Chapter 7: Equilibrium

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

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.

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

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 

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

1. 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 

2. 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] 

3. 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] 

4. 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 

5. 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 

5. Find out the value of KcKc  for each of the following equilibria from the value of KpKp :

1. 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)

2. 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)

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.

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.

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 

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 

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=74.79×1014LmolKc=74.79×1014Lmol

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.

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)N2(g)+3H2(g)⇌2NH3(g)

The given concentration of various species is

[N2]=1.5720molL−1[H2]=1.9220molL−1[N2]=1.5720molL−1[H2]=1.9220molL−1 

[NH3]=8.1320molL−1[NH3]=8.1320molL−1 

Now, reaction quotient QcQc  is:

QC=[NH3]2[N2][H2]3𝑄𝐶=[𝑁𝐻3]2[𝑁2][𝐻2]3 

QC=((8.13)20)2(1.5720)(1.9220)3𝑄𝐶=((8.13)20)2(1.5720)(1.9220)3 

QC=2.4×103𝑄𝐶=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

13. The equilibrium constant expression for a gas reaction is: 

Kc=[NH3]4 [O2]5[NO]4 [H2O]6Kc=[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:

4NO (g)+6H2O (g)⇌4NH3(g)+5O2(g)4NO (g)+6H2O (g)⇌4NH3(g)+5O2(g) 

14. One mole of H2OH2O  and one mole of COCO  are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with COCO  according to the equation: H2(g)+CO(g)⇌H2(g)+CO2(g)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)H2O(g) + CO(g)⇌H2(g) + CO2(g) 

110M110M00110M110M00 

1 – 0.410M1 – 0.410M0.410M0.410M1 – 0.410M1 – 0.410M0.410M0.410M 

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

Therefore, the equilibrium constant for the reaction:

Kc=[H2][CO2][H2O][CO]Kc=[H2][CO2][H2O][CO]

Kc=0.04×0.040.06×0.06Kc=0.04×0.040.06×0.06

Kc=0.444Kc=0.444

15.  At 700 K, equilibrium constant for the reaction H2(g)+I2(g)⇌2HI (g)H2(g)+I2(g)⇌2HI (g) is 54.8. If 0.5molL−10.5molL−1 of HI (g)HI (g)  is present at equilibrium at 700 K, what are the concentration of H2(g)H2(g)  and I2(g)I2(g)  assuming that we initially started with HI (g)HI (g)  and allowed it to reach equilibrium at 700 K?

Ans: It is given that equilibrium constant KcKc  for the reaction

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

Therefore, at equilibrium, the equilibrium constant K’cK’c  for the reaction

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

[HI]=0.5molL−1[HI]=0.5molL−1 will be: 154.8154.8 

Let the concentrations of hydrogen and iodine at equilibrium be xmolL−1𝑥molL−1 

[H2]=[I2]=xmolL−1[H2]=[I2]=𝑥molL−1 

Therefore, [H2] [I2][HI]2=K’c[H2] [I2][HI]2=K’c 

x×x(0.5)2=154.8𝑥×𝑥(0.5)2=154.8 

⇒x2=0.2554.8⇒x=0.06754⇒𝑥2=0.2554.8⇒𝑥=0.06754

x=0.068 molL –  1𝑥=0.068 molL –  1 (approximately)

Hence, at equilibrium, [H2]=[I2]=0.068 molL –  1[H2]=[I2]=0.068 molL –  1

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

2ICI (g)⇌I2(g)+Cl2(g)2ICI (g)⇌I2(g)+Cl2(g) ; Kc=0.14Kc=0.14 

Ans: For the given reaction:

2ICI (g)⇌I2(g)+Cl2(g)2ICI (g)⇌I2(g)+Cl2(g)

The initial concentration of ICIICI is 0.78M and for I2I2 and Cl2Cl2 is zero (0).

At equilibrium, the concentration for ICIICI is (0.78−2x)M(0.78−2𝑥)M and for I2I2 and Cl2Cl2is xM𝑥M.

Now, we can write:

[I2][Cl2][ICI]2=Kc[I2][Cl2][ICI]2=Kc

x×x(0.78−2x)2=0.14𝑥×𝑥(0.78−2𝑥)2=0.14

x0.78−2x=0.374𝑥0.78−2𝑥=0.374

x=0.292−0.748x𝑥=0.292−0.748𝑥

1.748x=0.2921.748𝑥=0.292

x=0.167𝑥=0.167

Therefore, at equilibrium

[H2][H2] = [I2][I2] = 0.1670.167

M[HI]=(0.78−2×0.167)M[HI]=(0.78−2×0.167)

M[HI]=0.446MM[HI]=0.446M

17. Kp=0.04atmKp=0.04atm  at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6C2H6  when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6(g)⇌C2H4(g)+H2(g)C2H6(g)⇌C2H4(g)+H2(g) 

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)C2H6(g)⇌C2H4(g)+H2(g)

The initial concentration of C2H6C2H6 is 4.0 atm and that of C2H4C2H4 and H2H2 is zero (0).

At equilibrium, the concentration of C2H6C2H6 is (4.0 – p) and that of C2H4C2H4 and H2H2is p. 

Then we can write,

⇒Kpp×p40−p=0.04⇒𝐾𝑝𝑝×𝑝40−𝑝=0.04

p2=0.16−0.04p𝑝2=0.16−0.04𝑝

p2+0.04p−0.16=0𝑝2+0.04𝑝−0.16=0

p=−0.04±(0.04)2−4×1×(−0.16)−−−−−−−−−−−−−−−−−−−−√2×1𝑝=−0.04±(0.04)2−4×1×(−0.16)2×1

p=−0.04±0.082𝑝=−0.04±0.082

If we take positive value:

p=0.762𝑝=0.762

p=0.38𝑝=0.38

So, at equilibrium:

[C2H6]−4−p=4−0.38=3.62 atm[𝐶2𝐻6]−4−𝑝=4−0.38=3.62 atm 

18.  A sample of pure PCl5PCl5  was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5PCl5  was found to be 0.5×10−1molL – 10.5×10−1molL – 1  . If value of KcKc  is 8.3×10−38.3×10−3  , what are the concentrations of PCl3PCl3  and Cl2Cl2  at equilibrium? PCl5(g)⇌PCl3(g)+Cl2(g)PCl5(g)⇌PCl3(g)+Cl2(g) 

Ans: Let the concentrations of both PCl3PCl3  and Cl2Cl2  at equilibrium be xmolL – 1𝑥molL – 1 . The given reaction is:

PCl5(g)⇌PCl3(g)+Cl2(g)PCl5(g)⇌PCl3(g)+Cl2(g)

At equilibrium the concentration of PCl5PCl5is 0.5×10−1molL – 10.5×10−1molL – 1

It is given that the value of equilibrium constant KcKc  is 8.3×10−38.3×10−3

The expression for equilibrium will be:

⇒[PCl3][Cl2][PCl5]=Kc⇒[PCl3][Cl2][PCl5]=Kc

⇒x×x0.5×10−1=8.3×10−3⇒𝑥×𝑥0.5×10−1=8.3×10−3

⇒x2=4.15×10−4⇒𝑥2=4.15×10−4

⇒x=2.04×10−2⇒𝑥=2.04×10−2

⇒x=0.0204⇒𝑥=0.0204

Therefore at equilibrium:

[PCl3]=[Cl2]=0.02molL−1[PCl3]=[Cl2]=0.02molL−1 

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 CO2CO2 .FeO (s)+CO (g)⇌Fe (s)+CO2(g)FeO (s)+CO (g)⇌Fe (s)+CO2(g) ; Kp=0.0265Kp=0.0265 at 1050 K.

What are the equilibrium partial pressures of COCO  and CO2CO2  at 1050 K if the initial partial pressures are: pCO=1.4atmpCO=1.4atm  and pCO2=0.80atmpCO2=0.80atm ?

Ans: For the given reaction:

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

The initial concentration of FeOFeO is 1.4 atm and the initial concentration of CO2CO2 is 0.80 atm. 

It is given in the that, Kp=0.265Kp=0.265.

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

Therefore, we can say that the pressure of COCO  will increase while the pressure of CO2CO2  will decrease. Now, let the increase in pressure of COCO  be equal to the  decrease in pressure of CO2CO2  be ‘p’. Then, we can write:

Qp=pCO2pCO=0.801.4=0.571Qp=pCO2pCO=0.801.4=0.571

Therefore, equilibrium partial of CO2CO2 will be, pCO2=0.080−0.339=0.461 atmpCO2=0.080−0.339=0.461 atm 

And, equilibrium partial pressure of COCO will be, pCO=1.4−0.339=1.739 atmpCO=1.4−0.339=1.739 atm 

20. Equilibrium constant, KcKc  for the reaction N2(g)+3H2(g)⇌2NH3(g)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−13.0 molL−1  N2N2 , 2.0 molL−12.0 molL−1  H2H2  and 0.5 molL−10.5 molL−1  NH3NH3 . 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)N2(g)+3H2(g)⇌2NH3(g)

At a particular time, the concentration of N2N2 is 3.0 molL−13.0 molL−1 , for H2H2 is 2.0 molL−12.0 molL−1 and for NH3NH3 is 0.5 molL−10.5 molL−1.

Now, we know that:

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

It is given that Kc=0.061Kc=0.061 

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

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

21. Bromine monochloride, BrClBrCl  decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g)⇌Br2(g)+Cl2(g)2BrCl (g)⇌Br2(g)+Cl2(g) for which Kc=32Kc=32 at 500 K. If initially pure BrClBrCl  is present at a concentration of 3.3×10−3 molL−13.3×10−3 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)2BrCl (g)⇌Br2(g)+Cl2(g)

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

Now, we can write.

⇒[Br2][Cl2][BrCl]=Kc⇒[Br2][Cl2][BrCl]=Kc

⇒x×x(3.3×10−3−2x)2=32⇒𝑥×𝑥(3.3×10−3−2𝑥)2=32

⇒x3.3×10−3−2x=5.66⇒𝑥3.3×10−3−2𝑥=5.66

⇒x=18.678×10−3−11.32x⇒𝑥=18.678×10−3−11.32𝑥

⇒12.32x=18.678×10−3⇒12.32𝑥=18.678×10−3

⇒x=1.5×10−3⇒𝑥=1.5×10−3

 Therefore, at equilibrium

[BrCl]=3.3×10−3−(2×1.5×10−3)[BrCl]=3.3×10−3−(2×1.5×10−3)

[BrCl]=3.3×10−3−3.0×10−3[BrCl]=3.3×10−3−3.0×10−3

[BrCl]=3.0×10−4 molL−1[BrCl]=3.0×10−4 molL−1

22. At 1127 K and 1 atm pressure, a gaseous mixture of COCO  and CO2CO2  in equilibrium with solid carbon has 90.55% COCO by mass: C (s)+CO2(g)⇌2CO (g)C (s)+CO2(g)⇌2CO (g). Calculate KcKc  for this reaction at the above temperature.

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

Mass of COCO = 90.55 gm

And, mass of CO2CO2 = (100 – 90.55) = 9.45 gm

Now, the number of moles of COCO , nCO=90.5528=3.234 mol𝑛𝐶𝑂=90.5528=3.234 mol 

Number of moles of CO2CO2, nCO2=9.4544=0.215 molnCO2=9.4544=0.215 mol 

Partial pressure of COCO , pCO=nCOnCO+nCO2×ptotal𝑝𝐶𝑂=𝑛𝐶𝑂𝑛𝐶𝑂+𝑛𝐶𝑂2×𝑝total 

pCO=3.2343.234+0.215×1𝑝𝐶𝑂=3.2343.234+0.215×1

pCO=0.938atm𝑝𝐶𝑂=0.938atm

Partial pressure of CO2CO2 ,

pCO2=nCO2nCO+nCO2×ptotal𝑝𝐶𝑂2=𝑛𝐶𝑂2𝑛𝐶𝑂+𝑛𝐶𝑂2×𝑝total

pCO2=0.2153.234+0.215×1𝑝𝐶𝑂2=0.2153.234+0.215×1

pCO2=0.062 atm𝑝𝐶𝑂2=0.062 atm

Therefore, 

For the given reaction:

Δn=2−1=1Δ𝑛=2−1=1 

We know that,

Kp=Kc(RT)Δn14.19Kp=Kc(RT)Δn14.19 

=Kc(0.082×1127)1=Kc(0.082×1127)1

Kc=14.190.082×1127Kc=14.190.082×1127

Kc=0.154Kc=0.154

23. Calculate:

1. ΔG∘ΔG∘ and
2. The equilibrium constant for the formation of NO2NO2  from NONO  and O2O2  at 298 K NO (g)+12O2(g)⇌NO2(g)NO (g)+12O2(g)⇌NO2(g) 

Where:

ΔfG∘(NO2)=52kJ/molΔfG∘(NO2)=52kJ/mol

ΔfG∘(NO)=87.0kJ/molΔfG∘(NO)=87.0kJ/mol

ΔfG∘(O2)=0kJ/molΔfG∘(O2)=0kJ/mol

Ans: For the given reaction: 

NO (g)+12O2(g)⇌NO2(g)NO (g)+12O2(g)⇌NO2(g)

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

ΔG∘=ΔG∘(Products)−ΔG∘(Reactants)ΔG∘=ΔG∘(Products)−ΔG∘(Reactants) 

ΔG∘=52.0−[87.0+0]ΔG∘=52.0−[87.0+0]

ΔG∘=−35.0 kJmol−1ΔG∘=−35.0 kJmol−1

2. We know that:

ΔG∘=RTlogKcΔG∘=RTlog⁡Kc

ΔG∘=2.303RTlogKcΔG∘=2.303RTlog⁡Kc

logKc=−35.0×10−3−2.303×8.314×298log⁡Kc=−35.0×10−3−2.303×8.314×298

logKc=6.134log⁡Kc=6.134

Kc=antilog (6.134)Kc=antilog (6.134)

Kc=1.36×106Kc=1.36×106

Hence, the equilibrium constant for the given reaction KcKc is 1.36×1061.36×106 .

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?

1. PCl5(g)⇌PCl3(g)+Cl2(g)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.

2. CaO (s)+CO2(g)⇌CaCO3(s)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.

3. 3Fe (s)+4H2O (g)⇌Fe3O4(s)+4H2(g)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.

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.

1. COCl2(g)⇌CO (g)+Cl2(g)COCl2(g)⇌CO (g)+Cl2(g) 
2. CH4(g)+2S2(g)⇌CS2(g)+2H2S (g)CH4(g)+2S2(g)⇌CS2(g)+2H2S (g) 
3. CO2(g)⇌C (s)+2CO (g)CO2(g)⇌C (s)+2CO (g) 
4. 2H2(g)+CO (g)⇌CH3OH (g)2H2(g)+CO (g)⇌CH3OH (g) 
5. CaCO3(s)⇌CaO (s)+CO2(g)CaCO3(s)⇌CaO (s)+CO2(g) 
6. 4NH3(g)+5O2(g)⇌4NO (g)+6H2O (g)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.

26. The equilibrium constant for the following reaction is 1.6×1051.6×105  at 1024 K. H2(g)+Br2(g)⇌2HBr (g)H2(g)+Br2(g)⇌2HBr (g) , Find the equilibrium pressure of all gases if 10.0 bar of HBrHBr  is introduced into a sealed container at 1024 K.

Ans: Given,

Kp𝐾𝑝 for the reaction i.e.,

H2(g)+Br2(g)↔2HBr(g)𝐻2(𝑔)+𝐵𝑟2(𝑔)↔2𝐻𝐵𝑟(𝑔) is 1.6×1051.6×105

Therefore, for the reaction 2HBr(g)↔H2(g)+Br2(g)2𝐻𝐵𝑟(𝑔)↔𝐻2(𝑔)+𝐵𝑟2(𝑔) the equilibrium constant will be, K′p=1Kp𝐾𝑝′=1𝐾𝑝

=11.6×10511.6×105

=6.25×10−66.25×10−6

Now, let pp be the pressure of both H2H2 and Br2Br2 at equilibrium.

pHBr×p2p2HBr=K′p𝑝𝐻𝐵𝑟×𝑝2𝑝2𝐻𝐵𝑟=𝐾𝑝′

p×p(10−2p)=6.25×10−6𝑝×𝑝(10−2𝑝)=6.25×10−6

p10−2p=2.5×10−3𝑝10−2𝑝=2.5×10−3

p=2.5×10−2−(5.0×10−)p𝑝=2.5×10−2−(5.0×10−)𝑝

p+(5.0×10−3)p=2.5×10−2𝑝+(5.0×10−3)𝑝=2.5×10−2

(1005×10−3)p=2.5×10−2(1005×10−3)𝑝=2.5×10−2

p=2.49×10−2𝑝=2.49×10−2 bar =2.5×10−2=2.5×10−2 (approximately) 

Therefore, at equilibrium,

[H2]=[Br2]=2.49×10−2 bar [H2]=[Br2]=2.49×10−2 bar 

[HBr]=10−2×(2.49×10−1) bar [HBr]=10−2×(2.49×10−1) bar 

=9.95bar=10 bar (approximately) 9.95bar=10 bar (approximately) 

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)CH4(g)+H2O (g)⇌CO (g)+3H2(g) 

1. Write an expression for KpKp  for the above reaction.

Ans: For the given equation:

Kp=pCO×p3H2pCH4×pH2O𝐾𝑝=𝑝𝐶𝑂×𝑝3𝐻2𝑝𝐶𝐻4×𝑝𝐻2𝑂 

2. How will the values of KpKp and composition of equilibrium mixture be affected by 
3. 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 KpKp 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 KpKp 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 KpKp is unaffected.

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

1. Addition of H2H2 

Ans: According to Le Chatelier’s principle, when H2H2  is added to a reaction, the equilibrium shifts in the forward direction.

2. Addition of CH3OHCH3OH  

Ans: The equilibrium will shift backwards with the addition of CH3OHCH3OH .

3. Removal of COCO  

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

4. Removal of CH3OHCH3OH 

Ans: When CH3OHCH3OH  is removed from the equation, the equilibrium shifts forward.

29. At 473 K, equilibrium constant KrKr  for decomposition of phosphorus pentachloride, PCl5PCl5  is 8.3×10−38.3×10−3  . If decomposition is depicted as: 

PCl5(g)⇌PCl3(g)+Cl2(g)PCl5(g)⇌PCl3(g)+Cl2(g) , ΔrH∘=124.0kJmol−1Δ𝑟𝐻∘=124.0kJmol−1 

1. Write an expression for KcKc  for the reaction. 

Ans: The expression for KcKc will be:

Kc=[PCl3] [Cl2][PCl5]Kc=[PCl3] [Cl2][PCl5] 

2. What is the value of KcKc  for the reverse reaction at the same temperature? 

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

Kcprime=1Kc𝐾𝑐prime=1𝐾𝑐

=18.3×10−3=1.2048×102=18.3×10−3=1.2048×102

=120−48=120−48

3. What would be the effect on KcKc  if 
4. More PCl5PCl5  is added?

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

2. Pressure is increased? 

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

3. The temperature has increased?

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

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 COCO  and H2H2 . In second stage, COCO  formed in first stage is reacted with more steam in water gas shift reaction:  CO (g)+H2O (g)⇌CO2(g)+H2(g)CO (g)+H2O (g)⇌CO2(g)+H2(g) If a reaction vessel at 400°C is charged with an equimolar mixture of COCO  and steam such that pCO=pH2O=4.0 bar𝑝CO=𝑝H2O=4.0 bar , what will be the partial pressure of H2H2  at equilibrium? Kp=10.1Kp=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)CO (g)+H2O (g)⇌CO2(g)+H2(g)

At equilibrium the concentration of COCO and H2OH2O will be (4.0−p)(4.0−𝑝).

It is given that, Kp=10.1Kp=10.1 

Now,

pco2×p2pco×pH2o=Kp𝑝𝑐𝑜2×𝑝2𝑝𝑐𝑜×𝑝𝐻2𝑜=𝐾𝑝

⇒p×p(4.0−p)(4.0−p)⇒𝑝×𝑝(4.0−𝑝)(4.0−𝑝)

⇒p40−p=3.178⇒𝑝40−𝑝=3.178

⇒p=12.712−3.178p⇒𝑝=12.712−3.178𝑝

⇒4.178p=12.712⇒4.178𝑝=12.712

⇒p=3.04⇒𝑝=3.04

Hence, at equilibrium, the partial pressure of H2H2  will be 3.04 bar.

31. Predict which of the following reaction will have appreciable concentration of reactants and products:

1. Cl2(g)⇌2Cl (g)Cl2(g)⇌2Cl (g) ; Kc=5×10−39𝐾𝑐=5×10−39 
2. Cl2(g)+2NO (g)⇌2NOCl (g)Cl2(g)+2NO (g)⇌2NOCl (g) ; Kc=3.7×108𝐾𝑐=3.7×108 
3. Cl2(g)+2NO2(g)⇌2NO2Cl (g)Cl2(g)+2NO2(g)⇌2NO2Cl (g) ; Kc=1.8𝐾𝑐=1.8 

Ans: If the value of KcKc  lies between 10−310−3  and 103103  , 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)3O2(g)⇌2O3(g) is 2.0×10−502.0×10−50 at 25°C. If the equilibrium concentration of O2O2  in air at 25°C is 1.6×10−21.6×10−2  , what is the concentration of O3O3 ?

Ans: The given reaction is:

3O2(g)⇌2O3(g)3O2(g)⇌2O3(g)

KC=[O3(g)]2[O2(g)]3𝐾𝐶=[𝑂3(𝑔)]2[𝑂2(𝑔)]3

 KC=2.0×10−50𝐾𝐶=2.0×10−50 and  [O2(g)]=1.6×10−2[𝑂2(𝑔)]=1.6×10−2

2.0×10−50=[O3(g)]2[1.6×10−2]32.0×10−50=[𝑂3(𝑔)]2[1.6×10−2]3 

⇒[O3(g)]2=2.0×10−50×(1.6×10−2)3⇒[𝑂3(𝑔)]2=2.0×10−50×(1.6×10−2)3 

⇒[O3(g)]2=8.192×10−50⇒[𝑂3(𝑔)]2=8.192×10−50 

⇒[O3(g)]=2.86×10−28M⇒[𝑂3(𝑔)]=2.86×10−28𝑀 

 Hence, the concentration of O3O3  is 2.86×10−28M2.86×10−28M .

33. The reaction CO (g)+3H2(g)⇌CH4(g)+H2O(g)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 COCO , 0.10 mol of H2H2  and 0.02 mol of H2OH2O  and an unknown amount of CH4CH4  in the flask. Determine the concentration of CH4CH4  in the mixture. The equilibrium constant, KcKc  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)CO (g)+3H2(g)⇌CH4(g)+H2O(g)

At equilibrium, the concentration of COCO  is 0.3 M , H2H2 is 0.1 M and H2OH2O  is 0.02 M 

Therefore,

[CH4(g)][H2O(g)][CO(g)][H2(g)]3=Kc[𝐶𝐻4(𝑔)][𝐻2𝑂(𝑔)][𝐶𝑂(𝑔)][𝐻2(𝑔)]3=𝐾𝑐 

⇒x×0.020.3×(0.1)3=3.90⇒𝑥×0.020.3×(0.1)3=3.90

⇒x=3.90×0.3×(0.1)30.02⇒𝑥=3.90×0.3×(0.1)30.02 

 = 0.001170.020.001170.02 

=0.0585M=0.0585𝑀 

=5.85×10−2M=5.85×10−2𝑀 

 Hence, the concentration of CH4CH4  at equilibrium is 5.85×10−2M5.85×10−2M. 

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−HNO2,CN−,HClO4,F−,OH−,CO32−,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. 

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

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

36. What will be the conjugate bases for the Brönsted acids: HF, H2SO4, HCO3HF, H2SO4, HCO3 

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

37. Write the conjugate acids for the following Brönsted bases: NH−2,NH3,HCOO−NH2−,NH3,HCOO− 

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

38. The species: H2O,HCO−3,HSO−4,NH3H2O,HCO3−,HSO4−,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.

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

1. OH−OH− 

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

2. F−F− 

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

3. H + H +  

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

4. BCl3BCl3 

Ans: BCl3BCl3  is a Lewis acid since it can accept a pair of electrons.

40. The concentration of hydrogen ion in a sample of soft drink is 3.8×10−3 M3.8×10−3 M . what is its pH?

Ans: Given, 

[H + ]=3.8×10−3 M[H + ]=3.8×10−3 M 

Therefore, the pH value of the soft drink:

pH=−log[H+]pH=−log⁡[𝐻+]

pH=−log(3.8×10−3)pH=−log⁡(3.8×10−3)

pH=−log3.8−log10−3pH=−log⁡3.8−log⁡10−3

pH=−log3.8+3pH=−log⁡3.8+3

pH=−0.58+3pH=−0.58+3

pH=2.42pH=2.42

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

Ans: Given the, pH=3.76pH=3.76 

It is known as:

pH=−log[H+]pH=−log⁡[H+]

log[H+]=−pHlog⁡[H+]=−pH

[H+]=antilog(−pH)[H+]=antilog(−pH)

[H+]=antilog(−3.76)[H+]=antilog(−3.76)

[H+]=1.74×10−4M[H+]=1.74×10−4M

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74×10−4 M1.74×10−4 M. 

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=2.08×10−6Kb=2.08×10−6

43. The ionization constant of phenol is 1.0×10−101.0×10−10 . 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+C6H5OH+H2O⇌C6H5O−+H3O+ 

The initial concentration of C6H5OHC6H5OH is 0.05 M. 

At equilibrium, the concentration of C6H5O−𝐶6𝐻5𝑂− and H3O+𝐻3𝑂+ will be x, then the concentration of C6H5OH𝐶6𝐻5𝑂𝐻 will be (0.05−x)(0.05−𝑥).

Ka=[C6H5O−][H3O+][C6H5OH]𝐾𝑎=[𝐶6𝐻5𝑂−][𝐻3𝑂+][𝐶6𝐻5𝑂𝐻] 

Ka=x×x0.05−x𝐾𝑎=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−−−−−−−−−−−−−√∴𝑥=1×10−10×0.05 

=5×10−12−−−−−−−−√=5×10−12 

=2.2×10−6M=[H3O+]=2.2×10−6M=[H3O+] 

Since[H3O+]=[C6H5O−]Since[H3O+]=[C6H5O−] 

[C6H5O−]=2.2×10−6M[C6H5O−]=2.2×10−6M 

Now, let α be the degree of ionization of phenol in the presence of 0.01 M C6H5ONaC6H5ONa

C6H5OH+H2O⇌C6H5O−+H3O+C6H5OH+H2O⇌C6H5O−+H3O+ 

The concentrations are as follows:

[C6H5OH]=0.05−0.05α[C6H5OH]=0.05−0.05𝛼

0.05 M[C6H5O−]=0.01+0.05α0.05 M[C6H5O−]=0.01+0.05𝛼

0.01 M[H3O+]=0.05α0.01 M[H3O+]=0.05𝛼

Ka=[C6H5O−][H3O+][C6H5OH]𝐾𝑎=[𝐶6𝐻5𝑂−][𝐻3𝑂+][𝐶6𝐻5𝑂𝐻] 

Ka=(0.01)(0.05α)0.05Ka=(0.01)(0.05𝛼)0.05

1.0×10−10=0.01α1.0×10−10=0.01𝛼

α=1×10−8𝛼=1×10−8

44. The first ionization constant of H2SH2S  is 9.1×10−89.1×10−8  . Calculate the concentration of HS−HS−  ion in its 0.1 M solution. How will this concentration be affected if the Ans:is 0.1 M in HClHCl  also? If the second dissociation constant of H2SH2S  is 1.2×10−131.2×10−13 , calculate the concentration of S2−S2− under both conditions.

Ans: To calculate the concentration of HS−HS− ion:

Case I (in the absence of HClHCl ):

Let the concentration of HS−HS−be x M:

H2S⇌H + +HS−H2S⇌H + +HS− 

The final concentration of H2SH2S will be (0.1−x)(0.1−𝑥) 

Then,

Kai=[H+][HS−][H2S]𝐾𝑎𝑖=[𝐻+][𝐻𝑆−][𝐻2𝑆] 

9.1×10−8=(x)(x)0.1−x9.1×10−8=(𝑥)(𝑥)0.1−𝑥 

(9.1×10−8)(0.1−x)=x2(9.1×10−8)(0.1−𝑥)=𝑥2

Taking,  0.1−xM;0.1M0.1−𝑥𝑀;0.1M we have (9.1×10−8)(0.1)=x2(9.1×10−8)(0.1)=𝑥2 

9.1×10−9=x29.1×10−9=𝑥2 

x=9.1×10−9−−−−−−−−√𝑥=9.1×10−9 

=9.54×10−5M=9.54×10−5M 

⇒[HS−]=9.54×10−5M⇒[𝐻𝑆−]=9.54×10−5M

Case II (in the presence of HClHCl):

In the presence of 0.1 M0.1 M of HClHCl, let [HS−][HS−] be y M.

 H2S⇌H + +HS−H2S⇌H + +HS−

The final concentration of H2SH2S will be (0.1−y)(0.1−𝑦) 

Now, 

Ka1=Ka1=

[HS−][H+][H2S][𝐻𝑆−][𝐻+][𝐻2𝑆]

Ka1=y(0.1+y)(0.1−y)𝐾𝑎1=𝑦(0.1+𝑦)(0.1−𝑦)

9.1×10−8=9.1×10−8=

y×0.10.1𝑦×0.10.1

y=9.1×10−8𝑦=9.1×10−8

The concentration of [HS−]=9.1×10−8[HS−]=9.1×10−8 

To calculate the concentration of [S2−][S2−] 

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

 HS−⇌H++S2−HS−⇌H++S2− 

[HS−]=9.54×10−5 M[HS−]=9.54×10−5 M 

Let [S2−][S2−] be XX .

Also, [H+]=9.54×10−5 M[H+]=9.54×10−5 M 

Ka2=[H+][S2−][HS−]𝐾𝑎2=[𝐻+][𝑆2−][𝐻𝑆−] 

Ka2=(9.54×10−5)(X)9.54×10−5𝐾𝑎2=(9.54×10−5)(𝑋)9.54×10−5 

1.2×10−13=X=[S2−]1.2×10−13=𝑋=[𝑆2−]

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

The concentration of S2−S2− be X’ MX’ M 

[HS−]=9.1×10−8 M[HS−]=9.1×10−8 M 

[H+]=0.1 M[H+]=0.1 M 

Then, 

Ka2=[H+][S2][HS−]𝐾𝑎2=[𝐻+][𝑆2][𝐻𝑆−]

1.2×10−13=(0.1)(X′)9.1×10−81.2×10−13=(0.1)(𝑋′)9.1×10−8 

10.92×10−21=0.1X′10.92×10−21=0.1𝑋′

10.92×10−210.1=X′10.92×10−210.1=𝑋′ 

X′=1.092×10−200.1𝑋′=1.092×10−200.1 

=1.092×10−19M=1.092×10−19M 

⇒Ka1=1.74×10−5⇒𝐾𝑎1=1.74×10−5

45. The ionization constant of acetic acid is 1.74×10−51.74×10−5 . 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+CH3COOH+H2O⇌CH3COO−+H3O+ 

The final concentration of CH3COO−CH3COO− and H3O + H3O +  is 0.05α0.05𝛼 . Then the final concentration of CH3COOHCH3COOH will be (0.05−0.05α)(0.05−0.05𝛼) .

Ka=(.05α)(.05α)(.05−0.05α)𝐾𝑎=(.05𝛼)(.05𝛼)(.05−0.05𝛼) 

=(.05α)(0.05α).05(1−α)=(.05𝛼)(0.05𝛼).05(1−𝛼) 

=.05α21−α=.05𝛼21−𝛼

1.74×10−5=.05α21−α1.74×10−5=.05𝛼21−𝛼 

1.74×10−5−1.74×10−5α=0.05α21.74×10−5−1.74×10−5𝛼=0.05𝛼2 

0.05α2+1.74×10−5α−1.74×10−50.05𝛼2+1.74×10−5𝛼−1.74×10−5 

D=b2−4ac𝐷=𝑏2−4𝑎𝑐

=(1.74×10−5)2−4(.05)(1.74×10−5)=(1.74×10−5)2−4(.05)(1.74×10−5) 

=3.02×10−25+.348×10−5=3.02×10−25+.348×10−5

α=Kac−−−√𝛼=𝐾𝑎𝑐 

α=1.74×10−5.05−−−−−−−−−−√𝛼=1.74×10−5.05 

=34.8×10−5×1010−−−−−−−−−−−−−−√=34.8×10−5×1010 

=3.48×10−6−−−−−−−−−√=3.48×10−6

[CH3COO−]=0.05×1.86×10−3[CH3COO−]=0.05×1.86×10−3 

=0.93×10−31000=0.93×10−31000 

=.000093=.000093

pH=−log[H+]𝑝𝐻=−log⁡[𝐻+] 

=−log(.093×10−2)=−log⁡(.093×10−2) 

∴pH=3.03∴𝑝𝐻=3.03

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

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 pKapKa .

Ans: Let the organic compound be HAHA.

HA⇌H++A−HA⇌H++A− 

Concentration of  HAHA is 0.01 M and its pH is 4.15.

−log[H+]=4.15−log⁡[𝐻+]=4.15 

[H+]=7.08×10−5[𝐻+]=7.08×10−5 

Now,Ka=[H+][A−][HA]Now,𝐾𝑎=[𝐻+][𝐴−][𝐻𝐴] 

[H+]=[A−]=7.08×10−5[𝐻+]=[𝐴−]=7.08×10−5 

[HA]=0.01[𝐻𝐴]=0.01

Now, 

Ka=[H+][A−][HA][H+]=[A−]=7.08×10−5[HA]=0.01Ka=[H+][A−][HA][H+]=[A−]=7.08×10−5[HA]=0.01

Then,

Ka=(7.08×10−5)(7.08×10−5)0.01𝐾𝑎=(7.08×10−5)(7.08×10−5)0.01 

Ka=5.01×10−7𝐾𝑎=5.01×10−7 

pKa=−logKa𝑝𝐾𝑎=−log⁡𝐾𝑎 

=−logKa=−log⁡𝐾𝑎 

=(5.01×10−7)=(5.01×10−7) 

pKa=6.3001𝑝𝐾𝑎=6.3001 

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

1. 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.

2. 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.

3.  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.

4. 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.

48. Calculate the pH of the following solutions: 

1. 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

2. 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

3. 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

4. 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=1.87pH=1.87

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=2.75Ka=2.75

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 .

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=2.75𝐾𝑎=2.75

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.

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 

1. 0.01M0.01M
2. 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⇒1.82×10−4×0.050.05=1.82×10−4

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?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.

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

Ans:

1. 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

2. 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

3. Human blood, 7.38:

pH=7.38=−log[H+]pH=7.38=−log⁡[H+]

∴[H+]=4.1710−8M∴[H+]=4.1710−8M

4. Human saliva, 6.4:

pH=6.4×pH=6.4× 

6.4=−log[H+]6.4=−log⁡[H+] 

[H+]=3.9810−7[H+]=3.9810−7

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+]

1. 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

2. 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

3. 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

4. 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

5. 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=1.5810−8M

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∴𝑝𝐻=12.70

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∴𝑝𝐻=13.495;13.50

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′=1.32×10−5𝛼′=1.32×10−5

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=2.02×10−4

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− 

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=2.325×10−5

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⇒𝐾𝑎=𝐾𝑤𝐾𝑎=10−146.6×10−6=1.51×10−9

63. Predict if the solutions of the following salts are neutral, acidic or basic:

NaCl,KBr,NaCN,NH4NO3,NaNO2NaCl,KBr,NaCN,NH4NO3,NaNO2 and KFKF

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.

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=7.94

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.

66. Calculate the pHpH of the resultant mixtures:

1. 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

2. 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.

3. 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=1.30

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−]=𝑠

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=2×0.65×10−4=1.30×10−4M𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦𝑜𝑓𝐴𝑔+=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=1.09×10−5𝑀

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=4.48×10−10M𝐼−=2𝑠=4.48×10−10𝑀

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=91.9𝑠𝑠′=6.5×10−5𝑀7.07×10−7𝑀=91.9

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?

(For cupric iodate Ksp=7.4×10−8)Ksp=7.4×10−8)

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.

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.

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)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.

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)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.

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=1.6×10−24Fes=6.3×10−18,MnS=2.5×10−13,ZnS=1.6×10−24

CdS=8.0×10−27𝐶𝑑𝑆=8.0×10−27

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.