NCERT Solutions for Class 11 Chemistry Part 1 Chapter 6 Equilibrium

 For the given reaction:

Δn=2−3=−1 

T=450K 

R=0.0831bar L bar K−1mol−1 

Kp=2.0×10^10bar−1 

We know that,

Kp=Kc(RT)Δn 

2.0×1010=Kc(0.0831×450)−1

Kc=2.0×1010(0.0831×450)

Kc=74.79×1014Lmol

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

Ans: The vapour pressure would first fall if the container's volume was rapidly increased. This is due to the fact that the amount of vapour remains constant while the volume increases rapidly. As a result, the same amount of vapour is spread over a wider surface area.

2. How do rates of evaporation and condensation change initially? 

Ans: The rate of evaporation is also constant because the temperature is constant. The density of the vapour phase reduces as the container volume increases. As a result, the rate of vapour particle collisions drops as well. As a result, the rate of condensation initially slows.

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

Ans: The rate of evaporation equals the rate of condensation when equilibrium is eventually restored. Only the volume changes in this situation, but the temperature remains fixed. Temperature, not volume, determines vapour pressure. As a result, the final vapour pressure will be the same as the system's initial vapour pressure.

The equilibrium constant for the given reaction will be:

Kc=[SO3]2 / [SO2]2[O2] =(1.90)2M2 / (0.60)2(0.821)M3

Kc=12.239M−1 (approximately)

Therefore, the equilibrium constant for the given reaction is 12.239M−1.

 

According to the question: The Kc for the forward reaction is 6.3×1014

Then, Kc for the reverse reaction will be:

K'c=1/ Kc 

K'c=1/ 6.3×1014

K'c=1.59×10−15

The equilibrium constant for the reverse reaction will be:

1.59×10−15.

 

For a pure substance which is both a solid and a liquid:

[Pure substance]=Number of molesVolume 

[Pure substance]=Mass / Molecular weightVolume

[Pure substance]=MassVolume × Molecular mass

[Pure substance]=DensityMolecular mass

The molecular mass and density of a pure substance (at a given temperature) are now always fixed and accounted for in the equilibrium constant. As a result, the equilibrium constant statement does not include the values of pure substances.

 

 Let the concentration of N2O at  equilibrium be x.

The initial concentration of N2 is 0.482mol.

The initial concentration of O2 is 0.933mol.

At equilibrium the concentration of N2 is (0.482-x)mol, the concentration of

O2is (0.933-x)mol and the concentration of N2O is x mol.

The reaction occurs in a 10L reaction vessel:

[N2]=0.482−x/10 , [O2]=0.933−x/2 /10 , [N2O]=x/10 

The value of equilibrium constant i.e.

Kc=2.0×10−37 is very small. Therefore, the amount of nitrogen and oxygen reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of nitrogen and oxygen. Then,

 

The given reaction is:

2NO (g)+Br2(g)⇌2NOBr (g)

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

NOBr  are formed from 0.0518 mol of NO. Again, 2 mol of NOBr  are formed from 1 mol of Br . Therefore, 0.0518 mol of NOBr are formed from 0.0581 /2 mol of

Br, or 0.0259 mol of NO. The amount of NO and Br present initially is as follows: 

[NO]=0.087mol and [Br2]=0.0437mol 

Therefore, the amount of NO present at equilibrium is:

[NO]=0.087−0.0518=0.0352mol And, the amount of Br  present at equilibrium is:

[Br2]=0.0437−0.0259=0.0178mol 

 

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: 

 

For the given reaction:

N2(g)+3H2(g)⇌2NH3(g)

The given concentration of various species is

[N2]=1.57/ 20molL−1  [H2]=1.92/ 20molL−1 

[NH3]=8.13/ 20molL−1 

Now, reaction quotient Qc  is:

QC=[NH3]^ /2[N2][H2]^3 

QC=((8.13)20)^2 /(1.5720)(1.9220)^3 

QC=2.4×103

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

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

 

Given,

Kp for the reaction i.e.,

H2(g)+Br2(g)↔2HBr(g) is 1.6×105

 

 The given reaction is: N2(g)+3H2(g)⇌2NH3(g)

At a particular time, the concentration of N2 is 3.0 molL−1 , for H2 is 2.0 molL−1 and for NH3 is 0.5 molL−1.

Now, we know that:

Qc=[NH3]2[N2][H2]3=(0.5)2(3.0)(2.0)3=0.0104

It is given that Kc=0.061 

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

Since Qc<Kc  , the reaction will proceed in the forward direction to reach equilibrium.

 

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)

The initial concentration of C2H6 is 4.0 atm and that of C2H4 and H2 is zero (0).

At equilibrium, the concentration of C2H6 is (4.0 – p) and that of C2H4 and

H2 is p. 

Then we can write,

⇒Kpp×p40−p=0.04

p2=0.16−0.04p

p2+0.04p−0.16=0

 

Let the concentrations of both PCl3  and Cl2  at equilibrium be xmolL - 1 . The given reaction is:

PCl5(g)⇌PCl3(g)+Cl(g)

At equilibrium the concentration of PCl5 is

0.5×10−1molL - 1.

It is given that the value of equilibrium constant Kc  is 8.3×10−3

The expression for equilibrium will be:

⇒[PCl3][Cl2] / [PCl5]=Kc

⇒x×x / 0.5×10−1=8.3×10^−3

⇒x2=4.15×10−4

⇒x=2.04×10−2

⇒x=0.0204

Therefore at equilibrium:

[PCl3]=[Cl2]=0.02molL−1 

 

 For the given reaction:

FeO (s)+CO (g)⇌Fe (s)+CO2(g)

The initial concentration of FeO is 1.4 atm and the initial concentration of

CO2 is 0.80 atm. 

It is given in the that, Kp=0.265.

 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  be equal to the  decrease in pressure of CO2  be ‘p’. Then, we can write:

Qp=pCO2pCO=0.801.4=0.571

Therefore, equilibrium partial of CO2 will be,

pCO2=0.080−0.339=0.461 atm 

And, equilibrium partial pressure of CO will be,

pCO=1.4−0.339=1.739 atm 

 

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

2BrCl (g)⇌Br2(g)+Cl2(g) 

At equilibrium the concentration of BrCl is(3.3×10−3−2x) 

Now, we can write.

⇒[Br2][Cl2][BrCl]=Kc

⇒x×x(3.3×10−3−2x)2=32

⇒x3.3×10−3−2x=5.66

⇒x=18.678×10−3−11.32x

⇒12.32x=18.678×10^−3

⇒x=1.5×10^−3

 Therefore, at equilibrium

[BrCl]=3.3×10−3−(2×1.5×10−3)

[BrCl]=3.3×10−3−3.0×10−3

[BrCl]=3.0×10−4 molL−1

 

Let, the total mass of the gas mixture = 100 gm

Mass of CO = 90.55 gm

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

Now, the number of moles of

nCO=90.5528=3.234 mol 

Number of moles of

nCO2=9.4544=0.215 mol 

Partial pressure of total pCO=nCOnCO+nCO2×ptotal 

pCO=3.2343.234+0.215×1

pCO=0.938atm

Partial pressure of CO2 ,

pCO2=nCO2nCO+nCO2×ptotal

pCO2=0.2153.234+0.215×1

pCO2=0.062 atm

Therefore, 

For the given reaction:

Δn=2−1=1 

We know that,

Kp=Kc(RT)Δn14.19 

=Kc(0.082×1127)1

Kc=14.190.082×1127

Kc=0.154

 

NO (g)+12O2(g)⇌NO2(g) 

Where: 

ΔfG∘(NO2)=52kJ/mol

ΔfG∘(NO)=87.0kJ/mol

ΔfG∘(O2)=0kJ/mol

Ans: For the given reaction: 

NO (g)+12O2(g)⇌NO2(g)

1. The ΔG∘ for the given reaction will be:

ΔG∘=ΔG∘(Products)−ΔG∘(Reactants) 

ΔG∘=52.0−[87.0+0]

ΔG∘=−35.0 kJmol−1

2. We know that:

ΔG∘=RTlog⁡Kc

ΔG∘=2.303RTlog⁡Kc

log⁡Kc=−35.0×10−^ / 3−2.303×8.314×298

log⁡Kc=6.13

Kc=antilog (6.134)

Kc=1.36×106

Hence, the equilibrium constant for the given reaction

Kc is 1.36×106 .

 

a. PCl5(g)⇌PCl3(g)+Cl2(g) 

Ans: The amount of moles produced by the reaction will rise. According to Le Chatelier's principle, as pressure is reduced, the equilibrium swings in the direction of a greater number of moles of gas. The number of moles of gaseous products in the reaction is greater than the number of moles of gaseous reactants. As a result, the reaction will be propelled ahead. As a result, there will be more moles of reaction products.

b. CaO (s)+CO2(g)⇌CaCO3(s) 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will decrease.

c. 3Fe (s)+4H2O (g)⇌Fe3O4(s)+4H2(g) 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will remain the same.

 

By increasing the pressure, the reactions in (i), (iii), (iv), (v), and (vi) will be changed. Because the number of moles of gaseous reactants is greater than the number of moles of gaseous products, the reaction in (iv) will proceed in the forward direction. Because the number of moles of gaseous reactants is smaller than that of gaseous products, the reactions in (i), (iii), (v), and (vi) will shift backward.

a. Write an expression for the above reaction.

Kp=pCO×p^3H2 / pCH4×pH2O

b. How will the values of Kp and composition of equilibrium mixture be affected by 

1. Increasing the pressure 

Ans: The equilibrium will shift in the backward direction, according to Le Chatelier's principle.

As the reaction shifts in the backward direction as the pressure is increased, the value of Kp decreases.

2. Increasing the temperature 

Ans: Because the reaction is endothermic, the equilibrium will shift forward, according to Le Chatelier's principle.

The value of Kp will grow as the temperature rises because the reaction will shift forward.

3. Using a catalyst?

Ans: The existence of a catalyst has no effect on the reaction's equilibrium. A catalyst does nothing but speed up a reaction. As a result, equilibrium will be soon achieved. Because the catalyst only changes the rate of reaction, the value of Kp is unaffected.

 

a. Addition of H2 

Ans: According to Le Chatelier's principle, when H2  is added to a reaction, the equilibrium shifts in the forward direction.

b. Addition of CH3OH  

Ans: The equilibrium will shift backwards with the addition of CH3OH .

c. Removal of CO

Ans: When CO  is removed from the equation, the equilibrium shifts backward.

d. Removal of CH3OH 

Ans: When CH3OH  is removed from the equation, the equilibrium shifts forward.

 

a. Write an expression for Kc  for the reaction. 

Ans: The expression for Kc will be:

Kc=[PCl3] [Cl2] / [PCl5] 

b. What is the value of Kc  for the reverse reaction at the same temperature? 

Ans: Value of Kc  for the reverse reaction at the same temperature is:

Kcprime=1 / Kc

=18.3×10^−3=1.2048×10^2

=120−48

c. What would be the effect on Kc  if 

1. More PCl5  is added?

Ans:  Because the temperature is constant in this scenario, Kc will remain constant.

2. Pressure is increased? 

Ans: At constant temperature, Kc remains constant. As a result, Kc would not alter in this scenario.

3. The temperature has increased?

Ans: In an endothermic process, the value of Kc rises as the temperature rises. Because the reaction is an endothermic reaction, the value of Kc will rise as the temperature rises.

 

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

CO (g)+H2O (g)⇌CO2(g)+H2(g)

At equilibrium the concentration of CO and H2O will b (4.0−p).

It is given that, Kp=10.1 

Now,

Pco2×p2 / pco×pH2o=Kp

⇒p×p / (4.0−p)(4.0−p)

⇒p40−p=3.178

⇒p=12.712−3.178p

⇒4.178p=12.712

⇒p=3.04

Hence, at equilibrium, the partial pressure of H2  will be 3.04 bar.

 

If the value of Kc  lies between 10^−3  and

10^3, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

 

The given reaction is:

 

Let the concentration of methane at equilibrium be x.

CO (g)+3H2(g)⇌CH4(g)+H2O(g)

At equilibrium, the concentration of CO  is 0.3 M , H2 is 0.1 M and

H2O  is 0.02 M 

 

A conjugate acid-base pair varies from one other by only one proton. In the table below, the conjugate acid-base for each species is listed. 

Lewis acids are those acids which can accept a pair of electrons. For example:

BF3, H + , NH4 +  are Lewis acids.

 

The table below lists the conjugate bases for the given Bronsted acids:

 

The table below lists the conjugate acids for the given Bronsted bases:

 

The table below lists the conjugate acids and conjugate bases for the given species.

 

a. OH− 

Ans: OH−is a Lewis base since it can donate its lone pair of electrons

b. F− 

Ans: F− is a Lewis base since it can donate a pair of electrons.

c. H +  

Ans: H +  is a Lewis acid since it can accept a pair of electrons.

d. BCl3 

Ans: BCl3  is a Lewis acid since it can accept a pair of electrons.

 

Given,  [H + ]=3.8×10−3 M 

Therefore, the pH value of the soft drink:

pH=−log⁡[H+]

pH=−log⁡(3.8×10−3)

pH=−log⁡3.8−log⁡10−3

pH=−log⁡3.8+3

pH=−0.58+3

pH=2.42

 

Given the, pH=3.76 

It is known as:

pH=−log⁡[H+]

log⁡[H+]=−pH

[H+]=antilog(−pH)

[H+]=antilog(−3.76)

[H+]=1.74×10−4M

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74×10−4 M.

 

 It is known as:

Kb=KwKa 

Given,

Ka of HF is 6.8×10−4

Hence,Kb of its conjugate base

Kb=KwKa

Kb=10−146.8×10−4

Kb=1.5×10−11

Given Ka of HCOOH is

10^−4 

Hence, Kb of its conjugate base

 

According to the reaction of ionization of phenol:

C6H5OH+H2O⇌C6H5O−+H3O+ 

The initial concentration of C6H5OH is 0.05 M. 

At equilibrium, the concentration of C6H5O− and H3O+ will be x, then the concentration of C6H5OH will be (0.05−x).

 

To calculate the concentration of HS− ion:

Case I (in the absence of HCl ):

Let the concentration of HS−be x M:

H2S⇌H + +HS− 

The final concentration of H2S will be (0.1−x) 

Then,

Kai=[H+][HS−] / [H2S] 

9.1×10^−8 = (x)(x) / 0.1−x 

(9.1×10−8)(0.1−x)=x2

Taking, 0.1−xM;0.1M we have (9.1×10−8)(0.1) = x^2 

9.1×10−9=x^2 

 x = √9.1×10−9 

=9.54×10−5M 

⇒[HS−]=9.54×10−5M

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let [HS−] be y M.

 H2S⇌H + +HS−

The final concentration of H2S will be (0.1−y) 

Now, 

Ka1= [HS−][H+] / [H2S]

Ka1=y(0.1+y)(0.1−y)

=9.1×10−8 = y×0.10.1

y=9.1×10−8

The concentration of [HS−]=9.1×10−8 

To calculate the concentration of [S2−] 

Case I (in the absence of 0.1 M of HCl):

 HS−⇌H++S2− 

[HS−]=9.54×10−5 M 

Let [S2−] be X .

Also,

[H+]=9.54×10^−5 M 

Ka2=[H+][S2−] / [HS−] 

Ka2=(9.54×10^−5)(X) / 9.54×10−5 

1.2×10−13=X=[S2−]

Case II (in the presence of 0.1 M of HCl):

The concentration of S2− be X' M 

[HS−]=9.1×10−8 M 

[H+]=0.1 M 

Then, 

Ka2=[H+][S2] / [HS−]

1.2×10−13=(0.1)(X′) / 9.1×10−8 

10.92×10−21=0.1X′

10.92×10−210.1=X′ 

X′=1.092×10−200.1 

=1.092×10−19M 

⇒Ka1=1.74×10−5

 

For the reaction:

CH3COOH+H2O⇌CH3COO−+H3O+ 

The final concentration of H3O +  is 0.05α . Then the final concentration of CH3COOH will be (0.05−0.05α) .

Ka=(.05α)(.05α) / (.05−0.05α) 

=(.05α)(0.05α) / .05(1−α) 

=.05α^2 / 1−α

1.74×10^−5=.05α^2 / 1−α 

1.74×10^−5−1.74×10^−5α=0.05α^2 

0.05α^2+1.74×10^−5α−1.74×10^−5 

D=b2−4ac

=(1.74×10−5)2−4(.05)(1.74×10−5) 

=3.02×10−25+.348×10−5

α = √K / ac 

α= √ 1.74×10^−5.05 

= √ 34.8×10^−5×10 / 10 

√=3.48×10^−6

[CH3COO−]=0.05×1.86×10−3 

=0.93×10^−3 / 1000 

=.000093

pH=−log⁡[H+] 

=−log⁡(.093×10−2) 

∴pH=3.03

 Hence, the concentration of acetate ion in the Ans:is 0.00093 M and its pH is 3.03.

 

Let the organic compound be

 

a. 0.003 M HCl 

Ans:

H2O+HCl⇌H3O++Cl− 

Since HCl is completely ionized.

[H3O+]=[HCl]⇒[H3O+]=0.003 

pH=−log⁡[H3O+]

pH=−log⁡(0.003)

pH=2.52

Hence, the pH of the solution is 2.52.

b. 0.005 M NaOH  

Ans:

NaOH(aq)⇌Na+(aq)+OH−(aq) 

[OH−]=[NaOH]⇒[OH−]=0.005 

pOH=−log⁡[OH−]

pOH=−log⁡(0.005)

pOH=2.30

pH=14−pOH

pH=14−2.30

pH=11.70

Hence, the pH of the solution is 11.70.

c. 0.002 M HBr 

Ans:

HBr+H2O⇌H3O++Br− 

[H3O+]=[HBr]⇒[H3O+]=0.002 

pH=−log⁡[H3O+]

pH=−log⁡(0.002)

pH=2.69

Hence, the pH of the solution is 2.69.

d. 0.002 M KOH 

Ans:

KOH(aq)⇌K+(aq)+OH−(aq) 

[OH−]=[KOH]

⇒[OH−]=0.002

pOH=−log⁡[OH−]

pOH=−log⁡(0.002)

pOH=2.69

pH=14−pOH

pH=14−2.69

pH=11.31

Hence, the pH of the solution is 11.31.

 

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

 

Degree of ionization,

α=0.132 

Concentration, c = 0.1 M

Thus, the concentration of

[H3O+] = 0.1×0.132 

[H3O+]=0.0132

pH=−log⁡[H+]

pH=−log⁡(0.0132)

pH=1.88

Now,

Ka=Cα2

Ka=0.1×(0.132)2

Ka=0.0017p

Ka=2.75

 

Ans: Degree of ionization,

α=0.132

Concentration,

c=0.1M

Thus, the concentration of

⇒0.1×0.132=0.0132  

pH=−log⁡[H+]=−log⁡(0.0132)=1.879:1.88

Now

Cα2 = 0.1×(0.132)^2

Ka=.0017p

Ka=2.75

 

a. 0.01M

b. 0.1 M in HCL

a. Human muscle fluid 6.83

pH=6.83pH=−log[pH=6.83pH=−log⁡[H+] 

∴6.83=−log⁡[H+] 

[H+]=1.4810−7M

b. Human stomach fluid, 1.2: 

pH=1.2 × 1.2=−log⁡[H+] 

∴[H+]=0.063

c. Human blood, 7.38:

pH=7.38=−log⁡[H+]

∴[H+]=4.1710−8M

d. Human saliva, 6.4:

pH=6.4× 6.4=−log⁡[H+] 

[H+]=3.9810−7

 

The hydrogen ion concentration in the given substances can be calculated by using the given relation:

pH=−log⁡[H+]

1. pH of milk = 6.8

Since,pH=−log⁡[H+]

6.8=−log⁡[H+]log

[H+]=−6.8 =anitlog(−6.8)

[H+]=anitlog⁡(−6.8)

=1.519−7M

2. pH of black coffee = 5.0

Since,

pH=×log⁡[H+]

5.0=−log⁡[H+]log

[H+]=−5.0

[H+]=anitlog⁡(−5.0)

=10−5M

3. pH of tomato juice = 4.2

Since,

pH=−log⁡[H+]

4.2=−log⁡[H+]log

[H+]=−4.2

[H+]=anitlog⁡(−4.2)

=6.3110−5M

4. pH of lemon juice = 2.2 

Since,

pH=− log

2.2=−log⁡[H+]log 

[H+]=−2.2 

[H+]=anitlog⁡(−2.2) 

=6.3110−3M

5. pH of egg white = 7.8

Since,

pH=−log⁡[H+]

7.8=−log⁡[H+]log

[H+]=−7.8

[H+]=anitlog⁡(−7.8)

=1.5810−8M

 

[KOHaq]=0.56115g/L

=2.805g/L

=2.805×156.11M=.05M

KOHaq→K(aq)++OH(aq)−

[OH−]=.05M

=[K+][H+][H−]=Kw

[H+]=Kw[OH−]

=10−140.05

=2×10−11M

∴pH=12.70

 

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

Then, concentration of Sr(OH)2 =19.23121.63M

=0.1581M

Sr⁡(OH)2(aq)→Sr(aq)2++2(OH−)(aq)

∴[Sr2+]=0.1581M

[OH−]=2×0.1581M=0.3126M

Now,

Kw=[OH−][H+] 

10−140.3126=[H+] 

⇒[H+]=3.2×10−14 

∴pH=13.495;13.50

 

 Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:

HA +H2O⇔H3O+ + A− 

(.05−0.0α)≈0.5.05α / .05^α

=(.05α)(.05α) / 0.05=.05α^2

α=Ka.05=1.63×10^−2

Then,

[H3O+]=.05α=.05×1.63×10−2=Kb.15×10^−4M

∴pH=3.09

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

[H3O+]=0.01

[A−]=005α′ 

[HA]=.05 

Ka=0.01×.05α′.05 

′1.32×10−5=.01×α′ 

α′=1.32×10^−5

 

c=0.1M 

pH=2.34 

−log⁡[H+]=pH 

−log⁡[H+]=2.34 

[H+]=4.5×10^−3

Also,

[H+]=cα

4.5×10−30.1=α

α=4.5×10^−3=.045 Then, 

=0.1×(45×10−3)^2

=202.5×10^−6

=2.02×10^−4

 

pH=3.44

We know that,

pH=−log⁡[H+]

∴[H+]=3.63×10−4

Then,

Kb=(3.63×10−4)20.02(∵ concentration =0.02M)

⇒Kb=6.6×10−6

Now,

Kb=KwKa

⇒Ka=Kw / Ka = 10^−14 / 6.6×10^−6 = 1.51×10^−9

 

(i) NaCl :

NaCl+H2O↔NaOH+HCl

Strong base     Strong base

Therefore, it is a neutral solution.

(ii) KBr:

KBr+H2O↔KOH+HBr

Strong base    Strong base

Therefore, it is a neutral solution.

iii) NaCN:

NaCN+H2O↔HCN+NaOH

Weak acid        Strong base

Therefore, it is a basic solution.

(iv) NH4NO3

NH4NO3+H2O↔NH4OH+HNO3

Weak acid               Strong base

Therefore, it is an acidic solution.

(v) NaNO2

NaNO2+H2O↔NaOH+HNO2

Strong base         Weak acid

Therefore, it is a basic solution.

(vi) KF

KF+H2O↔KOH+HF

Strong base     Weak acid

Therefore, it is a basic solution.

 

a. 10mL of 0.2MCa(OH)2+25mL of 0.1MHCl

Ans: Moles of

H3O+=25×0.1 / 1000=.0025mol

OH−=10×0.2×2 / 1000=.0040mol

Thus, excess of OH−=.0015mol

b. 10mL of 0.01MH2SO4+10mL of 0.01MCa(OH)2

Ans: Moles of H3O+=2×10×0.1 / 1000=.0002mol

Moles of OH−=2×10×0.1 / 1000=.0002mol

Since there is neither an excess of The solution is neutral. Hence, pH=7.

c. 10mL of 0.1MH2SO4+10mL of 0.1MKOH

Ans: Moles of H3O+=2×10×0.11000=.002mol

Moles of OH−=10×0.11000=0.001mol

Excess of H3O+=.001mol

Thus,

[H3O+]=.00120×10−3=10−320×10−3=.05

∴pH=−log(0.05)

∴pH=−log⁡(0.05)

=1.30

=1.30

 

1. Silver chromate: Ag2CrO4→2Ag++CrO42−

Then,

Ksp=[Ag+]2[CrO42−]

Let the solubility of Ag2CrO4 be

⇒[Ag+]=2s and [CrO42−]=s

Then,

Ksp=(2s)2⋅s=4s3 

⇒1.1×10−12=4s3 

.275×10−12=s3 

MolarityofAg+=2s=2x0.65x10^−4=1.30x10^−4M

Molarity of CrO42−=s=0.65×10^−4M

2. Barium Chromate:

BaCrO4→Ba2++CrO42−

Then,

Ksp=[Ba2+][CrO42−]

Let the solubility of BaCrO4 be [Ba2+]=s and [CrO42−]=s⇒Ksp=s2

⇒1.2×10−10=s2

M ⇒s=1.09×10^−5M

Molarity of Ba2+= Molarity of

CrO42−=s=1.09×10−5M

3. Ferric Hydroxide:

Fe(OH)3→Fe2+3OH− 

Ksp=[Fe2+][OH−]3

Let s be the solubility of Fe(OH)3

Molarity of OH−=3s=4.17×1010M

4. Lead Chloride:

PbCl2→Pb2++2Cl− 

KSP=[Pb2+][Cl−]2

KSP be the solubility of

[PB2+]=s and [Cl−]=2s 

 Thus, Ksp=s.(2s)2 

⇒1.6×10−5=4s3 

⇒0.4×10−5=s3 

4×10−6=s3⇒1.58×10−2M=S.1

Molarity of chloride

=2s=3.16×10−2M

5. Mercurous Iodide:

Hg2I2→Hg2++2I− 

Ksp=[Hg22+][I−]2

Let s be the solubility of

⇒[Hg22+]=s and [I−]=2s

Thus,

Ksp=s(2s)2⇒Ksp=4s3

4.5×10−29=4s3

1.125×10−29=s3 

⇒s=2.24×10−10M

Molarity of Hg22+=s=2.24×10−10M

Molarity of I−=2s=4.48×10−10M

 

Let s be the solubility of Ag2CrO4

Then, Ag2CrO4→Ag2++2CrO42−

Ksp=(2s)2⋅s=4s3 

1.1×10^−12=4s3 

s=6.5×10^−5M

Let s′ be the solubility of AgBr.

AgBr(s)↔Ag++Br− 

Ksp=s′2=5.0×10^−13 

∴s′=7.07×10^−7M

Therefore, the ratio of the molarities of their saturated solution is

ss′=6.5×10^−5M / 7.07×10^−7M = 91.9

 

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e. 0.001M. Then,

NalO3→Na++lO3− 0.001M0.001M 

Cu(ClO3)2→Cu2++2ClO3− 

0.001M0.001M

0.001M0.001M

Now, the solubility equilibrium for copper iodate can be written as:

Cu(lO3)2→Cu(aq)2++2lO3(aq)−

Ionic product of copper iodate:

=[Cu2+]×[lO3−]2 

=(0.001)(0.001)2 

=1×10−9

Since the ionic product (1×10−9) is less than (7.4×10−8), precipitation will not occur.

Let the maximum concentration of each solution be xmol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e. x/2.

If the concentrations of both solutions are equal to or less than 5.02×10^−9M, then there will be no precipitation of iron sulphide.

 

CaSO4(s)↔Ca2+(aq)+SO4(aq)2− 

Ksp=[Ca2+][SO42−]

Let the solubility of CaSO4 be s.

Then,

Ksp=s^2

9.1×10−6=s2 

s=3.02×10−3mol/L

Molecular mass of CaSO4=136g/mol

Solubility of CaSO4 in gram/L =3.02×10−3×136 

=0.41g/L

This means that we need 1L of water to dissolve 0.41g of CaSO4

Therefore, to dissolve 1g of CaSO4 we require

=1 / 0.41L=2.44L of water.