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**NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium- ** In this chapter, you will get the NCERT solutions for class 11 chemistry chapter 7 Equilibrium of chemical and physical processes and details about equilibrium's dynamic nature. There are some insights about the equilibrium constant, the law of mass action and what are the factors affecting the equilibrium state. It is a lengthy and also an important chapter of NCERT syllabus class 11 chemistry.

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- Equilibrium Class 11 Chemistry Chapter 7 Notes

In NCERT solutions for class 11 chemistry chapter 7 Equilibrium, there are 73 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 7 Equilibrium are prepared and solved by chemistry experts. These NCERT solutions will help you in the preparation of your class 11 final examination as well as in the various competitive exams like JEE Mains, NEET, BITSAT, etc. By referring to the NCERT solutions for class 11 , students can understand all the important concepts and practice questions well enough before their examination

**What is the Equilibrium? **

Equilibrium is the state of a process in which the properties like the concentration of the system, pressure, and temperature do not change with the passage of time. It can be established for both chemical and physical processes. At the stage of equilibrium, the rate of forwarding and reverse reactions are equal. The state of equilibrium can only be achieved if the reversible reaction is taking place in a closed system.

Important formulas of NCERT class 11 chemistry chapter 7 Equilibrium 1.Equilibrium constant, K 2. Concentration quotient, Q 3. 4. 5. |

**NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium **

** Question ****7.1(a) ** A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

What is the initial effect of the change on vapour pressure?

** Answer ** :

By increasing the volume of the container suddenly, initially, the vapour pressure would decrease. It is due to, the amount of vapour has remained the same but it is distributed in a larger volume.

** Question ****7.1(b) ** A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

How do rates of evaporation and condensation change initially?

** Answer ** :

Here the temperature is constant so that the rate of evaporation is also the same as before. On increasing the volume of the container, the density of vapour decreases due to which the rate of collision between vapour particles decreases. Hence the condensation rate also decreases initially.

** Question ****7.1(c) ** A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased .

What happens when equilibrium is restored finally and what will be the final vapour pressure?

** Answer ** :

When equilibrium is restored, the rate of evaporation is equal to the rate of condensation. The temperature is constant and the volume is changed. The vapour pressure is temperature-dependent, not volumes. So, that final vapour pressure is equal to the initial vapour pressure.

** Answer ** :

Followings are the given information values to solve the above problems-

The given chemical equation is -

The equilibrium constant for this reaction is expressed as;

(approx)

** Question ****7.3 ** At a certain temperature and total pressure of 10 ^{ 5 } Pa, iodine vapour contains 40% by volume of I atoms

Calculate Kp for the equilibrium.

** Answer ** :

It is given that total pressure (P _{ T } )is and partial pressure of atom = 40 % of P _{ T }

So, the partial pressure of atom = Pascal

The partial pressure of = 60% of P _{ T }

So, the partial pressure of = Pascal

Now, for the reaction

** Question ****7.4 ** Write the expression for the equilibrium constant, K _{ c } for each of the following reactions:

(i)

(ii)

(iii)

(iv)

(v)

** Answer ** :

The equilibrium constant for any reaction can be written as (concentration of products) / (concentration of reactants). And we considered constant values for the solids and liquids because their density per unit volume or mass per unit volume does not change.

Thus,

(i)

(ii)

(iii)

(iv)

(v)

** Question ****7.5 ** Find out the value of K _{ c } for each of the following equilibria from the value of K _{ p } :

(i)

(ii)

** Answer ** :

We know that the relation between and is expressed as;

............................(i)

here = (no. of moles of product) - (no. of moles of reactants)

R = 0.0831 bar L /mol/K, and

For (i)

= and Temp (T) = 500K

= 3 - 2 = 1

By putting the all values in eq (i) we get

For (ii)

= and temp(T) = 1073 K

= 2 - 1 = 1

Now, by putting all values in eq (i) we get,

** Question ****7.6 ** For the following equilibrium, K _{ c } = 6.3 × 10 ^{ 14 } at 1000 K

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K _{ c } , for the reverse reaction?

** Answer ** :

It is given that,

we know that for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:

** Question ****7.7 ** Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

** Answer ** :

For the pure liquids and solids, the molecular mass and the density at a particular temperature is always fixed and it is considered as a constant. Thus they can be ignored while writing the equilibrium constant expression

** Question ****7.8 ** Reaction between N _{ 2 } and O _{ 2- } takes place as follows:

If a mixture of 0.482 mol N _{ 2 } and 0.933 mol of O _{ 2 } is placed in a 10 L reaction vessel and allowed to form N _{ 2 } O at a temperature for which K _{ c } = 2.0 ×10 ^{ -37 } , determine the composition of equilibrium mixture.

** Answer ** :

It is given that,

Let the concentration of at equilibrium be . So,

initial conc 0.482 0.933 0 (in moles)

at equilibrium 0.482- 0.933 - (in moles)

The equilibrium constant is very small. So, we can assume 0.482- = 0.482 and 0.933 - = 0.933

We know that,

{dividing the moles by 10 to get concentration of ions)

So, the concentration of

** Question ****7.9 ** Nitric oxide reacts with Br _{ 2 } and gives nitrosyl bromide as per reaction given below:

When 0.087 mol of NO and 0.0437 mol of Br _{ 2 } are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br _{ 2 } .

** Answer ** :

The initial concentration of and is 0.087 mol and 0.0437 mol resp.

The given chemical reaction is-

Here, 2 mol of produces from 2 mol of . So, 0.0518 mol of is obtained from 0.518 mol of .

Again, From 1 mol of two mol of produced. So, to produce 0.518 mol of we need mol of .

Thus, the amount of present at equilibrium = 0.087 - 0.0518 = 0.0352 mol

and the amount of present at the equilibrium = 0.0437-0.0259 = 0.0178 mol

** Question ****7.10 ** At 450K, K _{ p } = 2.0 × 10 ^{ 10 } /bar for the given reaction at equilibrium.

What is K _{ c } at this temperature?

** Answer ** :

We have,

We know that the relation between and ;

Here = ( moles of product) - (moles of reactants)

So. here = 2-3 = -1

By applying the formula we get;

= 7.48

** Answer ** :

The initial pressure of is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of is ** 0.2 - 0.04 = 0.16 ** .

The given reaction is:

At equilibrium,

Therefore,

** Answer ** :

We have,

The concentration of species are-

We know the formula of

The reaction is not in equilibrium. Since , the equilibrium proceeds in reverse direction.

** Question ****7.13 ** The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.

** Answer ** :

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

Calculate the equilibrium constant for the reaction.

** Answer ** :

The given reaction is-

initial conc 1/10 1/10 0 0

At equilibrium 0.6/10 0.6/10 0.04 0.04

Now, the equilibrium constant for the reaction can be calculated as;

= 0.44 (approx)

** Question ****7.15 ** At 700 K, equilibrium constant for the reaction:

is 54.8. If 0.5 mol L ^{ -1 } of HI(g) is present at equilibrium at 700 K, what are the concentration of H _{ 2 } (g) and I _{ 2 } (g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

** Answer ** :

We have,

The equilibrium constant of the reaction =

moles of = 0.5 mol/L

The given reaction is-

So, the reverse equilibrium constant is

Suppose the concentration of hydrogen and iodine at equilibrium be

Therefore,

So, the value of =

** Answer ** :

The given reaction is:

Initial conc. 0.78 M 0 0

At equilibrium (0.78 - 2 ) M M M

The value of

Now we can write,

By solving this we can get the value of = 0.167

** Answer ** :

Suppose the pressure exerted by the hydrogen and ethene gas be at equilibrium.

the given reaction is-

initial pressure 4 atm 0 0

At equilibrium 4 -

Now,

By solving the quadratic equation we can get the value of = 0.38

Hence,at equilibrium,

= 4 - p = 4 -.038

= 3.62 atm

** Question ****7.18(i) ** Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

(i) Write the concentration ratio (reaction quotient), Q _{ c } , for this reaction (note: water is not in excess and is not a solvent in this reactio ** n) **

** Answer ** :

The given reaction is-

the concentration ratio (reaction quotient) of the given chemical reaction is-

** Question ****7.18(ii) ** Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

(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.

** Answer ** :

Let the volume of the mixture will be V.

initial conc. 1/V 0.18/V 0 0

At equilibrium (= 0.829/V)

So the equilibrium constant for the reaction can be calculated as;

(approx)

** Question ****7.18(iii) ** Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

(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 ** :

Let the volume of the mixture will be V.

initial conc. 1/V 0.5/V 0 0

At equilibrium (= 0.786/V)

Therefore the reaction quotient of the reaction-

Since equilibrium has not been reached.

** Answer ** :

We have,

concentration of = 0.05 mol/L

and

Suppose the concentrations of both and at equilibrium be mol/L. The given reaction is:

at equilibrium 0.05

it is given that the value of the equilibrium constant,

Now we can write the expression for equilibrium as:

(approx)

Hence the concentration of and is 0.0204 mol / L

** Answer ** :

We have,

the initial pressure of and are 1.4 atm and 0.80atm resp.

The given reaction is-

initially, 1.4 atm 0.80 atm

Since the reaction will proceed in the backward direction to attain equilibrium. ** The partial pressure of will increase = decrease in the partial pressure of = **

** **

** By solving ** the above equation we get the value of ** = 0.339 atm **

** Question ****7.21 ** Equilibrium constant, K _{ c } for the reaction

At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L ^{ -1 } N _{ 2 } , 2.0 mol L ^{ -1 } H _{ 2 } and 0.5 mol L ^{ -1 } NH _{ 3 } . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

** Answer ** :

The given reaction is:

at a particular time: 3.0molL ^{ -1 } 2.0 molL ^{ -1 } 0.5molL ^{ -1 }

Now, we know that,

^{}

It is given that K _{ c } = 0.061

Since, Qc Kc, the reaction mixture is not at equilibrium.

Again, , the reaction will proceed in the forward direction to attain the equilibrium.

** Question**** 7.22 ** Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

For which K _{ c } = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10 ^{ -3 } mol L ^{ -1 } , what is its molar concentration in the mixture at equilibrium?

** Answer ** :

Suppose the amount of bromine and chlorine formed at equilibrium. The given reaction is:

Initial Conc. ^{ } 0 0

at equilibrium - 2

Now, we can write,

By solving the above equation we get,

=

Hence, at equilibrium

M

** Question ****7.23 ** At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO _{ 2 } in equilibrium with soild carbon has 90.55% CO by mass

Calculate K _{ c } for this reaction at the above temperature .

** Answer ** :

Suppose the total mass of the gaseous mixture is 100 g.

Total pressure is 1 atm

Mass of = 90.55 g

And, mass of = (100 - 90.55) = 9.45 g

Now, number of moles of = 90.55/28 = 3.234 mol (mol. wt of = 28)

Number of moles of = 9.45/44 = 0.215 mol (mol. wt of = 44)

Partial pressure of ,

= 0.938 atm

Similarly partial pressure of ,

= 0.062 atm

Thus,

By using the relation

= 0.159 (approx)

** Question ****7.24 ** Calculate a) for the formation of NO2 from NO and O2 at 298K

where

** Answer ** :

Given data,

given chemical reaction-

for the reaction,

= (products) - (reactants)

= (52-87-0)

= -35 kJ/mol

(a)

(b)

(c)

** Answer ** :

According to Le Chatellier's principle, if the pressure is decreased, then the equilibrium will shift in the direction in which more number of moles of gases is present.

So,

- (a)The number of moles of reaction products will increase.
- (b)The number of moles of reaction products will decrease
- (c)The number of moles of reaction products remains the same.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

** Answer ** :

According to ** Le Chatellier's principle ** , if the pressure is increased, then the equilibrium will shift in the direction in which less number of moles of gases is present. So, as per this rule following given reactions are affected by the increasing pressure-

The reaction (i), (iii), and (vi)- all proceeds in the backward direction

Reaction(iv) will shift in the forward direction because the number of moles of gaseous reactants is more than that of products.

** Question ****7.27 ** The equilibrium constant for the following reaction is 1.6 ×10 ^{ 5 } at 1024K

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K .

** Answer ** :

Given that,

for the reaction =

Let the pressure of both and at equilibrium be .

initial conc. 10 0 0

at eq 10-2p p p

Now,

By solving the above equation we get,

= 0.00248 bar

Hence the pressure of and is 0.00248 bar and pressure of is 0.00496 bar

** Question ****7.28(a) ** Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

Write an expression for Kp for the above reaction.

** Answer ** :

the expression of ionisation constant ( Kp) for the reaction can be defined as the ratio of the product of concentration to the product of reactants.

** Question ****7.28(b) ** Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

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) According to Le Chatellier's principle, if pressure is increased, then the reaction will shift towards the less number of moles of gases. So, here the direction of equilibrium is backward and the value of decreases.

(ii) According to Le Chatellier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. The value of is increases.

(iii) The equilibrium of the reaction is not affected by the presence of the catalyst. It only increases the rate of reaction.

** Question ****7.29(a) ** Describe the effect of :

(a) addition of H _{ 2 }

_{ }

** Answer ** :

(a)According to Le Chatelliers principle, on the addition of dihydrogen, the number of mole of increases on the reactant side. Thus to attain the equilibrium again the reaction will move in the forward direction.

(b) According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

(c) If we remove the from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

(d) On removal of the equilibrium will shift in the forward direction.

** Question ****7.29(b) ** Describe the effect of :

addition of CH _{ 3 } OH

** Answer ** :

According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

** Question ****7.29(c) ** Describe the effect of :

removal of CO

** Answer ** :

If we remove the from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

(a) write an expression for Kc for the reaction.

(b) what is the value of K _{ c } for the reverse reaction at the same temperature ?

(c) what would be the effect on K _{ c } if

(i) more PCl _{ 5 } is added

(ii) pressure is increased

(iii) the temperature is increased ?

** Answer ** :

We have,

Equilibrium constant for the above reaction =

(a) Expression of for this reaction-

(b) The value of reverse equilibrium constant can be calculated as;

(c).i would remain the same because the temperature is constant in this case.

(c). ii If we increase the pressure, there is no change in because the temperature is constant in this case also.

(c). iii In an endothermic reaction, the value of increases with increase in temperature.

If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that = 4.0 bar, what will be the partial pressure of H _{ 2 } at equilibrium? Kp= 10.1 at 400°C

** Answer ** :

We have,

The partial pressure of and is 4 bar and the

Let be the partial pressure of and at equilibrium. The given reaction is-

Initial concentration 4 bar 4 bar 0 0

At equilibrium 4 - 4 -

Therefore, we can write,

By solving the above equation we get, = 3.04

Hence, the partial pressure of dihydrogen at equilibrium is 3.04 bar

** Question ****7.32 ** Predict which of the following reaction will have an appreciable concentration of reactants and products:

(a)

(b)

(c)

** Answer ** :

If the value of is in the range of to , then the reaction has appreciable concentrations of both reactants and products.

Therefore, the third reaction (c) will have an appreciable concentration of reactants and products.

** Answer ** :

We have,

equilibrium constant of the reaction =

the concentration of dioxygen =

the given reaction is-

Then we have,

(equilibrium constant)

Thus the concentration of dioxygen is

** Question ****7.34(a) ** The reaction,

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of , 0.10 mol of and 0.02 mol of and an unknown amount of CH _{ 4 } in the flask. Determine the concentration of CH _{ 4 } in the mixture. The equilibrium constant, K _{ c } for the reaction at the given temperature is 3.90.

** Answer ** :

Given that,

Total volume = 1L

0.3 mol of , 0.10 mol of dihydrogen( )and the 0.02 mol of water( )

the equilibrium constant = 3.90

Let be the concentration of methane at equilibrium. The given reaction is-

At equilibrium, 0.3 mol/L 0.1 mol/L 0.02 mol/L

Therefore,

Thus the concentration of methane at equilibrium is

** Question ****7.34(b) ** What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

** Answer ** :

A conjugate acid-base pair means that the species are differed by only one proton. for example; is an acid because it donates a proton to water. So , and these pairs are called conjugate acid-base pair.

Species | Conjugate acid-base |

(Base) | |

(Acid) | |

(base) | |

(acid) | |

(acid) | |

(acid) | |

(acid) |

** Question ****7.35 ** Which of the followings are Lewis acids?

** Answer ** :

** Lewis acid- **

Those species which can accept the pair of electrons are called Lewis acids. For example Boron trifluoride ( ), ammonium ion( ) and the hydrogen ion( ). Among them water molecule is a Lewis base, it can donate pair of electrons.

** Question ****7.36 ** What will be the conjugate bases for the Brönsted acids: ?

** Answer ** :

When Brönsted acids lose their proton then they become a conjugate base of that corresponding acids.

Followings are the conjugate base of Brönsted acid-

** Question ****7.38 ** Write the conjugate acids for the following Brönsted bases: .

** Answer ** :

When Brönsted base accepts a proton then they become a conjugate acid of that corresponding base.

Followings are the conjugate acid of Brönsted base-

** Question ****7.39 ** The species: can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

** Answer ** :

When acid or base accept or lose a proton, they form conjugate acid or base of that corresponding species.

Lists of the conjugate acid and conjugate base of the given species-

Species | Conjugate acid | Conjugate base |

** Question ****7.40 ** Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

(a)

(b)

(c) .

(d)

** Answer ** :

Species which donate pair of an electron are called Lewis base and which accepts pair of electrons are called acid.

(a) ^{ } is a Lewis base since it can donate its lone pair of electrons.

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

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

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

** Question ****7.41 ** The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10 ^{ -3 } M. what is its pH?

** Answer ** :

We have,

the concentration of Hydrogen ion sample is M

So,

** Question ****7.42 ** The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

** Answer ** :

We have,

The pH of a sample of vinegar is 3.76

=?

Therefore,

Taking antilog on both sides we get,

= antilog (-3.76)

Hence the concentration of hydrogen ion

** Answer ** :

We have,

IOnization constant of hydrogen fluoride, methanoic acid and hydrogen cyanide are and respectively.

It is known that,

...........................(i)

of the conjugate base

Similarly,

By using the equation (i)

of the conjugate base

Again, with the help of eq (i)

of the conjugate base

** Answer ** :

We have,

The ionization constant of phenol is , and

the concentration of phenol is 0.05 M

degree of ionisation = ?

Ionization of phenol;

At equilibrium,

the concentration of various species are-

As we see, the value of ionisation is very less. Also will be very small. Thus we can ignore .

Hence the concentration of phenolate ion is

Let be the degree of dissociation of phenol in the presence of 0.01 M of phenolate ion.

Concentration (1 - ) 0.05 0.05 0.05

So,

therefore,

The degree of dissociation is

** Answer ** :

We have,

1st ionisation constant of hydrogen sulphide is and the 2nd dissociation constant is

** Case 1st-(absence of hydrochloric acid) **

To calculate the concentration of

Let be the concentration of and the ionisation of hydrogen sulphide is;

0.1 M

At equilibrium, the concentration of various species are,

Since the dissociation constant is is very small. So, can be neglected.

the concentration of = 1- M

the concentration of and is M

So,

from here can be calculated and we get,

** Case 2nd (In presence of 0.1 M, HCl) **

Suppose is dissociated is .Then at equilibrium,

and the

So,

Thus the concentration of [ ] is

** Answer ** :

It is given,

The ionisation constant of acetic acid is and concentration is 0.05 M

The ionisation of acetic acid is;

Therefore,

So, the of the solution =

=

=

We know that,

** Answer ** :

We have,

pH of organic acid is 4.15 and its concentration is 0.01M

Suppose the organic acid be HA. The dissociation of organic acid can be written as;

(By taking antilog of -4.15)

Now,

[HA] = 0.01

Then,

Thus

** Question ****7.48 ** 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 ** :

Assuming the complete dissociation. So,

(a) The ionisation of hydrochloric acid is

Since it is fully ionised then

Therefore,

** of the solution ** =

= ** = 2.52 **

(b) The ionisation of

Therefore,

of the solution =

** of the solution is equal to (14 - 2.301 =11.70) **

** (c) ** The ionisation of

Therefore,

of the solution =

** of the solution is equal to (2.69) **

(d) The ionisation of

Therefore,

of the solution =

** of the solution is equal to (14 - 2.69 = 11.31) **

** Question ****7.49(a) ** Calculate the pH of the following solutions:

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

** Answer ** :

Here, 2 g of dissolves in water to give 2 litres of solution

So, the concentration of =

(the molar mass of is 221)

can be dissociated as

Therefore, (since K _{ w = ) }

So, the concentration of =

Thus

= 11.65(approx)

** Question ****7.49(b) ** Calculate the pH of the following solutions:

0.3 g of dissolved in water to give 500 mL of solution.

** Answer ** :

The calcium hydroxide ion dissociates into-

Molecular weight of = 74

the concentration of =

We know that,

=

Thus

= ** 14 - 1.79 = 12.21 **

** Question ****7.49(c) ** Calculate the pH of the following solutions:

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

** Answer ** :

dissociates into

So, the concentration of =

We know that ,

Now, =

** Question ****7.49(d) ** Calculate the pH of the following solutions:

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

** Answer ** :

We know that,

M _{ 1 } V _{ 1 } (before dilution) = M _{ 2 } V _{ 2 } (after dilution)

initially V _{ 1 = } 1mL and M _{ 1 } = 13.6 M

and V _{ 2 } = 1L and M _{ 2 } = ?

By putting all these values we get,

Thus

= 1.86 (approx)

** Answer ** :

We have,

Degree of ionization(a) = 0.132

Concentration of bromoacetic acid (C) = 0.1 M

Thus the concentration of

= 0.0132

Therefore

= 1.879

Now, we know that,

So,

(approx)

** Question ****7.51 ** The pH of 0.005M codeine solution is 9.95. Calculate its ionization constant and pK _{ b. }

** Answer ** :

We have,

C = 0.005 M

= 9.95 and = 14 - 9.95 = 4.05

we know that

By taking antilog on both sides we get.

concentration of =

C.a =

So, a =

We know that,

Thus

** Answer ** :

We have,

C = 0.001 M

=

Degree of inozation of aniline (a) = ?

Ionization constant of the conjugate acid ( ) = ?

We know that

=

= (0.001)

Thus

=

Then [Base] = C.a = ( )( )

=

Now,

= 14 - 6.187

= ** 7.813 **

It is known that,

So,

** This is ionization constant. **

0.01M

** Answer ** :

We have,

C = 0.05 M

By taking antilog on both sides we get,

from here we get the value of

After adding hydrochloric acid, the concentration of ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease.

(i) when 0.01 HCl is taken

Initial conc. 0.05 0 0

after dissociation 0.05 - 0.001+

As the dissociation is very small.

So we can write 0.001+ 0.001 and 0.05 - 0.05

Now,

So, the value of =

Now degree of dissociation = (amount dissociated) (amount taken)

=

=

0.1M in HCl ?

** Answer ** :

Let the amount of acetic acid is dissociated in this case

Initial conc. 0.05 0 0

after dissociation 0.05 - 0.1+

As the dissociation is very small.

So we can write 0.1+ 0.1 and 0.05 - 0.05

So, the value of =

Now the degree of dissociation = (amount dissociated) (amount is taken)

=

=

** Answer ** :

We have,

(Degree of ionization)

Concentration of dimethylamine = 0.02 M

If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization

(0.1 M) (0.1 M)

and also,

0.02-

(since the dissociation is very small)

Therefore,

Hence in the presece of 0.1 M of sodium hydroxide , 0.54% of dimethylamine get dissociated.

** Question ****7.55(a) ** Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human muscle-fluid, 6.83

** Answer ** :

We have 6.83

It is known that

Therefore,

By taking antilog on both sides we get,

** Question ****7.55(b) ** Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human stomach fluid, 1.2

** Answer ** :

We have 1.2

It is known that

Therefore,

By taking antilog on both sides we get,

** Question ****7.55(c) ** Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human blood, 7.38

** Answer ** :

we have 7.38

It is known that

Therefore,

By taking antilog on both sides we get,

** Question ****7.55(d) ** Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human saliva, 6.4.

** Answer ** :

we have 6.4

It is known that

Therefore,

By taking antilog on both sides we get,

** Answer ** :

We already know that can be calculated as-

to calculate the concentration of = antilog (- )

Thus, the hydrogen ion concentration of followings values are-

(i) of milk = 6.8

Since,

6.8 =

= -6.8

= anitlog(-6.8)

=

(ii) of black coffee = 5.0

Since,

5.0 =

= -5.0

= anitlog(-5.0)

=

(iii) of tomato juice = 4.2

Since,

4.2 =

= -4.2

= anitlog(-4.2)

=

(iv) of lemon juice = 2.2

Since,

2.2 =

= -2.2

= anitlog(-2.2)

=

(v) of egg white = 7.8

Since,

7.8 =

= -7.8

= anitlog(-7.8)

=

** Answer ** :

We have 0.562 g of potassium hydroxide ( ). On dissolving in water gives 200 mL of solution.

Therefore, concentration of =

= 2.805 g/L

It is a strong base. So, that it goes complete dissociation.

It is known that,

Therefore,

** Answer ** :

By given abova data, we know the solubility of at 298 K = 19.23 g/L

So, concentration of

= (Molecular weight of = 121.63 u)

= 0.1581 M

and the concentration of

Now

It is known that,

=

Therefore

** Answer ** :

Let the degree of ionization of propanoic acid be . Then Let suppose we can write propanoic acid to be HA,

It is known that,

We have

ionization constant of propanoic acid ( )= and the concentration is 0.005 M

By putting the values in above formula we get,

[Acid] = = C. =

Therefore,

If we add 0.01M hydrochloric acid then,

initial con C 0 0

at equi. C - C 0.01 +

Now, by using the formula of

The value of is calculated as ;

(Degree of ionisation)

** Answer ** :

We have,

Concentration of cyanic acid = 0.1 M

Therefore, the concentration of = antilog (-2.34)

=

It is known that,

=

Then Ionization constant ( ) =

** Answer ** :

We have,

Ionization constant of nitrous acid =

Concentration of sodium nitrite ( ) = 0.04 M

Degree of hydrolysis can be calculated as;

Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid ( )

Suppose moles of salt undergoes hydrolysis, then the concentration of-

, and

Therefore

from here we can calculate the value of ;

Now

Therefore the degree of hydrolysis

** Question ****7.62 ** A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

** Answer ** :

Given,

= 3.44

We know that

By taking antilog on both sides we get,

= antilog (- 3.44)

pyridinium hydrochloride completely ionised.

Then = (conc. of products)/ (conc, of reactants)

= (? Concentration is 0.02M)

Now,

(approx)

** Question ****7.63 ** Predict if the solutions of the following salts are neutral, acidic or basic:

** Answer ** :

Salts of strong acid and strong base are neutral in nature for example-

Salts of a strong base and weak acid are basic in nature for example-

Salts of strong acid and a weak base are acidic in nature for example-

** Answer ** :

We have,

Ionisation constant of chloroacetic acid( ) is

The concentration of acid = 0.1 M

Ionisation if acid, =

We know that,

....................(i)

As it completely ionised

Putting the values in eq (i)

Therefore, of the solution =

=

=

Now,

0.1 M (sod. chloroacetate) is basic due to hydrolysis-

For a salt of strong base+strong acid

** Question ****7.65 ** Ionic product of water at 310 K is 2.7 × 10 ^{ -14 } . What is the pH of neutral water at this temperature?

** Answer ** :

We have the ionic product of water at 310 K is

It is known that,

ionic product

SInce , therefore

at 310 K is

here we can calculate the value of concentration.

Thus,

Hence the of neutral water is 6.78

** Question ****7.66(a) ** Calculate the pH of the resultant mixtures:

10 mL of 0.2M Ca(OH) _{ 2 } + 25 mL of 0.1M HCl

** Answer ** :

Given that,

Vol. of 0.2 M = 10 mL

Vol. of 0.1 M HCl = 25 mL

therefore, by using the formula,

By substituting the value in these equations, we get;

Now,

since

= 14-1.221

= 12.78

** Question ****7.66(b) ** Calculate the pH of the resultant mixtures:

10 mL of 0.01M H _{ 2 } SO _{ 4 } + 10 mL of 0.01M Ca(OH) _{ 2 }

** Answer ** :

In this case, both the solutions have the same number of moles of and , therefore they both can get completely neutralised. Hence the = 7.0

** Question ****7.66(c) ** Calculate the pH of the resultant mixtures:

c) 10 mL of 0.1M H _{ 2 } SO _{ 4 } + 10 mL of 0.1M KOH

** Answer ** :

Given that,

Volume of 0.1 M = 10 mL, and

Volume of 0.1 M = 10 mL

So, by using the formula of,

By putting the values we get,

Hence,

** Answer ** :

Solubility product is the product of ionic concentrations in a saturated solution.

(i) silver chromate ( )

Ionization of silver chromate

Let " " be the solubility of

According to the table of =

(ii) Barium chromate ( )

Ionization of silver chromate

Let " " be the solubility of

According to the table of =

(iii) Ferric hydroxide ( )

Ionization of Ferric hydroxide

Let " " be the solubility of

According to the table of =

(iv)

Lead chloride ( )

Ionization of Lead chloride

Let " " be the solubility of

According to the table of =

So molarity of and molarity of

** Answer ** :

silver chromate ( )

Ionization of silver chromate

Let " " be the solubility of

of =

Ionization of Silver bromide ( )

of =

Now, the ratio of solubilities

** Answer ** :

We have,

solubility product ( ) of cupric iodate =

When equal volumes of sodium iodate and cupric chlorate are mixed together the molar concentration of both the solution becomes half (= 0.001)

Ionization of cupric iodate is;

0.001 M 0.001 M

So, can be calculated as;

Sinc eionic product is less than the so no precipitation occurs.

** Answer ** :

It is given that,

of buffer solution is 3.19. So, the concentration of ion can be calculated as;

= antilog (-3.19)

=

Ionization of benzoic acid;

It is given that

Therefore,

** Answer ** :

We have,

The solubility product of the

Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume.

Let be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half.

So,

The ionisation of ferrous sulphide;

Therefore, for no precipitation, ionic product = solubility product

By solving the above equation, we get

The maximum concentration of both the solution is M

** Answer ** :

We have,

The solubility product of calcium sulphate is .

given mass of calcium sulphate = 1g

Ionization of calcium sulphate;

Therefore,

Let the solubility of calcium sulphate be .

Then,

mol/L

Thus,

mass/ (mol. wt) volume Molarity

mass =

So, that to dissolve 1 g of calcium sulphate we need = of water.

** Answer ** :

We have,

the concentration of and the volume of the solution containing sulphur ion = 10 mL.

Volume of metal salts solution added = 5mL

Before mixing,

M

After mixing,

Volume = 15 mL

So, the concentration of

concentration of

So, the ionic product =

=

=

For the precipitation of the solution, the ionic product should be greater than the corresponding values.

of are and respectively.

Hence precipitation will take place in and metal salts.

** More About Equilibrium Class 11 NCERT Chemistry Chapter 7 Solutions**

After studying NCERT solutions for class 11 chemistry chapter 7 Equilibrium, you will be able to identify equilibrium's dynamic nature involved in chemical and physical processes; explain the law of equilibrium; explain characteristics of equilibria involved in chemical and physical processes; write expressions for equilibrium constants; explain factors affecting the equilibrium state of a reaction; classify substances as bases or acids according to Bronsted-Lowry, Arrhenius and Lewis concepts; also able to classify bases and acids as weak or strong in terms of their ionization constants and calculate solubility product constant.

**NCERT Solutions for Class 11 Chemistry **

Chapter 1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | Equilibrium |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 |

** NCERT Solutions for Class 11 Subject Wise **

** Benefits of NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium **

- The step-by-step solutions given in the NCERT solutions for class 11 chemistry chapter 7 Equilibrium will help you in understanding chapter easily.
- Revision will be easier because with the help of solutions you will always remember the concepts and get very good marks in your class.
- Homework will not be a problem now, all you need to do is check the detailed NCERT solutions for class 11 chemistry and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

**Also Check NCERT Books and NCERT Syllabus here:**

1. Where can I find complete solutions of NCERT class 11 Chemistry

Refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry

2. How are the ncert solutions beneficial for board exam

Most of the questions are asked directly from NCERT, hence it is must to do NCERT solutions. For more questions students can refer to NCERT exemplar problems.

3. Whether the unit Equilibrium is helpful in higher studies?

Yes, it is helpful to a large extent in the field of chemistry and chemical engineering.

4. What is the official website of NCERT

official website of NCERT: http://www.ncert.nic.in/

5. What is the weightage of NCERT chapter equilibrium for NEET exam

One problem from the chapter equilibrium can be expected for NEET exam. Follow NCERT book and NEET previous year papers for a good mark in the NEET exam.

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