Class 11 - Chemistry - Equilibrium

Question 7.1

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.

The volume of the container is suddenly increased.

  • What is the initial effect of the change on vapour pressure?
  • How do rates of evaporation and condensation change initially?
  • What happens when equilibrium is restored finally and what will be final vapour pressure?

Answer.

  • On increasing the volume of the container, initial vapour pressure decrease
  • because same amount of vapours are now distributed over a larger space.
  • Rate of evaporation increases initially because now more space is available.
  • In case of condensation, amount of vapours per unit volume decrease on increasing the volume.
  • Equilibrium will be restored when the rates of the forward and backward processes become equal.
  • Vapour pressure will remain unchanged as it depends on the temperature.

 

 

Question 7.2

What is the 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?

2 SO2 (g) + O2(g) --> 2SO3 (g)

Answer:

 Given equation is

2 SO2 (g) + O2(g) --> 2SO3 (g)

Class_11_Chemistry_Equilibrium_Equation1

Kc = 12.229 M-1 = 12.229 L mol-1.

 

Question 7.3

At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of 1 atoms

I2 (g) --> 2I (g)

Calculate Kp for the equilibrium.

Answer.

Given

Total pressure of equilibrium mixture = 105 Pa

Partial pressure of iodine atoms = (40/100) x (105 Pa) = 0.4 x 105 Pa

Partial pressure of iodine molecules = (60/100) x 105 Pa = 0.6 x 105 Pa

I2 (g) -->2I (g)

Class_11_Chemistry_Equilibrium_Equation2

Kp = 2.67 x 104 Pa

 

Question 7.4

Write the expression for the equilibrium constant, Kc for each of the following reactions:

  • 2NOCl (g) --> 2NO (g) + Cl2 (g)
  • 2Cu(NO3)2 (s) --> 2CuO (s) + 4NO2 (g) + O2 (g)
  • CH3COOC2H5 (aq) + H2O (l) --> CH3COOH (aq) + C2H5OH (aq)
  • Fe3+ (aq) + 3OH- (aq) --> Fe(OH)3 (s)
  • I2 (s) + 5F2 --> 2IF5

Answer.

 Class_11_Chemistry_Equilibrium_Equation3

 

Question 7.5

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

  • 2NOCl (g) --> 2NO (g) + Cl2 (g) ; Kp = 1.8 x 10-2 at 500 K
  • CaCO3 (s) --> CaO (s) + CO2 (g); Kp = 167 at 1073 K

Answer.

Relation between Kp and Kc is

Kp = Kc (RT) ∆n

  • R = 0.0831 bar L mol-1 K-1

Δn = 3 – 2 = 1

T = 500 K

Kp = 1.8 x 10-2

Using above relation,

1.8 x 10-2 = Kc (0.0831 x 500)1

Kc = 4.33 x 10-4

  • Δn = 2 – 1 = 1

T = 1073 K

Kp = 167

Put the values in the equation:

167 = Kc (0.0831 x 1073) Δn

Kc = 1.87

  

Question 7.6

For the following equilibrium, Kc = 6.3 x 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 reactions?

Answer.

For reverse reaction, Kc = 1/K

Kc = (1/6.3 x 1014 )= 1.59 x 10-15

 

Question 7.7

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

Answer.

Pure liquids and solids are ignored while writing the equilibrium constant expression because

molar concentration of pure solids or liquid are independent of the amount present.

Class_11_Chemistry_Equilibrium_Equation4

Molar mass and density of solid and pure liquid are fixed. Molar concentration is constant.

 

 

Question 7.8

Reaction between N2 and O2 takes place as follows:

2N2 (g) + O2 (g) --> 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 x 10-37, determine the composition of equilibrium mixture.

Answer.

Let x mole of N2 takes part in the reaction.

According to the equation given, (x/2) moles of O2 (g) will react to form x moles of N2O.

The molar concentration per litre of different species before and at the equilibrium point is

                                               2N2 (g)        +         O2 (g)                è                2N2O

Initial mole.litre                      0.482/10                0.933/10                                 Zero

Mole/litre at equi point            ( 0.482 – x)/10     (0.933 – x/2)/10                        x/10

 

The value of equilibrium constant is 2 x 10-37 is extremely small. This means only small amount of reactants have reacted.

This can be omitted.

Class_11_Chemistry_Equilibrium_Equation5

X2 = 43.352 x 10-40 = 6.6 x 10-20

X is extremely so it can be neglected.

In the equilibrium mixture, molar concentration of N2 = 0.0482 mol L-1

Molar concentration of O2 = 0.0933 mol L-1

Molar concentration of N2O = (0.1 X x) = (0.1 x 6.6 x 10-20 mol L-1)= 6.6 x 10-21 mol L-1.

 

 

Question 7.9

Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reacton given below:

2NO (g) + Br2 (g) -->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.

Answer.

2NO (g) + Br2 (g) --> 2NOBr (g)

2 mol     1 mol        2 mol

2 mol of NOBr are formed from 2 mol of NO.

0.0518 mol of NOBr are formed from 0.0581 mol of NO.

2 mol of NOBr are formed from 1 mol of Br.

0.0518 mol of NOBr are formed from (0.0581/2) mol of Br or 0.0259 mol of NO.

Initial amount of NO and Br are

[NO] = 0.087 mol

[Br2] = 0.0437 mol

Equilibrium amount of NO are

[NO] = (0.087 – 0.0518) = 0.0352 mol

Equilibrium amount of Br are

[Br2] = (0.0437 – 0.0259) = 0.0178 mol

 

 

Question 7.10

At  450K, Kp = 2 x 1010/ bar for the given reaction at equilibrium.

2SO2 (g) + O2 (g) --> 2SO3 (g)

What is the Kc at this temperature?

Answer.

Class_11_Chemistry_Equilibrium_Equation6

Kp = 2 x 1010 bar-1

R = 0.083 L bar K-1 mol-1

T = 450 K

Δng = 2 – 3 = -1

Putting all the values,

Kc = 7.47 x 1011 mol-1 L

 

Question 7.11

A sample of HI (g) is placed in flask at a pressure of 0.2 atm.

At equilibrium he partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

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

Answer.

pHI = 0.04 atm

pH2 = 0.08 atm

pI2 = 0.08 atm

Class_11_Chemistry_Equilibrium_Equation7

Putting all the values, Kp = 4.0

 

Question 7.12

A mixture of 1.57 mole 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) + 3H3 --> 2NH3 (g) is 1.7 x 102. Is the reaction mixture at equilibrium? If not, what os the direction of the net reaction?

Answer.

The above given reaction, concentration quotient, Qc is

Qc = 2.38 x 103

The equilibrium constant, Kc for the given reaction is 1.7 x 10-2.

Now, Qc not equal to Kc. This means that reaction is not in a state of equilibrium.

 Class_11_Chemistry_Equilibrium_Equation8

Qc = 2.38 x 103

 

Question 7.13

The equilibrium constant expression for a gas reaction is

Kc = [NH3]4 [O2]5/[NO]4 [H2O]6

Write the balanced chemical equation corresponding to this expression.

Answer.

The balanced chemical equation to the given expression can be

 4NO(g) + 6H2O(g) --> 4NH3(g) + 5O2(g)

  

Question 7.14

One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 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.

Answer.

Number of moles of water originally present = 1 mol

Percentage of water reacted = 40%

Number of moles of water reacted = (1) x (40/100) = 0.4 mol

Number of moles of water left = (1 – 0.4) = 0.6 mol

0.4 moles of water will react with 0.4 mol of carbon dioxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.

It can also be represented as

Class_11_Chemistry_Equilibrium_Equation9

Kc = 0.44

 

Question 7.15

At 700 K, equilibrium constant for the reaction:

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

Is 54.8. if 0.5 ml 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?

Answer.

Equilibrium constant for a reaction, Kc = 54.8.

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

[HI] = 0.5 mol L-1 will be (1/54.8).

Let, at equilibrium concentrations of hydrogen and iodide be x mol L-1.

[H2] = [I2] = x mol L-1

Class_11_Chemistry_Equilibrium_Equation10

x = 0.068 mol L-1.

 [H2] = 0.068 mol L-1.

[I2] = 0.068 mol L-1.

 

Question 7.16

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

Answer.

                            2ICl      -->     I2      +      Cl2

Initial conc.        0.78 M              0                0

At equilib.     (0.78 – 2x)             x                x

 

  (x X x)/  (0.78 – 2x)2          = 0.14

 

(x)/((0.78 – 2x)2   = 0.14

 

On solving

1.748 x = 0.748x

1.748 x = 0.292

x = 0.167

Hence at equilibrium,

[Cl2] = [I2] = 0.167M [ICl]

= (0.78 – 2 x 0.167) M

[HI] = 0.446 M

 

 

Question 7.17   

Kp = 0.04 atm at 899 K for the equilibrium shown below. What os the equilibrium concentration of C2H6

when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6 (g) --> C2H4 (g) + H2 (g)

Answer.

C2H(g) C2H4 (g) + H2 (g)

  4 atm                    0              0

(4 – p) atm          p atm       p atm

Class_11_Chemistry_Equilibrium_Equation11

0.04 = (p2/(4-p))

p2 = 0.04 ( 4 – p)

p2 + 0.04p – 0.16 = 0

On solving we get, p = 0.38

Equilibrium pressure or concentration of C2H6 = (4 – 0.38) = 3.62 atm

 

 

Question 7.18

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

CH3COOH (l) + C2H5OH (l) --> CH3COOC2H5 (l) + H2O (l)

  • Write the concentration ratio (reaction quotient), Qc for this reaction (note: water is not in excess and is not a solvent in this reaction)
  • 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.
  • 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 sometime.
  • Has equilibrium been reached?

Answer.

  • Concentration ratio for the reaction is

 Class_11_Chemistry_Equilibrium_Equation12

  • CH3COOH (l) + C2H5OH (l) --> CH3COOC2H5 (l) + H2O (l)

Initial molar conc.                 1 mol               0.18 mol                  0                       0

Molar conc. at equilibrium  (1-0.17)           (0.18 -0.171)          0.171 mol     0.171 mol

                                                    0.829 mol            0.009 mol

Applying the Law of Chemical equilibrium,

Class_11_Chemistry_Equilibrium_Equation13

Kc = (0.171) x (0.171)/(0.829 )  x (0.009) = 3.92

  •        CH3COOH (l) + C2H5OH (l) --> CH3COOC2H5 (l) + H2O (l)

Initial molar conc.                 1 mol               0.5 mol                 0.214                0.214

Molar conc. at equilibrium  (1-0.214)           (0.5 - 0.241)            0.786 mol           0.286 mol                                                                                                                                                             .                                                      

Class_11_Chemistry_Equilibrium_Equation12

=[(0.214 mol) x(0.21 mol)]/[(0.786 mol)x (0.286 mol)]

Qc = 0.204

Now, Qc is less than Kc. Hence, equilibrium has not been reached. Reactants are still being converted into products.

 

 

Question 7.19

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  x 10-1 mol L-1.

If value of Kc is 8.3 x 10-3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5 (g) --> PCl3 (g) + Cl2 (g)

Answer.

Let the initial molar concentration of PCl5 per L = x mol

Molar concentration of PCl5 at equilibrium = 0.05 mol

Moles of PCl5 decomposed = (x – 0.05) mol

Moles of PCl3 formed = (x – 0.05) mol

Moles of Cl2 formed = (x – 0.05) mol

Molar concentration of reactants and products are

                                                             PCl5 (g) --> PCl3 (g) + Cl2 (g)

Initial moles/L                                          x                 0                0

Mole/L at equilibrium point                   0.05          (0-0.05)     (x-0.05)

Equilibrium constant (Kc) = 8.3 x 10-3 = 0.0083

Using Law of Chemical equilibrium,

Class_11_Chemistry_Equilibrium_Equation14

(x – 0.05)2 = 0.083 x 0.05 = 4.15 x 10-4

(x – 0.05) = (4.15 x 10-4)1/2 = 2.037 x 10-2 = 0.02 moles

x = 0.05 + 0.02 = 0.07 mol

Molar concentration per litre of PCl3 at equilibrium = 0.07 – 0.05 = 0.02 mol

Molar concentration per litre of Cl2 at equilibrium = 0.07 – 0.05 = 0.02 mol

 

 

Question 7.20

One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

FeO (s) + CO (g) --> Fe (s) + CO2 (g) ; Kp = 0.265 atm at 1050 K

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?

Answer.

Given reaction is

FeO + CO --> Fe + CO2

Initial concentration of CO = 1.4 atm

Initial concentration of CO2 = 0.80 atm

Qp = pCO2/pCO = 0.80.1.4 = 0.571

Qp > Kp so the reaction will proceed in the backward direction.

Pressure of CO will increase while the pressure of CO2 will decrease.

Let the increase in pressure of CO = decrease in pressure of CO2 be p.

Kp = (pCO2/pCO)

0.265 = 0.80 – (p /1.4) + p

0.371 + 0.265p = 0.80 – p

1.265p = 0.429

P = 0.339 atm

Equilibrium partial of CO2, pCO = 0.80 – 0.339 = 0.461 atm

Equilibrium partial of CO, pCO = 1.4 + 0.339 = 1.739 atm

 

 

Question 7.21

Equilibrium constant, Kc for the reaction

N2 (g) + 3H2 (g) --> 2NH3 (g) at 500 K is 0.061

At a particular time, the analysis show 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?

Answer.

According to the given information,

N2 = [3.0]

H2 = [2.0]

NH3 = [0.50]

Class_11_Chemistry_Equilibrium_Equation15

Qc = 0.0104

Because Qc is not equal to Kc the reaction is not equilibrium.

Also, , the reaction proceeds in the forward direction to reach equilibrium.

Value of Qc is less than that of Kc, so the reaction is not in a state of equilibrium. It will proceed in a forward direction, till Qc = Kc.

 

Question 7.22

Bromine monochloride , BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl (g) --> Br2 (g) + Cl2 (g)

For which Kc = 32 at 500. If initially pure BrCl is present at a concentration of 3.3 x 10-3 mol L-1,

what is its molar concentration in the mixture at equilibrium?

Answer.

Let the number of moles that decompose BrCl in order to attain equilibrium = x

                                                       2BrCl (g) -->  Br2 (g) + Cl2 (g)

Initial Moles/L                              0.0033                  0           0

Moles/L at equilibrium point              0.0033 – x        x/2      x/2

According to the Law of Chemical equilibrium,

 Kc = [(Br2)(Cl2)]/[(BrCl)]2

32 = (x/2) x (x/2) /(0.0033 – x)2

 On taking square root, 5.656 = (x/2) /  (0.0033 – x) = 11.31

          

X = (0.037/12.31) = 0.003

Molar concentration of BrCl at equilibrium point = 0.0033 – 0.003

= 0.0003 mol L-1

= 3 x 10-4 mol L-1

 

 

Question 7.23

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

C (s) + CO2 (g) --> 2CO (g)

Calculate Kc for this reaction at the above temperature.

Answer.

Let the total mass of the gaseous mixture = 100 g

Mass of CO in the mixture = 90.55 g

Mass of CO2 in the mixture = (100 – 90.55) = 9.45 g

Number of moles of CO = 990.55/28) = 3.234 mol

Number of moles of CO2 = (9.45/44) = 0.215 mol

Class_11_Chemistry_Equilibrium_Diagram_1

 

Question 7.24

Calculate a) ΔG and b) the equilibrium constant for the formation of NO2 form NO and O2 at 298 K

NO (g) + ½ O2 (g) --> NO2 (g)

Where

Δf G (NO2) = 52.0 kJ / mol

Δf G(NO) = 87.0 kJ /mol

Δf G (O2) = 0 kJ /mol

Answer.

(a)

Class_11_Chemistry_Equilibrium_Equation17

 

Log Kc = 6.134

Kc = Antilog 6.134 = 1.36 x 106

 

Question 7.25

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?

  • PCl5 (g) --> PCl3 (g) + Cl2 (g)
  • CaO (s) + CO2 (g) --> CaCO3 (s)
  • 3Fe (s) + 4H2O (g) --> Fe3O4 (s) + 4H2 (g)

Answer.

  • Pressure will increase in forward reaction and number of moles of products will increase.
  • According to Le Chatelier Principle, if pressure is decreased equilibrium shifts in the direction in which number of moles of gases is more.
  • Pressure will increase in the backward reaction and number of moles of products will decrease.
  • Change in pressure will have no effect on the equilibrium constant and there will not be any change in the number of moles.

 

 

Question 7.26

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.

  • COCl2 (g) --> CO (g) + Cl2 (g)
  • CH4 (g) + 2S2 (g) -->CS2 (g) + 2H2S (g)
  • CO2 (g) + C (s) --> 2CO (g)
  • 2H2 (g) + CO (g) --> CH3OH (g)
  • CaCO3 (s) --> CaO (s) + CO2 (g)
  • 4NH3 (g) + 5O2 (g) --> 4NO (g) + 6 H2O (g)

Answer.

  • Increase in pressure will not affect equilibrium as np = nr = 3.
  • Increase in pressure will favour backward reaction as np (2) > nr (1).
  • Increase in pressure will favour backward reaction as np (10) > nr (9).
  • Increase in pressure will favour forward reaction as np (1) < nr (2).
  • Increase in pressure will favour backward reaction as np (2) > nr (1).
  • Increase in pressure will favour backward reaction as np (1) > nr (0).

 

 

Question 7.27

The equilibrium constant for the following reaction is 1.6 x 105 at 1024 K

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.

Answer.

Calculate Kp

Class_11_Chemistry_Equilibrium_Equation18

Δn = 2 – 2 = zero.

Calculate the partial pressure of gases.

                        H2 (g) + Br2 (g) è 2HBr (g)

Initial pressure      10            zero          zero

Equi. Pressure      (10-P)       (P/2) bar     (P/2) bar

Class_11_Chemistry_Equilibrium_Equation19

= 4 x 102

(20 – 2P) = 4 x 102 P

P(4 x 102 + 2) = 20

P = (20)/(400 + 2) = (20/402) = 0.50 bar

pH2 = 0.025 bar

pBr2 = 0.025 bar

pHBr = 10 – 0.05 = 9.95 bar

 

Question 7.28

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH4 (g) + H2O (g) --> CO (g) + 3H2 (g)

  • Write an expression for Kp for the above reaction.
  • How will the values of Kp and composition of equilibrium mixture be affected by
  • Increasing the pressure
  • Increasing the temperature
  • Using a catalyst

Answer.

Class_11_Chemistry_Equilibrium_Equation20

  • On increasing the pressure, number of moles per unit volume will increase.
  • In order to decrease this, pressure gets shifted in the backward direction. As more reactants will be formed, value of Kp will decrease.
  • With the increase in temperature, forward reaction will be favoured. Equilibrium gets shifted to the right and value of Kp will increase.
  • On adding catalyst, there will not be any change in the equilibrium.
  • It influences both forward and backward reactions to the same extent. Equilibrium will be attained more quickly.

 

 

Question 7.29

Describe the effect of

  • Addition of H2
  • Addition of CH3OH
  • Removal of CO
  • Removal of CH3OH

On the equilibrium of the reaction.

2H2 (g) + CO (g) --> CH3OH (g)

Answer.

  • Equilibrium will shift in the forward direction.
  • Equilibrium will shift in the backward direction.
  • Equilibrium will shift in the backward direction.
  • Equilibrium will shift in the forward direction.

 

 

Question 7.30

At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride. PCl5 is 8.3 x 10-3. If decomposition is depicted as

PCl5 (g) + --> PCl3 (g) + Cl2 (g) Δr H = 124.0 kJ mol-1

  1. Write an expression for Kc for the reaction.
  2. What is the value of Kc for the reverse reaction at the same temperature?
  3. What would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased?

Answer.

Class_11_Chemistry_Equilibrium_Equation21

 =1.2048 x 102

=120.48

  • On adding PCl5, value of Kc will remain constant because there is no change in temperature.

On increasing temperature, forward reaction will be favoured as it is endothermic in nature. Value of equilibrium constant will increase.

 

 

Question 7.31

Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam.

The first stage of two stage 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 + H2 (g)

If the reaction vessel at 400 o C is charged with an equimolar mixture of CO and steam such that pCO = pH2O = 4.0 bar,

what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400o C.

Answer.

Let the partial pressure of H2 at equilibrium is p bar.

                                         CO (g) + H2O (g) --> CO2 + H2 (g)

Initial pressure             4 bar          4 bar            0          0

Final pressure              (4-p)             (4-p)            p           p

 

 

(PCO2 x PH2)/ [(PCO x PH2O)] = Kp

​⇒ (p x p)/ [(4.0−p) (4.0−p)] = 10.1

⇒ (p)/ (4.0-p) = 3.178

=> p= 12.712 – 3.17 p

4.17p = 12.712

0r p= (12.712)/ (4.178)

=3.04

So, partial pressure of H2 is 3.04 bar at equilibrium.

 

Question 7.32

Predict which of the following reaction will have appreciable concentration of reactants and products:

  1. Cl2 (g) --> 2Cl (g) Kc = 5 x 10-39
  2. Cl2 (g) + 2 NO (g) --> 2NOCl (g) Kc = 3.7 x 108
  3. Cl2 (g) + 2NO2 (g) --> 2NO2Cl (g) Kc = 1.8

Answer.

  • Value of Kc is very small, which means molar concentration of products is very small as compared to the reactants.
  • Value of Kc is quite large which means molar concentration of products is very large as compared to that of the reactants.
  • Value of kc is 1.8, which means both reactant and products have appreciable concentration.

  

Question 7.33

The value of Kc for the reaction 3O2 (g) --> 2O3 (g) is 2.0 x 10-50 at 25o C.

If the equilibrium concentration of O2 in air at 25o C is 1.6 x 10-2, what is the concentration of O3?

Answer.

3O-2 (g) --> 2O3 (g)

Class_11_Chemistry_Equilibrium_Equation23

[O3]2 = (2.0 x 10-50 x (1.6 x 10-2)3

[O3]2 = 8.192 x 10-56

[O3] = 2.86 x 10-28 M

 

 

Question 7.34

The reaction CO (g) + 3H2 (g) --> CH4 (g) + H2O (g)

Is at equilibrium at 1300 K in a 1 L flask. It also contain 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.

Answer.

Using the given data,

Class_11_Chemistry_Equilibrium_Equation24

[CH4] = 3.9 x 0.30 x (0.01/0.02) = 5.85 x 10-2 M

 

 

Question 7.35

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

HNO2, CN-, HClO4, F-, OH-, CO32- and S2-

Answer.

Conjugate acid – HCN, H2O, HCO3-, HS-

Conjugate base – NO2-, ClO4-, O2-

 

Question 7.36

Which of the following are Lewis acids? H2O, BF3, H+ and NH4+.

Answer.

BF3, H+ ions are Lewis acids.

 

 

Question 7.37

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

Answer.

Conjugate bases are F-, HSO4-, CO3- respectively.

 

 

Question 7.38

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

Answer.

NH3, NH4+, HCOOH respectively.

  

Question 7.39

The species: H2O, HCO3-, HSO4- and NH3 can act both as Bronsted acids and bases.

For each case give the corresponding conjugate acid and base.

Answer.     

Conjugate acid          Species        Conjugate base

H2O+                          H2O             OH-

H2CO3                        HCO3-         CO32-

H2SO4                        HSO4-          SO42-

NH4+                           NH3                   NH2-

 

 

Question 7.40

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

  • OH-
  • F-
  • H+
  • BCl3

Answer.

  • OH- ions can donate an electron pair and act as Lewis base.
  • F- ions can donate an electron pair and act as Lewis base.
  • H is a Lewis acid since it can accept a paiir of electrons.

  • BCl3 can accept an electron pair since B-atom is electron deficient (it is Lewis acid).

 

 

Question 7.41

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

Answer.

pH = -log [H+]

pH = - log (3.8 x 10-3)

pH = 3 – 0.5798 = 2.4202 = 2.42

 

 

Question 7.42

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

Answer.

pH = - log[H+]

Log [H+] = -pH = -3.76 = 4.24

[H+] = Antilog 4.24 = 1.738 x 10-4 = 1.74 x 10-4 M

 

 

Question 7.43

The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10-4, 1.8 x 10-4 and 4.8 x 10-9 respectively.

Calculate the ionization constant of the corresponding conjugate base.

Answer.

For F-,

Kb = Kw/Ka

= 10-14/(6.8 x 10-4)

= 1.47 x 10-11

= 1.5 x 10-11

 

For HCOO-,

Kb = 10-14/(1.8 x 10-4) = 5.6 x 1011

 

For CN-,

Kb = 10-14/ (4.8 x 10-9) = 2.08 x 10-6

 

 

Question 7.44

The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol?

What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Answer.

   

The balanced chemical equation for the ionization of phenol is:

PhOH +H2O ⇌PhO- + H3O+

The initial concentrations of phenol, phenoxide ion and hydronium ion are 0.05, 0 and 0 respectively.

Corresponding equilibrium concentrations are 0.05−x, x and x respectively.

Ka = [PhO-] [H3O+]/ [PhOH]

(1 x 10-10) = (x X x)/ (0.05)

x is very small (as the ionization constant is very small) . Hence, x can be ignored in the denominator.

1×10-10 = (x X x)/ (0.05)

x=2.2×10-6 M = [H3O+] = [PhO-]

Let α be the degree of dissociation of phenol in presence of 0.01 M phenoxide.

The dissociation of phenoxide is as represented:

PhONa → PhO- + Na+

Also,

PhOH+H2O⇌PhO- + H3O+

[PhOH]=0.05−0.05α≃0.05 M (as the degree of dissociation is very small.)

[PhO-] = 0.01+0.05α≃0.01 M

[H3O+] = 0.05α

Ka = [PhO-] [H3O+]/ [PhOH]

1×10-10 = (0.01 x 0.05α)/ (0.05)

α = 1 x 10-8

 

 

Question 7.45

The first ionization constant of H2S is 9.1 x 10-8, calculate the concentration of HS- ion in its 0.1 M solution.

How will this concentration be affected if the solution is 0.1 M in HCl also?

If the second dissociation constant of H2S is 1.2 x 10-13, calculate the concentration of S2- under both conditions.

Answer.

For HS-

                                  H2S        -->        H+     +     HS-

Initial                             0.1 M

After dissociation          (0.1 – x)                  x            x

Ka = x X (x/0.1) = 9.1 x 10-8

x = 9.54 x 10-5

In the presence if 0.1 M HCl, let H2S dissociated is y.

So, at equilibrium [H2S] = 0.1 – y ≈ 0.1

[H+] = 0.1 + y ≈ 0.1

[HS-] = y M

Ka = (0.1 x y) /(0.1) =9.1 x 10-8 M

 Ka =([H+]2[S2-])/([H2S])

For [S2-]

Ka = Ka1x Ka2 = 9.1 x 10-8 x 1.2 x 10-13 = 1.092 x 10-20

In the absence of 0.1 M HCl

[H+] = 2[S2-]

If [S2-] = x, [H+] = 2x

(2x)2 x/(0.1) = 1.093 x 1020

4x3 = 1.092 x 10-21 = 273 x 10-24

3 log x = log 273 – 24 = 2.4362 – 24

Log x = 0.8127 – 8 = 8.8127

X = antilog 8.8127 = 273 x 10-24

x = 6.5 x 10-8 M

 

In the presence of 0.1 M HCl, let [S2-] = y

[H2S] = 0.1 – y ≈ 0.1 M

[H+] = 0.1 + y ≈ 0.1 M

Ka = (0.1 )2 x (y/0.1) = 1.09 x 10-10 M

  

Question 7.46

The ionization constant of acetic acid is 1.74 x 10-5 , calculate the degree of dissociation of actic acid in its 0.05 M solution.

Calculate the concentration of S2- under both conditions.

Answer.

Class_11_Chemistry_Equilibrium_Equation25

[H+] = 9.33 x 10-4 M

[CH3COO-] = [H+] = 9.33 x 10-4 M

pH = -log (9.33 x 10-4) = (4 – 0.9699) =(4 – 0.97) = 3.03

 

 

Question 7.47

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 ionization constant of the acid and its pKa.

Answer.

Class_11_Chemistry_Equilibrium_Equation26

[H+] =7.08 x 10-5 M

[A-] = [H+] = 7.08 x 10-5M

Ka = [H+] [A-]/[HA]

Putting all the values

Ka = 5.0 x 10-7

pKa = -log Ka = - log(5 x 10-7) = 7 – 0.699 = 6.301

 

 

Question 7.48

Assuming complete dissociation, calculate the pH of the following solutions.

  • 0.003 M HCl
  • 0.005 M NaOH
  • 0.002 M HBr
  • 0.002 M KOH

Answer.

  • HCl + H2O H+ + Cl-

[H+] = [HCl] = 3 x 10-3 M

pH = -log (3 x 10-3) = 2.52

  • NaOH(aq) Na+ + OH-

[OH-] = NaOH

[OH-] = 5 x 10-3 M

pOH = -log[OH-] = -log(0.005]

pOH = 2.30

Therefore, pH = 14 - 2.30

=11.70

  • HBr + H2O H3O+ + Br-

[H3O+]  = [HBr]

[H3O+] = 2 x 10-3 M

pH = -log ( 2 x 10-3) = 2.69

  • KOH(aq) K+ + OH-

[OH-] = [KOH]

[OH-] = 2 x 10-3 M

pOH = -log[OH-]

=2.69

Therefore, pH = 14 - 2.69

=11.31

 

Question 7.49

Calculate  the pH of the following solutions:

  1. 2 g of TlOH dissolved in water to gove 2 litre of solution
  2. 3 g of Ca(OH)2 dissolved in water to give 500 mL of solution
  3. 3 g of NaOH dissolved in water to give 200 mL of solution
  4. 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution

Answer.

  • Molar conc of TlOH = [2g /(204 + 16 + 1)] 1/2L = 4.52 x 10-3 M

[OH-] = [TlOH] = 4.52 x 10-3 M

[H+] = 10-14/(4.52 x 10-3) = 2.21 x 10-12 M

pH = - log (2.21 x 10-12) = 12 – (0.3424) = 11.66

  • Molar conc. of Ca(OH)2 = [0.3/(40 + 34)] x (1/0.5) = 8.11 x 10-3 M

Ca(OH)2 --> Ca2+ + 2OH-

[OH-] = 2[Ca(OH)2] = 2 x (8.11 x 10-3) = 3 – 1.2101 = 1.79

pH = 14 – 1.79 = 12.21

  • Molar conc. of NaOH = (0.3/40) x (1/1.2) = 3.75 x 10-2 M

[OH-] = 3.75 x 10-2 M

pOH = - log (3.75 x 10-2) = (2 – 0.0574) = 1.43

pH = 14 – 1.43 = 12.57

  • M1V1 = M2V2

13.6 M x 1 mL = M2 x 1000

M2 = 1.36 x 10-2 M) 2 – 0.1206

[H+] = [HCl] = 1.36 x 10-2 M

pH = -log (1.36  x 10-2) = 2 – 0.1335 = 1.87

 

 

Question 7.50

The degree of ionization of a 1.0 M bromoacetic acid solution is 0.132.

Calculate the pH of the solution and the pKa of bromoacetic acid.

Answer.

                          CH2(Br) COOH --> CH2 (Br) COO- + H+

Initial conc.                   C                             0                0

Conc. at equil.         C – Ca                        C a              C a

Class_11_Chemistry_Equilibrium_Equation27

Ka = 0.1 x (0.132)2 = 1.74 x 10-3

pKa = - log (1.74 x 10-3) = 3 – 0.2405 = 2.76

[H+] = C a = 0.1 x 0.132 = 1.32 x 10-2 M

pH = -log (1.32 x 10-2) = 2 – 0.1206 = 1.88

 

 

Question 7.51

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

Answer.

Cod + H2O --> Cod H+ + OH-

pH = 9.95

pOH = 14 – 9.95 = 4.05

-log [OH-] = 4.05

Log [OH-] = -4.05 = 5.95

[OH-] = 8.913 x 10-5 M

Kb = [Cod H+] [OH-]/[Cod]

Kb =[OH-]2 /[Cod] = (8.91 x 10-5 )2/(5 x 10-3) = 1.588 x 10-6

pKb = - log(1.588 x 10-6) = 6 – 0.1987 = 5.8

 

 

Question 7.52

What is the pH of the 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.

Answer.

Kb = 4.27 x 10-10

C = 0.001 M

Kb = c a2

4.27 x 10-10 = 0.001 x a2

4270 x 10-10 = a2

4270 x 10-5 = a = 6.53 x 10-4

a = 6.53 x 10-4

[OH]  = ca = 0.001 x 65.34 x 10-5 = 0.065 x 10-5

pOH = - log (0.056 x 10-5)

2.34 x 10-5 is ionization constant = 6.187

pH = 7.813

Ka x Kb = Kw

Ka = 10-14/(4.27 x 10-10) = 2.34 x 10-5

 

 

Question 7.53

Calculate the degree of ionization 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.

Answer.

pKa = 4.74

-log Ka = 4.74

Ka = antilog (-4.74) = 1.820 x 10-5

Ca2 = 1.82 x 10-5

 

a2 = 1.82 x 10-5/C

a2 = 3.74 x 10-4

a = 1.9 x 10-2

                                   CH3COOH --> CH3 COO- + H+

Initial molar conc.               C                   0              0

Equim. Conc.                   C (1 – a)          Ca           Ca

Class_11_Chemistry_Equilibrium_Equation28

[H+] = 0.01 M

1.82 x 10-5 = Ca/C x 100

a = 1.82 x 10-3

[H+] = 0.1 M

1.82 x 10-5 = (Ca/C) x 100

a = 1.82 x 10-4

 

 

Question 7.54

The ionization constant of dimethylamine is 5.4 x 10-4. Calculate its  degree of ionization in its 0.02 M solution.

What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer.

Ka= 5.4 x 10-4

C = 0.02 M

Ka = Ca2

a2 = (Ka/C) = 5.4 x 10-4 /(0.02)

a2 = 0.164

(CH3)2NH + H2O (CH3)2 NH2+  + OH-

                         

Initial Conc.                C                          0                 0

At equil. Conc.       C(1 – a)                    Ca              Ca

But [OH-] = 0.1 M

Ka = Ca x [OH-]

5.4 x 10-4 = Ca x 0.1/C(1 – a)

% of ionization = 100 a = 100 x 0.0054 = 0.54

 

 

Question 7.55

Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

  • Human muscle fluid 6.83
  • Human stomach fluid 1.2
  • Human blood 7.38
  • Human saliva 6.4

Answer.

  • pH = 6.83

[H+] = antilog (-6.83) = 1.48 x 10-7 M

  • pH = 1.2

[H+] = antilog (-1.2) = 6.30 x 10-2 M

  • pH = 7.38

[H+] = antilog (-7.38) = 4.17 x 10-8 M

  • pH = 6.4

[H+] = antilog (-6.4) = 3.98 x 10-7 M

 

 

Question 7.56

pH of milk, black coffee, tomato juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively.

Calculate corresponding hydrogen ion concentration in each.

Answer.

Milk, pH = 6.8

[H+] = antilog (-pH) = antilog (-6.8) = 1.58 x 10-7 M

Black coffee, pH = 5.0

[H+] = antilog (-pH) = antilog (-5.0) = 1 x 10-5 M

Tomato juice, pH = 4.2

[H+] = antilog(-pH) = antilog (-4.2) = 6.31 x 10-5 M

Lemon juice, pH = 2.2

[H+] = antilog (-pH) = antilog (-2.2) = 6.31 x 10-3 M

Egg white, pH = 7.8

[H+] = antilog (-pH) = antilog (-7.8) = 1.58 x 108 M

 

 

Question 7.57

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.

Calculate the concentrations of potassium , hydrogen and hydroxyl ions. What is its pH?

Answer.

Molar mass of KOH = 56 g mol-1

Number of moles of KOH = (0.561/56) = 0.01 mol

Concentration of KOH = (0.01/0.2) = 0.05 mol/L

KOH --> K+ + OH-

[K+] = [OH-] = 0.05 M

[H+] = Kw /[OH-] = 1.0 x 10-14/0.05 = 2 x 10-13

pH = - log(2.0 x 10-13) = 12.70

 

 

Question 7.58

The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution.

Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH.

What will be its degree of ionization if the solution is 0.01 M in HCl also?

Answer.

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

Solubility of Sr(OH)2 in mole L-1 = (19.23/121.6) = 0.1581 M

If complete dissociation takes places,

Sr(OH)2  --> Sr2+ + 2 OH-

[Sr2+] = 0.1581 M

[OH-] = 2 x 0.1581 = 0.3162 M

pOH = - log (0.3162) = 0.5

pH = 14 – 0.5 = 13.5

 

 

Question 7.59

The ionization constant of propanoic acid is 1.32 x 10-5.

Calculate the degree of ionization of the acidin its 0.05 M solution and also its pH.

What will be its degree of ionization if the solution is 0.01 M in HCl also?

Answer.

Ka = 1.32 x 10-5

C = 0.05 M

Class_11_Chemistry_Equilibrium_Equation29

a = 1.62 x 10-2

[H+] = Ca = 0.05 x 1.62 x 10-2 = 8.1 x 10-4

pH = -log(8.1 x 10-4)

pH = 3.09

In 0.01 M HCl, [H+] = 0.01 M

                                    C2H5COOH --> C2H5COO- + H+

Initial molar conc.               C                    0               0

Equi. Molar conc.          C(1 – a)              Ca             Ca

Using law of Chemical equation,

Class_11_Chemistry_Equilibrium_Equation30

1.32 x 10-5 = 0.01 a

a = 1.32 x 10-3

 

 

Question 7.60

The pH 0.1 M solution of cyanic acid (HCNO) is 2.34.

Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer.

HCNO -->H+ + CNO-

pH = 2.34

log [H+ ] = -2.34

[H+] = antilog (-2.34) = 4.57 x 10-3 M

[CNO-] = [H+] = 4.57 x 10-3 M

Class_11_Chemistry_Equilibrium_Equation31

a = 0.0457

 

Ka =cα2

=0.1×(45×10−3)2

=202.5×10−6

=2.02×10−4

 

Question 7.61

The ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer.

Class_11_Chemistry_Equilibrium_Equation32

Thus, the degree of hydrolysis is

= (y/0.04) = (0.093×10−5)/ (0.04)

=2.325×10−5

 

 

Question 7.62

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

Answer.

Pyrimidine hydrochloride is a salt of weak base and strong acid.

pH = (7)- (1/2) [log C + pKb)]

Kb = dissociation constant of pyridine

3.44 = 7 – ½ (log 0.02 + pKb)

-3.56 = -1/2 (-170 + pKb)

-7-12 = 1.70 - pKb

pKb = 1.70 + 7.12 = 8.82

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

 

 

Question 7.63

Predict if the solutions of the following salts ar4e neutral, acidic or basic: NaCl, KBr, NaCN, NH4 NO3, NaNO2 and KF.

Answer.

NaCN, NaNO3, KF solutions are basic as they are the salts of strong base and weak acid.

NaCl, KBr solutions are neutral as they are salts of strong acid and strong base.

NH4NO3 solution is acidic as it is a salt of strong acid and weak base.

 

 

Question 7.64

The ionization constant of chloroacetic acid is 1.35 x 10-3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Answer.

pH of an acid solution

[H+] =√ (Ka x C)

Ka = 1.35 x 10-3

C = 0.1 M

[H+] =√ (1.35 x 10-3 x 0.1)

=√ (1.35 x 10-4)

[H+] = 1.16 x 10-2

pH = - log [H+] = - log (1.16 x 10-2) = 2 – 0.064 = 1.936

pH of 0.1 M sodium salt solution.

Sodium salt of chloroacetic acid is alt of weak acid and strong base.

Class_11_Chemistry_Equilibrium_Equation33

pKa = - log Ka = - log (1.35 x 10-3) = (-0.1303 – 3) = 2.8697

pKw = - log 10-14 = 14

C = 0.1 M

logC = log (0.1) = -1

pH = ½ [14 + 2.8697 – 1] = 7.935

 

 

Question 7.65

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

Answer.

We know that

[H+] = =√ (Kw) = =√ (2.7 x 10-14)

=1.643 x 10-7 M

pH = - log [H+] = -log (1.643 x 10-7)

= 7 – 0.2156 = 6.78

 

Question 7.66

Calculate the pH of the resultant mixtures:

  1. 10 mL of 0.2 M Ca(OH)2 + 25 mL of 0.1 M HCl
  2. 10 mL of 0.01 M H2SO4 + 10 mL of 0.01 M Ca(OH)2
  3. 10 mL of 0.1 M H2SO4 + 10 mL of 0.1 M KOH

Answer.

  • 10 mL of 0.2 M Ca(OH)2

= 10 x 0.2 millimoles = 2 millimoles of Ca(OH)2

25 mL of 0.1 M HCl = 2.5 x 0.1 millimoles = 2.5 millimoles of HCl

 

1 millimoles of Ca(OH)2  reacts with 2 millimoles of HCl

Therefore, 2.5 millimoles of HCl reacts with 1.25 millimoles of Ca(OH)2

 

Remaining Ca(OH)2 = 2 – 1.25 = 0.75 millimoles

(HCl is the limiting reactant)

Total volume of solution = 10 + 25 mL = 35 mL

 

Molarity of Ca(OH)2  in the mixture solution = 0.75/35 M = 0.0214 M

[OH-] = 2 x 0.0214 M = 0.0428 M = 4.28 x 10-2 M

 

pOH = - log(4.28 x 10-2) = 2 – 0.6314 = 1.3686 = 1.37 (approx)

pH comes out to be 14 – 1.37 = 12.63

 

  • 10 mL of 0.01 M H2SO4

= 10 x 0.01 millimole = 0.1 millimole

10 mL of 0.01 M Ca(OH)2 = 10 x 0.01 millimole = 0.1 millimole

 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 is neutral.

(c) 10 mL of 0.1 M H2SO4 = 1  millimole

10 mL of 0.1 M KOH = 1 millimole

 

1 millimole of KOH will react with 0.5 millimole of H2SO4.

Remaining H2SO4 = 1 – 0.5 = 0.5 millimole

Volume of reaction mixture = 10 + 10 = 20 mL

Molarity of H2SO4 = 0.5/20 = 2.5 x 10-2 M

[H+] = 2 x 2.5 x 10-2 = 5 x 10-2

pH = - log(5 x 10-2) = 2 – 0.699 = 1.3

 

 

Question 7.67

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride

and mercurous iodide at 298 K from their solubility product constants given in Table 7.9.

Determine also the molarities of individual ions.

Answer.

Silver chromate, Ag2CrO4

Ag2 CrO4 --> 2Ag+ (aq) + CrO42- (aq)

Ksp = 4S3

Ksp = 1.1 x 10-12

S = 6.50 x 10-5 M

[Ag+] = 2 x 6.50 x 10-5 = 1.30 x 10-4 M

[CrO42-] = 6.50 x 10-5 M

 

Barium chromate, BaCrO4

BaCrO4 (s) --> Ba2+ (aq) + CrO42- (aq)

Ksp = S2 = 1.2 x 10-10

S = 1.09 x 10-10 M

[Ba2+] = [CrO42-] = 1.09 x 10-10 M

 

Ferric hydroxide, Fe(OH)3

Fe(OH)3 --> Fe3+ + 3OH-

Ksp = 27S4 = 1 x 10-38

S = 1.38 x 10-10 M

[Fe3+] = 1.38 x 10-10 M

[OH-] = 3 x 1.38 x 10-10 = 4.16 x 10-10 M

 

Lead chloride, PbCl2

PbCl2 --> Pb2+ + 2Cl-

Ksp = 4S3 =1.6 x 10-5

S = 1.59 x 10-2 M

[Pb2+] = 1.59 x 10-2 M

[Cl-] = 2 x 1.59 x 10-2 = 3.18 x 10-2 M

 

Mercurous iodide, Hg2I2

Hg2I2 --> Hg2+ + 2I-

Ksp = 4S3 = 4.5 x 10-29

S = 2.24 x 10-10 M

[Hg2+] = 2.24 x 10-10 M

[I-] = 2 x 2.24 x 10-10 = 4.48 x 10-10 M

 

 

 

Question 7.68

The solubility product constant of Ag2CrO4 and AgBr are 1.1 x 10-12 and 5.0 x 10-13 respectively.

Calculate the ratio of the molarities of their saturated solutions.

Answer.

Ksp (Ag2 CrO4) > Ksp (AgBr)

[Which means Ag2CrO4 is more soluble]

Ag2 CrO4 2 Ag + (aq) + CrO2-4(aq)

S = (1.1 x 10-12)(1/3)/(4)

=6.50 x 10-5 M

Ksp = 4S3

 

AgBr(s) --> Ag+ (aq)  + CrO42- (aq)

Ksp= s’2

S’ = (5 x 10-13)1/2

S’ = 7.07 x 10-7 M

Ratio of molarities = S/S’ = 91.94

 

 

Question 7.69

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together.

Will it lead to precipitation of copper iodate? (For cupric iodate ksp = 7.4 x 10-8)

Answer.

On mixing  equal volumes of sodium iodate and copper chlorate, molar concentrations of both the solutes

would be reduced to half.

NaIO3 --> Na2+ + IO3-

0.001 M           0.001 M

Cu(ClO3)2 --> Cu2+ + 2ClO3-

0.001 M                0.001 M

After mixing, [IO3-] =  [NaIO3] = 0.001 M

[Cu2+] = [Cu(IO3)2] = 0.001 M

Solubility equilibrium for copper iodate will be written as

Cu(IO3)2 (s) --> Cu2+ (aq) + 2IO3- (aq)

Ionic product of copper iodate

[Cu2+][IO3]2 = (0.001)(0.001)2

=1 x 10-9

Ionic product (1 x 10-9) is less than Ksp (7.4 x 10-8), so no precipitation will take place.

  

Question 7.70

The ionization constant of benzoic acid is 6.46 x 10-5 and Ksp for silver benzoate is 2.5 x 10-13.

How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer.

Let us consider S is the molar solubility of silver benzoate in water

Class_11_Chemistry_Equilibrium_Equation35

 

 

Silver benzoate is 3.32 times more soluble in buffer of pH 3.19 than in pure water.

 

Question 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 x 10-18)

Answer.

Let concentration of FeSO4 and Na2S is x mol L-1.

Mixing equal volumes of in both the solutions,

Class_11_Chemistry_Equilibrium_Equation34

FeS --> Fe2+ + S2-

Ksp = [Fe2+] [S2-]

Putting the values,

6.3 x 10-18 = (x/2) x (x/2)

X2 = 4 x 6.3 x 10-18 = 25.2 x 10-18

x = 5.02 x 10-19 mol L-1

Maximum concentration of FeSO4 and Na2S which will not precipitate iron sulphide is 5.02 x 10-9 mol L-1.

 

 

Question 7.72

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10-6)

Answer.

Reaction comes out to be

CaSO4 (s) --> Ca2+ (aq) + SO42-

Let S is the solubility of CaSO4 in mol L-1

Ksp = [Ca2+] [SO42+] = S2-

S = √ Ksp  = √ 9.1 x 10-6 = 3.02 x 10-3 mol L-1

s = 3.02 x 10-3 x 136 g L-1

s = 0.411 g L-1

Molar mass of CaSO4 = 136 g mol-1

1 L of water is required for dissolving 0.411 g.

For dissolving 1 g = water required = (1/0.411) L = 2.43 L

 

 

Question 7.73

The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 x  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 these solutions precipitation will take place?

Answer.

Precipitation takes place when ionic product is greater than solubility product.

As 10 mL of solution contain S2- is mixed with 5 mL of metal salt solution, so after mixing

[S2-] = 1 x 10-19 x (10/15) = 6.67 x 10-20

[Fe2+] = [Mn2+] = [Zn2+] = [Cd2+] = (5/15) x 0.04

= 1.33 x 10-2 M

So, the ionic product of [M2+] [S2-] = 1.33 x 10-2 x 6.67 x 10-20 = 8.87 x 10-22

Ksp of ZnS = 1.6 x 10-24

Ksp of CdS = 8 x 10-27

In these, ionic product exceeds solubility product. Precipitation will take place.

Share this with your friends  

Download PDF


You can check our 5-step learning process


.