NCERT Solutions For Class 11 Chemistry Chapter 7

NCERT Solutions For Class 11 Chemistry Chapter 7 PDF Free Download

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium is the finest study material that provides answers to the questions in the textbooks along with answers to important questions from previous year question papers and sample papers.

NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium provide exemplary problems and questions in different formats like MCQ’s, Short and long answer type questions, fill in the blank and match the following question for the benefit of class 11 Science students.

Class 11  NCERT Solutions for chemistry chapter 7 Equilibrium

This solution helps students to test their knowledge about the topic equilibrium and understand the different concepts related to it. The chapter equilibrium included in class 11 chemistry syllabus is an important concept and students need to be thorough with it.

As such, the solutions for NCERT chapter 7 that we are offering here consists of the main questions covered in the NCERT textbook which is also relevant for class 11 studies in CBSE schools. BYJU’s solutions for class 11 chemistry chapter 7 – equilibrium has been prepared and solved by our experienced subject experts as per latest CBSE syllabus 2018-19 and thus it will help the students to solve the CBSE class 11 chemistry paper without much difficulty. The set of NCERT solutions for class 11 chemistry chapter 7 equilibrium is available in the form of a PDF for easy access and free download.

Along with the structured questions, the NCERT solutions come with detailed explanations to help students learn and understand equilibrium concepts in a more seamless manner. If you have been looking for the most detailed, accurate and free solutions for class 11 NCERT chemistry, you are in the right place. Students who use these NCERT solutions will definitely have some advantage over others and it will also help them to score better marks in their exams.

Subtopics of Class 11 Chemistry Chapter 7 Equilibrium

  1. Solid-liquid Equilibrium
    1. Liquid-vapour Equilibrium
    2. Solid – Vapour Equilibrium
    3. Equilibrium Involving Dissolution Of Solid Or Gases In Liquids
    4. General Characteristics Of Equilibria Involving Physical Processes
  2. Equilibrium In Chemical Processes – Dynamic Equilibrium
  3. Law Of Chemical Equilibrium And Equilibrium Constant
  4. Homogeneous Equilibria
    1. Equilibrium Constant In Gaseous Systems
  5. Heterogeneous Equilibria
  6. Applications Of Equilibrium Constants
    1. Predicting The Extent Of A Reaction
    2. Predicting The Direction Of The Reaction
    3. Calculating Equilibrium Concentrations
  7. Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G
  8. Factors Affecting Equilibria
    1. Effect Of Concentration Change
    2. Effect Of Pressure Change
    3. Effect Of Inert Gas Addition
    4. Effect Of Temperature Change
    5. Effect Of A Catalyst
  9. Ionic Equilibrium In Solution
  10. Acids, Bases And Salts
    1. Arrhenius Concept Of Acids And Bases
    2. The Bronsted-lowry Acids And Bases
    3. Lewis Acids And Bases
  11. Ionization Of Acids And Bases
    1. The Ionization Constant Of Water And Its Ionic Product
    2. The Ph Scale
    3. Ionization Constants Of Weak Acids
    4. Ionization Of Weak Bases
    5. The Relation Between Ka And Kb
    6. Di- And Polybasic Acids And Di- And Polyacidic Bases
    7. Factors Affecting Acid Strength
    8. Common Ion Effect In The Ionization Of Acids And Bases
    9. Hydrolysis Of Salts And The Ph Of Their Solutions
  12. Buffer Solutions
  13. Solubility Equilibria Of Sparingly Soluble Salts
    1. Solubility Product Constant
    2. Common Ion Effect On Solubility Of Ionic Salts.

 Class 11 Physics Chapter 4 Equilibrium  Important questions

Q.1. At a fixed temperature a liquid is in equilibrium with its vapour in a closed vessel. Suddenly, the volume of the vessel got increased.

I)What will be the final vapour pressure and  what will happen when equilibrium is restored finally?

II)Write down, how initially the rates of evaporation and condensation got changed? 

III)Write down the effect observed when there was a change in vapour pressure.


(I)Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

(II)On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(III)On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.


Q.2.Find out Kc for the given reaction in equilibrium state

: [SO2]= 0.6 M, [O2] = 0.82 M and [SO3] = 1.9 M ?

\(2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g)\)


As per the question,

\(2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g)\) (Given)


\(K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\ =\frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}}\\ \\ =12.229M^{-1}\)(approximately)

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


Q.3. At a definite temperature and a total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms

\(I_{2} (g) \leftrightarrow 2I (g)\)

Find Kp for the equilibrium.


Partial pressure of Iodine atoms (I)

\(p_{I}=\frac{40}{100}\times p_{total}\\ \\ =\frac{40}{100}\times 10^{5}\\ \\ =4\times 10^{4}Pa\)

Partial pressure of I2 molecules,

\(p_{I}=\frac{60}{100}\times p_{total}\\ \\ =\frac{60}{100}\times 10^{5}\\ \\ =6\times 10^{4}Pa\)

Now, for the given reaction,

\(K_{p}=\frac{(p_{I})^{2}}{p_{I_{2}}} =\frac{(4\times 10^{4})^{2}Pa^{2}}{6\times 10^{4}Pa}\\ \\ =2.67\times 10^{4}Pa\)


Q.4. For the given reaction , find expression for the equilibrium constant

(i)\(2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g)\)

(ii)\(2Cu(NO_{3})_{2}(s)\leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)\)

(iii)\(CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)\leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)\)

(iv)\(Fe^{3+}(aq)+3OH^{-}(aq)\leftrightarrow Fe(OH)_{3}(s)\)

(v)\(I_{2}(s)+5F_{2}\leftrightarrow 2IF_{5}\)



(ii)\(K_{C}=\frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}}\\ \\ =[NO_{2(g)}]^{4}[O_{2(g)}]\)

(iii)\(K_{C}=\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]}\\ \\ =\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}\)

(iv)\(K_{C}=\frac{Fe(OH)_{3(s)}}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\\ \\ =\frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\)

(v)\(K_{C}=\frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}}\\ \\ =\frac{[IF_{5}]^{2}}{[F_{2}]^{5}}\)


Q.5.Find the value of , Kc for each of the following equilibria from the given value of Kp:

(i)\(2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g);\: \: \: K_{p}=1.8\times 10^{-2}\: \: at\: 500K\)

(ii)\(CaCO_{3}(s)\leftrightarrow CaO(s)+CO_{2}(g);\: \: \: K_{p}=167\: \: at\: 1073K\)


The relation between Kp and Kc is given as:

\(K_{p}=K_{c}(RT)^{\Delta n}\)

(a) Given,

R = 0.0831 barLmol–1K–1

\(\Delta n = 3-2 =1\)

T = 500 K

Kp=\(1.8\times 10^{-2}\)


Kp = Kc (RT) ∆n

\(\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \Rightarrow K_{c}=\frac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\ =4.33\times 10^{-4}(approximately)\)

(b) Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol–1K–1

T = 1073 K

Kp= 167


Kp = Kc (RT) ∆n

\(\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\frac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)\)


Q.6. For the following equilibrium, \(K_{c}=6.3\times 10^{14}\: \: \: at\: 1000K\\ \\ NO(g)+O_{3(g)}\leftrightarrow NO_{2(g)}+O_{2(g)}\)

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


For the reverse reaction, \(K_{c}=\frac{1}{K_{c }}\\ \\ =\frac{1}{6.3\times 10^{14}}\\ \\ =1.59\times 10^{-15}\)


Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression?


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

Mole concentration= \(\frac{Number\: of\: moles}{Volume}\\ \\ \frac{Mass/molecular\:mass}{Volume}\\ \\ =\frac{Mass}{Volume\,\times\, Molecular \:mass}\\ \\ =\frac{Density}{Molecular\: mass}\)

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

\(∴\)Molar concentration are constatnt.


Q.8. When  oxygen and nitrogen react with each other, then the following reaction takes place:

\(2N_{2}(g)+O_{2}\leftrightarrow 2N_{2}O(g)\)

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 =\(2.0\times 10^{-37}\), determine the composition of equilibrium solution.


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

\([N_{2}]=\frac{0.482-x}{10}\cdot [O_{2}]=\frac{0.933-\frac{x}{2}}{10},[N_{2}O]=\frac{x}{10}\)

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

\([N_{2}]=\frac{0.482}{10}=0.0482molL^{-1} \: \:and\:\: [O_{2}]=\frac{0.933}{10}=0.0933molL^{-1}\)


\(K_{c}=\frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]}\\ \\ \Rightarrow 2.0\times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)}\\ \\ \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37}\times (0.0482)^{2}\times (0.0933)\\ \\ \Rightarrow x^{2}=43.35\times 10^{-40}\\ \\ \Rightarrow x=6.6\times 10^{-20}\) \([N_{2}O]=\frac{x}{10}=\frac{6.6\times 10^{-20}}{10}\\ \\ =6.6\times 10^{-21}\)


Q.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.


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 \(\frac{0.0518}{2}\) 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


Q.10. At 450 K, Kp= \(2.0 \times 10^{10}\)/bar for the given reaction at equilibrium.

2SO2(g)+O2(g) ⇒ 2SO3(g)

What is Kc at this temperature?


For the given reaction,

∆n = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K–1 mol–1

Kp=\(2.0 \times 10^{10}bar^{-1}\)

We know that,

\(K_{p}=K_{c}(RT)\Delta n\\ \\ \Rightarrow 2.0\times 10^{10}bar^{-1}=K_{c}(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}\\ \\ \Rightarrow K_{c}=\frac{2.0\times 10^{10}bar^{-1}}{(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}}\\ \\ =(2.0\times 10^{10}bar^{-1})(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)\\ \\ =74.79\times 10^{10}L\, mol^{-1}\\ \\ =7.48\times 10^{11}L\, mol^{-1}\\ \\ =7.48\times 10^{11}\, M^{-1}\)


Q.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)


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               2.15

2                      2

=0.08atm         =0.08atm


\(K_{p=}\frac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\ =\frac{0.08\times 0.08}{(0.04)^{2}}\\ \\ =\frac{0.0064}{0.0016}\\ \\ =4.0\)

Hence, the value of Kp for the given equilibrium is 4.0.


Q.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 \(1.7\times 10^{2}\)

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


The given reaction is:

N2(g)+3H2(g) 2NH3(g)

The given concentration of various species is

[N2]= \(\frac{1.57}{20}mol\, L^{-1}\) [H2]= \(\frac{1.92}{20}mol\, L^{-1}\) [NH3]= \(\frac{8.31}{20}mol\, L^{-1}\)

Now, reaction quotient Qc is:

\(Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}}\\ \\ =2.4\times 10^{3}\)

Since, ?? ≠ ??, the reaction mixture is not at equilibrium.

Again, ?? > ??. Hence, the reaction will proceed in the reverse direction.


Q.13. The equilibrium constant expression for a gas reaction is,


Write the balanced chemical equation corresponding to this expression.


The balanced chemical equation corresponding to the given expression can be written as:



Q.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.


The given reaction is:

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

Initial conc.                    1     M             1     M                            0               0

10                    10

At  equilibrium   1-0.6  M        1-0.6  M                                0.6  M               0.6  M

10                  10                                       10                       10

=0.04 M           =0.04M                                =0.06 M            =0.06M

Therefore, the equilibrium constant for the reaction,

Kc=\(K_{c}=\frac{[H_{2}][CO_{2}]}{[H_{2}O][CO]}\\ \\ =\frac{0.06\times 0.06}{0.04\times 0.04}\\ \\ =\frac{0.0036}{0.0016}\\ \\ =2.25(approximately)\)


Q.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?


It is given that equilibrium constant ?? for the reaction

H2(g)+I2(g)                              2HI(g)   is  54.8.

Therefore, at equilibrium, the equilibrium constant Kc for the reaction

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

[HI]=0.5 molL-1 will be 1/54.8.

Let the concentrations of hydrogen and iodine at equilibrium be x molL–1

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

Therefore, \(\frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{‘}_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.5)^{2}}=\frac{1}{54.8}\\ \\ \Rightarrow x^{2}=\frac{0.25}{54.8}\\ \\ \Rightarrow x=0.06754\\ \\ x=0.068molL^{-1}(approximately)\)

Hence, at equilibrium, [H2]=[I2]=0.068 mol L-1.


Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl(g) ⇒ I2(g)+Cl2(g) ;  Kc =0.14



The given reaction is:

2ICl(g)                                           I2(g)      +   Cl2(g)

Initial conc.                 0.78 M                                           0             0

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

Now, we can write, \(\frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x}{0.78-2x}=0.374\\ \\ \Rightarrow x=0.292-0.748x\\ \\ \Rightarrow 1.748x=0.292\\ \\ \Rightarrow x=0.167\)

Hence, at equilibrium,

[H2]=[I2]=0.167 M

[HI]= \((0.78-2\times 0.167)M\\ \\ =0.446M\)


Q.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)


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


C2H6(g)                                         C2H4(g)    +     H2(g)

Initial conc.                 4.0 M                                           0                      0

At equilibrium          (4.0-p)                                            p                     p

We can write,

\(\frac{p_{c_{2}H_{4}}\times p_{H_{2}}}{p_{c_{2}H_{6}}}=K_{P}\\ \\ \Rightarrow \frac{p\times p}{40-p}=0.04\\ \\ \Rightarrow p^{2}=0.16-0.04p\\ \\ \Rightarrow p^{2}+0.04p-0.16=0\)


\(p=\frac{-0.04\pm \sqrt{(0.04)^{2}-4\times 1\times (-0.16)}}{2\times 1}\\ \\ =\frac{-0.04\pm 0.80}{2}\\ \\ =\frac{0.76}{2}\: \: \: \: \: \: \: \: \: (Taking \:positive\: value)\\ \\ =0.38\)

Hence, at equilibrium,

[C2H6] – 4 – p = 4 – 0.38

= 3.62 atm


Q.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?


(i)Reaction quotient,


(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

CH3COOH(l)+c2H5OH(l)                          CH3COOC2H5(l)+H2O(l)

Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, equilibrium constant for the given reaction is:

\(K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.009}{V}}=3.919\\ \\ =3.92(approximately)\)

(iii)Let the volume of the reaction mixture be V.


Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, the reaction quotient is,

\(K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.286}{V}}=0.2037\\ \\ =0.204(approximately)\)

Since Qc<Kc , equilibrium has not been reached.


Q.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 \(0.5\times 10^{-1}\)mol L–1. If value of Kc is \(8.3\times 10^{-3}\), what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5(g) ⇒ PCl3(g)   +   Cl2(g)


Consider the conc. Of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:

PCl5(g)PCl3(g      +      Cl2(g)

At equilibrium \(0.5\times 10^{-10}molL^{-1}\)         x mol L-1      x mol L-1

It is given that the value of equilibrium constant , Kc is \(8.3\times 10^{-10}molL^{-3}\)

Now we can write the expression for equilibrium as:

\(\frac{[PCl_{2}][Cl_{2}]}{[PCl_{3}]}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{0.5\times 10^{-10}}=8.3\times 10^{-3}\\ \\ \Rightarrow x^{2}=4.15\times 10^{-4}\\ \\ \Rightarrow x=2.04\times 10^{-2}\\ \\ =0.0204\\ \\ =0.02(approximately)\)

Therefore, at equilibrium,

[PCl3]=[Cl2]=0.02mol L-1


Q.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 ???2= 0.80 atm?


For the given reaction,

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

Initialy,              1.4 atm                                                      0.80 atm


Qp=\(\frac{p_{CO_{2}}}{p_{CO}}\\ \\ =\frac{0.80}{1.4}\\ \\ =0.571\)

Since Qp > Kp , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then, we can write,

\(K_{p}=\frac{p_{CO_{2}}}{p_{CO}}\\ \\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \\ \Rightarrow 0.371+0.265p=0.80-p\\ \\ \Rightarrow 1.265p=0.429\\ \\ \Rightarrow p=0.339 atm\)

Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm

And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm


Q.21. A reaction is given:

N2(g)+3H2(g) ⇒ 2NH3

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.


N2(g)        +        3H2(g)                                  2NH3

At a particular time:  3.0 mol L-1          2.0mol L-1                              0.5 mol L-1


Qc=\(\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\\ \\ =0.0104\)

It is given that Kc=0.061

\(∵\)   \(Qc \neq K_{c}\), the reaction is not at equilibrium.

\(∵\)    \(Qc < K_{c}\), the reaction preceeds in the forward direction to reach at equilibrium.


Q.22.Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇒ Br2(g)   +   Cl2(g)

For which Kc= 42 at 600 K.

If initially pure BrCl is present at a concentration of \(5.5 \times 10^{-5}\)molL-1 , what is its molar concentration in the mixture at equilibrium?


Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

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

Initial conc. \(5.5\times 10^{-5}\)                   0                         0

At equilibrium \(5.5\times 10^{-5}-2x\)         x                          x

Now, we can write,

\(\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(5.5\times 10^{-5}-2x)^{2}}=42\\ \\ \Rightarrow \frac{x}{5.5\times 10^{-5}-2x}=6.48\\ \\ \Rightarrow x=35.64\times 10^{-5}-12.96x\\ \\ \Rightarrow 13.96x=35.64\times 10^{-5}\\ \\ \Rightarrow x=\frac{35.64}{13.96}\times 10^{-5}=2.55\times 10^{-5}\)

So\(,\)at equilibrium

\([BrCl]=5.5\times 10^{-5}-(2\times 2.55\times 10^{-5})\\ \\ =5.5\times 10^{-5}-5.1\times 10^{-5}\\ \\ =0.4\times 10^{-5}\\ \\ =4.0\times 10^{-6}molL^{-1}\)


Q.23. Find out Kc for the given reaction at temperature 1127K where the pressure is 1 atm. A solution of CO and CO2 is in equilibrium with carbon(solid). It has 93.55% CO by mass.


C(s)+CO2 (g) ⇒ 2CO (g)


Let us assume that the solution is of 100g in total.

Given, mass of CO = 93.55 g

Now, the mass of CO2 =(100 – 93.55)=6.45 g

Now, number of moles of CO, \(n_{CO}=\frac{93.5}{28}=3.34\: mol\)

Number of moles of CO2, \(n_{CO_{2}}=\frac{6.45}{44}=0.146\: mol\)

Partial pressure of CO,

PCO= \(\frac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{3.34}{3.34+0.146}\times 1=0.958\, atm\)

Partial pressure of CO2, \(P_{CO_{2}}=\frac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{0.146}{3.34+0.146}\times 1=\, atm\)

Therefore, Kp= \(\frac{[CO]^{2}}{[CO_{2}]}\\ \\ =\frac{(0.938)^{2}}{0.062}\\ \\ =14.19\)

For the given reaction,

∆n = 2 – 1 = 1

We know that,

Kp=Kc(RT) \(\Delta n\)

\(\Rightarrow 14.19=K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}=\frac{14.19}{0.082\times 1127}\\ \\ =0.154(approximately)\)


Q.24.Find out

(I)The equilibrium constant for the formation of NO2 from NO and O2 at 298 K and

(II) ∆G°

\(NO(g)+\frac{1}{2}O_{2}(g)\leftrightarrow NO_{2}(g)\)  


fG° (NO2) = 52.0 kJ/mol

 ∆fG° (NO) = 87.0 kJ/mol

 ∆fG° (O2) = 0 kJ/mol


(I)We know that,

∆G° = RT log Kc

∆G° = 2.303 RT log Kc

\(K_{c}=\frac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ =6.134\\ \\ ∴ K_{c}=antilog(6.134)\\ \\ =1.36\times 10^{6}\)

Therefore, the equilibrium constant for the given reaction Kc is \(1.36\times 10^{6}\)


Q.25. When each of the following equilibria is subjected to a decrease in pressure by increasing the volume, does the number of moles of reaction products increase, decrease or remain same?

(I) PCl5(g) ⇒ PCl3 +Cl2 (g)

(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)

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


(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.


Q.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)


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


Q.27. The equilibrium constant for the following reaction is \(1.6 \times 10^{5}\)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.



Given, ?? for the reaction i.e., H2 (g) + Br2 (g)2HBr (g) is \(1.6 \times 10^{5}\).

Therefore, for the reaction   2HBr (g)H2 (g) + Br2 (g)  the equilibrium constant will be,

Kp= \(\frac{1}{K_{p}}\\ \\ =\frac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}\)

Now, let p be the pressure of both H2 and Br2 at equilibrium.

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

Initial conc.                   10                                                         0                         0

At equilibrium           10-2p                                                       p                         p


Now, we can write,

\(\frac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\ \frac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \frac{p}{10-2p}=2.5\times 10^{-3}\\ \\ p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})=2.5\times 10^{-2}\\ \\ p=2.49\times 10^{-2}bar=2.5\times 10^{-2}bar (approximately)\)

Therefore, at equilibrium,

[H2]=[Br2]= \(2.49\times 10^{-2}bar\) [HBr]= \(10-2\times (2.49\times 10^{-2})bar\\ \\ =9.95 bar=10 bar(approximately)\)


Q.28.Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

\(CH_{4}(g)+H_{2}O(g)\leftrightarrow CO(g)+3H_{2}(g)\)

(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?


(I)For the given reaction,

Kp=\(\frac{p_{CO}\times p_{H_{2}}^{3}}{p_{CH_{4}}\times p_{H_{2}O}}\)

(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.


Q.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)


(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.


Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is \( 8.3 \times 10^{-3} \). If decomposition is depicted as,

\(PCl_{5}(g)\leftrightarrow PCl_{3}(g)+Cl_{2}(g)\)

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?



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

\(K_{c}^{‘} = \frac{1}{K_{c}}\\ \\ =\frac{1}{8.3 \times 10^{-3}} = 1.2048 \times 10^{2}\\ \\ =120.48\)

(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.


Q.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,

\(CO(g)+H_{2}O(g)\leftrightarrow CO_{2}(g)+H_{2}(g)\)

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


Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

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

Initial conc.           4.0 bar          4.0 bar                            0                     0

At equilibrium      4.0-p             4.0-p                                p                     p

Given Kp = 10.1

\(\frac{P_{CO_{2}}\times P_{H_{2}}}{P_{CO}\times P_{H_{2}O}} = K_{P}\\ \\ \Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1\\ \\ \Rightarrow \frac{p}{4.0-p} =3.178\\ \\ \Rightarrow p =12.712-3.178p\\ \\ 4.178p = 12.712\\ \\ p =\frac{12.712}{4.178}\\ \\ p=3.04\)

So, partial pressure of H2 is 3.04 bar at equilibrium.
Q.32. Predict which of the following reaction will have appreciable concentration of reactants and products:

(a)\(Cl_{2}(g)\leftrightarrow 2Cl(g);K_{c}=5 \times 10^{-39}\)

(b)\(Cl_{2}(g)+2NO(g)\leftrightarrow 2NOCl(g);K_{c}=3.7 \times 10^{8}\)

(c)\(Cl_{2}(g)+2NO_{2}(g)\leftrightarrow 2NO_{2}Cl(g);K_{c}=1.8\)


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.


Q.33. The value of Kc for the reaction 3O2 (g) ⇒ 2O3 (g) is \(2.0\times 10^{-50} \) at 25°C. If the equilibrium concentration of O2 in air at 25°C is \( 1.6 \times 10^{-2} \) , what is the concentration of O3?



3O2 (g)2O3 (g)

Then, Kc = \( \frac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}\)

Given that Kc = \( 2.0 \times 10^{-50}\) and [O2(g)] = \(1.6 \times 10^{-2}\)


\( 2.0 \times 10^{-50} = \frac { [O_{3}(g)]^{2} }{ [1.6 \times 10^{ -2 }]^{3}}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 8.192 \times 10^{-56}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.86 \times 10^{-28} M\)

So, the conc. of O3 is \(2.86 \times 10^{-28}M\).


Q.34. The reaction, \(CO_{(g)} \; + \; 3H_{2(g)} \; \rightarrow \; CH_{4(g)} \; + \; H_{2}O_{(g)}\) at 1300K is at equilibrium in a 1L container. It has 0.30 mol of CO, 0.10 mol of \(H_{2}\) and 0.02 mol of \(H_{2}O\) and y amount of \(CH_{4}\) in the container. Find the concentration of \(CH_{4}\) in the mixture.

The equilibrium constant, \(K_{c}\) is 3.90 at the given temp.


Let the concentration of \(CH_{4}\)  at equilibrium be y.

\(CO_{(g)} \; + \; 3H_{2(g)} \; \rightarrow \; CH_{4(g)} \; + \; H_{2}O_{(g)}\)

At equilibrium,

For CO – \(\frac{0.3}{1} \; = \; 0.3M\)

For \(H_{2}\) – \(\frac{0.1}{1} \; = \; 0.1M\)

For \(H_{2}O\) – \(\frac{0.02}{1} \; = \; 0.02M\) \(K_{c}\) = 3.90



\(\frac{[CH_{4(g)}][H_{2}O_{(g)}]}{[CO_{(g)}][H_{2(g)}]^{3}} \; = \; K_{c}\) \(\frac{y \times 0.02}{0.3 \times (0.1)^{3}} \; = \; 3.9\) \(y \; = \; \frac{3.9 \times 0.3 \times (0.1)^{3}}{0.02}\) \(y \; = \; \frac{0.00117}{0.02}\) \(y \; = \; 0.0585M\) \(y \; = \; 5.85 \times 10^{-2}M\)


Therefore, the concentration of \(CH_{4}\) at equilibrium is \(5.85 \times 10^{-2}M\)


Q.35. What is conjugate acid-base pair? Find the conjugate acid/base of the given species:

(i) \(HNO_{2}\)

(ii) \(CN^{-}\)

(iii) \(HClO_{4}\)

(iv) \(F^{-}\)

(v) \(OH^{-}\)

(vi) \(CO_{3}^{2-}\)

(vii) \(S^{-}\)


A conjugate acid-base pair is a pair that has a difference of only one proton.

The conjugate acid-base pair of the following are as follows:

(i) \(HNO_{2}\) – \(NO_{2}^{-}\) (Base)

(ii) \(CN^{-}\) – HCN (Acid)

(iii) \(HClO_{4}\) – \(ClO_{4}^{-}\) (Base)

(iv) \(F^{-}\) – HF (Acid)

(v) \(OH^{-}\) – \(H_{2}O\) (Acid)/ \(O^{2-}\) (Base)

(vi) \(CO_{3}^{2-}\) – \(HCO_{3}^{-}\) (Acid)

(vii) \(S^{-}\) – \(HS^{-}\) (Acid)


Q.36. From the compounds given below which are Lewis acids?

(i) \(H_{2}O\)

(ii) \(BF_{3}\)

(iii) \(H^{+}\)

(iv) \(NH_{4}^{+}\)


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

(i) \(H_{2}O\) – Not Lewis acid

(ii) \(BF_{3}\) – Lewis acid

(iii) \(H^{+}\) – Lewis acid

(iv) \(NH_{4}^{+}\) – Lewis acid


Q.37. From the compounds given below which will be the conjugate base for the Bronsted acids?

(i) HF

(ii) \(H_{2}SO_{4}\)

(iii) \(HCO_{3}\)


The following shows the conjugate bases for the Bronsted acids:

(i) HF – \(F^{-}\)

(ii) \(H_{2}SO_{4}\) – \(HSO_{4}^{-}\)

(iii) \(HCO_{3}\) – \(CO_{3}^{2-}\)


Q.38. For the Brönsted bases given below find their conjugate acids.

  1. NH3
  2. \(HCOO^{-}\)
  3. \(NH_{2}^{-}\)


Brönsted base Conjugate acid
1 NH3 \(NH_{4}^{+}\)
2 \(HCOO^{-}\) HCOOH
3 \(NH_{2}^{-}\) NH3



Q.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


Species Conjugate base Conjugate acid
1 \(HCO_{3}^{-}\) \(CO_{3}^{2-}\) H2CO3
2 \(HSO_{4}^{-}\) \(SO_{4}^{2-}\) H2SO4
3 NH3 \(NH_{2}^{-}\) \(NH_{4}^{+}\)
4 H2O \(OH^{-}\) H3O+



Q.40. Classify the species given below into bases and acids and also show that these species act as base/acid:

  1. BCl3
  2. \(H^{+}\)
  3. \(OH^{-}\)
  4. \(F^{-}\)


  1. BCl3:

It is a Lewis acid as it has tendency to accept a pair of electrons.

  1. \(H^{+}\)

It is a Lewis acid as it has tendency to accept a pair of electrons.

  1. \(OH^{-}\)

It is a Lewis base as it has tendency to lose a pair of electrons.

  1. \(F^{-}\)

It is a Lewis base as it has tendency to lose its lone pair of electrons.


Q.41.A sample soft drink is taken, whose hydrogen ion concentration is \(2.5\times 10^{-4}M\).Find out pH.


\(pH=-log[H^{+}]\\ \\ =-log(2.5\times 10^{-4})\\ \\ =-log\: 2.5\, -\, log\: 10^{-4}\\ \\ =-log\, 2.5+4\\ \\ =-0.398+4\\ \\ =3.602\)


Q.42.A sample of white vinegar is taken , whose pH is 2.36. Find out the hydrogen ion concentration in the sample.


\(pH = -log [H^{+}]\\ \\ \Rightarrow log[H^{+}]=-pH\\ \\ \Rightarrow [H^{+}]=antilog(-pH)\\ \\ =antilog(-2.36)\\ \\ =0.004365\\ \\ =4.37\times 10^{-3}\) \(∴\) \(4.37\times 10^{-3}\) is the concentration of white vinegar sample.


Q.43. Ionization constant for the following acids are given:

HF = \(5.7\times 10^{-5}\) at 298K

HCOOH = \(1.7\times 10^{-3}\) at 298K

HCN = \(3.7\times 10^{-8}\) at 298K

Find out the conjugate bases for the above acids.


For F,  \(K_{b} = \frac{K_{w}}{K_{a}}=\frac{10^{-14}}{(5.7\times 10^{-5})}=1.75\times 10^{-9}\)

For HCOO, \(K_{b}=\frac{10^{-14}}{(1.7\times 10^{-3})}=5.88\times 10^{-11}\)

For CN = \(K_{b}=\frac{10^{-14}}{(3.7\times 10^{-8})}=\times 10^{-11}=2.70\times 10^{-6}\)


Q.44.Phenol has ionization constant of \(1.0\times 10^{-8}\). In a 0.06M of phenol solution calculate the presence of phenolate ion. Find out the degree of ionization if 0.02M of sodium phenolate is given.


C6H5OH                            C6H5O + H+

Initial                            0.06M

After dissociation       0.06-x                               x               x

\(∴ K_{a}=\frac{x\times x}{0.06-x}=1.0\times 10^{-8}\\ \\ \Rightarrow \frac{x^{2}}{0.06}=1.0\times 10^{-8}\\ \\ \Rightarrow x^{2} = 6\times 10^{-10}\\ \\ \Rightarrow x=2.4\times 10^{5}M\)

In presence of 0.02  sodium phenolate(C6H5Na), suppose y is the amount of phenol dissociated, then at equilibrium

[C6H5OH] = 0.06 – y \(\simeq\) 0.06,

[C6H5O] = 0.02 + y \(\simeq\)0.01M,

[H+]=y M

\(∴ K_{a}=\frac{(0.02)(y)}{0.06}=1.0\times 10^{-8}\\ \\ \Rightarrow y=\frac{1.0\times 0.06}{(0.02)}\times 10^{-8}\\ \\ \Rightarrow y=6\times 10^{-8}\) \(∴\) degree of ionization = \(\alpha =\frac{y}{c}=\frac{6\times 10^{-8}}{6\times 10^{-2}}( Here c=0.06= 6\times 10^{-2})=10^{-6} \)

So, \(\alpha = 10^{-6}\)


Q.45 Given, \(9.1\times 10^{-8}\) is the initial(first) ionization constant of the gas H2S.Find out concentration of the ion HS  in 0.1M solution of H2S .  Find the changes in concentration if the concentration is 0.1M in HCl. Find the concentration of S2- under both conditions, if \(1.2\times 10^{-13}\) is the second dissociation constant of H2S.


To calculate [HS]

H2S                           H+  +  HS

Intial                          0.1 M

After dissociation    0.1-x                          x          x

\(\simeq 0.1\) \(K_{a}=\frac{x\times x}{0.1}=9.1\times 10^{-8}\\ \\ \Rightarrow x^{2}=9.1\times 10^{-9}\\ \\ \Rightarrow x=9.54\times 10^{-5}\)

In the presence of 0.1 M HCl, suppose H2S dissociated is y. Then at equilibrium, [H2S] =0.1-y \(\simeq 0.1\),

[H+]=0.1 + y \(\simeq 0.1\),

[HS] = y M

\(K_{a}=\frac{0.1\times y}{0.1}=9.1\times 10^{-8}\\ \\ y=\frac{9.1\times 0.1}{0.1}\times 10^{-8}\\ \\ y=9.1\times 10^{-8}\)

To calculate [S2-]                         

H2S                           H+  +  HS  , HS                           H+  +  S2-

For the overall reaction,

H2S                           2H+  +  S2-

\(K_{a}=K_{a_{1}}\times K_{a_{2}}=9.1\times 10^{-8}\times 1.2\times 10^{-13}=1.092\times 10^{-20}\) \(K_{a}=\frac{[H^{+}]^{2}[S^{2-}]}{[H_{2}S]}\)

In the absence of 0.1M HCl,

[H+]= 2[S2-]

Hence, if [S2-] = x , [H+] =2x

\(∴ \frac{(2x)^{2}}{0.1}=1.092\times 10^{-20}\\ \\ \Rightarrow 4x^{3}=1.092\times 10^{-21}\\ \\ \Rightarrow x^{3}=\frac{1.092}{4}\times 10^{-21}=273\times 10^{-24}\\ \\ \Rightarrow log\, x^{3}=log\, 273-log \, 10^{-24}=2.4362-24\\ \\ \Rightarrow 3logx=2.4362-24\\ \\ \Rightarrow log\, x=\frac{2.4362}{3}-\frac{24}{3}\\ \\ \Rightarrow x=0.8127-8=-7.1873\\ \\ \Rightarrow x=Antilog\,-7.1873 =6.497\times 10^{-8}=6.5\times 10^{-8}\)

In presence of 0.1M HCl,

Suppose [S2-]=y, then



\(K_{a}=\frac{(0.1)^{2}\times y}{0.1}=1.09\times 10^{-20}\\ \\ y=1.09\times 10^{-19}M\)


Q.46.Given,the ionization constant of acetic acid is \(1.74\times 10^{-5}\). Find the degree of dissociation of acetic acid in its 0.05 M solution. Find  the concentration of acetate ion in the solution and its pH.



\(K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=\frac{[H^{+}]^{2}}{[CH_{3}COOH]}\\ \\ \Rightarrow [H^{+}]=\sqrt{K_{a}[CH_{3}COOH]}=\sqrt{(1.74\times 10^{-5})(5\times 10^{-2})}=9.33\times 10^{-4}M\\ \\ [CH_{3}COO^{-}]=[H^{+}]=9.33\times 10^{-4}M\\ \\ pH=-log(9.33\times 1.0^{-4})=4-0.9699=4-0.97=3.03\)


Q.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.


HA                         H+   +   A



[H+] = \(7.08\times 10^{-5}\, M\) [A] = [H+] = \(7.08\times 10^{-5}\, M\) \(K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}=\frac{(7.08\times 10^{-5})(7.08\times 10^{-5})}{10^{-2}}=5.0\times 10^{-7}\\ \\ p_{K_{a}}=-log\, K_{a} = -log\, (5.0\times 10^{-7})=7-0.699=6.301\)


Q.48.Consider complete dissociation, find out the pH of the following :

(I)0.004 M HCl

(II)0.003 M NaOH

(III)0.002 M HBr

(IV)0.002 M KOH


(I)HCl + aq                     H+  +  Cl

\(∴ [H^{+}]=[HCl]=4\times 10^{-3}M\\ \\ pH=-log(4\times 10^{-3})=2.398\)

(II)NaOH + aq                      Na+ + OH

\(∴ [OH^{-}]=3\times 10^{-3}M\\ \\ [H^{+}]=\frac{10^{-14}}{(3\times 10^{-3})}=3\times 10^{-12}M\\ \\ pH=-log(3\times 10^{-12})=11.52\)

(III)HBr + aq                      H+  +  Br

\(∴ [H^{+}]=2\times 10^{-3}M\\ \\ pH=-log(2\times 10^{-3}M)=2.70\)

(IV)KOH + aq                      K+  +  OH

\(∴ [OH^{+}]=2\times 10^{-3}M\\ \\ [H^{+}]=\frac{10^{-14}}{(2\times 10^{-3})}=5\times 10^{-12}\\ \\ pH=-log(5\times 10^{-12})=11.30\)


Q.49.Find out the pH of the following solution:

(I)2g of TIOH dissolved in water to give 2 litre of the solution

(II)0.3g of Ca(OH)2 dissolved in water to given 500mL of the solution

(III)0.3g of NaOH dissolved in water to give 200mL of the solution

(IV)1 mL of 13.6 M HCl is diluted with water to given 1 litre of the solution


(I)Molar conc. Of TlOH = \(\frac{2g}{(204+16+1)g\: mol^{-1}}\times \frac{1}{2L}=4.52\times 10^{-3}M\\ \\ [OH_{-}]=[TlOH]=4.52\times 10^{-3}M\\ \\ [H^{+}]=\frac{10^{-14}}{(4.52\times 10^{-3})}=2.21\times 10^{-12}M\\ \\ ∴ pH=-log(2.21\times 10^{-12})=12-(0.3424)=11.66\)

(II)Molar conc. Of Ca(OH)2=\(\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.5L}=8.11\times 10^{-3}M\\ \\ [OH_{-}]=2[Ca(OH)_{2}]=2\times (8.11\times 10^{-3})M=16.22\times 10^{-3}M\\ \\ pOH=-log(16.22\times 10^{-3})=3-1.2101=1.79\\ \\ pH=14-1.79=12.21\)

(III)Molar conc. of NaOH = \(\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.2L}=3.75\times 10^{-2}M\\ \\ [OH_{-}]=3.75\times 10^{-2}M\\ \\ pOH=-log(3.75\times 10^{-2})=2-0.0574=1.43\\ \\ pH=14-1.43=12.57\)

(IV)M1V1 = M2V2

\(∴ 13.6M\times \times 1mL=M_{2}\times 1000mL\\ \\ ∴ M_{2}=1.36\times 10^{-2}M\\ \\ [H^{+}]=[HCl]=1.36\times 10^{-2}M\\ \\ pH=-log(1.36\times 10^{-2})=2-0.1335\simeq 1.87\)


Q.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.


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

Initial conc.                                  C                                     0                  0

Conc. at eqm.                         C – Cα                                   Cα                     Cα

\(K_{a}=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}=\frac{C\alpha ^{2}}{1-\alpha }\simeq C\alpha ^{2}=0.1\times (0.132)^{2}=1.74\times 10^{-3}\\ \\ p_{K_{a}}=-log(1.74\times 10^{-3})=3-0.2405=2.76\\ \\ [H^{+}]=C\alpha =0.1\times 0.132=1.32\times 10^{-2}M\\ \\ pH=-log(1.32\times 10^{-2})=2-0.1206=1.88\)


Q.51. 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.


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

Initial conc.                                  C                                     0                  0

Conc. at eqm.                         C – Cα                                   Cα                     Cα

\(K_{a}=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}=\frac{C\alpha ^{2}}{1-\alpha }\simeq C\alpha ^{2}=0.1\times (0.132)^{2}=1.74\times 10^{-3}\\ \\ p_{K_{a}}=-log(1.74\times 10^{-3})=3-0.2405=2.76\\ \\ [H^{+}]=C\alpha =0.1\times 0.132=1.32\times 10^{-2}M\\ \\ pH=-log(1.32\times 10^{-2})=2-0.1206=1.88\)


Q.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.


Kb = \(4.27 \times 10^{-10}\)

c = 0.001M

pH =?

α =?

\(K_{b}=c\alpha ^{2}\\ \\ 4.27 \times 10^{-10}=0.001\times \alpha ^{2}\\ \\ 4270\times 10^{-10}=\alpha ^{2}\\ \\ 65.34\times 10^{-5}=\alpha =6.53\times 10^{-4}\\ \\ Then,[anion]=c\alpha =0.001\times 65.34\times10^{-5}=0.065\times10^{-5}\) \(pOH=-log(0.065\times 10^{-5})\\ \\ =6.187\\ \\ pH=7.813\\ \\ Now,\\ \\ K_{a}\times K_{b}=K_{w}\\ \\ K_{a}=\frac{10^{-14}}{4.27\times 10^{-10}}\\ \\ =2.34\times 10^{-5}\) \(∴ 2.34\times 10^{-5}\) is the ionization constant.


Q.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

(I)0.01 M

(II)0.1 M in HCl?





\(K_{a}=1.82\times 10^{-5}\\ \\ K_{a}=c\alpha ^{2}\\ \\ \alpha =\sqrt{\frac{K_{a}}{c}}\\ \\ \alpha =\sqrt{\frac{1.82\times 10^{-5}}{5\times 10^{-2}}}=1.908\times 10^{-2}\)

When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case 1: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

CH3COOH                           H+  +  CH3COO

Initial conc.                          0.05M                            0               0

After dissociation              0.05-x                         0.01+x            x

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

\(K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\ ∴ =\frac{(0.01)x}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.01}\\ \\ x=1.82\times 10^{-3}\times 0.05M\)


\(\alpha =\frac{Amoun \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-3}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-3}\)

Case 2: When 0.1 M HCl is taken.

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

[CH3COOH]=0.05 – X; 0.05 M


[H+]=0.1+X ; 0.1M

\(K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\ ∴ K_{a}=\frac{(0.1)X}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.1}\\ \\ x=1.82\times 10^{-4}\times 0.05M\)


\(\alpha =\frac{Amoun \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-4}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-4}\)


Q.54. The ionization constant of dimethylamine is\(5.4 \times × 10^{-4}\). 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?


\(K_{b}=5.4\times 10^{-4}\\ \\ c=0.02M\\ \\ Then,\alpha =\sqrt{\frac{K_{b}}{c}}\\ \\ =\sqrt{\frac{5.4\times 10^{-4}}{0.02}}=0.1643\)

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

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

0.1M        0.1M


(CH3)2 NH   +  H2O                           (CH3)2 NH2+   +  OH

(0.02-x)                                                     x                    x

;0.02M                                                                          ;0.1M

Then,[ (CH3)2 NH+2]=x

[OH ]=x+0.1;0.1

\(\Rightarrow K_{b}=\frac{[(CH_{3})_{2}NH_{2}^{+}][OH^{-}]}{[(CH_{3})_{2}NH]}\\ \\ 5.4\times 10^{-4}=\frac{x\times 0.1}{0.02}\\ \\ x=0.0054\)

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.


Q.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(I)Human saliva, 6.4

(II)Human stomach fluid, 1.2 

(III)Human muscle-fluid, 6.83

(IV)Human blood, 7.38


(I)Human saliva, 6.4:

pH = 6.4

6.4 = – log [H+] [H+] = \(3.98\times 10^{-7} \)

(II)Human stomach fluid, 1.2:

pH =1.2

1.2 = – log [H+] \(∴\) [H+] = 0.063

(III)Human muscle fluid 6.83:

pH = 6.83

pH = – log [H+]

6.83 = – log [H+] [H+] =\(1.48\times 10^{-7}\, M\)

(IV) Human blood, 7.38:

pH = 7.38 = – log [H+] [H+] = \(4.17\times 10^{-8}\, M\)


Q.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.


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)

= \(1.5\times 10^{-7} M\)

(II)pH ofblack coffee = 5.0

Since, pH = –log [H+]

5.0 = –log [H+] log

[H+] = –5.0

[H+] = anitlog(–5.0)

= \( 10^{-5} M\)

(III)pH of tomato= 4.2

Since, pH = –log [H+]

4.2 = –log [H+] log

[H+] = –4.2

[H+] = anitlog(–4.2)

= \(6.31\times 10^{-5} M\)

(IV)pH of lemon juice= 2.2

Since, pH = –log [H+]

2.2 = –log [H+] log

[H+] = –2.2

[H+] = anitlog(–2.2)

= \(6.31\times 10^{-3} M\)

(V)pH of egg white= 7.8

Since, pH = –log [H+]

7.8 = –log [H+] log

[H+] = –7.8

[H+] = anitlog(–7.8)

= \(1.58\times 10^{-8} M\)


Q.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?


[KOHaq]= \(\frac{0.561}{\frac{1}{5}}g/L\\ \\ =2.805g/L\\ \\ =2.805\times \frac{1}{56.11}M\\ \\ =0.05M\)

KOH(aq)                            K+(aq)  +  OH(aq)

[OH]=0.05M=[K+] [H+][H]=Kw

[H+]=\(\frac{K_{w}}{[OH^{-}]}\) \(=\frac{10^{-14}}{0.05}=2\times 10^{-13}M\\ \\ ∴ pH=12.70\)


Q.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.


Solubility of Sr(OH)2 = 19.23 g/L

Then, concentration of Sr(OH)2

=\(\frac{19.23}{121.63}M\\ \\ =0.1581M\)

Sr(OH)2(aq)                         Sr2+(aq) + 2(OH)(aq)

\(∴\)[ Sr2+]=0.1581M

[OH]= \(2\times 0.1581M=0.3126\)


Kw=[OH][H+] \(\frac{10^{-14}}{0.3126}=[H^{+}]\\ \\ \Rightarrow [H^{+}]=3.2\times 10^{-14}\\ \\ ∴ pH=13.495;13.50\)


Q.59. The ionization constant of propanoic acid is \(1.32 \times × 10^{-15}\). 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?


Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA               +    H2O                                                H3O+       +       A

(0.05-0.0α) \(\approx\)0.05                                       0.05α              0.05α

\(K_{\alpha }=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\\ \\ =\frac{(0.05\alpha )(0.05\alpha )}{0.05}=0.05\alpha ^{2}\) \(\alpha =\sqrt{\frac{K_{\alpha }}{0.05}}=1.63\times 10^{-2}\)

Then,[ H3O+ ]= 0.05α =\(0.05\times 1.63\times 10^{-2}=K_{b}\cdot 15\times 10^{-4}M\\ \\ ∴ pH=3.09\)

In the presence of 0.1M of HCl, let α´ be the degree of ionization.

Then,[ H3O+ ]=0.01

[A]=0.05 α´


\(K_{\alpha }=\frac{0.01\times 0.05 {\alpha}’}{0.05}\\ \\ 1.32\times 10^{-5}=0.01\times {\alpha }’\\ \\ {\alpha }’=1.32\times 10^{-3}\)



Q.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.


c = 0.1 M

pH = 2.34

-log [H+] = pH

-log [H+] = 2.34

[H+]=\(4.5\times 10^{-3}\)



\(4.5\times 10^{-3}=0.1\times \alpha \\ \\ \frac{4.5\times 10^{-3}}{0.1}=\alpha \\ \\ \alpha =4.5\times 10^{-3}=0.045\)


\(K_{a}=c\alpha ^{2}\\ \\ =0.1\times (45\times 10^{-3})^{2}\\ \\ =202.5\times 10^{-6}\\ \\ =2.02\times 10^{-4}\)


Q.61. for nitrous acid Ka = \(4.5\times 10^{-4}\). Calculate degree of hydrolysis and pH for 0.04M of sodium nitrite.


Sodium nitrite is a salt of NaOH (strong base) and HNO2 (weak acid).

\(NO_{2}^{-} + H_{2}O \leftrightarrow HNO_{2} + OH^{-}\) \(K_{h} = \frac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}\) \(\Rightarrow \;\frac{K_{w}}{K_{a}} = \frac{10^{-14}}{4.5\times 10^{-4}} = 22\times 10^{-10}\)

Let, y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:

\([NO_{2}^{-}] = 0.04 – y; 0.04\) \([HNO_{2}] = y\) \([OH^{-}] = y\) \(K_{h} = \frac{y^{2}}{0.04} = 0.22\times 10^{-10}\) \(y^{2} = 0.0088\times 10^{-10}\) \(y = 0.093\times 10^{-5}\) \(∴ [OH^{-}]= 0.093\times 10^{-5}M\) \([H_{3}O^{+}] = \frac{10^{-10}}{0.093\times 10^{-5}} = 10.75\times 10^{-9}M\)

Thus, \(pH = -\log (10.75\times 10^{-9})\)

= 7.96

Thus, the degree of hydrolysis is

\(= \frac{y}{0.04} = \frac{0.093\times 10^{-5}}{0.04}\) \(= 2.325\times 10^{-5}\)


Q.62. 0.02M solution of pyridinium hydrochloride (C5H6ClN)  is having pH = 3.44. Determine the ionization constant of C5H5N (pyridine).


pH = 3.44

As we know,

\(pH = \log [H^{+}]\) \(∴ [H^{+}] = 3.63\times 10^{-4}\)

Now, \(K_{h} = \frac{3.63\times 10^{-4}}{0.02}\); (Given that concentration = 0.02M)

\(\Rightarrow K_{h} = 6.6\times 10^{-6}\)

As we know that,

\(K_{h} = \frac{K_{w}}{K_{a}}\) \(K_{a} = \frac{K_{w}}{K_{h}} = \frac{10^{-14}}{6.6\times 10^{-6}}\)

= \(1.51\times 10^{-9}\)


Q.63. Few salts are given below;

  1. KBr
  2. NH4NO3
  3. KF
  4. NaNO2
  5. NaCN
  6. NaCl

Determine the nature of solution of these salts i.e.  Is it acidic or basic or neutral?


  1. KBr


KBr + H2O \(\leftrightarrow\) KOH (Strong base)+ HBr (Strong acid)

Thus, it is a neutral solution.


  1. NH4NO3


NH4NO3 + H2O \(\leftrightarrow\) NH4OH(Weak base) + HNO2 (Strong acid)

Thus, it is an acidic solution.


  1. KF


KF + H2O \(\leftrightarrow\) KOH (Strong base) + HF (weak acid)

Thus, it is a basic solution.

  1. NaNO2


NaNO2 + H2O \(\leftrightarrow\) NH4OH(Strong base) + HNO2(Weak acid)

Thus, it is a basic solution.


  1. NaCN


NaCN + H2O \(\leftrightarrow\) HCN (Weak acid) + NaOH (Strong base)

Thus, it is a basic solution.


  1. NaCl


NaCl + H2O \(\leftrightarrow\) NaOH (Strong base) + HCL (Strong acid)

Thus, it is a neutral solution.


Q.64. Find the pH of 0.1M acid and its 0.1M NaCl solution. The Ka for chloroacetic acid is \(1.35\times 10^{-3}\).


The Ka for chloroacetic acid (ClCH2COOH) is \(1.35\times 10^{-3}\).

\(\Rightarrow K_{a} = c\alpha ^{2}\) \(∴ \alpha = \sqrt{\frac{K_{a}}{c}}\)

= \(\sqrt{\frac{1.35\times 10^{-3}}{0.1}}\); (given concentration = 0.1M)

\(\alpha = \sqrt{1.35\times 10^{-2}}\)

= 0.116

\(∴ [H^{+}] = c\alpha = 0.1*0.116 = 0.0116\)
  • pH = \(-\log [H^{+}]\) = 1.94

ClCH2COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH2COOH

\(ClCH_{2}COO^{-} + H_{2}O \leftrightarrow ClCH_{2}COOH + OH^{-}\) \(K_{h} = \frac{[ClCH_{2}COO][OH^{-}]}{[ClCH_{2}COO^{-}]}\)

Now, \(K_{h} = \frac{K_{w}}{K_{a}}\) \(K_{h} = \frac{10^{-14}}{1.35\times 10^{-3}}\) \(= 0.740\times 10^{-11}\)

Also, \(K_{h} = \frac{y^2}{0.1}\) \(\Rightarrow \; 0.740\times 10^{-11} = \frac{y^2}{0.1}\) \(\Rightarrow \; 0.0740\times 10^{-11} = y^2\) \(y = 0.86\times 10^{-6}\) \([OH^{-}] = 0.86\times 10^{-6}\) \(∴ [H^{+}] =\frac{K_{w}}{0.86\times 10^{-6}}\) \(=\frac{10^{-14}}{0.86\times 10^{-6}}\) \([H^{+}] = 1.162\times 10^{-3}\)

pH = \(- \log [H^{+}]\)

= 7.94


Q.65. Determine the pH of neutral water at 310K temperature. Ionic product of H2O is \(2.7\times 10^{-14}\).


Ionic Product,

\(K_{w} = [H^{+}][OH^{-}]\)

Assuming, \([H^{+}]\) = y

As, \([H^{+}] = [OH^{-}]\), Kw = y2.

Kw at 310K is \(2.7\times 10^{-14}\).

\(∴ 2.7\times 10^{-14} = y^{2}\)
  • y = \(1.64\times 10^{-7}\)
  • \([H^{+}] = 1.64\times 10^{-7}\)
  • pH = \(-\log [H^{+}]\)

= \(-\log [1.64\times 10^{-7}]\)

= 6.78

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



Q.66. Find out the pH of resultant mixture;

i) n10 ml of 0.02M H2SO4 + 10 ml of 0.02M Ca(OH)2

ii) 10 ml of 0.1M H2SO4 + 10 ml of 0.1M KOH

iii) 10 ml of 0.2M Ca(OH)2 + 25 ml of 0.1M HCl


i) Moles of \(OH^{-}\)

= \(\frac{2*10*0.02}{1000} = 0.0004 mol\)

Moles of \(H_{3}O^{+}\)

= \(\frac{2*10*0.02}{1000} = 0.0004 mol\)


ii) Moles of \(OH^{-}\)

= \(\frac{2*10*0.1}{1000} = 0.002 mol\)

Moles of \(H_{3}O^{+}\)

= \(\frac{2*10*0.1}{1000} = 0.001 mol\)

Here, the \(H_{3}O^{+}\) is in excess is 0.01 mol

So, \([H_{3}O^{+}] = \frac{0.001}{20\times 10^{-3}} = \frac{10^{-3}}{20\times 10^{-3}} = 0.5\)

Thus, pH = -log(0.05)

= 1.3

As the solution id neutral pH = 7.


iii) Moles of \(OH^{-}\)

= \(\frac{2*10*0.2}{1000} = 0.004 mol\)

Moles of \(H_{3}O^{+}\)

= \(\frac{25*0.1}{1000} = 0.0025 mol\)

Here, the \(OH^{-}\) is in excess is 0.0015 mol

So, \([OH^{-}] = \frac{0.0015}{35\times 10^{-3}} = 0.0428\)

Thus, pH = -log(OH)

= 1.36

pH = 14 – 1.36 = 12.63

As the solution id neutral pH = 12.63.


Q.67. Calculate the solubilities of

a) barium chromate

b) ferric hydroxide

c) lead chloride

d) mercurous iodide

e) silver chromate

At 300K from their solubility product constant. Also calculate the molarities of the individual ions.


a) Barium Chromate

BaCrO4 à Ba2+ + \(CrO_{4}^{2-}\)

Now, Ksp = [Ba2+]\([CrO_{4}^{2-}]\)

Aumming the solubility of BaCrO4 is ‘x’.


[Ba2+] = x and  \(CrO_{4}^{2-}\) = x

  • Ksp = x2
  • \(1.2\times 10^{-10} = x^{2}\)
  • x = \(1.09\times 10^{-10}M\)

Molarity of Ba2+ = Molarity of \(CrO_{4}^{2-}\) = x = \(1.09\times 10^{-10}M\)

b) Ferric Hydroxide

Fe(OH)3 à Fe3+ + \(OH^{-}\)

Now, Ksp = [Fe3+]\([OH^{-}]^{3}\)

Aumming the solubility of Fe(OH)3 is ‘x’.


[Fe3+] = x and  \(OH^{-}\) = 3x

  • Ksp = x(3x)3

= x*27x3

Ksp = 27x4

  • \(1.0\times 10^{-38} = 27x^{4}\)
  • x = \(0.00037\times 10^{-36}M\)

Molarity of Fe3+ = x = \(1.39\times 10^{-10}M\)

Molarity of \(OH^{-}\) = 3x = \(4.17\times 10^{-10}M\)

c) Lead Chloride

PbCl2 à Pb2+ + \(2Cl^{-}\)

Now, Ksp = [Pb2+]\([Cl^{-}]\)

Aumming the solubility of PbCl2 is ‘x’.


[Pb2+] = x and  \(Cl^{-}\) = 2x

  • Ksp = x(2x)2

= x*4x2

Ksp = 4x3

  • \(1.6\times 10^{-5} = 4x^{3}\)
  • x = \(1.58\times 10^{-2}M\)

Molarity of Pb2+ = x = \(1.58\times 10^{-2}M\)

Molarity of \(Cl^{-}\) = 2x = \(3.16\times 10^{-2}M\)

d) Mercurous iodide

Hg2I2 à Hg2+ + \(2I^{-}\)

Now, Ksp = [Hg2+]2\([I^{-}]^{2}\)

Aumming the solubility of Hg2I2 is ‘x’.


[Hg2+] = x and  \(I^{-}\) = 2x

  • Ksp = x(2x)2

= x*4x2

Ksp = 4x3

  • \(4.5\times 10^{-29} = 4x^{3}\)
  • x = \(2.24\times 10^{-10}M\)

Molarity of Hg2+ = x = \(2.24\times 10^{-10}M\)

Molarity of \(I^{-}\) = 2x = \(4.48\times 10^{-10}M\)

e) Silver Chromate

Ag2CrO4 à 2Ag2+ + \(CrO_{4}^{2-}\)

Now, Ksp = [Ag2+]2\([CrO_{4}^{2-}]\)

Aumming the solubility of Ag2CrO4 is ‘x’.


[Ag2+] = 2x and \(CrO_{4}^{2-}\)  = x

  • Ksp = (2x)2*x
  • \(1.1\times 10^{-12} = 4x^{3}\)
  • x = \(0.65\times 10^{-4}M\)

Molarity of Ag2+ = 2x = \(1.3\times 10^{-4}M\)

Molarity of \(CrO_{4}^{2-}\) = x = \(0.65\times 10^{-4}M\)


Q-68. Determine the ratio of molarities to their saturated solutions for the following:

Ag2CrO4 and AgBr

The solubility product constant of Ag2CrO4 and AgBr are \(1.1\times 10^{-12}\; and\; 5.0\times 10^{-13}\) respectively.


Ag2CrO4 à 2Ag2+ + \(CrO_{4}^{-}\)

Now, Ksp = [Ag2+]2\([CrO_{4}^{-}]\)

Asuming the solubility of Ag2CrO4 is ‘x’.


[Ag2+] = 2x and \(CrO_{4}^{-}\)  = x

  • Ksp = (2x)2*x
  • \(1.1\times 10^{-12} = 4x^{3}\)
  • x = \(0.65\times 10^{-4}M\)

Assuming the solubility of AgBr is y.

AgBr(s) à Ag2+ + \(2CrO_{4}^{-}\)

  • Ksp = (y)2
  • \(5.0\times 10^{-13} = y^{2}\)

y = \(7.07\times 10^{-7}M\)

The ratio of molarities to their saturated solution is:

\(\frac{x}{y} = \frac{0.65\times10^{-4}M}{7.07\times10^{-7}M} = 91.9\)


Q.69. Cupric chlorate and sodium iodate having equal volume of 0.002M. Will the precipitation of copper iodate will occur or not?


Cupric chlorate and sodium iodate having equal volume are mixed together, then molar concentration of cupric chlorate and sodium iodate will reduce to half.

So, molar concentration of cupric chlorate and sodium iodate in mixture is 0.001M.

Na(IO3)2 à Na+ + \(IO_{3}^{-}\)

0.0001M                                0.001M

Cu(ClO3)2 à Cu2+ + \(2CIO_{3}^{-}\)

0.001M                                   0.001M

The Solubility for Cu(IO3)2 ⇒ Cu2+ (aq) + 2IO3–  (aq)

Now, the ionic product of the copper iodate is:

= [Cu2+] \([IO_{3}^{-}]^{2}\)

= (0.001)(0.001)2

= \(1.0\times 10^{-9}M\)

As the value of Ksp is more than Ionic product.

Thus, the precipitation will not occur.


Q.70. For benzoic acid the ionization constant is \(6.46\times 10^{-5}M\) and for silver benzoate Ksp is \(2.5\times 10^{-5}M\). Give relation between the solubility of silver benzoate in buffer of pH = 3.19 and its solubility in water.


Here, pH = 3.19

[H3O+] = \(6.46\times 10^{-5}M\)

C6H5COOH + H2O à \(C6H5COO^{-}\) + H3O

\(K_{a}\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{C_{6}H_{5}COOH}\) \(K_{a}\frac{[C_{6}H_{5}COOH]}{C_{6}H_{5}COO^{-}} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46\times 10^{-4}}{6.46\times 10^{-5}} = 10\)

Assuming the solubility of silver benzoate (C6H5COOAg) is y mol/L.

Now,  [Ag+] = y

[C6H5COOH] = \([C6H5COO^{-}]\) = y

10\([C6H5COO^{-}]\) + \([C6H5COO^{-}]\) = y

\([C6H5COO^{-}]\) = y/11

Ksp[Ag+]\([C6H5COO^{-}]\) = y

\(2.5\times 10^{13} = y\frac{y}{11}\)

y = \(1.66\times 10^{-6}\) mol/L

Hence, solubility of C6H5COOAg in buffer of pH = 3.19 is \(1.66\times 10^{-6}\) mol/L.

For, water:

Assuming the solubility of silver benzoate (C6H5COOAg) is x mol/L.

Now,  [Ag+] = x M

Ksp = [Ag+]\([C6H5COO^{-}]\)

Ksp = (y)2

y = \(\sqrt{K_{sp}} = \sqrt{2.5\times 10^{-13}} = 5\times 10^{-7} mol/L\) \(∴ \frac{y}{x} = \frac{1.66\times 10^{-6}}{5\times 10^{-7}} = 3.32\)

Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.


Q.71. Calculate the maximum concentration of equimolar solutions of  FeSO4 and Na2SO4 so that when they are mixed in equal volume than there is no precipitation of FeS? (Ksp for Fes is \(6.3\times 10^{-18}\))


Assuming the maximum concentration of each solution is y mol/L

On mixing the solutions the volume of the concentration of each solution is reduced to half.

After mixing the maximum concentration of each solution is y/2 mol/L.

Thus, [FeSO4] = [Na2S] = y/2 M

So, [Fe2+] = [FeSO4] = y/2 M

FeS(s) \(\leftrightarrow\) \(Fe^{2+}_{(aq)} + S^{2-}_{(aq)}\)

Ksp = [Fe2+][S2-] \(6.3\times 10^{-18} = (\frac{y}{2})(\frac{y}{2})\) \(\frac{y^{2}}{4} = 6.3\times 10^{-18}\)

Thus, y = \(5.02\times 10^{-9}\)

Thus, if the concentration of FeSO4 and Na2SO4 are equal to or less than that of \(5.02\times 10^{-9}M\), then there won’t be precipitation of FeS.


Q.72. Find the minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K?

Ksp for CaSO4 is \(9.1\times 10^{-6}\)


\(CaSO_{4(s)}\leftrightarrow Ca^{2+}_{(aq)} + SO_{4}^{2-}(aq)\)

Ksp = \([Ca^{2+}][SO_{4}^{2-}]\)

Assuming the solubility of calcium sulphate is x.

So, Ksp = x2

\(∴ \;9.1\times 10^{-6} = x^{2}\) \(∴ \;x = 3.02\times 10^{-3} mol/L\)

Now, molecular mass os calcium sulphate is 136g/mol.

Solubility in calcium sulphate in g/mol is

\(= 3.02\times 10^{-3} \times 136\)

= 0.41 g/L

i.e. 1 litre H2O will be required to dissolve 0.41g of calcium sulphate.

Thus, minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K is

= \(\frac{1}{0.41} L = 2.44 L\)


Q.73. The concentration of S2- in 0.1M HCl solution saturated with H2S is \(1.0\times 10^{19}M\). 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 = \(2.5\times 10^{-13}\)

For ZnS, Ksp = \(1.6\times 10^{-24}\)

For CdS, Ksp = \(8.0\times 10^{-27}\)

For FeS, Ksp = \(6.3\times 10^{-18}\)



If the ionic product exceeds the Ksp value, then only precipitation can take place.

Before mixing:

[S2-] = Ksp = \(1.0\times 10^{-19}M\) [M2+] = 0.04M

Volume = 10mL                                                        Volume = 5mL

After mixing:

[S2-] = ? and                                                                           [M2+] = ?

Total volume = (10 + 5) = 15mL                                         Volume = 15mL

[S2-] = \(\frac{1.0\times 10^{-19}\times 10}{15} = 6.67\times 10^{-20}M\) [M2+] = \(\frac{0.04\times 5}{15} = 1.33\times 10^{-2}M\)


Now, the ionic product = [M2+][S2-]

= \((1.33\times 10^{-2})(6.67\times 10^{-20})\)

= \(8.87\times 10^{-22}\)

Here, the ionic product of CdS and ZnS exceeds its corresponding Ksp value.

Thus, precipitation will occur in  ZnCl2 and CdCl2 solutions.

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