Chapter 7 Equilibrium Questions and Answers: NCERT Solutions for Class 11th Chemistry (Partnership Chemistry )

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Class 11 Chemistry (Partnership Chemistry ) Chapter 7: Equilibrium - Questions and Answers of NCERT Book Solutions.

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1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container suddenly increased. a. 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.
b. 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.
c. 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.

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2. What is KcKc for the following equilibrium when the equilibrium concentration of each substance

Ans. is: [SO2]=0.60M[SO2]=0.60M , [O2]=0.82M[O2]=0.82M and [SO3]=1.90M[SO3]=1.90M ? 2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)
Ans: The equilibrium constant for the given reaction will be:
Kc=[SO3]2[SO2]2[O2]Kc=[SO3]2[SO2]2[O2]
=(1.90)2M2(0.60)2(0.821)M3=(1.90)2M2(0.60)2(0.821)M3
Kc=12.239M−1Kc=12.239M−1 (approximately)
Therefore, the equilibrium constant for the given reaction is 12.239M−112.239M−1.

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3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of iodine atoms. I2(g)⇌2I (g)I2(g)⇌2I (g) . Calculate KPKP for the equilibrium.

Ans: Partial pressure for II atoms:
p1=40100×ptotalp1=40100×ptotal
= 40100×10540100×105
= 4×104Pa4×104Pa
Partial pressure for I2I2 molecules:
pl2=60100×ptotalpl2=60100×ptotal
=60100×105=60100×105
=6×104Pa=6×104Pa
Now for, the given reaction:
Kp=(pI)2p12Kp=(pI)2p12
=(4×104)Pa26×104Pa=(4×104)Pa26×104Pa
=2.67×104Pa=2.67×104Pa

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4. Write the expression for the equilibrium constant, KcKc for each of the following reactions:

Ans.
i.2NOCl (g)⇌2NO (g)+Cl2(g)2NOCl (g)⇌2NO (g)+Cl2(g)
Ans: The expression for the equilibrium constant is:
Kc=[NO]2[Cl2][NOCl]2Kc=[NO]2[Cl2][NOCl]2
ii. 2Cu(NO3)2(s)⇌2CuO (s)+4NO2(g)+O2(g)2Cu(NO3)2(s)⇌2CuO (s)+4NO2(g)+O2(g)
Ans: The expression is:
Kc=[CuO]2[NO2]4[O2][Cu(NO3)2]2Kc=[CuO]2[NO2]4[O2][Cu(NO3)2]2
Kc=[NO2]4[O2]Kc=[NO2]4[O2]
iii.CH3COOC2H5(aq)+H2O (l)⇌CH3COOH (aq)+C2H5OH (aq)CH3COOC2H5(aq)+H2O (l)⇌CH3COOH (aq)+C2H5OH (aq)
Ans: The expression will be:
Kc=[CH3COOH] [C2H5OH][CH3COOC2H5] [H2O]Kc=[CH3COOH] [C2H5OH][CH3COOC2H5] [H2O]
Kc=[CH3COOH] [C2H5OH][CH3COOC2H5]Kc=[CH3COOH] [C2H5OH][CH3COOC2H5]
iv. Fe3 + (aq)+3OH−(aq)⇌Fe(OH)3(s)Fe3 + (aq)+3OH−(aq)⇌Fe(OH)3(s)
Ans: The expression will be:
Kc=[Fe(OH)3][Fe3+][OH−]3Kc=[Fe(OH)3][Fe3+][OH−]3
Kc=1[Fe3+][OH−]3Kc=1[Fe3+][OH−]3
v. I2(s)+5F2⇌2IF5I2(s)+5F2⇌2IF5
Ans: The expression will be:
Kc=[IF5]2[I2][F2]2Kc=[IF5]2[I2][F2]2
Kc=[IF5]2[F2]2Kc=[IF5]2[F2]2

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5. Find out the value of KcKc for each of the following equilibria from the value of KpKp : 2NOCl (g)⇌2NO (g)+Cl2(g)2NOCl (g)⇌2NO (g)+Cl2(g) ; Kp=1.8×10−2Kp=1.8×10−2 at 500K

i. 2NOCl (g)⇌2NO (g)+Cl2(g)2NOCl (g)⇌2NO (g)+Cl2(g) ; Kp=1.8×10−2Kp=1.8×10−2 at 500K Ans: The relationship between KpKp and KcKc is given as: Kp=Kc(RT)ΔnKp=Kc(RT)Δn Δn=3−2=1Δn=3−2=1 R=0.0831 barLmol−1K−1R=0.0831 barLmol−1K−1 T = 500K Kp=1.8×10−2Kp=1.8×10−2 Now, Kp=Kc(RT)ΔnKp=Kc(RT)Δn 1.8×10−2=Kc(0.0831×500)11.8×10−2=Kc(0.0831×500)1 Kc=1.8×10−20.0831×500Kc=1.8×10−20.0831×500 Kc=4.33×10−4Kc=4.33×10−4 (Approximately) ii.CaCO3(s)⇌CaO (s)+CO2(g)CaCO3(s)⇌CaO (s)+CO2(g) ; Kp=167Kp=167 at 1073K Ans: Here, Δn=2−1=1Δn=2−1=1 R=0.0831 barLmol−1K−1R=0.0831 barLmol−1K−1 T = 1073K Kp=167Kp=167 Kp=Kc(RT)ΔnKp=Kc(RT)Δn 167=Kc(0.0831×1073)1167=Kc(0.0831×1073)1 Kc=1670.0831×1073Kc=1670.0831×1073 Kc=1.87Kc=1.87 (Approximately)

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6. For the following equilibrium, Kc=6.3×1014Kc=6.3×1014 at 1000K NO (g)+O3(g)⇌NO2(g)+O2(g)NO (g)+O3(g)⇌NO2(g)+O2(g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is KcKc for the reverse reaction?

Ans: According to the question:
The KcKc for the forward reaction is 6.3×10146.3×1014
Then, KcKcfor the reverse reaction will be:
K'c=1KcK'c=1Kc
K'c=16.3×1014K'c=16.3×1014
K'c=1.59×10−15K'c=1.59×10−15
The equilibrium constant for the reverse reaction will be: 1.59×10−151.59×10−15.

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

Ans: For a pure substance which is both a solid and a liquid:
[Pure substance]=Number of molesVolume[Pure substance]=Number of molesVolume
[Pure substance]=Mass / Molecular weightVolume[Pure substance]=Mass / Molecular weightVolume
[Pure substance]=MassVolume × Molecular mass[Pure substance]=MassVolume × Molecular mass
[Pure substance]=DensityMolecular 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.

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8. Reaction between nitrogen and oxygen takes place as follows: 2N2(g)+O2(g)⇌2N2O (g)2N2(g)+O2(g)⇌2N2O (g) If a mixture of 0.482 mol of N2N2 and 0.933 mol of O2O2 is placed in a 10 L reaction vessel and allowed to form N2ON2O at a temperature for which Kc=2.0×10−37Kc=2.0×10−37 , determine the composition of equilibrium mixture.

Ans: Let the concentration of N2ON2O at equilibrium be x.
The initial concentration of N2N2 is 0.482mol.
The initial concentration of O2O2 is 0.933mol.
At equilibrium the concentration of N2N2 is (0.482-x)mol, the concentration of O2O2is (0.933-x)mol and the concentration of N2ON2O is x mol.
The reaction occurs in a 10L reaction vessel:
[N2]=0.482−x10[N2]=0.482−x10 , [O2]=0.933−x/210[O2]=0.933−x/210 , [N2O]=x10[N2O]=x10
The value of equilibrium constant i.e. Kc=2.0×10−37Kc=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,
[N2]=0.48210=0.0482 molL−1[N2]=0.48210=0.0482 molL−1 and [O2]=0.93310=0.0933 molL - 1[O2]=0.93310=0.0933 molL - 1
Now, according to the equation:
Kc=[N2O]2[N2]2[O2]Kc=[N2O]2[N2]2[O2]
2.0×10−37=(x2)2(0.0482)2(0.0933)2.0×10−37=(x2)2(0.0482)2(0.0933)
x2100=2.0×10−37×(0.0482)2(0.0933)x2100=2.0×10−37×(0.0482)2(0.0933)
x2=43.35×10−40x2=43.35×10−40
x=6.6×10−20x=6.6×10−20
[N2O]=x10=6.6×10−2010[N2O]=x10=6.6×10−2010
[N2O]=6.6×10−21[N2O]=6.6×10−21

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9. Nitric oxide reacts with Br2Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g)+Br2(g)⇌2NOBr (g)2NO (g)+Br2(g)⇌2NOBr (g) When 0.087 mol of NONO and 0.0437 mol of Br2Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBrNOBr is obtained at equilibrium. Calculate equilibrium amount of NONO and Br2Br2 .

Ans: The given reaction is:
2NO (g)+Br2(g)⇌2NOBr (g)2NO (g)+Br2(g)⇌2NOBr (g)
Now, 2 mol of NOBrNOBr are formed from 2 mol of NONO . Therefore, 0.0518 mol of NOBrNOBr are formed from 0.0518 mol of NONO. Again, 2 mol of NOBrNOBr are formed from 1 mol of BrBr .
Therefore, 0.0518 mol of NOBrNOBr are formed from 0.058120.05812 mol of BrBr, or 0.0259 mol of NONO.
The amount of NONO and BrBr present initially is as follows:
[NO]=0.087mol[NO]=0.087mol and [Br2]=0.0437mol[Br2]=0.0437mol
Therefore, the amount of NONO present at equilibrium is:
[NO]=0.087−0.0518=0.0352mol[NO]=0.087−0.0518=0.0352mol
And, the amount of BrBr present at equilibrium is:
[Br2]=0.0437−0.0259=0.0178mol[Br2]=0.0437−0.0259=0.0178mol

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10. At 450K, Kp=2.0×1010bar - 1Kp=2.0×1010bar - 1 for the given reaction reaction at equilibrium. 2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g) , What is KcKc at this temperature?

Ans: For the given reaction:
Δn=2−3=−1Δn=2−3=−1
T=450KT=450K
R=0.0831bar L bar K−1mol−1R=0.0831bar L bar K−1mol−1
Kp=2.0×1010bar−1Kp=2.0×1010bar−1
We know that,
Kp=Kc(RT)ΔnKp=Kc(RT)Δn
2.0×1010=Kc(0.0831×450)−12.0×1010=Kc(0.0831×450)−1
Kc=2.0×1010(0.0831×450)Kc=2.0×1010(0.0831×450)
Kc=74.79×1014Lmol

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11. A sample of HI (g)HI (g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI (g)HI (g) is 0.04 atm. What is KpKp for the given equilibrium? 2HI (g)⇌H2(g)+I2(g)2HI (g)⇌H2(g)+I2(g)

Ans: The initial concentration of HIHI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.
Therefore, a decrease in the pressure of HIHI is 0.2 – 0.04 = 0.16. The given reaction is:
At equilibrium the pressure of HIHI is 0.04 atm, the pressure for H2=0.162=0.08atmH2=0.162=0.08atm and the pressure for O2=0.162=0.08atmO2=0.162=0.08atm.
Therefore, the value of KpKp at equilibrium is 4.0.

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12. A mixture of 1.57 mol of N2N2 , 1.92 mol of H2H2 and 8.13 mol of NH3NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, KcKc for the reaction is 1.7×1021.7×102 . The reaction is: N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g) . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans: For the given reaction: N2(g)+3H2(g)⇌2NH3(g)
The given concentration of various species is
[N2]=1.5720molL−1[H2]=1.9220molL−1
[NH3]=8.1320molL−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

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13. The equilibrium constant expression for a gas reaction is:
Kc=[NH3]4 [O2]5[NO]4 [H2O]6
Write the balanced chemical equation corresponding to this expression.

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

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14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation: H2(g)+CO(g)⇌H2(g)+CO2(g) Calculate the equilibrium constant for the reaction.

Ans: The given reaction is:
H2O(g) + CO(g)⇌H2(g) + CO2(g)
110M110M00
1 - 0.410M1 - 0.410M0.410M0.410M
= 0.06M = 0.06M = 0.04M = 0.04M
Therefore, the equilibrium constant for the reaction:
Kc=[H2][CO2][H2O][CO]
Kc=0.04×0.040.06×0.06
Kc=0.444

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15. At 700 K, equilibrium constant for the reaction H2(g)+I2(g)⇌2HI (g) is 54.8. If 0.5molL−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?

Ans: It is given that equilibrium constant Kc for the reaction
H2(g)+I2(g)⇌2HI (g) is 54.8
Therefore, at equilibrium, the equilibrium constant K'c for the reaction
2HI (g)⇌H2(g)+I2(g)
[HI]=0.5molL−1 will be: 154.8
Let the concentrations of hydrogen and iodine at equilibrium be xmolL−1
[H2]=[I2]=xmolL−1
Therefore, [H2] [I2][HI]2=K'c
x×x(0.5)2=154.8
⇒x2=0.2554.8⇒x=0.06754
x=0.068 molL - 1 (approximately)
Hence, at equilibrium, [H2]=[I2]=0.068 molL - 1

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16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M?
2ICI (g)⇌I2(g)+Cl2(g) ; Kc=0.14

Ans: For the given reaction:
2ICI (g)⇌I2(g)+Cl2(g)
The initial concentration of ICI is 0.78M and for I2 and Cl2 is zero (0).
At equilibrium, the concentration for ICI is (0.78−2x)M and for I2 and Cl2is xM.
Now, we can write:
[I2][Cl2][ICI]2=Kc
x×x(0.78−2x)2=0.14
x0.78−2x=0.374
x=0.292−0.748x
1.748x=0.292
x=0.167
Therefore, at equilibrium
[H2] = [I2] = 0.167
M[HI]=(0.78−2×0.167)
M[HI]=0.446M

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17. Kp=0.04atm 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?

Ans: 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 H2is p.
Then we can write,
⇒Kpp×p40−p=0.04
p2=0.16−0.04p
p2+0.04p−0.16=0
p=−0.04±(0.04)2−4×1×(−0.16)−−−−−−−−−−−−−−−−−−−−√2×1
p=−0.04±0.082
If we take positive value:
p=0.762
p=0.38
So, at equilibrium:
[C2H6]−4−p=4−0.38=3.62 atm

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18. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5×10−1molL - 1 . If value of Kc is 8.3×10−3 , what are the concentrations of PCl3 and Cl2 at equilibrium?

Ans: Let the concentrations of both PCl3 and Cl2 at equilibrium be xmolL - 1 . The given reaction is:
PCl5(g)⇌PCl3(g)+Cl2(g)
At equilibrium the concentration of PCl5is 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×x0.5×10−1=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

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19. 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=0.0265 at 1050 K.
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO=1.4atm and pCO2=0.80atm ?

Ans: 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

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20. Equilibrium constant, Kc for the reaction N2(g)+3H2(g)⇌2NH3(g) at 500 K os 0.061. At a particular time, the analysis shows that composition of the reaction mixture is 3.0 molL−1 N2 , 2.0 molL−1 H2 and 0.5 molL−1 NH3 . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Ans: 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 >

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21. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g)⇌Br2(g)+Cl2(g) for which Kc=32 at 500 K. If initially pure BrCl is present at a concentration of 3.3×10−3 molL−1 , what is its molar concentration in the mixture at equilibrium?

Ans: 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.0
⇒[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

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22. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass: C (s)+CO2(g)⇌2CO (g). Calculate Kc for this reaction at the above temperature.

Ans: 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 CO , nCO=90.5528=3.234 mol
Number of moles of CO2, nCO2=9.4544=0.215 mol
Partial pressure of CO , 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

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23. Calculate:
ΔG∘ and
The equilibrium constant for the formation of NO2 from NO and O2 at 298 K 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)
The ΔG∘ for the given reaction will be:
ΔG∘=ΔG∘(Products)−ΔG∘(Reactants)
ΔG∘=52.0−[87.0+0]
ΔG∘=−35.0 kJmol−1
We know that:
ΔG∘=RTlogKc
ΔG∘=2.303RTlogKc
logKc=−35.0×10−3−2.303×8.314×298
logKc=6.134
Kc=antilog (6.134)
Kc=1.36×106
Hence, the equilibrium constant for the given reaction Kc is 1.36×106 .

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24. Does the number of moles of reaction products increase, decrease or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

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.
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.
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.

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25. 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)

Ans: 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.

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26. The equilibrium constant for the following reaction is 1.6×105 at 1024 K. H2(g)+Br2(g)⇌2HBr (g) , Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Ans: Given,
Kp for the reaction i.e.,
H2(g)+Br2(g)↔2HBr(g) is 1.6×105
Therefore, for the reaction 2HBr(g)↔H2(g)+Br2(g) the equilibrium constant will be, K′p=1Kp
=11.6×105
=6.25×10−6
Now, let p be the pressure of both H2 and Br2 at equilibrium.
pHBr×p2p2HBr=K′p
p×p(10−2p)=6.25×10−6
p10−2p=2.5×10−3
p=2.5×10−2−(5.0×10−)p
p+(5.0×10−3)p=2.5×10−2
(1005×10−3)p=2.5×10−2
p=2.49×10−2 bar =2.5×10−2 (approximately)
Therefore, at equilibrium,
[H2]=[Br2]=2.49×10−2 bar
[HBr]=10−2×(2.49×10−1) bar =9.95bar=10 bar (approximately)

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27. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH4(g)+H2O (g)⇌CO (g)+3H2(g)

a. Write an expression for Kp for the above reaction. Ans: For the given equation: Kp=pCO×p3H2pCH4×pH2O b.How will the values of Kp and composition of equilibrium mixture be affected by I.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. II.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. III.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.

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28. Describe the effect of the following on the equilibrium of the reaction: 2H2(g)+CO (g)⇌CH3OH (g)

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.

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

PCl5(g)⇌PCl3(g)+Cl2(g) , ΔrH∘=124.0kJmol−1 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=1Kc =18.3×10−3=1.2048×102 =120−48 c. What would be the effect on Kc if I.More PCl5 is added? Ans: Because the temperature is constant in this scenario, Kc will remain constant. II.Pressure is increased? Ans: At constant temperature, Kc remains constant. As a result, Kc would not alter in this scenario. 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.

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30. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2 . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction: CO (g)+H2O (g)⇌CO2(g)+H2(g) If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pCO=pH2O=4.0 bar , what will be the partial pressure of H2 at equilibrium? Kp=10.1 at 400°C.

Ans: 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 be (4.0−p).
It is given that, Kp=10.1
Now,
pco2×p2pco×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.

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31. Predict which of the following reaction will have appreciable concentration of reactants and products:
Cl2(g)⇌2Cl (g) ; Kc=5×10−39
Cl2(g)+2NO (g)⇌2NOCl (g) ; Kc=3.7×108
Cl2(g)+2NO2(g)⇌2NO2Cl (g) ; Kc=1.8

Ans: 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.
32. The value of Kc for the reaction 3O2(g)⇌2O3(g) is 2.0×10−50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6×10−2 , what is the concentration of O3 ?
Ans: The given reaction is:
3O2(g)⇌2O3(g)
KC=[O3(g)]2[O2(g)]3
KC=2.0×10−50 and [O2(g)]=1.6×10−2
2.0×10−50=[O3(g)]2[1.6×10−2]3
⇒[O3(g)]2=2.0×10−50×(1.6×10−2)3
⇒[O3(g)]2=8.192×10−50
⇒[O3(g)]=2.86×10−28M Hence, the concentration of O3 is 2.86×10−28M .

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33. The reaction CO (g)+3H2(g)⇌CH4(g)+H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO , 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Ans: 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
Therefore,
[CH4(g)][H2O(g)][CO(g)][H2(g)]3=Kc
⇒x×0.020.3×(0.1)3=3.90
⇒x=3.90×0.3×(0.1)30.02
= 0.001170.02
=0.0585M
=5.85×10−2M
Hence, the concentration of CH4 at equilibrium is 5.85×10−2M.

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34. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2,CN−,HClO4,F−,OH−,CO2−3,S−

Ans: 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.

Species	Conjugate Acid-Base
HNO2	NO3− (base)
CN−	HCN (acid)
HClO4	ClO4− (base)
F−	HF (acid)
OH−	H2O (acid) / O2− (base)
CO2−3	HCO3− (acid)
S−	HS− (acid)

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35. Which of the followings are Lewis acids? H2O, BF3, H + , NH4+.

Ans: Lewis acids are those acids which can accept a pair of electrons. For example: BF3, H + , NH4 + are Lewis acids.

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36. What will be the conjugate bases for the Brönsted acids: HF, H2SO4, HCO3

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

Bronsted Acid	Conjugate Base
HF	F−
H2SO4	HSO4−
HCO3−	CO2−3

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37. Write the conjugate acids for the following Brönsted bases: NH−2,NH3,HCOO−

Bronsted Base	Conjugate Acid
NH2−	NH3
NH3	NH4−
HCOO−	HCOOH

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38. The species: H2O,HCO−3,HSO−4,NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

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

Species	Conjugate Acid	Conjugate Base
H2O	H3O +	OH
HCO−3	H2CO3	CO2−3
HSO−4	H2SO4	SO2−4
NH3	NH + 4	NH−2

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39. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

Ans.
i.OH−
Ans: OH−is a Lewis base since it can donate its lone pair of electrons
ii. F−
Ans: F− is a Lewis base since it can donate a pair of electrons.
iii.H +
Ans: H + is a Lewis acid since it can accept a pair of electrons.
iv.BCl3
Ans: BCl3 is a Lewis acid since it can accept a pair of electrons.

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

Ans: 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=−log3.8−log10−3
pH=−log3.8+3
pH=−0.58+3 pH=2.42

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

Ans: 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 .
Kb=1.5×10−11
Given Ka of HCOOH is 1.8×10−4
Hence, Kb of its
conjugate base HCOO−:
Kb
=KwKaKb
=10−141.8×10−4
Kb=5.6×10−11
Given Ka of HCN is 4.8×10−9
Hence, Kb of its
conjugate base CN−:
Kb=KwKa
Kb=10−144.8×10−9 Kb=2.08×10−6

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42. The ionization constant of HF, HCOOH, HCNHF, HCOOH, HCN at 298 K are 6.8×10−46.8×10−4 , 1.8×10−41.8×10−4 and 4.8×10−94.8×10−9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Ans: It is known as:
Kb=KwKaKb=KwKa
Given, KaKa of HFHF is 6.8×10−46.8×10−4
Hence, KbKb of its conjugate base F−F−:
Kb=KwKaKb=KwKa
Kb=10−146.8×10−4Kb=10−146.8×10−4
Kb=1.5×10−11Kb=1.5×10−11
Given KaKa of HCOOHHCOOH is 1.81.8××10−410−4
Hence, KbKb of its
conjugate base HCOO−HCOO−:
KbKb
=KwKaKbKwKaKb
=10−141.8×10−4=10−141.8×10−4
Kb=5.6×10−11Kb=5.6×10−11
Given KaKa of HCNHCN is 4.84.8××10−910−9
Hence, KbKb of its
conjugate base CN−CN−:
Kb=KwKaKb=KwKa
Kb=10−144.8×10−9Kb=10−144.8×10−9
Kb=2.08×10−

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43. The ionization constant of phenol is 1.0×10−10 . What is the concentration of phenolate ion in 0.05 M Ans:of phenol? What will be its degree of ionization if the Ans:is also 0.01M in sodium phenolate?

Ans: 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).
Ka=[C6H5O−][H3O+][C6H5OH]
Ka=x×x0.05−x
As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.
∴x=1×10−10×0.05−−−−−−−−−−−−−√
=5×10−12−−−−−−−−√
=2.2×10−6M=[H3O+]
\ Since[H3O+]=[C6H5O−]
[C6H5O−]=2.2×10−6M
Now, let α be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa
C6H5OH+H2O⇌C6H5O−+H3O+
The concentrations are as follows:
[C6H5OH]=0.05−0.05α
0.05 M[C6H5O−]=0.01+0.05α
0.01 M[H3O+]=0.05α
Ka=[C6H5O−][H3O+][C6H5OH]
Ka=(0.01)(0.05α)0.05
1.0×10−10=0.01α α=1×10−

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

Ans: 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)=x2
9.1×10−9=x2
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

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45. The ionization constant of acetic acid is 1.74×10−5 . Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the Ans:and its pH.

Ans: For the reaction:
CH3COOH+H2O⇌CH3COO−+H3O+
The final concentration of CH3COO− and 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)
=.05α21−α
1.74×10−5=.05α21−α
1.74×10−5−1.74×10−5α=0.05α2
0.05α2+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=Kac−−−√
α=1.74×10−5.05−−−−−−−−−−√
=34.8×10−5×1010−−−−−−−−−−−−−−√
=3.48×10−6−−−−−−−−−√
[CH3COO−]=0.05×1.86×10−3
=0.93×10−31000
=.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.

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

Ans: Let the organic compound be HA.
HA⇌H++A−
Concentration of HA is 0.01 M and its pH is 4.15.
−log[H+]=4.15
[H+]=7.08×10−5
Now,Ka=[H+][A−][HA]
[H+]=[A−]=7.08×10−5
[HA]=0.01
Now,
Ka=[H+][A−][HA][H+]=[A−]=7.08×10−5[HA]=0.01
Then,
Ka=(7.08×10−5)(7.08×10−5)0.01
Ka=5.01×10−7
pKa=−logKa
=−logKa
=(5.01×10−7) pKa=6.3001

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47. Assuming complete dissociation, calculate the pH of the following solutions: 0.003 M HClHCl

Ans: H2O+HCl⇌H3O++Cl−H2O+HCl⇌H3O++Cl−
Since HClHCl is completely ionized.
[H3O+]=[HCl]⇒[H3O+]=0.003[H3O+]=[HCl]⇒[H3O+]=0.003
pH=−log[H3O+]pH=−log⁡[H3O+]
pH=−log(0.003)pH=−log⁡(0.003)
pH=2.52pH=2.52
Hence, the pH of the solution is
2.52.
ii. 0.005 M NaOHNaOH
Ans: NaOH(aq)⇌Na+(aq)+OH−(aq)NaOH(aq)⇌Na+(aq)+OH−(aq)
[OH−]=[NaOH]⇒[OH−]=0.005[OH−]=[NaOH]⇒[OH−]=0.005
pOH=−log[OH−]pOH=−log⁡[OH−]
pOH=−log(0.005)pOH=−log⁡(0.005)
pOH=2.30pOH=2.30
pH=14−pOHpH=14−pOH
pH=14−2.30pH=14−2.30
pH=11.70pH=11.70
Hence, the pH of the solution is
11.70.
iii. 0.002 M HBrHBr
Ans: HBr+H2O⇌H3O++Br−HBr+H2O⇌H3O++Br−
[H3O+]=[HBr]⇒[H3O+]=0.002[H3O+]=[HBr]⇒[H3O+]=0.002
pH=−log[H3O+]pH=−log⁡[H3O+]
pH=−log(0.002)pH=−log⁡(0.002)
pH=2.69pH=2.69
Hence, the pH of the solution is
2.69.
iv.0.002 M KOHKOH
Ans: KOH(aq)⇌K+(aq)+OH−(aq)KOH(aq)⇌K+(aq)+OH−(aq)
[OH−]=[KOH][OH−]=[KOH]
⇒[OH−]=0.002⇒[OH−]=0.002
pOH=−log[OH−]pOH=−log⁡[OH−]
pOH=−log(0.002)pOH=−log⁡(0.002)
pOH=2.69pOH=2.69
pH=14−pOHpH=14−pOH
pH=14−2.69pH=14−2.69
pH=11.31pH=11.31
Hence, the pH of the solution is 11.31.

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48. Calculate the pH of the following solutions

Ans. a. 2 g of TlOHTlOH dissolved in water to give 2 litre of solution.
Ans: For 2g of TlOHTlOH dissolved in water to give 2 L of solution:
[TIOH(aq)]=22g/L[TIOH(aq)]=22g/L
=22×1221M=22×1221M
=1221M=1221M
TlOH (aq)→Tl+(aq)+OH−(aq)TlOH (aq)→Tl+(aq)+OH−(aq)
TIOH(aq)→Tl+(aq)+OH−(aq)TIOH(aq)→Tl(aq)++OH(aq)−
[OH−(aq)]=[TIOH(aq)]=1221M[OH(aq)−]=[TIOH(aq)]=1221M
Kw=[H+][OH−]Kw=[H+][OH−]
10−14=[H+](1221)10−14=[H+](1221)
221×10−14=[H+]221×10−14=[H+]
⇒pH=−log[H+]=−log(221×10−14)⇒pH=−log⁡[H+]=−log⁡(221×10−14)
=−log(2.21×10−12)=−log⁡(2.21×10−12)
=11.65=11.65
0.3 g of Ca(OH)2Ca(OH)2 dissolved in water to give 500 mL of solution.
Ans: Molecular weight of Ca(OH)2Ca(OH)2 is 74g.
So, 0.3g Ca(OH)2Ca(OH)2 = 0.374 mole0.374 mole
Molar Concentration = Moles in SolutionVolume in SolutionMolar Concentration = Moles in SolutionVolume in Solution
Molar concentration of Ca(OH)2Ca(OH)2 = 0.374 mole0.5 L=8.11×10−3 M0.374 mole0.5 L=8.11×10−3 M
Ca(OH)2→Ca+2+2OH−Ca(OH)2→Ca+2+2OH−
[OH−]=2[Ca(OH)2][OH−]=2[Ca(OH)2]
[OH−]=2×(8.11×10−1) M[OH−]=2×(8.11×10−1) M
[OH−]=16.22×10−1M[OH−]=16.22×10−1M
pOH=−log(16.22×10−1)pOH=−log⁡(16.22×10−1)
pOH=3−1.2101pOH=3−1.2101
pOH=1.79pOH=1.79
pH=14−1.79pH=14−1.79
pH=12.21pH=12.21
0.3 g of NaOHNaOH
dissolved in water to give 200 mL of solution.
Ans: 1 mole of NaOHNaOH = 40g
Since, 40 gm of NaOHNaOH = 1 mole
of NaOHNaOH
So, 0.3 gm of NaOH=NaOH=\frac{0.3}{40}$
Molar Concentration=Moles in SolutionVolume in SolutionMoles in SolutionVolume in Solution
Molar concentration = 0.340Moles0.2L=3.75×10−1M0.340Moles0.2L=3.75×10−1M
[OH−]=3.75×10−2 M[OH−]=3.75×10−2 M
pOH=−log(3.75×10−2)pOH=−log⁡(3.75×10−2)
pOH=1.43pOH=1.43
pH=14−1.43pH=14−1.43
pH=12.57pH=12.57
1mL of 13.6 M HClHCl is diluted with water to give 1 litre of solution.
Ans: For this question:
M1V1=M2V2M1V1=M2V2 >
13.6×1 mL=M2×1000 mL13.6×1 mL=M2×1000 mL
13.6×10−3=M2×1 L13.6×10−3=M2×1 L
M2=1.36×10−2M2=1.36×10−2
[H+]=1.36×10−2[H+]=1.36×10−2
pH=−log(1.36×10−2)pH=−log⁡(1.36×10−2)
pH=−0.1335+2pH=−0.1335+2 pH=1.87

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49. The degree of ionization of a 0.1M bromoacetic acid is 0.132. Calculate the pH of pKapKa of bromoacetic acid.

Ans: Degree of ionization, α=0.132α=0.132
Concentration, c = 0.1 M
Thus, the concentration of
H3O+H3O+=c.αc.α
[H3O+][H3O+] = 0.1×0.1320.1×0.132
[H3O+]=0.0132[H3O+]=0.0132
pH=−log[H+]pH=−log⁡[H+]
pH=−log(0.0132)pH=−log⁡(0.0132)
pH=1.88pH=1.88
Now,
Ka=Cα2Ka=Cα2
Ka=0.1×(0.132)2Ka=0.1×(0.132)2
Ka=0.0017pKa=0.0017p Ka=2.75

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50. 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.

Ans: Kb=4.2710−10Kb=4.2710−10
c=0.001Mc=0.001M
pH=?pH=?
α=?α=?
ka=Cα2ka=Cα2
4.27×10−10=0.001×α24.27×10−10=0.001×α2
4270×10−10=α24270×10−10=α2
65.34×10−5=α=6.53×10−465.34×10−5=α=6.53×10−4
Then, concentration of anion:
[anion]=c.α=0.01×65.34×10−5[anion]=c.α=0.01×65.34×10−5
[anion]=0.65×10−5[anion]=0.65×10−5
pOH=−log(.065×10−5)pOH=−log⁡(.065×10−5)
=6.187=6.187
pH=7.813pH=7.813
Ka×Kb=KwKa×Kb=Kw
∴4.27×10−10×Ka=Kw∴4.27×10−10×Ka=Kw
Ka=10−144.27×10−10Ka=10−144.27×10−10
=2.34×10−5=2.34×10−5
Thus, the ionization constant of the conjugate acid of aniline is 2.34×10−52.34×10−5 .

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51. The degree of ionization of a 0.1M0.1M bromoacetic acid is 0.132. Calculate the pHpH and the pKapKa of bromoacetic acid.

Ans: Degree of ionization, α=0.132α=0.132
Concentration, c=0.1Mc=0.1M
Thus, the concentration of H3O+=H3O+=c. αα
⇒0.1×0.132=0.0132⇒0.1×0.132=0.0132
pH=−log[H+]=−log(0.0132)=1.879:1.88pH=−log⁡[H+]=−log⁡(0.0132)=1.879:1.88
Now
KaKa = Cα2Cα2 = 0.1×(0.132)20.1×(0.132)2
Ka=.0017pKa=.0017p
Ka=2.75

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52. What is the pH of 0.001M0.001M 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.

Ans:
Kb=4.27×10−10Kb=4.27×10−10
c=0.001Mc=0.001M
pH=?pH=?
α=?α=?
ka=Cα2ka=Cα2
4.27×10−10=0.001×α24.27×10−10=0.001×α2
4270×10−10=α24270×10−10=α2
65.34×10−5=α=6.53×10−465.34×10−5=α=6.53×10−4
Then,[anion]=Then,[anion]=\text{c}\alpha =0.01\times 65.34\times 10^{-5}$
=0.65×10−5=0.65×10−5
pOH=−log(.065×10−5)pOH=−log⁡(.065×10−5)
=6.187=6.187
pH=7.813pH=7.813
Now,
Ka×Kb=KwKa×Kb=Kw
∴4.27×10−10×Ka=Kw∴4.27×10−10×Ka=Kw
Ka=10−144.27×10−10Ka=10−144.27×10−10
=2.34×10−5=2.34×10−5
Thus, the ionization constant of the conjugate acid of aniline is 2.34×10−52.34×10−5.

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53. Calculate the degree of ionization of 0.05M0.05M acetic acid if its pKapKa value is 4.74. How is the degree of dissociation affected when it's also contains 0.01M0.01M
0.1M0.1M in HClHCl ?

Ans:
c=0.05Mc=0.05M
pKa=4.74pKa=4.74
pKa=−logKapKa=−log⁡Ka
Ka=1.82×10−5Ka=1.82×10−5
c=0.05Mc=0.05M
pKa=4.74pKa=4.74
pKa=−log(Ka)pKa=−log⁡(Ka)
Ka=1.82×10−5Ka=1.82×10−5
Ka=cα2α=Kac−−−√Ka=cα2α=Kac
α=1.82×10−55×10−2−−−−−−−−−−√=1.908×10−2α=1.82×10−55×10−2=1.908×10−2
Ka=cα2α=Kac−−−√Ka=cα2α=Kac
α=1.82×10−55×10−2−−−−−−−−−−√=1.908×10−2α=1.82×10−55×10−2=1.908×10−2
When HCl is added to the solution, the concentration of H+H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease. Case I: When 0.01MHCl0.01MHCl is taken.
Let xx be the amount of acetic acid dissociated after the addition of HCl.
CH3COOHH++CH3COO−CH3COOHH++CH3COO−
Initial conc. 0.05M Afterdissociation 0.05−x0.01+xxInitial conc. 0.05M Afterdissociation 0.05−x0.01+xx
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05−x0.05−x and 0.01+x0.01+x can be taken as 0.050.05 and 0.010.01 respectively.
Ka=[CH3COO−][H+][CH3COOH]Ka=[CH3COO−][H+][CH3COOH]
∴Ka=(0.01)x0.05∴Ka=(0.01)x0.05
x=1.82×10−5×0.050.01x=1.82×10−5×0.050.01
x=1.82×10−3×0.05Mx=1.82×10−3×0.05M
Ka=[CH3COO−][H+][CH3COOH]Ka=[CH3COO−][H+][CH3COOH]
∴Ka=(0.01)x0.05∴Ka=(0.01)x0.05
x=1.82×10−5×0.050.01x=1.82×10−5×0.050.01
x=1.82×10−3×0.05Mx=1.82×10−3×0.05M
Now,
α= Amount of acid dissociated Amount of acid taken α= Amount of acid dissociated Amount of acid taken
⇒1.82×10−5×0.050.05=1.82×10−3⇒1.82×10−5×0.050.05=1.82×10−3
Case II: When 0.1MHCl0.1MHCl is taken.
Let the amount of acetic acid dissociated in this case be XX. As we have done in the first case, the concentrations of various species involved in the reaction are:
[CH3COOH]=0.05−X;0.05M[CH3COOH]=0.05−X;0.05M
[CH3COO−]=X[CH3COO−]=X
[H+]=0.1+X;0.1M[H+]=0.1+X;0.1M
Ka=[CH3COO−][H+][CH3COOH]Ka=[CH3COO−][H+][CH3COOH]
∴Ka=(0.1)x0.05∴Ka=(0.1)x0.05
[CH3COOH]=0.05−X;0.05M[CH3COO−]=X[CH3COOH]=0.05−X;0.05M[CH3COO−]=X
[H+]=0.1+X;0.1M[H+]=0.1+X;0.1M
Ka=[CH3COO−][H+][CH3COOH]Ka=[CH3COO−][H+][CH3COOH]∴Ka=(0.1)x0.05∴Ka=(0.1)x0.05
x=1.82×10−5×0.050.1x=1.82×10−5×0.050.1 x=1.82×10−4×0.05Mx=1.82×10−4×0.05M Now,
α= Amount of acid dissociated Amount of acid taken α= Amount of acid dissociated Amount of acid taken
⇒1.82×10−4×0.050.05=1.82×10−4

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54. The ionization constant of dimethylamine is 5.410−45.410−4. Calculate its degree of ionization in its 0.02M0.02M solution. What percentage of dimethylamine is ionized if the Ans:is also 0.1M0.1M in NaOH?

Ans:Kb=5.4×10−4Kb=5.4×10−4
c=0.02Mc=0.02M
c=0.02Mc=0.02M Then, α=Kbc−−−√α=Kbc =5.4×10−40.02−−−−−−−−−√=5.4×10−40.02 =0.1643=0.1643
Now, if 0.1M0.1M of NaOHNaOH is added to the solution, then NaOHNaOH (being a strong base) undergoes complete ionization.
(image will be uploaded soon)
And,
(CH3)2NH+H2O↔(CH3)2NH+2+OH(CH3)2NH+H2O↔(CH3)2NH2++OH
(0.02−x)x(0.02−x)x
;0.02M;0.02M
Then, [(CH3)2NH+2]=x Then, [(CH3)2NH2+]=x
[OH−]=x+0.1;0.1[OH−]=x+0.1;0.1
⇒Kb=[(CH3)2NH+2][OH−][(CH3)2NH]⇒Kb=[(CH3)2NH2+][OH−][(CH3)2NH]
5.4×10−4=x×0.10.025.4×10−4=x×0.10.02
x=0.0054x=0.0054
It means that in the presence of 0.1MNaOH,0.54%0.1MNaOH,0.54% of dimethylamine will get dissociated.

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

Ans:
Human muscle fluid 6.83
pH=6.83pH=−log[H+]pH=6.83pH=−log⁡[H+]
∴6.83=−log[H+]∴6.83=−log⁡[H+]
[H+]=1.4810−7M[H+]=1.4810−7M
Human stomach fluid, 1.2:
pH=1.2×pH=1.2×
1.2=−log[H+]1.2=−log⁡[H+]
∴[H+]=0.063∴[H+]=0.063
Human blood, 7.38:
pH=7.38=−log[H+]pH=7.38=−log⁡[H+]
∴[H+]=4.1710−8M∴[H+]=4.1710−8M
Human saliva, 6.4:
pH=6.4×pH=6.4×
6.4=−log[H+]6.4=−log⁡[H+]
[H+]=3.9810−7

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56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,5.0,4.2,2.26.8,5.0,4.2,2.2 and 7.87.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Ans: The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH=−log[H+]pH=−log⁡[H+]
pHpH of milk =6.8=6.8
Since, pH=−log[H+]pH=−log⁡[H+]
6.8=−log[H+]log6.8=−log⁡[H+]log
[H+]=−6.8[H+]=−6.8
[H+]=anitlog(−6.8)[H+]=anitlog⁡(−6.8)
=1.519−7M=1.519−7M
pHpH of black coffee =5.0=5.0
Since, pH=×log[H+]pH=×log⁡[H+]
5.0=−log[H+]log5.0=−log⁡[H+]log
[H+]=−5.0[H+]=−5.0
[H+]=anitlog(−5.0)[H+]=anitlog⁡(−5.0)
=10−5M=10−5M
pHpH of tomato juice =4.2=4.2
Since, pH=−log[H+]pH=−log⁡[H+]
4.2=−log[H+]log4.2=−log⁡[H+]log
[H+]=−4.2[H+]=−4.2
[H+]=anitlog(−4.2)[H+]=anitlog⁡(−4.2)
=6.3110−5M=6.3110−5M
pHpH of lemon juice =2.2=2.2
Since, pH=−pH=− log [H+][H+]
2.2=−log[H+]2.2=−log⁡[H+]log
[H+]=−2.2[H+]=−2.2
[H+]=anitlog(−2.2)[H+]=anitlog⁡(−2.2)
=6.3110−3M=6.3110−3M
pHpH of egg white =7.8=7.8
Since, pH=−log[H+]pH=−log⁡[H+]
7.8=−log[H+]log7.8=−log⁡[H+]log
[H+]=−7.8[H+]=−7.8
[H+]=anitlog(−7.8)[H+]=anitlog⁡(−7.8)
=1.5810−8M

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57. If 0.561g0.561g of KOHKOH is dissolved in water to give 200mL200mL at 298K298K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pHpH ?

Ans: [KOHaq]=0.56115g/L[KOHaq]=0.56115g/L
=2.805g/L=2.805g/L
=2.805×156.11M=.05M=2.805×156.11M=.05M
KOHaq→K+(aq)+OH−(aq)KOHaq→K(aq)++OH(aq)−
[OH−]=.05M[OH−]=.05M
=[K+][H+][H−]=Kw=[K+][H+][H−]=Kw
[H+]=Kw[OH−][H+]=Kw[OH−]
=10−140.05=10−140.05
=2×10−11M=2×10−11M
∴pH=12.70

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58. The solubility of Sr(OH)2Sr(OH)2 at 298K298K is 19.23g/L19.23g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pHpH of the solution.

Ans:Solubility of Sr(OH)2=19.23g/LSr(OH)2=19.23g/L
Then, concentration of Sr(OH)2Sr(OH)2
=19.23121.63M=19.23121.63M
=0.1581M=0.1581M
Sr(OH)2(aq)→Sr2+(aq)+2(OH−)(aq)Sr⁡(OH)2(aq)→Sr(aq)2++2(OH−)(aq)
∴[Sr2+]=0.1581M∴[Sr2+]=0.1581M
[OH−]=2×0.1581M=0.3126M[OH−]=2×0.1581M=0.3126M
Now,
Kw=[OH−][H+]Kw=[OH−][H+]
10−140.3126=[H+]10−140.3126=[H+]
⇒[H+]=3.2×10−14⇒[H+]=3.2×10−14
∴pH=13.495;13.50

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59. The ionization constant of propanoic acid is 1.32×10−5.1.32×10−5. Calculate the degree of ionization of the acid in its 0.05M0.05M Ans:and also its pHpH. What will be its degree of ionization if the Ans:is 0.01M0.01M in HCl also?

Ans: Let the degree of ionization of propanoic acid be αα. Then, representing propionic acid as HA, we have:
HA +H2O⇔H3O+ + A−HA +H2O⇔H3O+ + A−
(.05−0.0α)≈0.5.05α.05α(.05−0.0α)≈0.5.05α.05α
Ka=[H3O+][A−][HA]Ka=[H3O+][A−][HA]
=(.05α)(.05α)0.05=.05α2=(.05α)(.05α)0.05=.05α2
α=Ka.05−−−√=1.63×10−2α=Ka.05=1.63×10−2
Then, [H3O+]=.05α=.05×1.63×10−2=Kb.15×10−4M[H3O+]=.05α=.05×1.63×10−2=Kb.15×10−4M
∴pH=3.09∴pH=3.09
In the presence of 0.1M0.1M of HClHCl, let αα ' be the degree of ionization.
Then, [H3O+]=0.01[H3O+]=0.01
[A−]=005α′[A−]=005α′
[HA]=.05[HA]=.05
Ka=0.01×.05α′.05Ka=0.01×.05α′.05
1.32×10−5=.01×α′1.32×10−5=.01×α′
α′=1.32×10−5

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60. The pHpHof 0.1M0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Ans: c=0.1Mc=0.1M
pH=2.34pH=2.34
−log[H+]=pH−log⁡[H+]=pH
−log[H+]=2.34−log⁡[H+]=2.34
[H+]=4.5×10−3[H+]=4.5×10−3
Also,
[H+]=cα[H+]=cα 4.5×10−3=0.1×α4.5×10−3=0.1×α 4.5×10−30.1=α4.5×10−30.1=α α=4.5×10−3=.045α=4.5×10−3=.045 Then,
=0.1×(45×10−3)2=0.1×(45×10−3)2
=202.5×10−6=202.5×10−6 =2.02×10−4

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61. The ionization constant of nitrous acid is 4.510−44.510−4. Calculate the pHpH of 0.04M0.04M sodium nitrite and also its degree of hydrolysis.

Ans: NaNO2NaNO2 is the salt of a strong base (NaOH)(NaOH) and a weak acid (HNO2)(HNO2).
NO−2+H2O↔HNO2+OH−NO2−+H2O↔HNO2+OH− 0
Kb=[HNO2][OH−][NO−2]Kb=[HNO2][OH−][NO2−]
⇒KwKa=10−144.5×10−14=.22×10−10⇒KwKa=10−144.5×10−14=.22×10−10
Now, If xx moles of the salt undergo hydrolysis, then the concentration of various species present in the solution, it will be:
[NO−2]=.04−x;0.04[NO2−]=.04−x;0.04
[HNO2]=x[HNO2]=x
[OH−]=x[OH−]=x
Kb=x20.04=0.22×10−10Kb=x20.04=0.22×10−10
x2=.0088×10−10x2=.0088×10−10
x=.093×10−5x=.093×10−5
∴[OH−]=0.093×10−5M∴[OH−]=0.093×10−5M
[H3O+]=10−14.093×10−5=10.75×10−9M[H3O+]=10−14.093×10−5=10.75×10−9M
⇒pH=−log(10.75×10−9)⇒pH=−log⁡(10.75×10−9)
= 7.96
Therefore, degree of hydrolysis
=x0.04=.093×10−5.04=x0.04=.093×10−5.04
=2.325×10−5

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62. A 0.02M0.02M solution of pyridinium hydrochloride has pH=3.44pH=3.44. Calculate the ionization constant of pyridine

Ans: pH=3.44pH=3.44
We know that,
pH=−log[H+]pH=−log⁡[H+] ∴[H+]=3.63×10−4∴[H+]=3.63×10−4
Then, Kb=(3.63×10−4)20.02(∵Kb=(3.63×10−4)20.02(∵ concentration =0.02M)=0.02M)
⇒Kb=6.6×10−6⇒Kb=6.6×10−6
Now, Kb=KwKaKb=KwKa
⇒Ka=KwKa=10−146.6×10−6=1.51×10−9

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63. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl,KBr,NaCN,NH4NO3,NaNO2NaCl,KBr,NaCN,NH4NO3,NaNO2 and KF

Ans:
(i) NaClNaCl :
NaCl+H2O↔NaOH+HClNaCl+H2O↔NaOH+HCl
Strong base Strong base
Therefore, it is a neutral solution.
(ii) KBr:
KBr+H2O↔KOH+HBrKBr+H2O↔KOH+HBr
Strong base Strong base
Therefore, it is a neutral solution.
iii) NaCN:NaCN:
NaCN+H2O↔HCN+NaOHNaCN+H2O↔HCN+NaOH
Weak acid Strong base
Therefore, it is a basic solution.
(iv) NH4NO3NH4NO3
NH4NO3+H2O↔NH4OH+HNO3NH4NO3+H2O↔NH4OH+HNO3
Weak acid Strong base
Therefore, it is an acidic solution.
(v) NaNO2NaNO2
NaNO2+H2O↔NaOH+HNO2NaNO2+H2O↔NaOH+HNO2
Strong base Weak acid
Therefore, it is a basic solution.
(vi) KF
KF+H2O↔KOH+HFKF+H2O↔KOH+HF
Strong base Weak acid
Therefore, it is a basic solution.

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64. The ionization constant of chloro acetic acid is 1.3510−3.1.3510−3. What will be the pHpH of 0.1M0.1M acid and its 0.1M0.1M sodium salt solution?

Ans: It is given that KaKa for ClCH2COOHClCH2COOH is 1.3510−31.3510−3
⇒Kb=cα2⇒Kb=cα2 ∴α=Kac−−−√∴α=Kac =1.35×10−30.1−−−−−−−−−−√(∴=1.35×10−30.1(∴ concentration of acid =0.1M)=0.1M)
⇒Kb−cα2⇒Kb−cα2
∴α=Kac−−−√∴α=Kac
=1.35×10−30.1−−−−−−−√(∴ concentration of acid =0.1M)=1.35×10−30.1(∴ concentration of acid =0.1M)
α=1.35×10−3−−−−−−−−−√α=1.35×10−3
=0.116=0.116
∴[H+]=cα=0.1×0.116∴[H+]=cα=0.1×0.116
⇒pH=−log[H+]=1.94⇒pH=−log⁡[H+]=1.94
ClCH2COONaClCH2COONa is the salt of a weak acid i.e., ClCH2COOHClCH2COOH and a strong base i.e., NaOHNaOH.
ClCH2COO−+H2O↔ClCH2COOH+OH−ClCH2COO−+H2O↔ClCH2COOH+OH−
Kb=[ClCH2COOH][OH−][ClCH2COO−]Kb=[ClCH2COOH][OH−][ClCH2COO−]
Kb=KwKaKb=KwKa
Kb=10−141.35×10−3Kb=10−141.35×10−3
=0.740×10−11=0.740×10−11
Also, Kb=x20.1Kb=x20.1 (where xx is the concentration of OH−OH−and ClCH2COOH)ClCH2COOH)
0.740×10−11=x20.10.740×10−11=x20.1
0.074×10−11=x20.074×10−11=x2
⇒x2=0.74×10−12⇒x2=0.74×10−12
x=0.86×10−6x=0.86×10−6
[OH−]=0.86×10−6[OH−]=0.86×10−6
∴[H+]=Kw0.86×10−6∴[H+]=Kw0.86×10−6
=10−140.86×10−6=10−140.86×10−6
[H+]=1.162×10−8[H+]=1.162×10−8
pH=−log[H+]pH=−log⁡[H+]
=7.94

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65. Ionic product of water at 310K310K is 2.710−142.710−14. What is the pHpH of neutral water at this temperature?

Ans: Ionic product,
Kw=[H+][OH−]Kw=[H+][OH−]
Let[H+]=xLet⁡[H+]=x
Since [H+]=[OH−],Kw=x2. Since [H+]=[OH−],Kw=x2.
⇒Kw at 310K is 2.7×10−14⇒Kw at 310K is 2.7×10−14
∴2.7×10−14=x2∴2.7×10−14=x2
⇒x=1.64×10−7⇒x=1.64×10−7
⇒[H+]=1.64×10−7⇒[H+]=1.64×10−7
⇒pH=−log[H+]⇒pH=−log⁡[H+]
=−log[1.64×10−7]=−log⁡[1.64×10−7]
=6.78=6.78
Hence, the pHpH of neutral water is 6.78.

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66. Calculate the pHpH of the resultant mixtures:

Ans. a. 10mL10mL of 0.2MCa(OH)2+25mL0.2MCa(OH)2+25mL of 0.1MHCl0.1MHCl
Ans: Moles of H3O+=25×0.11000=.0025molH3O+=25×0.11000=.0025mol
Moles of OH−=10×0.2×21000=.0040molOH−=10×0.2×21000=.0040mol
Thus, excess of OH−=.0015molOH−=.0015mol
b. 10mL10mL of 0.01MH2SO4+10mL0.01MH2SO4+10mL of 0.01MCa(OH)20.01MCa(OH)2
Ans: Moles of H3O+=2×10×0.11000=.0002molH3O+=2×10×0.11000=.0002mol
Moles of OH−=2×10×0.11000=.0002molOH−=2×10×0.11000=.0002mol
Since there is neither an excess of H3O+H3O+or OH−OH−
The solution is neutral. Hence, pH=7pH=7.
c.10mL10mL of 0.1MH2SO4+10mL0.1MH2SO4+10mL of 0.1MKOH0.1MKOH
Ans: Moles of H3O+=2×10×0.11000=.002molH3O+=2×10×0.11000=.002mol
Moles of OH−=10×0.11000=0.001molOH−=10×0.11000=0.001mol
Excess of H3O+=.001molH3O+=.001mol
Thus, [H3O+]=.00120×10−3=10−320×10−3=.05[H3O+]=.00120×10−3=10−320×10−3=.05 ∴pH=−log(0.05)∴pH=−log⁡(0.05) =1.30

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67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercuries iodide at 298K298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Ans:
1. Silver chromate:Ag2CrO4→2Ag++CrO2−4Ag2CrO4→2Ag++CrO42−
Then,
Ksp=[Ag+]2[CrO2−4]Ksp=[Ag+]2[CrO42−]
Let the solubility of Ag2CrO4Ag2CrO4 be ss.
⇒[Ag+]=2s and [CrO2−4]=s⇒[Ag+]=2s and [CrO42−]=sM
Then,
Ksp=(2s)2⋅s=4s3Ksp=(2s)2⋅s=4s3
⇒1.1×10−12=4s3⇒1.1×10−12=4s3
.275×10−12=s3.275×10−12=s3
s=0.65×10−4Ms=0.65×10−4M
MolarityofAg+=2s=2x0.65x10−4=1.30x10−4MMolarityofAg+=2s=2x0.65x10−4=1.30x10−4M
Molarity of CrO2−4=s=0.65×10−4MCrO42−=s=0.65×10−4M
2. Barium Chromate:
BaCrO4→Ba2++CrO2−4BaCrO4→Ba2++CrO42−
Then, Ksp=[Ba2+][CrO2−4]Ksp=[Ba2+][CrO42−]
Let the solubility of BaCrO4BaCrO4 be ss.
So, [Ba2+]=s[Ba2+]=s and [CrO2−4]=s⇒Ksp=s2[CrO42−]=s⇒Ksp=s2
⇒1.2×10−10=s2⇒1.2×10−10=s2
⇒s=1.09×10−5M⇒s=1.09×10−5
Molarity of Ba2+=Ba2+= Molarity of CrO2−4=s=1.09×10−5MCrO42−=s=1.09×10−5M
3.Ferric Hydroxide:
Fe(OH)3→Fe2+3OH−Fe(OH)3→Fe2+3OH−
Ksp=[Fe2+][OH−]3Ksp=[Fe2+][OH−]3
Let ss be the solubility of Fe(OH)3Fe(OH)3
Molarity of OH−=3s=4.17×1010MOH−=3s=4.17×1010M
4.Lead Chloride:
PbCl2→Pb2++2Cl−PbCl2→Pb2++2Cl−
KSP=[Pb2+][Cl−]2KSP=[Pb2+][Cl−]2
Let KSPKSP be the solubility of PbCl2PbCl2.
[PB2+]=s and [Cl−]=2s[PB2+]=s and [Cl−]=2s
Thus, Ksp=s.(2s)2 Thus, Ksp=s.(2s)2
=4s3=4s3
⇒1.6×10−5=4s3⇒1.6×10−5=4s3
⇒0.4×10−5=s3⇒0.4×10−5=s3
4×10−6=s3⇒1.58×10−2M=S.14×10−6=s3⇒1.58×10−2M=S.1
Molarity of PB2+=s=1.58×10−2MPB2+=s=1.58×10−2M
Molarity of chloride =2s=3.16×10−2M=2s=3.16×10−2M
5. Mercurous Iodide:
Hg2I2→Hg2++2I−Hg2I2→Hg2++2I−
Ksp=[Hg2+2][I−]2Ksp=[Hg22+][I−]2
Let ss be the solubility of Hg2I2Hg2I2.
⇒[Hg2+2]=s and [I−]=2s⇒[Hg22+]=s and [I−]=2s
Thus, Ksp=s(2s)2⇒Ksp=4s3Ksp=s(2s)2⇒Ksp=4s3
4.5×10−29=4s34.5×10−29=4s3
1.125×10−29=s31.125×10−29=s3
⇒s=2.24×10−10M⇒s=2.24×10−10M
Molarity of Hg2+2=s=2.24×10−10MHg22+=s=2.24×10−10M
Molarity of I−=2s=4.48×10−10M

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68. The solubility product constant of Ag2CrO4Ag2CrO4 and AgBrAgBr are 1.1×10−121.1×10−12 and 5.0×10135.0×1013 respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans: Let ss be the solubility of Ag2CrO4Ag2CrO4
Then, Ag2CrO4→Ag2++2CrO2−4Ag2CrO4→Ag2++2CrO42−
Ksp=(2s)2⋅s=4s3Ksp=(2s)2⋅s=4s3
1.1×10−12=4s31.1×10−12=4s3
s=6.5×10−5Ms=6.5×10−5M
Let s′s′ be the solubility of AgBrAgBr.
AgBr(s)↔Ag++Br−AgBr(s)↔Ag++Br−
Ksp=s′2=5.0×10−13Ksp=s′2=5.0×10−13
∴s′=7.07×10−7M∴s′=7.07×10−7M
Therefore, the ratio of the molarities of their saturated solution is
ss′=6.5×10−5M7.07×10−7M=91.9

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69. Equal volumes of 0.002M0.002M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?

Ans: 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., ,
0.001M0.001M. Then,
NalO3→Na++lO−3NalO3→Na++lO3−
0.001M0.001M0.001M0.001M
Cu(ClO3)2→Cu2++2ClO−3Cu(ClO3)2→Cu2++2ClO3−
0.001M0.001M0.001M0.001M
Now, the solubility equilibrium for copper iodate can be written as:
Cu(lO3)2→Cu2+(aq)+2lO−3(aq)Cu(lO3)2→Cu(aq)2++2lO3(aq)−
Ionic product of copper iodate:
=[Cu2+]×[lO−3]2=[Cu2+]×[lO3−]2
=(0.001)(0.001)2=(0.001)(0.001)2
=1×10−9=1×10−9
Since the ionic product (1×10−9)(1×10−9) is less than Ksp(7.4×10−8)Ksp(7.4×10−8), precipitation will not occur.

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70. The ionization constant of benzoic acid is 6.46×10−56.46×10−5 and KspKsp for silver benzoate is 2.5×10−132.5×10−13. How many times is silver benzoate more soluble in a buffer of pH=3.19pH=3.19 compared to its solubility in pure water?

Ans: Since pH=3.19, Since pH=3.19,
[H3O+]=6.46×10−4M[H3O+]=6.46×10−4M
C6H5COOH+H2O↔C6H5COO−+H3OC6H5COOH+H2O↔C6H5COO−+H3O
Ka=[C6H5COO−][H3O+][C6H5COOH]Ka=[C6H5COO−][H3O+][C6H5COOH]
[C6H5COOH][C6H5COO−]=[H3O+]Ka=6.46×10−46.46×10−5=10[C6H5COOH][C6H5COO−]=[H3O+]Ka=6.46×10−46.46×10−5=10
Let the solubility of C6H5COOAgC6H5COOAg be xmol/Lxmol/L.
Then,
[Ag+]=x[Ag+]=x
[C6H5COOH]+[C6H5COO−]=x[C6H5COOH]+[C6H5COO−]=x
10[C6H5COO−]+[C6H5COO−]=x10[C6H5COO−]+[C6H5COO−]=x
[C6H5COO−]=x11[C6H5COO−]=x11
Ksp[Ag+][C6H5COO−]Ksp[Ag+][C6H5COO−]
2.5×10−13=x(x11)2.5×10−13=x(x11)
x=1.66×10−6mol/Lx=1.66×10−6mol/L
Thus, the solubility of silver benzoate in a pH3.19pH3.19 solution is 1.66×10−6mol/L1.66×10−6mol/L. Now, let the solubility of C6H5COOAgC6H5COOAg be x′mol/Lx′mol/L.
Then, [Ag+]=x′M[Ag+]=x′M and [C6H5COO−]=x′M[C6H5COO−]=x′M
Ksp=[Ag+][C6H5COO−]Ksp=[Ag+][C6H5COO−]
Ksp=(x′)2Ksp=(x′)2
x′=Ksp−−−√=2.5×10−13−−−−−−−−−√=5×10−7mol/Lx′=Ksp=2.5×10−13=5×10−7mol/L
∴xx′=1.66×10−65×10−7=3.32∴xx′=1.66×10−65×10−7=3.32
Hence, C6H5COOAgC6H5COOAg is approximately 3.317 times more soluble in a low pHpH solution.

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71. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide?
(For iron sulphide, Ksp=6.310−18)

Ans: Let the maximum concentration of each solution be xmol/Lxmol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2x/2.
∴[FeSO4]=[Na2S]=x2M∴[FeSO4]=[Na2S]=x2M
Then, [Fe2+]=[FeSO4]=x2M[Fe2+]=[FeSO4]=x2M
Also, [S2−]=[Na2S]=x2M[S2−]=[Na2S]=x2M
FeS(x)↔Fe2+(aq)+S2−(aq)FeS(x)↔Fe(aq)2++S(aq)2−
Ksp=[Fe2+][S2−]Ksp=[Fe2+][S2−]
6.3×10−18=(x2)(x2)6.3×10−18=(x2)(x2)
x24=6.3×10−18x24=6.3×10−18
⇒x=5.02×10−9⇒x=5.02×10−9
If the concentrations of both solutions are equal to or less than 5.02×10−9M5.02×10−9M, then there will be no precipitation of iron sulphide.

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72. What is the minimum volume of water required to dissolve 1g1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp Ksp is 9.110−6)

Ans: CaSO4(s)↔Ca2+(aq)+SO2−4(aq)CaSO4(s)↔Ca2+(aq)+SO4(aq)2−
Ksp=[Ca2+][SO2−4]Ksp=[Ca2+][SO42−]
Let the solubility of CaSO4CaSO4 be ss.
Then, Ksp=s2Ksp=s2
9.1×10−6=s29.1×10−6=s2
s=3.02×10−3mol/Ls=3.02×10−3mol/L
Molecular mass of CaSO4=136g/molCaSO4=136g/mol
Solubility of CaSO4CaSO4 in gram/L
=3.02×10−3×136=3.02×10−3×136
=0.41g/L=0.41g/L
This means that we need 1L1L of water to dissolve 0.41g0.41g of CaSO4CaSO4
Therefore, to dissolve 1g1g of CaSO4CaSO4 we require =10.41L=2.44L=10.41L=2.44L of water.

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73. The concentration of sulphide ion in 0.1MHCl0.1MHCl solution saturated with hydrogen sulphide is 1.0 10−19M10−19M. If 10mL10mL of this is added to 5mL5mL of 0.04M0.04M solution of the following:
FeSO4,MnCl2,ZnCl2FeSO4,MnCl2,ZnCl2 and CdCl2CdCl2. in which of these solutions precipitation will take place?
Givěn KspKsp for Fes=6.3×10−18,MnS=2.5×10−13,ZnS=1.6×10−24

Ans: For precipitation to take place, it is required that the calculated ionic product exceeds the KspKsp value.
Before mixing:
[S2−]=1.0×10−19M[M2+]=0.04M[S2−]=1.0×10−19M[M2+]=0.04M
volume =10mL volume =5mL volume =10mL volume =5mL
After mixing:
[S2−]=?[M2+]=?[S2−]=?[M2+]=?
volume =(10+5)=15mL volume =15mL volume =(10+5)=15mL volume =15mL
[S2−]=1.0×10−19×1015=6.67×10−20M[S2−]=1.0×10−19×1015=6.67×10−20M
[M2+]=0.04×515=1.33×10−2M[M2+]=0.04×515=1.33×10−2M
Ionic product =[M2+][S2−] Ionic product =[M2+][S2−]
=(1.33×10−2)(6.67×10−20)=(1.33×10−2)(6.67×10−20)
=8.87×10−22=8.87×10−22
This ionic product exceeds the Kspof ZnsKspof Zns and CdSCdS. Therefore, precipitation will occur in CdCl2CdCl2 and ZnCl2ZnCl2 solutions.

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Last Updated on: Mar 13, 2024