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Question-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?

Solution:
a. The vapour pressure will decrease due to increase in volume.
b. If the volume of the container is increased but the surface area remains the same then the rate of evaporation remains the same while the rate of condensation decreases.
c. When equilibrium is restored finally, the vapour pressure will have the same value as vapour pressure depends only on the temperature and not on the volume of the vessel.

Question-2

What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M?
2SO2(g) + O2(g) f 2SO3(g)

Solution:

Kc = = = 12.23

Kc = 12.23.

Question-3

At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms I2(g) f 2I(g) Calculate Kp for the equilibrium.
Solution:
po = 105 Pa
pI2(g) = 2.6 × 109 Pa (Experimentally)
Kp = Pa = 2.67 × 104 Pa.

Question-4

Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl(g) f 2NO(g) + Cl2(g) (ii) I2(S) + 5F2 f 2IF5

Solution:
(i) Kc = (ii) Kc =

Question-5

Find out the value of Kc for each of the following equilibrium from the value of kp:
(i) 2NOCl(g) f 2NO(g) + Cl2(g); Kp = 1.8 × 10-2 at 500 k
(ii) CaCO3(S) f CaO(S) + CO2(g); Kp = 167 at 1073 K

Solution:
(a) 2NOCl(g) 2NO(g) + Cl2(g)
Kp = 1.8 × 10-2 at 500 K

For the reaction,
Δn = (2+1) – 2 = 1
Kp = Kc (RT)n

KC = = = 4.4
× 10-4


(b) CaCO3(s) CaO(s) + CO2(g)
Kp = 167 at 1073K

Kc =

The concentration of CaO(s) and CaCO3(s) may be assumed to be constant i.e.
[CaO(s)] – 1 and [CaCO3(s)] = 1 and hence neglected so that Kc = [CO2(g)]

Since the partial pressure of a gaseous component is proportional to its concentration
Kp = p[CO2(g)]

Kc = = = 1.90.

Question-6

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

Solution:
Kc for the reverse reaction is inverse of Kc for forward reaction. Kc (reverse) = = 1.6 × 10-15.

Question-7

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Solution:
Concentration of solids and liquids are constant and taken as unity
[Solids] = 1; [liquids] = 1.

Therefore pure liquids and solids can be ignored while writing the equilibrium constant expression.

Question-8

Reaction between N2 and O2- takes place as follows:
2N2(g) + O2(g) f 2N2O(g)

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10-37, determine the composition of equilibrium mixture.

Solution:
2N2(g) + O2(g) 2N2O(g)

Initial moles: 0.482 + 0.933

Moles at equilibrium: (0.482 – x) (0.933 – x/2) x

Since the value of K is very small (2.0 × 10-37) it indicates that the equilibrium is very much in favour of reactants and thus the value of x is very small. Therefore, the approximate concentrations at equilibrium may be written as

[N2] = = 0.0482 mol L-1

[O2] = = 0.0933 mol L-1

[N2O] = or 0.1 x mol L-1

Kc = = = 2.0 × 10-37

x = 6.6 × 10-20

[N2O] = 0.1 × 6.6 × 10-20 = 6.6 × 10-21 mol L-1

[N2] = 0.0482 mol L-1

[O2] = 0.0933 mol L-1.

Question-9

2NO(g) + Br2(g) f 2NOBr(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 NO and Br2.

 


Solution:
Amount of NO Br, formed at equilibrium = 0.0518 mol.
Initial amount of NO = 0.087 mol
Initial amount of Br2 = 0.0437 mol.
The reaction is
2NO + Br2(g)
2NOBr 2NOBr 2NOBr

2 mols of NOBr is formed from 2 mols of NO
Therefore 0.0518 mol of NOBr is formed from 0.0518 mol of NO
So the amount of NO at the equilibrium = 0.087 – 0.0518 = 0.0352 mol.

Amount of Br2 at the equilibrium

According to equation
Number of moles of Br2, reacting with 0.0518 mol of NoBr = = 0.0259 mol.
Therefore Amount of Br2 at the equilibrium = 0.0437 – 0.0259 = 0.0178 mol.

Question-10

2HI(g) f H2(g) + I2(g)

Solution:

2HI(g) H2(g) + I2(g)

Initial Pressure 0.2 atm.

At equilibrium 0.2 p p/2 p/2

(0.04 atm) 0.8 atm 0.8 atm

Kp = = = 400.

Question-11

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) f 2NH3(g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Solution:
[H2] = M = 0.096 M

[N2] = M = 0.079 M

[NH3] = M = 0.407 M

Qc = = = 2370
since Qc > Kc (equilibrium constant) the reaction is said to proceed in the reverse direction.

Question-12

The equilibrium constant expression for a gas reaction is,
Ke = replace [H2O] by [H2O]6 replace [O2] by [O2]5
Write the balanced chemical equation corresponding to this expression.

Solution:

Ke =
Hence chemical equation is
4NO + 6H2O 4NH3 + 5O2.

Question-13

Calculate the equilibrium constant for the reaction.
 

Solution:

The equilibrium reaction involved is

H2O (g) + CO (g) H2 (g) + CO2 (g) 

Initial moles of H2O and CO taken = 1 mole each.
% dissociation of H2O = 40%
Number of moles of H2O dissociated =40/100 ×1= 0 .4 moles
Number of moles of CO at equilibrium = 1- 0.4 = 0.6
Number of moles of H2O at equilibrium = 1 - 0.4 = 0.6
No. of moles of H2 formed = 0.4
No. of moles of CO2 formed = 0.4
Therefore [H2O] = 0.6/10 =0.06mol/L
[CO] = 0.6/10 = 0.06mol/L
[H2] = 0.4/10 = 0.04mol/L
[CO2] = 0.4/10 = 0.04mol/L
Hence,
             
              = 4/9 = 0.44

Question-14

At 700 K, equilibrium constant for the reactions:
H2(g) + I2(g) f 2HI(g) is 54.8. If 0.5 mol L-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Solution:
For the equilibrium,
H2(g) +I2(g) 2HI(g) 
The value of equilibrium constant K is 54.8.
Hence, if we start with HI and allow the reaction to reach equilibrium then the equilibrium constant K' = 1/K = 1/54.8
K' = [H2][I2] /[HI]2 = 0.0182

Let the concentration of both H2 and I2 be x
Then K' = 0.0182 = x2/ (0.5)2 
       x2 = 0.0182 × 0.25 = 0.00456
       x   = 0.067mol/litre. 

Question-15

2ICl(g) I2(g) + Cl2(g) ; Kc = 0.14

 


Solution:
2Icl(g) I2(g) + Cl2(g)

Initial Concentration 0.78M - -

At equilibrium (0.78-x)M x/2M x/2M

Kc =

0.14 = =

x = 0.33M

[ICl] = 0.78 – 0.33 = 0.45M

[I2] = = 0.165M

[Cl2] = = 0.165M.

Question-16

The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as:
CH3COOH + C2H5OH CH3COOC2H5 + H2O

(i) Write the concentration ratio, Q, for this reaction. Note that water is not in excess and is not a solvent in this reaction.

(ii) At 293 K, if one starts with 1.0 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.50 mol of ethanol and 1.0 mol of acetic acid and maintaining at 293 K, 0.214 mol of ethyl acetate is formed after some time. Has equilibrium been reached?

Solution:
(i)

(ii) At equilibrium, concentration of acetic acid is = 1.0 - 0.171 = 0.829 mol
and concentration of ethanol is                          = 0.180-0.171
                                                                      = 0.009
                               K = (0.171)/(0.829 × 0.009) = 22.19

(iii) Initial conc. of ethanol = 0.5mol
Initial conc. of acetic acid = 1.0mol
After some time, conc. of ethanol = 0.5 - 0.214 = 0.286 mol
                   conc. of acetic acid = 1.0 - 0.214 = 0.786 mol
             Hence, Q = (0.214)/(0.786 × 0.286)
                           = 0.9520
Since the value of Q is lesser than that of K, the reaction must be proceeding in the forward direction readily.

Question-17

A sample of pure PCl5 was introduced into an evacuated vessel at 473K. After equilibrium was attained, concentration of Pcl5 was found to be 0.5 × 10-1 mol L-1 . If value of Kc is 8.3 × 10-3, what are the concentrations of PCl3 and Cl2 at equilibrium?

Solution:
PCl5 PCl3 + Cl2
Let the concentration of PCl5 and Cl2 be x at equilibrium.
Kc =
Or 8.3
× 10-3 =
Or 4.15 × 10-4 = x2
Or x = 2
× 10-2 = 0.02 mol L-1
Hence the concentration of PCl5 and Cl2 each at equilibrium = 0.02 mol L-1.

Question-18

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: PCO = 1.4 atm and PCO2 = 0.80 atm?

 


Solution:
FeO(s) + CO(g) Fe(s) + CO2(g)

Kp = 0.265 atm at 1050 K

FeO(s) + CO(g) Fe(s) + CO2(g)

Initaial pressure 1.4 atm 0.8 atm

Equilibrium pressure (1.4 – x) atm (0.8 + x) atm

Kp = = 0.265 atm

0.8 + x = 0.265 (1.4 – x) = 0.371 – 0.265 x

x + 0.265x = 0.371 – 0.8

1.265x = -0.429

x = -0.34 atm

Equilibrium pressure of CO = 1.4 – (-0.34) = 1.74 atm

Equilibrium pressure of CO2 = 0.8 + (-0.34) = 0.46 atm.

Question-19

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 3.00 mol L-1 of N2, 2.00 mol L-1 of H2 and 0.500 mol L-1 of NH3. Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Solution:
N2(g) + 3H2(g) 2NH3(g)
Kc = == = = 0.01
Since Qc < Kc the reaction is not at equilibrium. The reaction proceeds to form more product.

Question-20

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.

 


Solution:
2BrCl(g) Br2(g) + Cl2(g)
3.30 × 10-3 0 0
KC = 32 at 500K
From the equation 2 mols of BrCl on decomposition gives = 1 mol of Br and 1 mol of Cl2
Therefore, 3.30
× 10-3 mol L-1 of BrCl gives =
                                                                 = 1.65
× 10-3 mol of Br2 and Cl2
Let x be the molar concentration of BrCl at the equilibrium
Molar concentration of Br2 = 1.65
× 10-3 mol.
Molar concentration of I2 = 1.65
× 10-3 mol
Therefore Kc =
               32 =
               x2 =
               x2 =
                 x = = 3
× 10-4 mol L-1
Therefore Molar concentration of BrCl = 3
× 10-4 mol L-1.

Question-21

Calculate the Kc for this reaction at the above temperature.

 


Solution:
Let total mass of the gaseous mixture be = 100g

Mass of CO = 90.55 g

Mass of CO2 = 9.45g

Moles of CO = = 3.234

Moles of CO2 = = 0.215

Total number of moles of gases = 3.234 + 0.215 = 3.449

PCO = × 1 = 0.938 atm

× 1 = 0.0623 atm

Kp = = 14.12

Kp = Kc(RT)Δ n = Kc(RT) Q Δ n = (2-1) = 1

Kc = = 0.153.

Question-22

Δ fG0 (O2) = 0 kJ/mol

 


Solution:
NO(g) + ½ O2(g) NO2(g)

Δ fG0 (NO2) = 52.0 kJ/mol-1

Δ fG0 (NO) = 87.0 kJ/mol-1

Δ fG0 (O2) = 0 kJ/mol-1 T = 298 K

Δ G° = Δ fG0 (NO2)[Δ fG0 (NO) + Δ fG0 (O2)]

= Δ fG0 (NO2)Δ fG0 (NO)

= (52-87) kJ mol-1

We have, Δ G° = -2.303 RT log Kc

-35 kJ mol-1 = -2.303 × 8.314 JK-1 × 298 × log Kc

log Kc =

=

=

log Kc = 6.134

Kc = 1.361 × 106.

Question-23

c) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

 


Solution:
a) No. of moles of the products increases.

b) No. of moles of the products decreases.

c) No. of moles of the products decreases.

Question-24

(i) CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
(ii) CO2(g) + C(s)
2CO(g)
(iii) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

 


Solution:
The following reactions will get effected by increase of pressure

(ii) CO2(g) + C(s)  2CO(g)

The increase of pressure will shift the equilibrium to left hand direction.

(iii) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

The increase of pressure will shift the equilibrium to left hand direction.

Question-25

The equilibrium constant for the following reaction is 1.6 × 105 at 1024K H2(g) + Br2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Solution:
Kp = Kc =
Let (pH2)eq = (pBr2)eq = x bar
1.6
× 105 =
x2 =

x =
or x = 0.25 × 10-1
or 2.5 × 10-2
Therefore, (pH2)eq = 2.5
× 10-2
(pBr2)eq = 2.5
× 10-2 bar
and (pHBr)eq = 10.0 bar.

Question-26

Hydrogen gas is obtained from natural gas by partial oxidation with steam as per folowing endothermic reaction.

(a) Write an expression for Kp for the above reaction.
(b) How will the value ok Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catelyst?

Solution:


(a) Kp =
 

(b)

 

(i) The value of Kp remains unchanged on increasing the pressure. When pressure is increased then according to Le-Chatelier’s principle the equilibrium shifts in the direction where there is less number of moles of gases i. e., backward direction in case of the given reaction.

(ii) In case of endothermic reactions the value of Kp increases with increase in temperature. With increase in temperature, the equilibrium shifts in the endothermic directions i. e., forward direction in case of the given reaction.

(iii) Kp will remain undisturbed. Equilibrium composition will remain unchanged. However, in the presence of catalyst, the equilibrium would be attained quickly.

Question-27

2H2(g) + CO(g) CH3OH(g)

 


Solution:
2H2(g) + CO(g) CH3OH(g)

a) Effect of addition of H2
This will favour forward reaction i. e., formation of CH3OH because increase in the concentration of the reactants will result in the formation of more products.

b) Addition of CH3OH
This will shift the equilibrium towards backward direction because when the product concentration increases, backward reaction is favoured.

c) Removal of CO
This favours the increase in formation of CO i. e., backward reaction is favoured.

d) Removal of CH3OH
This will result in the increased production of CH3OH i. e., forward reaction is favoured.

All these explanations are according to Le Chatelier’s principles.

Question-28

(a) Write an expression of 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) the pressure is increased (iii) the temperature is increased?

 


Solution:
PCl5(g) PCl3(g) + Cl2(g)
(a) Kc =

(b) Reverse reaction

PCl3(g) + Cl2(g) PCl5(g)
Kc = = =120.48

(c) (i) More PCl5 is added: The equilibrium will be shifted to the right. This will continue till the entire extra amount of PCl5 has been used up.

(ii) The pressure is increased:

PCl5(g) PCl3(g) + Cl2(g)
1 mole        1 mole      1 mole

Since backward reaction takes place with decrease in number of moles, an increase in pressure will favour combination of PCl3 and Cl2 molecules to produce back PCl5.

(iii) The temperature is increased: The reaction is endothermic. Therefore, an increase in temperature will shift the equilibrium to the right.

Question-29

c) Cl2(g) + 2NO2(g) 2NO2Cl(g) Kc = 1.8

 


Solution:
a) Will have appreciable concentration of reactants.

b) Will have appreciable concentration of products.

c) Will have appreciable concentration of reactants and products.

Question-30

The value of Kc for the reaction 3O2(g) 2O3(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?

Solution:
3O2(g) 2O3(g)

Kc = 2 × 10-5 T = 25° C = 298 K

Kc =

[O3]2 = Kc × [O2]3

= 2 × 10-50 × (1.6 × 10-2)3

= 2 × 10-50 × 10-6 × 4.096

= 8.192 × 10-56

[O3] = 81.92 × 10-12 = 2.86 × 10-28.M

Question-31

The reaction, CO(g) + 3H2(g) f CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Solution:
CO(g) + 3H2(g) CH4(g) + H2O(g)

Kc =

3.9 =

[CH4] = = 0.0585 mol.

Question-32

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 is added to 5 ml of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of the solutions will precipitation take place. (Please consult table 8).

Solution:
Ksp FeS    .........................
 
= 6.3 x 10-13
Ksp MnS    ........................
 
= 2.5 x 10-13
Ksp ZnS ..............................
 
= 1.6 x 10-24
Ksp CdS   ......................
 
= 8.0 x 10-27
FeSO4  Fe2+ + SO42-
 
 
H2S   ..........................
 
= 2H+ = S2-
Fe2+ + S2- FeS
 
 
The total volume after mixing
 
= (10 + 5) = 15 ml
[Fe2+]  ........................
 
= 1.33 x 10-2
[S2-]   .........................
 
= 0.6 x 10-19
[Fe2+] [S2-]  ................. = 0.798 x 10-21

Similarly [Zn2+] [S2-] and [Cd2+] [S2-] are also 0.798 x 10-21

Since in the case of ZnCl2 and CdCl2 Solutions, the [Zn2+] [S2-] and [Cd2+] [S2-] > Ksp of ZnS and Cds respectively.

Therefore ZnCl2 and CdCl2 give precipitation of their respective sulphides.

Question-33

Which of the followings are Lewis acids?
H2O, BF3 , H+ and NH

Solution:
BF3, H+, NH4+.

Question-34

What will be the conjugate bases for the Bronsted acids: HF, H2SO4 and HCO3?

Solution:
The conjugate base of HF is F- that of H2SO4 is HSO4- and that of HCO3- is CO32-.

Question-35

Write the conjugate acids for the following Bronsted bases: NH2-, NH3 and HCOO-

Solution:

Bronsted Bases

Conjugate Acids

NH2-

NH3

NH3

NH4+

HCOO-

HCOOH

 

Question-36

The spieces: H2O, HCO3-, HSO4- and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Solution:

Bronsted Acid/Base

Conjugate Acids

Conjugate Base

H2O

H3O+

OH-

HCO3-

H2CO3

CO32-

HSO4-

H2SO4

SO42-

NH3

NH4+

NH2-

 

Question-37

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.

Solution:
a) OH- ion is a Lewis base as it has lone pairs of electrons on oxygen which it can donate

b) F- ion has four lone pairs of electrons. Hence, it can act as a Lewis base by donating any one of these lone pair.

c) H+ acts as a Lewis acid as it has a vacant orbital and hence can accept a pair of electrons from Lewis bases.

d) BCl3 acts as a Lewis acid. In BCl3, there are only six electrons in the valence shell of boron. It can accept a pair of electrons from bases such as NH3 to complete its octet.

 

Question-38

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

Solution:
PH = -log [H3O+]

= - log (3.8 × 10-3)

= - log 3.8 + (-log 10-3)

= -log 3.8 + 3 log 10

pH = 2.42

Question-39

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

Solution:
 

 pH = 3.76

pH = -log [H3O+]

3.76 = -log [H3O+]

log [H3O+] = -3.76

log [H3O+] = 4.2400

[H3O+] = antilog (4.2400) = 1.7 x 10-4

Question-40

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.

Solution:
Given data: Kα of acetic acid = 1.74 × 10-5

α = ?

Concentration of acetic acid = 0.05 M

α = 0.0187

[CH3COO-] = Cα = 0.05 × 0.0187 mol L-1 = 0.00094 mol L-1

[CH3COO-] = 9.4 × 10-4 mol L-1.

[H3O+] = [CH3COO] = 0.05M = 5 × 10-2 M

pH = -log [H3O+]

= -log (5 × 10-2)

= - log5 + 2 log 10

= -0.6990 + 2 = 1.301

pH = 1.301.

Question-41

The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 × 10-4, 1.8 × 10-4 and 4.8 × 10-4 respectively. Calculate the ionisation constants of the corresponding conjugate base.

Solution:

We know that Ka x Kb = Kw = 1 x 10 -14 Kb = Kw/Ka.

i) Ionisation constant of HF (Ka) = 6.8 x 10-4

Ionisation constant of F- (Kb) = Kw/Ka = 1.5 x 10-11

ii) Ionisation constant of HCOOH (Ka) = 1.8 x 10-4

Ionisation constant of HCOO- (Kb) = 5.6 x 10-11

iii) Ionisation constant of HCN (Ka) = 4.8 x 10-9

Ionisation constant of CN- (Kb) = 2.08 x 10-6.

Question-42

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

Solution:

Ionisation constant of phenol

= 1.0 × 10-10

Degree of ionisation

= α =

 

= =
= = 4.47
× 10-5

[Phenolate ion]   ..................




PH   ..........................

= cα =
=
= 2.2
× 10-6
= -log (2.2
× 10-6 M/M)
= 5.65

Degree of ionisation of the solution is also 0.01M


Degree of ionisation

=
=

=
= 1
× 10-4

 

Question-43

The first ionization constant of H2S is 9.1 × 10-8.    10-8. Calculate the concentration of HS- ion in its 0.1 M solution and how will this concentration be effected 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.

Solution:
H2S H+ + HS- Ka1 = 9.1 × 10-8
HS- H+ + S2- Ka2 = 1.2
× 10-13
HCl H+ + Cl-
In the case of HCl, the dissociation is complete.
So [H+] = 0.1 M.
Let [H+] from H2S be x and also x be the [HS-].

In the solution of 0.1M HCl

[H+]  ..........................

= 0.1 + x

[HS]   ..........................

= X

[H2S] ..........................

= [0.1 – x]

Ka ..........................

=

Ka1    ..........................

= 9.1 × 10-8

Since x is very small

 

9.1 × 10-18   ..........................

=

x    ..........................

= = 9.1 × 10-8

In 0.1 M HCl solution, concentration of [HS-] = 9.1 × 10-8 M / litre

2nd dissociation

 

HS- H+ + S2-

 

Ka2    ..........................

= 1.2 × 10-13

Ka2   .......................... 

=

1.2 × 10-13  ..................

=

S2-    ..........................

= 1.09 × 10-20 M / Litre

 

Question-44

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

Solution:

a) 0.003 M HCl

 

HCl + H2O H3O+ + Cl

 

0.003 M

 

pH ..........................

= -log [H3O]+

 

= -log(3 × 10-3)

 

= (3-log 3)

 

= (3-0.4771)

 

= 2.5229 = 2.52

 

 

b) 0.005 M NaOH

 

NaOH Na+(aq) + OH-(aq)

 

0.005M

 

[OH-]  ..............................

= 0.005 M = 5 × 10-3 M

[H3O+]   ..........................

= =

 

= 0.2 × 10-11 = 2 × 10-12

PH ..........................

= - log [H3O+]

 

= - log (2 × 10-12)

 

= [12 – log 2]

 

= [12 - 0.3010 ] = 11.70

 

 

c) 0.002 M HBr

 

HBr + H2O Br-(aq) + H3O+(aq)

0.002 M

 

[H3O+] ..........................

= 0.002 M = 2 × 10-3 M

 pH ................................

= - log[H3O+]

 

= - log(2 × 10-3)

 

= (3 – log2) = (3-0.3010)

 

= 2.699 = 2.70

 

 

d) 0.002 KOH

 

KOH K+(aq) = OH-(aq)
0.002 M

 

[OH-]     ..........................

= 0.002 M = 2 × 10-3 M

[H3O+] ...........................

= = = 0.5 × 10-11 M

pH       ............................

= -log[H3O+]

 

= -log [5 × 10-12]

 

= (12 – log 5)

 

= 12 – 0.699

 

= 11.3

 

Question-45

d) 1 ml of 13.6 M HCl is diluted with water to give 1 litre of solution.

 


Solution:
 

a) Molar mass of TlOH = 221.4

nTlOH = 2/221.4 = 9.03 × 103

[OH-] = n TlOH_/ Vlit = 9.03 × 10-3/ 2 = 4.51 × 10-3 M

pOH = log (4.51 × 10-3 ) = 3log 4.51 = 30.6542 = 2.35

pH + pOH = 14 ; pH = 14 2.35 = 11.65

b) Molar mass of Ca(OH)2 = 74

n Ca(OH)2 = 0.3 / 74 = 4.05 × 103

[OH] = 4.05 × 103/ 0.5 = 0.0081 M

2[OH] = 0.0081 × 2 = 16.2 × 10-3 (Ca(OH)2 contains 2 OH groups.)

pOH = log (16.2 × 10-3) = 3 – log 16.2 = 1.79

pH = 14 – 1.79 = 12.21

(c) n NaOH = 0.3 / 40 = 7.5 × 10-3M

[OH-] = 7.5×103/0.2 = 0.0375

pOH = log(0.0375) = log(3.75 × 102) = 2 log3.75 = 1.43

pH = 14  - 1.43 = 12.57

(d) To get molarity(concentration) we can use V1M1 = V2M2

1 × 13.6 = 1000 × M2

M2 = 1× 13.6/1000 = 0.0136 = [H+]

pH =log(1.36 × 10-1) = 1 log 1.36 = 1.87

Question-46

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

Solution:
Given Data:

C = 0.1 M

α = 0.132

[H3O+] = Cα = 0.1 × 0.132 = 0.0132

pH = - log[H3O+]

= - log 0.0132

= - log (1.32 × 10-2)

= - log 1.32 + 2 log 10

= -0.1206 + 2

pH = 1.88

α =

Squaring both sides,

α 2 =

Ka = Cα2 / 1−α = 0.1 × (0.132)2 / 10.132

Ka = 2.01 × 10-3

pKa = - log Ka

pKa = - log (2.01 × 10-3) = 3 log 2.01 = 2.70

pKa = 2.77

Question-47

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

Solution:
 

pH = 9.95 ; [H+] = antilog (-9.95) = 1.12 X 10-10M

[OH-] = 1X10-14/1.12 X 10-10 = 8.91 X 10-5M

Kb = [M+] [OH-] / [MOH] = (8.91 X 10-5)2 /0.005 = 1.59 X 10-6

pKb = -log Kb = -log(1.59 X 10-6) = 5.80

Question-48

What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Solution:
pH = 0.001M

From the table, the ionization constant of aniline is

Kb = 4.3 × 10-10

α b = =

α b = 0.002.

Kb × Kα = Kw

Kα = = 2.33 × 10-5.

Kα = 2.33 × 10-5.

Question-49

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

Solution:

a) pH   ................................

= 6.83

log [H3O+] ..........................

= -pH = -6.83

[H3O+] ...............................

= antilog(- 6.83)

 

= 10-7 × 100.17

[H3O+] ...............................

= 1.48 × 10-7 M

 

 

b) pH .................................

= 1.2

log [H3O+]  .........................

= -pH = -1.2

[H3O+] ...............................

= antilog(-1.2)

 

= 10-2 × 100.8

 

= 6.31 × 10-2 M

 

 

c) pH   ................................

= 7.38

log [H3O+] ..........................

= -pH = -7.38

[H3O+]  ...............................

= Antilog(-7.38)

[H3O+] ...............................

= 10-7.38

 

= 10-8 × 10.62

 

= 4.17 × 10-8

d) pH    ...............................

= 6.4

log [H3O+]   .......................

= -pH = -6.4

[H3O+]  .............................

= Antilog(-6.4)

 

= 3.98 × 10-7 M

 

Question-50

The pH of milk, black office, tomoto 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.

Solution:

i) pH

= 6.8

log[H3O+

= -pH = -6.8

[H3O+]

= Antilog [-6.8]

 

= 1.5 × 10-7 M

   

ii) pH = 5.0

 

log[H3O+]

= -pH =-5.0

[H3O+]

= Antilog(-5.0) = 10-5 M

   

iii) pH

= 4.2

log [H3O+]

= -pH = -4.2

[H3O+

= Antilog(-4.2)

 

= 6.31 × 10-5 M

   

iv) pH

= 2.2

log[H3O+]

= -pH = -2.2

[H3O+]

= Antilog(-2.2)

 

= 6.31 × 10-3

   

v) pH

= 7.8

log[H3O+]

= -pH = - 7.8

[H3O+]

= Antilog (-7.8)
= 1.58
× 10-8 M

 

Question-51

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298K. For Calcium sulphate Ksp is 9.1 × 10—6.

Solution:
CaSO4 Ca2+ + SO42-
Let the solubility of CaSO4 be x moles/litre. Then the solution will contain x moles of Ca2+ and x moles SO42- ions respectively per litre. Hence, the solubility product, Ksp of CaSO4 is

Ksp    .......................

= [Ca2+] [SO42-]

 

= x × (x) = x2

We know Ksp   .................

= 9.1 x 10-6

Hence x2   .......................

= 9.1 × 10-6

Or x    .............................

= 3.01 × 10-3 mol L-1

1 mole of CaSO4 .............

= 40 + 32 + 64 = 136g

3.01 × 10-3 of CaSO4 ......

= 136 × 3.01 × 10-3

 

= 136 × 0.00301

 

= 0.409 g

0.41 g needs 1 L of water for complete dissolution                                                            
1 g needs = = 2.38 L.

Question-52

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

Solution:
Solubility of Sr(OH)2 at 298 K = 19.23 g/L
Sr(OH)2 Sr2+(aq) + 2OH-
Mol. mass of Sr(OH)2 = 87 + 34 = 121

Molar concentration of Sr(OH)2

=
= = 0.1581 M
= 0.16

Sr(OH)2 .........................   

= Sr2+(aq) + 2OH-

0.16M 2(0.16) ......................

= 32 M

[Sr2+]    .........................

= 0.1581 M

[OH-]    .........................

= 2 × 0.158

 

= 0.316 M


[H3O+]  .........................

= = =
=

pH   ......................... 

= - log H3O+

 

= - log 10-12 – log 32

 

= 12 + 1.51

pH .........................

= 13.51

 

Question-53

The ionization constant of propionic acid is 1.32 × 10-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01 in HCl also.

Solution:

Ionisation constant of propionic acid

= 1.32 × 10-5

Concentration of solution

= 0.05M

Let the degree of ionisation is α

α..................................................

  = =

 

= 1.63 × 10-2

[H+]  ..................................

= cα

 

= 0.05 × 1.63 × 10-2

 

= 0.0815 × 10-2

 

= 815 × 10-6

pH .....................................

= -log(815 × 106)

 

= - (log815 + log 10-6)

 

= 6-log 815

 

= 6-2.9112

 

= 3.088

 

= 3.09

In the case of molarity of HCl is 0.01 M, it is assumed that it is fully dissociated. Let [H+] be x from the ionisation of propionic acid. This is (x) also the concentration of propionate ion.

[H+] = 0.01 + X ; [Propionate] 

= x and [HP] = (C-x)

Ka .....................................

= [H+] [P-] / [HP]

 

=

Since x is very small, Ka .......

=

Or x ....................................

= C. Ka / 0.01

 

= 0.05 × 1.32 × 10-5 / 0.01

x ..........................................

= 1.60 × 10-5

Degree of ionisation in 0.1 M HCl = x/c

   
 

= 1.60 × 10-5 / 0.05

 

= 1.32 × 10_3

 

Question-54

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.

Solution:

PH ...................................

= 2.34

- log [H3O+]  ....................

= 2.34

or [H3O+]  .......................

= Antilog [-2.34]

 

= 4.57 × 10-3

For weak acid [H+]  ..........

= c α = 4.57 × 10-3

C    ................................

= 0.1M

α    .................................

= = 0.0457

Ka .................................

= α 2 / C

 

=

Ionisation constant Ka

= 2.09 × 10-4

 

Question-55

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.

Solution:

Ka .................................

= 4.5 × 10-4

Hydrolysis constant Kh

= = = × 10-10

Let the degree of hydrolysis

= h

Kh ................................

=

Since h is small 1-h is negligible

 

Concentration of NaNO3

= 0.04 M

............................

= 0.04 × h2

0.180h2 ......................

= 10-10

h2 ..............................

=

h = .............

= 2.36 × 10-5

PH..................................

= 7.0 + (pKa + log c)

 

= 7.0 + (log 0.04 – log(4.5 × 10-4

 

= 7.0 + [-1.40 + 3.35]

 

= 7.0 + [1.95] = 7.0 + 0.97 = 7.97

 

Question-56

A 0.02 M solution of pyridinium hydrochloride has pH = 3.44 M NaOH solution. Calculate ionization constant of Pyridine.

Solution:

Since pyridinium hydrochloride is a salt of weak base and strong acid pH = 7 (log C + pkb), where kb is the dissociation constant of pyridine

 

3.44 = 7 (log 0.02 + pkb)

3.56 = (1.70 + pkb)

7.12 = 1.70 pkb

Pkb = 1.70 + 7.12 = 8.82

Kb = antilog(8.82) = 1.513 X 10-9

 

Question-57

NaCl, KBr , NaCN ,NH4NO3, NaNO2 and KF.

 


Solution:
Nacl and NaNO2 solutions are neutral. KBr , NaCN and KF solutions are basic. NH4NO3 solution is acidic.

Question-58

The ionisation constant of chloro acetic acid is 1.35 × 10-3. What will be pH of 0.1 M acid its 0.1 M salt solution?

Solution:
Sodium acetate is a salt of strong base and weak acid.

PH ...............................

= - [logKw + log Ka – log c]

Kw  ..............................

= 1.0 × 10-14, Ka = 1.35 × 10-3 , c = 0.1 M

PH   .............................

= - [log-14 + log (1.35 + 10-3) – log(0.1)

 

= --14-[-3+0.1303 + 1]

 

= - [-14-3+0.1303 + 1]

 

= - [-17 + 1.313]
= 7.94

pH of its sat solution ...........

= 7.94.

 

Question-59

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

Solution:

Ionic product of water

= 2.7 × 10-14

H2O H+ + OH-14
           
x      x

 

[H+] [OH-] .........................

= 2.7 × 10-14

x2 ....................................

= 2.7 × 10_14

x   ....................................

= = 1.64 × 10_7

pH ...................................

= -log [H+]

 

= 7 – log 1.64

 

= 7 – 0.2148

 

= +6.7696

pH ..................................

= 6.7852.

 

Question-60

Calculate the pH of the resulting mixture of
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

Solution:

a) n Ca(OH)2 = 0.2 ×0.01 = 0.002

.nOH-. = 2..× n Ca(OH)2..= 2.× 0.002 = 0.004

n H+ = nHCl = 0.025× 0.1 = 0.0025

H+ is completely consumed.

.nOH-.(left) = 0.004 – 0.0025 = 0.0015

[OH-] = = 0.0429 M

pOH = -log(0.0429) = 2-0.6325 = 1.3675

pH = 14-pOH = 14-1.3675 = 12.6325


...b) nH2SO4.= .01 ×..01 = 0.0001

nH+...= 2..× .. ) nH2SO4.= 0.0002.....

The base viz., Ca(OH)2 has same volume and concentration.

Hence nH+ = nOH- = 0.0002
pH of the mixture = 7

c) nH2SO4 = .01 ×..0.1 = 0.001

nH+ = 2 nH2SO4 =2 ×.0.001 = 0.002

nOH- = 0.01× 0.1 = 0.001

nH+ (left) = 0.002 -0.001 = 0.001

[H+] = 0.001 / 0.020 = 0.05 M

pH = -log(5 ×10-2) = 2-log 5 = 1.30

Question-61

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility products constants. Determine also the molarities of individual ions.

Solution:
i) Silver chromate Ag2 CrO4, Ksp of Ag2CrO4
                                            = 1.1
× 10-12
Let the solubility of Ag2CrO4 be equal to S,

Ag2CrO4 2Ag+ + CrO

1.1 × 10-12    .................

= (2s)2 (s)

or 1.1 × 10-12   ..............

= 4s3

or s3   ..........................

=

or s ............................

=

Molarity of CrO  .....

= 0.65 × 10-4 M

Molarity of Ag+ .............

= 2 × 0.65 × 10-4 M

 

= 1.30 × 10-4 M

   

(ii) Barium chromate : BaCrO4

BaCrO4 Ba2+ + CrO

Ksp of BaCrO4 .................

= 1.2 × 10-10

1.2 × 10-10   ...................

= s × s

or 1.2 × 10-10  ................

= s2

or s ...............................

=

 

= 1.1 × 10-5 M

Molarity of Ba2+ and CrO42-

= 1.1 × 10-5 M each

   

(iii) Ferric hydroxide: Fe(OH)3

 

Fe(OH)3 Fe3+ + 3OH-

 

Ksp of Fe(OH)3  ...................

= 1.0 × 10_38

Or 1.0 × 10-38 ...................

= s × 3(s)3

Or 1.0 × 10-38 = 3s4

Or s4 =

Or .................................

=

 

= 1.39 × 10-10 M

Molarity of Fe3+ ...............

= 1.39 × 10-10 M

Molarity of OH-   ..............

= 3 × 1.39 × 10-10

 .............

= 4.17 × 10-10 M

   

iv) Lead chloride : PbCl2

 

PbCl2 Pb2+ + 2Cl-

 

Ksp of PBCl2 ..................

= 1.6 × 10_5

1.6 × 10-5 .....................

= 2s3

or s3 ...........................

= 0.8 × 10-5

or s    ...........................

= (0.8 × 10-5)1/3

 

= (8 × 10-6)1/3

 

= 1.59 × 10-2 M

Molarity of Pb2+

= 1.59 × 10-2

Molarity of Cl

= 2 × 1.59 × 10-2

 

= 3 .18 × 10-2 M

   

v) Mercurous iodide : Hg2I2

 

Hg2I2 2Hg2I2

 

Ksp of Hg2I2 .....................

= 4.5 × 10-29

4.5 × 10-29    .....................

= 2s2 × 2s2

or 4.5 × 10-29  .................

= 4s4

or s4  ..............................

= × 10-29

or s   ...............................

=

 

= 2.24 × 10-10 M.

Molarity of Hg22+ and I- each = 4.48 × 10-10 M.

Question-62

The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10-12 and 5.0 × 10-13 respectively. Calculate the ratio of the molarities in their saturated solution.

Solution:
Ag2CrO4 dissolves in water and the equilibrium in the saturated solution is
Ag2CrO4(s) = 2Ag+(aq) + CrO42-(aq)
Let the solubility of Ag2CrO4 be S1

[Ag+(aq)]   .......................

= 2s1

and [CrO42-(aq)]  ...............

= s1

Ksp   ...........................

= [Ag+(aq)]2 [CrO42-(aq)

 

= (2s1)2 × s = 4s13

or s1 ........................

= =

 

= 0.65 × 10-4

Similalry AgBr

= Ag(aq) + Br- (aq)

Let the solubility of AgBr be s2

[Ag+(aq)]   ...................

= s2 and [Be-(aq)] = s2

Ksp ....................

= [Ag(aq)] [Be-(aq)]

  

= s2 × s2 = s12

or s1 .........................

= (Ksp)1/2 = (5.0 × 10-13)1/2

 

= (0.5 × 10-12)1/2

 

= 0.707 × 10-6 mol L-1

   

Raid of the molarities of silver chromate to silver bromide comes to = 0.65 × 10-4 ; 0.707 × 10-6 .

Question-63

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 copper iodate Ksp = 7.4 × 10-8.

Solution:
NaIO3  Na+ + IO3-
Cu(ClO4)2 Cu2+ + 2ClO4-
Cu2+ + 2IO3- Cu(IO3)2

Molarity of NaIO3 and Cu(ClO4) each is 0.002 M.

They ionize completely
So [Cu2+] [IO3-] = 0.002 × 0.002

                          = 4 × 10_6

Since the products of [Cu2+] [IO3-] > KSP of Copper Iodate (7.4 × 10-8)
Copper iodate will be precipitated.

Question-64

What is the maximum concentration of equimolar solution of ferrous sulphate and sodium sulphide. For iron sulphide Ksp = 6.3 × 10-18.

Solution:

FeSO4 ..............................

= Fe2+ + SO42-

Na2S    .............................

= 2Na+ + S2-

Fe2+ + S2- FeS

 

Solubility product of iron sulphide

= 6.3 × 10-18

Let Fe2+  ........................

= [x] and S2- = [x]

[x] [x]      ......................

= 6.3 × 10-18

x2  ...............................

= 6.3 × 10-18

x    ................................

= (6.3 × 10-18)1/2

 

= 2.51 × 10-9

Hence the highest molarity for the solution is 5.02 × 10-9 M.

 





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