.01 M H2SO4 = 10 × 0.01 millimole = 0. 1 millimole 10 mL of 0. 01 M Ca(OH)2 = 10 × 0. 01 millimole = 0. 1 millimole Ca(OH)2 + H2SO4 CaSO4 + 2 H2O 1 mole of Ca(OH)2 reacts with 1 mole of H2SO4 ∴ 0.1 millimole of Ca(OH)2 will react completely with 0.1 millimole

of H2SO4. Hence, solution will be neutral with pH = 7. 0 (c) 10 mL of 0.1 M H2SO4 = 1 millimole 10 mL of 0. 1 M KOH = 1 millimole 2 KOH + H2SO4 ––––→ K2SO4 + 2 H2O 1 millimole of KOH will react with 0·5 millimole of H2SO4 ∴ H2SO4 left = 1– 0.5 = 0.5 millimole Volume of reaction mixture = 10 + 10 = 20 mL

∴ Molarity of H2SO4 in the mixture solution =0 520

2 5 10 2. .= × − M

[H+] = 2 × 2.5 × 10–2 = 5 × 10–2 pH = – log (5 × 10–2) = 2 – 0.699 = 1.3 67. Determine the solubilities of silver chromate, barium chromate, ferric

hydroxide, lead chloride and mercurous iodide at 298 K. Given Ksp values: Ag2CrO4 = 1.1×10–12, BaCrO4 = 1.2×10–10, Fe(OH)3= 1.0×10–38, PbCl2 = 1.6×10–5, Hg2I2 = 4.5 ×10–29 Determine also the molarities of individual ions. Ans. Ag2CrO4 2Ag+ + CrO4

2–, Ksp = (2s)2(s) = 4s3

S 2S S ∴ S3 = Ksp/4 = (1.1 × 10 –12) /4 = 0.275 × 10–12 = 2.75 × 10–13

3 log S = log(2.75 × 10–13) = –13 + 0.4393 = –12. 5607 or log S = – 4.1869 = 5 8131. Taking Antilog S = 6.5 × 10–5mol L–1

[Ag+] = 2 × 6.5× 10–5 = 13.0 × 10–5 M = 1.30 ×10–4M [CrO4

2–] = 6. 5 × 10–5 M For BaCrO4

BaCrO4 (s) Ba2+ + CrO4–2

Equilibrium


244

S S S

Ksp = s2 = 1.2 × 10–10, ∴ = × −s 1 2 10 10. = 1.1 × 10–5 M = [Ba2+] = [CrO4

–2] For Fe(OH)3, Fe(OH)3 Fe3+ + 3OH–1

S S 3S Ksp = S × (3S)3 = 27 S4

∴ S = = × −( / ) ( / )Ksp 27 1 10 274 384

Taking log

log S = log 1 1027

38 1 4×

− /

= − = − −−1

410 27 1

438 2738[log log ] [ log ]

= − − = − = −14

38 1 43136 14

39 43 9 8578[ . ] [ . ] .

∴ log S = – 9.8578 Taking Antilog S = 1.35 × 10–10 M Hence [Fe]3+ = 1.35 × 10–10 M; [OH–] = 3 ×1.35× 10–10 = 4.04×10–10 M For PbCl2 PbCl2 Pb2+ + 2Cl–

S S 2S

Ksp = 4S3 ∴ S K /4sp= = × −( / ) ( . )4 1 6 103 53

log S = 1/3 [log (1.6 × 10–5) – log 4]

log S =log . log log . . .1 6 5 10 4

30 204 5 0 6020

35 398

3− −

=− −

=−

log S = – 1.799 Taking Antilog S = 1.59 × 10–2 M Hence [Pb2+] = 1.59 × 10–2M; [Cl–] = 2×1.59 ×10–2 = 3.18 × 10–2 M For Hg2I2 Hg2I2 Hg2

+2 + 2I–

S S 2S

Equilibrium


245

Ksp = (S)(2S)2 = 4S3

⇒ =×

= ×−

−S M4 5 104

2 24 1029

3 10. .

[Hg2+2] = S = 2.24 × 10–10 M

[I–] = 2S = 4.48 × 10–10 M 68. The solubility product constants of Ag2CrO4 and AgBr are 1.1×10–12 and

5.0×10–13 respectively. Calculate the ratio of molarities of their saturated solutions.

Ans. Ag2CrO4 2Ag+ + CrO42–,

Ksp = (2S)2(S) = 4 S3, ∴ S3 = Ksp/4 = (1.1× 10–12)/4 = 0. 275 × 10–12 = 2.75 × 10–13

31og S = log (2.75 × 10–13) = –13 + 0.4393 = –12.5607 or log S = – 4.1869 = 5 8131. ∴ S = 6.5 × 10–5 mol L–1

AgBr Ag+ + Br–

S S S Ksp = S2 = 5 × 10–13

∴ S = × −5 10 13 = 7.071 × 10–7

∴ ratio of [Ag2CrO4] : [AgBr] = 6.5 × 10–5 : 7.071 × 10–7 = 91.9 Silver chromate is more soluble than silver bromide. 69. Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate

are mixed together. Will it lead to precipitation of copper iodate? For cupric iodate, Ksp = 7.4×10–8.

Ans. 2 NaIO3 + CuCrO4––––→ Na2CrO4 + Cu(IO3)2

After mixing, [NaIO3] = [IO3–] = (2 × 10–3) / 2 = 10–3 M

[CuCrO4] = [Cu2+] = (2 × 10–3) / 2 = 10–3 M Ionic product of Cu(IO3)2 = [Cu2+] [IO3

–]2 = (10–3) (10–3)2 = 10–9

As ionic product is less than Ksp, no precipitation will occur. 70. The ionisation constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is

2.5 ×10–13. How many times is silver benzoate more soluble in a buffer of pH 3·19 compared to its solubility in pure water ?

Ans. C6H5COOAg ––––→ C6H5COO–+ Ag+

Solubility in water. Suppose solubility in water = x mol L–1. Then [C6H5COO–] = [Ag+] = x mol L–1

Equilibrium


246

∴ x2 = Ksp or x K= = × −sp 2 5 10 13.

= 5 × 10–7 mol L–1 Solubility in buffer of pH = 3.19 pH = 3.19 means – log [H +] = 3.19 or, log [H+] = –3.19 = 4 81. or

[H+] = 6.457×10–4 M C6H5COO– ions now combine with the H+ ions to form benzoic acid but [H+]

remains almost constant because we have buffer solution. Now C6H5COOH C6H5COO –+ H+.

K

C H COO HC H COOH

or [C H COOH][C H COO ]

6 5

6 5

6 5

6 5a =

− +

−[ ][ ]

[ ]

= =××

=+ −

−[ ] .

.HKa

6 457 106 46 10

104

5 …(i)

Suppose solubility in the buffer solution is y mol L–1. Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionised), we have

y = [Ag+] = [C6H5COO–] + [C6H5COOH] = [C6H5COO–] + 10 [C6H5COO–] = 11 [C6H5COO–] ...(using eqn. (i)) [C6H5COO–] = y/11 ∴ Ksp = [C6H5COO–] [Ag+] i.e. 2.5 × 10–3 = y/11× y or y2 = 2.75 ×10–12 or y = 1.66 × 10–6

yx=

×

×=

−

−1 66 10

5 103 32

6

7. .

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

Ans. Suppose the concentration of each of FeSO4 and Na2S is x mol L–1. Then after mixing equal volumes,

[FeSO4] = [Na2S] = x/2 M, i.e., [Fe2+] = [S2–] = x/2 M Ksp for FeS = [Fe2+] [S2–], i.e., 6.3×10–18 = (x/2) × (x/2) or x2 = 25·2 × 10–18, or x = 5.02 ×10–9 M.

Equilibrium


247

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

Ans. CaSO4 (s) Ca2+ (aq) + SO42– (aq)

If S is the solubility of CaSO4 in moles L–1, then Ksp = [Ca2+] × [SO42–] = S2

or S = (Ksp)1/2 = (9.1× 10–6)1/2 = 3.02×10–3 mol L–1

Solubility in grams per litre = 3.02 × 10–3 × 136 g L–1 = 0.411 gL–1

(Molar mass of CaSO4 = 136 g mol–1) Thus, for dissolving 0.411 g, water required = 1 L ∴ For dissolving 1 g, water required = 1/(0. 411) L = 2.43 L. 73. The concentration of sulphide ion in 0.1 M HCl solution saturated with

hydrogen sulphide is 1.0 ×10–19 M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of FeSO4, MnCl2, ZnCl2 and CdCl2, in which solutions precipitation will take place ? Given Ksp for FeS = 6.3 × 10–18, MnS = 2.5 × 10–13, ZnS = 1.6× 10–24 and CdS = 8.0 × 10–27.

Ans. Precipitation will take place in the solution for which ionic product is greater than solubility product. As 10 mL of solution containing S2– ion is mixed with 5 mL of metal salt solution, after mixing [S2–] = 1.0×10–19× (10 /15) = 6.67 ×10–20

[Fe2+] = [Mn2+] = [Zn2+] = [Cd2+] = (5/15)× 0. 04 = 1.33×10–2 M ∴ Ionic product for each of these will be = [M2+] [S2– ] = (1.33 × 10–2) (6.67 × 10–20) = 8.87× 10–22

As this is greater than the solubility product of ZnS and CdS, therefore, ZnCl2 and CdCl2 solutions be precipitated.

b/9w øãç 9ø Ù {½ i - studymate · equilibrium e z øãç ø Ù ò 222 ans.for the given...

Documents