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 class 11 chemistry. In NCERT solutions for class 11 chemistry chapter 7 Equilibrium, there are 73 questions in the exercise. The solutions of NCERT 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.
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 BronstedLowry, 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.
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. 
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 temperaturedependent, 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:
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
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:
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:
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 = 0.04370.0259 = 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 = 23 = 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:
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:
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:
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:
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
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:
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)
= (52870)
= 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
Answer:
Given that,
for the reaction =
Let the pressure of both and at equilibrium be .
initial conc. 10 0 0
at eq 102p 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
(ii) increasing the temperature
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 :
_{ }
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 :
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 :
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
(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.
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:
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,
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 acidbase pair? Find the conjugate acid/base for the following species:
Answer:
A conjugate acidbase 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 acidbase pair.
Species 
Conjugate acidbase 

(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:
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.
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)
=
=
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:
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:
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:
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:
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
= 141.221
= 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