Equilibrium

Class 11 > Chemistry > NCERT-I > Equilibrium

Equilibrium

Book Solutions

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?

Answer

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Exercise

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)↔2SO3(g)

Answer

As per the question,

2SO2(g)+O2(g)↔2SO3(g) (Given)

Kc=[SO3]2[SO2]2[O2]=(1.9)2M2(0.6)2(0.82)M3=12.229M−1(approximately)

Hence, K for the equilibrium is 12.229 M–1.

Exercise

3

At a definite temperature and a total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms. 

I2(g)↔2I(g)  

Find Kp for the equilibrium.

Answer

Exercise

4

For the given reaction, find expression for the equilibrium constant

(i)2NOCl (g) ↔ 2NO (g) + Cl2 (g)

(ii)2Cu(NO3)2 (s) ↔ 2CuO (s) + 4NO2 (g) +O2(g)

(iii)CH3COOC2H5 (aq) +H2O (l) ↔ CH3COOH (aq) +C2H5OH (aq)

(iv)Fe3+ (aq) + 3OH− (aq) ↔ Fe(OH)3 (s)

(v)I2 (s) + 5F2  (g) ↔ 2IF5 (s)

Answer

Exercise

5

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

(i)2NOCl (g) ↔ 2NO (g) + Cl2 (g); Kp = 1.8×10−2 at 500K

(ii)CaCO3 (s) ↔ CaO (s) + CO2 (g); Kp = 167 at 1073K

Answer


Exercise

6

For the following equilibrium, Kc = 6.3×1014 at 1000K

NO (g) + O3 (g) ↔ NO2 (g) +O2 (g)

Both the reverse and forward reactions in the equilibrium are elementary bimolecular reactions. Calculate Kc' for the reverse reaction?

Answer

For the reverse reaction

Kc'=1Kc=16.3×1014=1.59×10−15

Exercise

7

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

Answer

This is because molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration= NumberofmolesVolumeMass/molecularmassVolume=MassVolume×Molecularmass=DensityMolecularmass

Though density of solid and pure liquid is fixed and molar mass is also fixed .

∴Molar concentration are constatnt.

Exercise

8

When oxygen and nitrogen react with each other, then the following reaction takes place: 

2N2 (g) +O2 ↔ 2N2O (g)

If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen 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 solution.

Answer

Let the concentration of N2O at equilibrium be x.

The given reaction is:

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

Initial conc.     0.482 mol                     0.933 mol                                  0

At equilibrium(0.482-x)mol            (1.933-x)mol                              x mol

[N2]=0.482−x10⋅[O2]=0.933−x210,[N2O]=x10

The value of equilibrium constant  is extremely small. This means that only small amounts . Then,

[N2]=0.48210=0.0482molL−1and[O2]=0.93310=0.0933molL−1



Exercise

9

Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction is 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 a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Answer

The given reaction is:

2NO(g) +   Br2(g)     ⇒         2NOBr(g)

2mol          1mol                    2mol

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

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

Therefore, 0.0518 mol of NOBr are formed from 0.05182 mol or Br, or 0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259 = 0.0178 mol

Exercise

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?

Answer

Exercise

11

A sample of HI(g) is placed in 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?

2HI (g) ⇒ H2 (g) + I2 (g)

Answer

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.

Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

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

Initial conc.               0.2 atm                                     0                     0

At equilibrium           0.4 atm                                     0.16/2             0.16/2

                                                                                = 0.08atm         = 0.08atm

Therefore,


Exercise

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 1.7 ×102

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer

Exercise

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 corresponding to the given expression can be written as:

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

Exercise

14

One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 60% 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

Exercise

15

At 700 K, equilibrium constant for the reaction H2 (g) + I2 (g) ⇒ 2HI (g) is 54.8. If 0.5 molL–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


Exercise

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 ICl(g) ⇌  I2(g) + Cl2(g) ; KC = 0.14

Answer

Exercise

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 to equilibrium?

C2H6(g) ⇒ C2H4 (g) + H2 (g)

Answer

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,

Exercise

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)

(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 sometime. Has equilibrium been reached?

Answer


Exercise

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 L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5 (g) ⇒ PCl3 (g)+ Cl2 (g)

Answer

Exercise

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 give 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 partial pressures are: pCO = 1.4 atm and pCO2 = 0.80 atm?

Answer


Exercise

21

A reaction is given:

N2 (g) + 3H2 (g) ⇒ NH3 (g)

For the above equation, Equilibrium constant = 0.061 at 500 K

At a specific time, from the analysis we can conclude that composition of the reaction mixture is, 2.0 mol L–1 H2 , 3.0 mol L –1 N2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.

Answer

Exercise

22

Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇒ Br2(g)   +   Cl2(g)

For which Kc= 32 at 600 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 molL–1, what is its molar concentration in the mixture at equilibrium?

Answer


Exercise

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 be 100 g.

Mass of CO = 90.55 g

And, mass of CO2 = (100 – 90.55) = 9.45 g



Exercise

24

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

NO(g)+12O2(g)↔NO2(g), Where;

∆fG° (NO2) = 52.0 kJ/mol

 ∆fG° (NO) = 87.0 kJ/mol

 ∆fG° (O2) = 0 kJ/mol

Answer

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 + 0}

= – 35.0 kJ mol–1

(b) We know that,

∆G° = RT log Kc

∆G° = 2.303 RT log Kc

Kc=−35.0×10−3−2.303×8.314×298=6.134∴Kc=antilog(6.134)=1.36×106

Therefore, the equilibrium constant for the given reaction Kc is 1.36×106

Exercise

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?

(I) PCl5(g) ⇒ PCl3 +Cl2 (g)

(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)

(III) 3Fe (s) + 4H2O (g) ⇒ Fe3O4 (s) +4H2 (g)

Answer

(I) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(II) The number of moles of reaction products will decrease.

(III) The number of moles of reaction products remains the same.  

Exercise

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) ⇒ 2CO (g)

(IV)2H2 (g) +CO  (g) ⇒ CH3OH(g)

(V)CaCO3 (s) ⇒ Cao (s) + CO2 (g)

(VI)4NH3 (g) +5O2 (g) ⇒ 4NO (g) + 6H2O (g)

Answer

When pressure is increased:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.

Since, the number of moles of gaseous reactants is more than that of gaseous products; the reaction given in (iv) will proceed in the forward direction.

Since, the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction.


Exercise

27

The equilibrium constant for the following reaction is 1.6 ×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


Exercise

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)

(I)Write as expression for Kp for the above reaction.

(II) How will the values of Kp and composition of equilibrium mixture be affected by

 (i)Increasing the pressure

(ii)Increasing the temperature

(iii)Using a catalyst?

Answer

(I)For the given reaction,

Kp=pCO×p3H2pCH4×pH2O

(II) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Exercise

29

Describe the effect of:

I) Removal of CO

II) Addition of H2

III) Removal of CH3OH on the equilibrium of the reaction:

IV) Addition of CH3OH

2H2 (g)+CO (g) ⇒ CH3OH (g) 

Answer

(I) On removing CO, the equilibrium will shift in the backward direction.

(II) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(III) On removing CH3OH, the equilibrium will shift in the forward direction.

(IV) On addition of CH3OH, the equilibrium will shift in the backward direction.


Exercise

30

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

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

∆rH° = 124.0 kJmol–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) The temperature is increased?

Answer

(a) Kc=[PCl3(g)][Cl2(g)][PCl3(g)]

(b) Value of Kc for the reverse reaction at the same temperature is:

K‘c=1Kc=18.3×10−3=1.2048×102=120.48

(c) (i) Kc would remain the same because in this case, the temperature remains the same.

(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change. (iii) In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.

Exercise

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(g)+H2(g)

If a reaction vessel at 400°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 400°C

Answer

Exercise

32

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

(a)Cl2(g)↔2Cl(g);Kc=5×10−39

(b)Cl2(g)+2NO(g)↔2NOCl(g);Kc=3.7×108

(c)Cl2(g)+2NO2(g)↔2NO2Cl(g);Kc=1.8

Answer

If the value of Kc lies between 10–3 and 103 , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.
Exercise

33

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

Answer

Exercise

34

The reaction, CO (g) + 3H2 (g) → 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 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

Let the concentration of methane at equilibrium be x.


Exercise

35

What is conjugate acid-base pair? Find the conjugate acid/base of the given species:

(i) HNO2

(ii) CN−

(iii) HClO4

(iv) F−

(v) OH−

(vi) CO32−

(vii) S−

Answer

Exercise

36

From the compounds given below which are Lewis acids?

(i) H2O

(ii) BF3

(iii) H+

(iv) NH+4

Answer

Lewis acids are the acids which can accept a pair of electrons.

(i) H2O – Not Lewis acid

(ii) BF3 – Lewis acid

(iii) H+ – Lewis acid

(iv) NH+4 – Lewis acid

Exercise

37

From the compounds given below which will be the conjugate base for the Bronsted acids?

(i) HF

(ii) H2SO4

(iii) HCO3

Answer

The following shows the conjugate bases for the Bronsted acids:

(i) HF – F−

(ii) H2SO4 – HSO−4

(iii) HCO3 – CO2−3

Exercise

38

For the bronsted bases given below find their conjugate acids.

  1. NH3
  2. HCOO−
  3. NH2−

Answer

Exercise

39

The species given below can act as both Brönsted bases as well as Brönsted acids. For each of them give their conjugate acid and base.

  1. HCO−3
  2. HSO−4
  3. NH3
  4. H2O

Answer

Exercise

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

Answer

(a) OH– is a Lewis base since it can donate its lone pair of electrons.

(b) F– is a Lewis base since it can donate a pair of electrons.

(c) H+ is a Lewis acid since it can accept a pair of electrons.

(d) BCl3 is a Lewis acid since it can accept a pair of electrons.

Exercise

41

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

Answer

Exercise

42

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

Answer

Exercise

43

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer


Exercise

44

The ionization 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 ionization if the solution is also 0.01M in sodium phenolate?

Answer


Exercise

45

The first ionization constant of H2­­S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

Answer




Exercise

46

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.

Answer


Exercise

47

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

Answer


Exercise

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

 

Answer




Exercise

49

Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution.

b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.

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

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

Answer


Exercise

50

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

Answer

Exercise

51

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

C = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)

Answer

C = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)


Exercise

52

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.

Answer

Kb = 4.27 × 10–10

c = 0.001M

pH =?

α =?


Exercise

53

Calculate the degree of ionization of 0.05M 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



Exercise

54

The ionization constant of dimethyl amine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethyl amine is ionized if the solution is also 0.1 M in NaOH?

Answer


Exercise

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.

Answer

(a) Human muscle fluid 6.83:

pH = 6.83

pH = – log [H+]

∴6.83 = – log [H+]

[H+] =1.48 × 10–7 M

(b) Human stomach fluid, 1.2:

pH =1.2

1.2 = – log [H+]

∴[H+] = 0.063

(c) Human blood, 7.38:

pH = 7.38 = – log [H+]

∴ [H+] = 4.17 × 10–8 M

(d) Human saliva, 6.4:

pH = 6.4

6.4 = – log [H+]

[H+] = 3.98 × 10–7

Exercise

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.

Answer

The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = –log [H+]

(I)pH of milk = 6.8

Since, pH = –log [H+]

6.8 = –log [H+] log

[H+] = –6.8

[H+] = anitlog(–6.8)

= 1.5×10−7M

(II)pH ofblack coffee = 5.0

Since, pH = –log [H+]

5.0 = –log [H+] log

[H+] = –5.0

[H+] = anitlog(–5.0)

= 10−5M

(III)pH of tomato= 4.2

Since, pH = –log [H+]

4.2 = –log [H+] log

[H+] = –4.2

[H+] = anitlog(–4.2)

= 6.31×10−5M

(IV)pH of lemon juice= 2.2

Since, pH = –log [H+]

2.2 = –log [H+] log

[H+] = –2.2

[H+] = anitlog(–2.2)

= 6.31×10−3M

(V)pH of egg white= 7.8

Since, pH = –log [H+]

7.8 = –log [H+] log

[H+] = –7.8

[H+] = anitlog(–7.8)

= 1.58×10−8M

Exercise

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

Exercise

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.

Answer


Exercise

59

The ionization constant of propanoic 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 0.01M in HCl also?

Answer


Exercise

60

The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer


Exercise

61

The ionization 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.

Answer

Exercise

62

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

Answer

Exercise

63

Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF

Answer


Exercise

64

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

Answer



Exercise

65

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

Answer

Ionic Product,

Kw=[H+][OH−]

Assuming, [H+] = y

As, [H+]=[OH−], Kw = y2.

Kw at 310K is 2.7×10−14.

∴2.7×10−14=y2

  • y = 1.64×10−7
  • [H+]=1.64×10−7
  • pH = −log[H+]

= −log[1.64×10−7]

= 6.78

Thus, the pH of neutral water at 310K temperature is 6.78.


Exercise

66

Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

Answer

Exercise

67

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Answer





Exercise

68

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 of their saturated solutions.

Answer

Exercise

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 × 10–8).

Answer

Exercise

70

The ionization 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?

Answer



Exercise

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).

Answer

Exercise

72

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

Answer

Exercise

73

The concentration of S2- in 0.1M HCl solution saturated with H2S is 1.0×1019 M. If 10mL of this added to 5mL of 0.04M solution given below:

  1. MnCl2
  2. ZnCl2
  3. CdCl2
  4. FeSO4

In which of the above solution the precipitation takes place?

For MnS, Ksp = 2.5×10−13

For ZnS, Ksp = 1.6×10−24

For CdS, Ksp = 8.0×10−27

For FeS, Ksp = 6.3×10−18

Answer

Exercise

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