Q.1. At a fixed temperature a liquid is in equilibrium with its vapour in a closed vessel. Suddenly, the volume of the vessel got increased.
I)What will be the final vapour pressure and what will happen when equilibrium is restored finally?
II)Write down, how initially the rates of evaporation and condensation got changed?
III)Write down the effect observed when there was a change in vapour pressure.
Ans.
(I)Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.
(II)On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(III)On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.
Q.2.Find out Kc for the given reaction in equilibrium state
: [SO2]= 0.6 M, [O2] = 0.82 M and [SO3] = 1.9 M ?
2SO2(g)+O2(g)↔2SO3(g)
Ans.
As per the question,
2SO2(g)+O2(g)↔2SO3(g) (Given)
Kc=[SO3]2[SO2]2[O2]=(1.9)2M2(0.6)2(0.82)M3=12.229M−1(approximately)
Hence, K for the equilibrium is 12.229 M–1.
Q.3. At a definite temperature and a total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms
I2(g)↔2I(g)
Find Kp for the equilibrium.
Ans.
Partial pressure of Iodine atoms (I)
pI=40100×ptotal=40100×105=4×104Pa
Partial pressure of I2 molecules,
pI=60100×ptotal=60100×105=6×104Pa
Now, for the given reaction,
Kp=(pI)2pI2=(4×104)2Pa26×104Pa=2.67×104Pa
Q.4. For the given reaction , find expression for the equilibrium constant
(i)2NOCl(g)↔2NO(g)+Cl2(g)
(ii)2Cu(NO3)2(s)↔2CuO(s)+4NO2(g)+O2(g)
(iii)CH3COOC2H5(aq)+H2O(1)↔CH3COOH(aq)+C2H5OH(aq)
(iv)Fe3+(aq)+3OH−(aq)↔Fe(OH)3(s)
(v)I2(s)+5F2↔2IF5
Ans.
KC=[NOg]2[Cl2(g)][NOCl(g)]2
(ii)KC=[CuO(s)]2[NO2(g)]4[O2(g)][Cu(NO3)2(g)]2=[NO2(g)]4[O2(g)]
(iii)KC=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)][H2O(l)]=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)]
(iv)KC=Fe(OH)3(s)[Fe3+(aq)][OH−(aq)]3=1[Fe3+(aq)][OH−(aq)]3
(v)KC=[IF5]2[I2(s)][F2]5=[IF5]2[F2]5
Q.5.Find the value of , Kc for each of the following equilibria from the given value of Kp:
(i)2NOCl(g)↔2NO(g)+Cl2(g);Kp=1.8×10−2at500K
(ii)CaCO3(s)↔CaO(s)+CO2(g);Kp=167at1073K
Ans.
The relation between Kp and Kc is given as:
Kp=Kc(RT)Δn
(a) Given,
R = 0.0831 barLmol–1K–1
Δn=3−2=1
T = 500 K
Kp=1.8×10−2
Now,
Kp = Kc (RT) ∆n
⇒1.8×10−2=Kc(0.0831×500)1⇒Kc=1.8×10−20.0831×500=4.33×10−4(approximately)
(b) Here,
∆n =2 – 1 = 1
R = 0.0831 barLmol–1K–1
T = 1073 K
Kp= 167
Now,
Kp = Kc (RT) ∆n
⇒167=Kc(0.0831×1073)Δn⇒Kc=1670.0831×1073=1.87(approximately)
Q.6. For the following equilibrium, Kc=6.3×1014at1000KNO(g)+O3(g)↔NO2(g)+O2(g)
Both the reverse and forward reactions in the equilibrium are elementary bimolecular reactions. Calculate Kc, for the reverse reaction?
Ans.
For the reverse reaction, Kc=1Kc=16.3×1014=1.59×10−15
Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression?
Ans.
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Mole concentration= NumberofmolesVolumeMass/molecularmassVolume=MassVolume×Molecularmass=DensityMolecularmass
Though density of solid and pure liquid is fixed and molar mass is also fixed .
∴Molar concentration are constatnt.
Q.8. When oxygen and nitrogen react with each other, then the following reaction takes place:
2N2(g)+O2↔2N2O(g)
If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc =2.0×10−37, determine the composition of equilibrium solution.
Ans.
Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g) + O2(g) 2N2O(g)
Initial conc. 0.482 mol 0.933 mol 0
At equilibrium(0.482-x)mol (1.933-x)mol x mol
[N2]=0.482−x10⋅[O2]=0.933−x210,[N2O]=x10
The value of equilibrium constant is extremely small. This means that only small amounts . Then,
[N2]=0.48210=0.0482molL−1and[O2]=0.93310=0.0933molL−1
Now,
Kc=[N2O(g)]2[N2(g)][O2(g)]⇒2.0×10−37=(x10)2(0.0482)2(0.0933)⇒x2100=2.0×10−37×(0.0482)2×(0.0933)⇒x2=43.35×10−40⇒x=6.6×10−20 [N2O]=x10=6.6×10−2010=6.6×10−21
Q.9. Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction is given below:
2NO(g)+Br2(g) ⇒ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Ans.
The given reaction is:
2NO(g)+Br2(g) 2NOBr(g)
2mol 1mol 2mol
Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.
Again, 2 mol of NOBr are formed from 1 mol of Br.
Therefore, 0.0518 mol of NOBr are formed from 0.05182 mol or Br, or 0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br2] = 0.0437 mol
Therefore, the amount of NO present at equilibrium is:
[NO] = 0.087 – 0.0518 = 0.0352 mol
And, the amount of Br present at equilibrium is:
[Br2] = 0.0437 – 0.0259 = 0.0178 mol
Q.10. At 450 K, Kp= 2.0×1010/bar for the given reaction at equilibrium.
2SO2(g)+O2(g) ⇒ 2SO3(g)
What is Kc at this temperature?
Ans.)
For the given reaction,
∆n = 2 – 3 = – 1
T = 450 K
R = 0.0831 bar L bar K–1 mol–1
Kp=2.0×1010bar−1
We know that,
Kp=Kc(RT)Δn⇒2.0×1010bar−1=Kc(0.0831LbarK−1mol−1×450K)−1⇒Kc=2.0×1010bar−1(0.0831LbarK−1mol−1×450K)−1=(2.0×1010bar−1)(0.0831LbarK−1mol−1×450K)=74.79×1010Lmol−1=7.48×1011Lmol−1=7.48×1011M−1
Q.11. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI(g) ⇒ H2(g)+I2(g)
Ans.
The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.
Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:
2HI(g) H2(g) + I2(g)
Initial conc. 0.2 atm 0 0
At equilibrium 0.4 atm 0.16 2.15
2 2
=0.08atm =0.08atm
Therefore,
Kp=pH2×pI2p2HI=0.08×0.08(0.04)2=0.00640.0016=4.0
Hence, the value of Kp for the given equilibrium is 4.0.
Q.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction
N2(g)+3H2(g) ⇒ 2NH3(g) is 1.7×102
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Ans.
The given reaction is:
N2(g)+3H2(g) ⇒ 2NH3(g)
The given concentration of various species is
[N2]= 1.5720molL−1 [H2]= 1.9220molL−1 [NH3]= 8.3120molL−1
Now, reaction quotient Qc is:
Q=[NH3]2[N2][H2]3=((8.13)20)2(1.5720)(1.9220)3=2.4×103
Since, ?? ≠ ??, the reaction mixture is not at equilibrium.
Again, ?? > ??. Hence, the reaction will proceed in the reverse direction.
Q.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:
4NO(g)+6H2o(g) ⇒ 4NH3(g)+5O2(g)
Q.14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 60% of water (by mass) reacts with CO according to the equation,
H2O(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)
Initial conc. 1 M 1 M 0 0
10 10
At equilibrium 1-0.6 M 1-0.6 M 0.6 M 0.6 M
10 10 10 10
=0.04 M =0.04M =0.06 M =0.06M
Therefore, the equilibrium constant for the reaction,
Kc=Kc=[H2][CO2][H2O][CO]=0.06×0.060.04×0.04=0.00360.0016=2.25(approximately)
Q.15. At 700 K, equilibrium constant for the reaction
H2(g)+I2(g) ⇒ 2HI(g)
is 54.8. If 0.5 molL–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 ?? for the reaction
H2(g)+I2(g) 2HI(g) is 54.8.
Therefore, at equilibrium, the equilibrium constant K’c for the reaction
H2(g)+I2(g) 2HI(g)
[HI]=0.5 molL-1 will be 1/54.8.
Let the concentrations of hydrogen and iodine at equilibrium be x molL–1
[H2]=[I2]=x mol L-1
Therefore, [H2][I2][HI]2=K‘c⇒x×x(0.5)2=154.8⇒x2=0.2554.8⇒x=0.06754x=0.068molL−1(approximately)
Hence, at equilibrium, [H2]=[I2]=0.068 mol L-1.
Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl(g) ⇒ I2(g)+Cl2(g) ; Kc =0.14
Ans.
The given reaction is:
2ICl(g) I2(g) + Cl2(g)
Initial conc. 0.78 M 0 0
At equilibrium (0.78-2x) M x M x M
Now, we can write, [I2][Cl2][IC]2=Kc⇒x×x(0.78−2x)2=0.14⇒x2(0.78−2x)2=0.14⇒x0.78−2x=0.374⇒x=0.292−0.748x⇒1.748x=0.292⇒x=0.167
Hence, at equilibrium,
[H2]=[I2]=0.167 M
[HI]= (0.78−2×0.167)M=0.446M
Q.17. Kp = 0.04 atm 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?
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)
Initial conc. 4.0 M 0 0
At equilibrium (4.0-p) p p
We can write,
pc2H4×pH2pc2H6=KP⇒p×p40−p=0.04⇒p2=0.16−0.04p⇒p2+0.04p−0.16=0
Now,
p=−0.04±(0.04)2−4×1×(−0.16)√2×1=−0.04±0.802=0.762(Takingpositivevalue)=0.38
Hence, at equilibrium,
[C2H6] – 4 – p = 4 – 0.38
= 3.62 atm
Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)
(i)Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii)At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Ans.
(i)Reaction quotient,
Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]
(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.
The given reaction is:
CH3COOH(l)+c2H5OH(l) CH3COOC2H5(l)+H2O(l)
Initial conc. 0 0
At equilibrium
= =
Therefore, equilibrium constant for the given reaction is:
Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.171V×0.171V0.829V×0.009V=3.919=3.92(approximately)
(iii)Let the volume of the reaction mixture be V.
CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)
Initial conc. 0 0
At equilibrium
= =
Therefore, the reaction quotient is,
Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.214V×0.214V0.786V×0.286V=0.2037=0.204(approximately)
Since Qc<Kc , equilibrium has not been reached.
Q.19. 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 0.5×10−1mol L–1. If value of Kc is 8.3×10−3, what are the concentrations of PCl3 and Cl2 at equilibrium?
PCl5(g) ⇒ PCl3(g) + Cl2(g)
Ans.
Consider the conc. Of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:
PCl5(g) ⇒ PCl3(g + Cl2(g)
At equilibrium 0.5×10−10molL−1 x mol L-1 x mol L-1
It is given that the value of equilibrium constant , Kc is 8.3×10−10molL−3
Now we can write the expression for equilibrium as:
[PCl2][Cl2][PCl3]=Kc⇒x×x0.5×10−10=8.3×10−3⇒x2=4.15×10−4⇒x=2.04×10−2=0.0204=0.02(approximately)
Therefore, at equilibrium,
[PCl3]=[Cl2]=0.02mol L-1
Q.20. 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= 0.265 at 1050 K.
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and ???2= 0.80 atm?
Ans.
For the given reaction,
FeO(g) + CO(g) Fe(s) + CO2(g)
Initialy, 1.4 atm 0.80 atm
Qp=pCO2pCO=0.801.4=0.571
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 = decrease in pressure of CO2 be p. Then, we can write,
Kp=pCO2pCO⇒0.265=0.80−p1.4+p⇒0.371+0.265p=0.80−p⇒1.265p=0.429⇒p=0.339atm
Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm
And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm
Q.21. A reaction is given:
N2(g)+3H2(g) ⇒ 2NH3
For the above equation, Equilibrium constant = 0.061 at 500 K
At a specific time, from the analysis we can conclude that composition of the reaction mixture is, 2.0 mol L–1 H2 , 3.0 mol L –1 N2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.
Ans.
N2(g) + 3H2(g) 2NH3
At a particular time: 3.0 mol L-1 2.0mol L-1 0.5 mol L-1
So,
Qc=[NH3]2[N2][H2]3=(0.5)2(3.0)(2.0)3=0.0104
It is given that Kc=0.061
∵ Qc≠Kc, the reaction is not at equilibrium.
∵ Qc<Kc, the reaction preceeds in the forward direction to reach at equilibrium.
Q.22.Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇒ Br2(g) + Cl2(g)
For which Kc= 42 at 600 K.
If initially pure BrCl is present at a concentration of 5.5×10−5molL-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)
Initial conc. 5.5×10−5 0 0
At equilibrium 5.5×10−5−2x x x
Now, we can write,
[Br2][Cl2][BrCl]2=Kc⇒x×x(5.5×10−5−2x)2=42⇒x5.5×10−5−2x=6.48⇒x=35.64×10−5−12.96x⇒13.96x=35.64×10−5⇒x=35.6413.96×10−5=2.55×10−5
So,at equilibrium
[BrCl]=5.5×10−5−(2×2.55×10−5)=5.5×10−5−5.1×10−5=0.4×10−5=4.0×10−6molL−1
Q.23. Find out Kc for the given reaction at temperature 1127K where the pressure is 1 atm. A solution of CO and CO2 is in equilibrium with carbon(solid). It has 93.55% CO by mass.
C(s)+CO2 (g) ⇒ 2CO (g)
Ans.)
Let us assume that the solution is of 100g in total.
Given, mass of CO = 93.55 g
Now, the mass of CO2 =(100 – 93.55)=6.45 g
Now, number of moles of CO, nCO=93.528=3.34mol
Number of moles of CO2, nCO2=6.4544=0.146mol
Partial pressure of CO,
PCO= nCOnCO+nCO2×ptotal=3.343.34+0.146×1=0.958atm
Partial pressure of CO2, PCO2=nCO2nCO+nCO2×ptotal=0.1463.34+0.146×1=atm
Therefore, Kp= [CO]2[CO2]=(0.938)20.062=14.19
For the given reaction,
∆n = 2 – 1 = 1
We know that,
Kp=Kc(RT) Δn
⇒14.19=Kc(0.082×1127)1⇒Kc=14.190.082×1127=0.154(approximately)
Q.24.Find out
(I)The equilibrium constant for the formation of NO2 from NO and O2 at 298 K and
(II) ∆G°
NO(g)+12O2(g)↔NO2(g)
Where;
∆fG° (NO2) = 52.0 kJ/mol
∆fG° (NO) = 87.0 kJ/mol
∆fG° (O2) = 0 kJ/mol
Ans.)
(I)We know that,
∆G° = RT log Kc
∆G° = 2.303 RT log Kc
Kc=−35.0×10−3−2.303×8.314×298=6.134∴Kc=antilog(6.134)=1.36×106
Therefore, the equilibrium constant for the given reaction Kc is 1.36×106
Q.25. When each of the following equilibria is subjected to a decrease in pressure by increasing the volume, does the number of moles of reaction products increase, decrease or remain same?
(I) PCl5(g) ⇒ PCl3 +Cl2 (g)
(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)
(III) 3Fe (s) + 4H2O (g) ⇒ Fe3O4 (s) +4H2 (g)
Ans.
(I) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(II) The number of moles of reaction products will decrease.
(III) The number of moles of reaction products remains the same.
Q.26. 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.)
When pressure is increased:
The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.
Since, the number of moles of gaseous reactants is more than that of gaseous products; the reaction given in (iv) will proceed in the forward direction
Since, the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction
Q.27. The equilibrium constant for the following reaction is 1.6×105at 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, ?? 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.
2HBr (g) H2 (g) + Br2 (g)
Initial conc. 10 0 0
At equilibrium 10-2p p p
Now, we can write,
pHBr×p2p2HBr=K‘pp×p(10−2p)2=6.25×10−6p10−2p=2.5×10−3p=2.5×10−2−(5.0×10−3)pp+(5.0×10−3)p=2.5×10−2(1005×10−3)=2.5×10−2p=2.49×10−2bar=2.5×10−2bar(approximately)
Therefore, at equilibrium,
[H2]=[Br2]= 2.49×10−2bar [HBr]= 10−2×(2.49×10−2)bar=9.95bar=10bar(approximately)
Q.28.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)
(I)Write as expression for Kp for the above reaction.
(II) How will the values of Kp and composition of equilibrium mixture be affected by
(i)Increasing the pressure
(ii)Increasing the temperature
(iii)Using a catalyst?
Ans.)
(I)For the given reaction,
Kp=pCO×p3H2pCH4×pH2O
(II) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.
Q.29. Describe the effect of:
I) Removal of CO
II) Addition of H2
III) Removal of CH3OH on the equilibrium of the reaction:
IV) Addition of CH3OH
2H2 (g)+CO (g) ⇒ CH3OH (g)
Ans.)
(I) On removing CO, the equilibrium will shift in the backward direction.
(II) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.
(III) On removing CH3OH, the equilibrium will shift in the forward direction.
(IV) On addition of CH3OH, the equilibrium will shift in the backward direction.
Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3×10−3. If decomposition is depicted as,
PCl5(g)↔PCl3(g)+Cl2(g)
∆rH° = 124.0 kJmol–1
a) Write an expression for Kc for the reaction.
b) What is the value of Kc for the reverse reaction at the same temperature?
c) What would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased?
(iii) The temperature is increased?
Ans.)
(a)Kc=[PCl3(g)][Cl2(g)][PCl3(g)]
(b)Value of Kc for the reverse reaction at the same temperature is:
K‘c=1Kc=18.3×10−3=1.2048×102=120.48
(c)(i)Kc would remain the same because in this case, the temperature remains the same.
(ii)Kc is constant at constant temperature. Thus, in this case, Kc would not change. (iii)In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.
Q.31. 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 ⇒ CO2(g) + H2(g)
Initial conc. 4.0 bar 4.0 bar 0 0
At equilibrium 4.0-p 4.0-p p p
Given Kp = 10.1
PCO2×PH2PCO×PH2O=KP⇒p×p(4.0−p)(4.0−p)=10.1⇒p4.0−p=3.178⇒p=12.712−3.178p4.178p=12.712p=12.7124.178p=3.04
So, partial pressure of H2 is 3.04 bar at equilibrium.
Q.32. Predict which of the following reaction will have appreciable concentration of reactants and products:
(a)Cl2(g)↔2Cl(g);Kc=5×10−39
(b)Cl2(g)+2NO(g)↔2NOCl(g);Kc=3.7×108
(c)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.
Q.33. 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.)
Given,
3O2 (g) ⇒ 2O3 (g)
Then, Kc = [O3(g)]2[O2(g)]3
Given that Kc = 2.0×10−50 and [O2(g)] = 1.6×10−2
Then,
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=8.192×10−56⇒[O3(g)]2=2.86×10−28M
So, the conc. of O3 is 2.86×10−28M.
Q.34. The reaction, CO(g)+3H2(g)→CH4(g)+H2O(g) at 1300K is at equilibrium in a 1L container. It has 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and y amount of CH4 in the container. Find the concentration of CH4 in the mixture.
The equilibrium constant, Kc is 3.90 at the given temp.
Ans.)
Let the concentration of CH4 at equilibrium be y.
\(CO_{(g)} \; + \; 3H_{2(g)} \; \rightarrow \; CH_{4(g)} \; + \; H_{2}O_{(g)}\)
At equilibrium,
For CO – 0.31=0.3M
For H2 – 0.11=0.1M
For H2O – 0.021=0.02M Kc = 3.90
Therefore,
[CH4(g)][H2O(g)][CO(g)][H2(g)]3=Kc y×0.020.3×(0.1)3=3.9 y=3.9×0.3×(0.1)30.02 y=0.001170.02 y=0.0585M y=5.85×10−2M
Therefore, the concentration of CH4 at equilibrium is 5.85×10−2M
Q.35. What is conjugate acid-base pair? Find the conjugate acid/base of the given species:
(i) HNO2
(ii) CN−
(iii) HClO4
(iv) F−
(v) OH−
(vi) CO2−3
(vii) S−
Ans.)
A conjugate acid-base pair is a pair that has a difference of only one proton.
The conjugate acid-base pair of the following are as follows:
(i) HNO2 – NO−2 (Base)
(ii) CN− – HCN (Acid)
(iii) HClO4 – ClO−4 (Base)
(iv) F− – HF (Acid)
(v) OH− – H2O (Acid)/ O2− (Base)
(vi) CO2−3 – HCO−3 (Acid)
(vii) S− – HS− (Acid)
Q.36. From the compounds given below which are Lewis acids?
(i) H2O
(ii) BF3
(iii) H+
(iv) NH+4
Ans.)
Lewis acids are the acids which can accept a pair of electrons.
(i) H2O – Not Lewis acid
(ii) BF3 – Lewis acid
(iii) H+ – Lewis acid
(iv) NH+4 – Lewis acid
Q.37. From the compounds given below which will be the conjugate base for the Bronsted acids?
(i) HF
(ii) H2SO4
(iii) HCO3
Ans.)
The following shows the conjugate bases for the Bronsted acids:
(i) HF – F−
(ii) H2SO4 – HSO−4
(iii) HCO3 – CO2−3
Q.38. For the Brönsted bases given below find their conjugate acids.
NH3
HCOO−
NH−2
Ans.)
Brönsted base Conjugate acid
1 NH3 NH+4
2 HCOO− HCOOH
3 NH−2 NH3
Q.39. The species given below can act as both Brönsted bases as well as Brönsted acids. For each of them give their conjugate acid and base.
HCO−3
HSO−4
NH3
H2O
Ans.)
Species Conjugate base Conjugate acid
1 HCO−3 CO2−3 H2CO3
2 HSO−4 SO2−4 H2SO4
3 NH3 NH−2 NH+4
4 H2O OH− H3O+
Q.40. Classify the species given below into bases and acids and also show that these species act as base/acid:
BCl3
H+
OH−
F−
Ans.)
BCl3:
It is a Lewis acid as it has tendency to accept a pair of electrons.
H+
It is a Lewis acid as it has tendency to accept a pair of electrons.
OH−
It is a Lewis base as it has tendency to lose a pair of electrons.
F−
It is a Lewis base as it has tendency to lose its lone pair of electrons.
Q.41.A sample soft drink is taken, whose hydrogen ion concentration is 2.5×10−4M.Find out pH.
Ans.)
pH=−log[H+]=−log(2.5×10−4)=−log2.5−log10−4=−log2.5+4=−0.398+4=3.602
Q.42.A sample of white vinegar is taken , whose pH is 2.36. Find out the hydrogen ion concentration in the sample.
Ans.)
pH=−log[H+]⇒log[H+]=−pH⇒[H+]=antilog(−pH)=antilog(−2.36)=0.004365=4.37×10−3 ∴ 4.37×10−3 is the concentration of white vinegar sample.
Q.43. Ionization constant for the following acids are given:
HF = 5.7×10−5 at 298K
HCOOH = 1.7×10−3 at 298K
HCN = 3.7×10−8 at 298K
Find out the conjugate bases for the above acids.
Ans.)
For F–, Kb=KwKa=10−14(5.7×10−5)=1.75×10−9
For HCOO–, Kb=10−14(1.7×10−3)=5.88×10−11
For CN– = Kb=10−14(3.7×10−8)=×10−11=2.70×10−6
Q.44.Phenol has ionization constant of 1.0×10−8. In a 0.06M of phenol solution calculate the presence of phenolate ion. Find out the degree of ionization if 0.02M of sodium phenolate is given.
Ans.)
C6H5OH C6H5O– + H+
Initial 0.06M
After dissociation 0.06-x x x
∴Ka=x×x0.06−x=1.0×10−8⇒x20.06=1.0×10−8⇒x2=6×10−10⇒x=2.4×105M
In presence of 0.02 sodium phenolate(C6H5Na), suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.06 – y ≃ 0.06,
[C6H5O–] = 0.02 + y ≃0.01M,
[H+]=y M
∴Ka=(0.02)(y)0.06=1.0×10−8⇒y=1.0×0.06(0.02)×10−8⇒y=6×10−8 ∴ degree of ionization = α=yc=6×10−86×10−2(Herec=0.06=6×10−2)=10−6
So, α=10−6
Q.45 Given, 9.1×10−8 is the initial(first) ionization constant of the gas H2S.Find out concentration of the ion HS– in 0.1M solution of H2S . Find the changes in concentration if the concentration is 0.1M in HCl. Find the concentration of S2- under both conditions, if 1.2×10−13 is the second dissociation constant of H2S.
Ans.)
To calculate [HS–]
H2S H+ + HS–
Intial 0.1 M
After dissociation 0.1-x x x
≃0.1 Ka=x×x0.1=9.1×10−8⇒x2=9.1×10−9⇒x=9.54×10−5
In the presence of 0.1 M HCl, suppose H2S dissociated is y. Then at equilibrium, [H2S] =0.1-y ≃0.1,
[H+]=0.1 + y ≃0.1,
[HS–] = y M
Ka=0.1×y0.1=9.1×10−8y=9.1×0.10.1×10−8y=9.1×10−8
Ka2
Ka1
To calculate [S2-]
H2S H+ + HS– , HS– H+ + S2-
For the overall reaction,
H2S 2H+ + S2-
Ka=Ka1×Ka2=9.1×10−8×1.2×10−13=1.092×10−20 Ka=[H+]2[S2−][H2S]
In the absence of 0.1M HCl,
[H+]= 2[S2-]
Hence, if [S2-] = x , [H+] =2x
∴(2x)20.1=1.092×10−20⇒4x3=1.092×10−21⇒x3=1.0924×10−21=273×10−24⇒logx3=log273−log10−24=2.4362−24⇒3logx=2.4362−24⇒logx=2.43623−243⇒x=0.8127−8=−7.1873⇒x=Antilog−7.1873=6.497×10−8=6.5×10−8
In presence of 0.1M HCl,
Suppose [S2-]=y, then
[H2S]=0.1-y≃0.1M,
[H+]=0.1+y≃0.1M
Ka=(0.1)2×y0.1=1.09×10−20y=1.09×10−19M
Q.46.Given,the ionization constant of acetic acid is 1.74×10−5. Find the degree of dissociation of acetic acid in its 0.05 M solution. Find the concentration of acetate ion in the solution and its pH.
Ans.)
CH3COOH ⇒ CH3COO– + H+
Ka=[CH3COO−][H+][CH3COOH]=[H+]2[CH3COOH]⇒[H+]=Ka[CH3COOH]−−−−−−−−−−−−−√=(1.74×10−5)(5×10−2)−−−−−−−−−−−−−−−−−−−√=9.33×10−4M[CH3COO−]=[H+]=9.33×10−4MpH=−log(9.33×1.0−4)=4−0.9699=4−0.97=3.03
Q.47. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Ans.)
HA H+ + A–
pH=-log[H+]
log[H+]=-4.15
[H+] = 7.08×10−5M [A–] = [H+] = 7.08×10−5M Ka=[H+][A−][HA]=(7.08×10−5)(7.08×10−5)10−2=5.0×10−7pKa=−logKa=−log(5.0×10−7)=7−0.699=6.301
Q.48.Consider complete dissociation, find out the pH of the following :
(I)0.004 M HCl
(II)0.003 M NaOH
(III)0.002 M HBr
(IV)0.002 M KOH
Ans.)
(I)HCl + aq H+ + Cl–
∴[H+]=[HCl]=4×10−3MpH=−log(4×10−3)=2.398
(II)NaOH + aq Na+ + OH–
∴[OH−]=3×10−3M[H+]=10−14(3×10−3)=3×10−12MpH=−log(3×10−12)=11.52
(III)HBr + aq H+ + Br–
∴[H+]=2×10−3MpH=−log(2×10−3M)=2.70
(IV)KOH + aq K+ + OH–
∴[OH+]=2×10−3M[H+]=10−14(2×10−3)=5×10−12pH=−log(5×10−12)=11.30
Q.49.Find out the pH of the following solution:
(I)2g of TIOH dissolved in water to give 2 litre of the solution
(II)0.3g of Ca(OH)2 dissolved in water to given 500mL of the solution
(III)0.3g of NaOH dissolved in water to give 200mL of the solution
(IV)1 mL of 13.6 M HCl is diluted with water to given 1 litre of the solution
Ans.)
(I)Molar conc. Of TlOH = 2g(204+16+1)gmol−1×12L=4.52×10−3M[OH−]=[TlOH]=4.52×10−3M[H+]=10−14(4.52×10−3)=2.21×10−12M∴pH=−log(2.21×10−12)=12−(0.3424)=11.66
(II)Molar conc. Of Ca(OH)2=0.3g(40+34)gmol−1×10.5L=8.11×10−3M[OH−]=2[Ca(OH)2]=2×(8.11×10−3)M=16.22×10−3MpOH=−log(16.22×10−3)=3−1.2101=1.79pH=14−1.79=12.21
(III)Molar conc. of NaOH = 0.3g(40+34)gmol−1×10.2L=3.75×10−2M[OH−]=3.75×10−2MpOH=−log(3.75×10−2)=2−0.0574=1.43pH=14−1.43=12.57
(IV)M1V1 = M2V2
∴13.6M××1mL=M2×1000mL∴M2=1.36×10−2M[H+]=[HCl]=1.36×10−2MpH=−log(1.36×10−2)=2−0.1335≃1.87
Q.50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans.)
CH2(Br)COOH CH2(Br)COO– + H+
Initial conc. C 0 0
Conc. at eqm. C – Cα Cα Cα
Ka=Cα⋅CαC(1−α)=Cα21−α≃Cα2=0.1×(0.132)2=1.74×10−3pKa=−log(1.74×10−3)=3−0.2405=2.76[H+]=Cα=0.1×0.132=1.32×10−2MpH=−log(1.32×10−2)=2−0.1206=1.88
Q.51. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans.)
CH2(Br)COOH CH2(Br)COO– + H+
Initial conc. C 0 0
Conc. at eqm. C – Cα Cα Cα
Ka=Cα⋅CαC(1−α)=Cα21−α≃Cα2=0.1×(0.132)2=1.74×10−3pKa=−log(1.74×10−3)=3−0.2405=2.76[H+]=Cα=0.1×0.132=1.32×10−2MpH=−log(1.32×10−2)=2−0.1206=1.88
Q.52. 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 = 4.27×10−10
c = 0.001M
pH =?
α =?
Kb=cα24.27×10−10=0.001×α24270×10−10=α265.34×10−5=α=6.53×10−4Then,[anion]=cα=0.001×65.34×10−5=0.065×10−5 pOH=−log(0.065×10−5)=6.187pH=7.813Now,Ka×Kb=KwKa=10−144.27×10−10=2.34×10−5 ∴2.34×10−5 is the ionization constant.
Q.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains
(I)0.01 M
(II)0.1 M in HCl?
Ans.)
c=0.05M
pKa=4.74
pKa=-log(Ka)
Ka=1.82×10−5Ka=cα2α=Kac−−−√α=1.82×10−55×10−2−−−−−−−√=1.908×10−2
When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case 1: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
CH3COOH H+ + CH3COO–
Initial conc. 0.05M 0 0
After dissociation 0.05-x 0.01+x x
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.
Ka=[CH3COO−][H+][CH3COOH]∴=(0.01)x0.05x=1.82×10−5×0.050.01x=1.82×10−3×0.05M
Now,
α=AmounofaciddissociationAmountofacidtaken=1.82×10−3×0.050.05=1.82×10−3
Case 2: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
[CH3COOH]=0.05 – X; 0.05 M
[CH3COO–]=X
[H+]=0.1+X ; 0.1M
Ka=[CH3COO−][H+][CH3COOH]∴Ka=(0.1)X0.05x=1.82×10−5×0.050.1x=1.82×10−4×0.05M
Now,
α=AmounofaciddissociationAmountofacidtaken=1.82×10−4×0.050.05=1.82×10−4
Q.54. The ionization constant of dimethylamine is5.4××10−4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?
Ans.)
Kb=5.4×10−4c=0.02MThen,α=Kbc−−−√=5.4×10−40.02−−−−−−√=0.1643
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
NaOH(aq) Na+(aq) + OH–(aq)
0.1M 0.1M
And,
(CH3)2 NH + H2O (CH3)2 NH2+ + OH
(0.02-x) x x
;0.02M ;0.1M
Then,[ (CH3)2 NH+2]=x
[OH– ]=x+0.1;0.1
⇒Kb=[(CH3)2NH+2][OH−][(CH3)2NH]5.4×10−4=x×0.10.02x=0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.
Q.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(I)Human saliva, 6.4
(II)Human stomach fluid, 1.2
(III)Human muscle-fluid, 6.83
(IV)Human blood, 7.38
Ans.)
(I)Human saliva, 6.4:
pH = 6.4
6.4 = – log [H+] [H+] = 3.98×10−7
(II)Human stomach fluid, 1.2:
pH =1.2
1.2 = – log [H+] ∴ [H+] = 0.063
(III)Human muscle fluid 6.83:
pH = 6.83
pH = – log [H+]
6.83 = – log [H+] [H+] =1.48×10−7M
(IV) Human blood, 7.38:
pH = 7.38 = – log [H+] [H+] = 4.17×10−8M
Q.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 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+]
(I)pH of milk = 6.8
Since, pH = –log [H+]
6.8 = –log [H+] log
[H+] = –6.8
[H+] = anitlog(–6.8)
= 1.5×10−7M
(II)pH ofblack coffee = 5.0
Since, pH = –log [H+]
5.0 = –log [H+] log
[H+] = –5.0
[H+] = anitlog(–5.0)
= 10−5M
(III)pH of tomato= 4.2
Since, pH = –log [H+]
4.2 = –log [H+] log
[H+] = –4.2
[H+] = anitlog(–4.2)
= 6.31×10−5M
(IV)pH of lemon juice= 2.2
Since, pH = –log [H+]
2.2 = –log [H+] log
[H+] = –2.2
[H+] = anitlog(–2.2)
= 6.31×10−3M
(V)pH of egg white= 7.8
Since, pH = –log [H+]
7.8 = –log [H+] log
[H+] = –7.8
[H+] = anitlog(–7.8)
= 1.58×10−8M
Q.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Ans.)
[KOHaq]= 0.56115g/L=2.805g/L=2.805×156.11M=0.05M
KOH(aq) K+(aq) + OH(aq)–
[OH–]=0.05M=[K+] [H+][H–]=Kw
[H+]=Kw[OH−] =10−140.05=2×10−13M∴pH=12.70
Q.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Ans.)
Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2
=19.23121.63M=0.1581M
Sr(OH)2(aq) Sr2+(aq) + 2(OH–)(aq)
∴[ Sr2+]=0.1581M
[OH–]= 2×0.1581M=0.3126
Now,
Kw=[OH–][H+] 10−140.3126=[H+]⇒[H+]=3.2×10−14∴pH=13.495;13.50
Q.59. The ionization constant of propanoic acid is 1.32××10−15. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Ans.)
Let the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
HA + H2O H3O+ + A–
(0.05-0.0α) ≈0.05 0.05α 0.05α
Kα=[H3O+][A−][HA]=(0.05α)(0.05α)0.05=0.05α2 α=Kα0.05−−−√=1.63×10−2
Then,[ H3O+ ]= 0.05α =0.05×1.63×10−2=Kb⋅15×10−4M∴pH=3.09
In the presence of 0.1M of HCl, let 뱫 be the degree of ionization.
Then,[ H3O+ ]=0.01
[A–]=0.05 α´
[HA]=0.05
Kα=0.01×0.05α′0.051.32×10−5=0.01×α′α′=1.32×10−3
Q.60. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Ans.)
c = 0.1 M
pH = 2.34
-log [H+] = pH
-log [H+] = 2.34
[H+]=4.5×10−3
Also,
[H+]=cα
4.5×10−3=0.1×α4.5×10−30.1=αα=4.5×10−3=0.045
Then,
Ka=cα2=0.1×(45×10−3)2=202.5×10−6=2.02×10−4
Q.61. for nitrous acid Ka = 4.5×10−4. Calculate degree of hydrolysis and pH for 0.04M of sodium nitrite.
Ans.)
Sodium nitrite is a salt of NaOH (strong base) and HNO2 (weak acid).
NO−2+H2O↔HNO2+OH− Kh=[HNO2][OH−][NO−2] ⇒KwKa=10−144.5×10−4=22×10−10
Let, y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:
[NO−2]=0.04–y;0.04 [HNO2]=y [OH−]=y Kh=y20.04=0.22×10−10 y2=0.0088×10−10 y=0.093×10−5 ∴[OH−]=0.093×10−5M [H3O+]=10−100.093×10−5=10.75×10−9M
Thus, pH=−log(10.75×10−9)
= 7.96
Thus, the degree of hydrolysis is
=y0.04=0.093×10−50.04 =2.325×10−5
Q.62. 0.02M solution of pyridinium hydrochloride (C5H6ClN) is having pH = 3.44. Determine the ionization constant of C5H5N (pyridine).
Ans.)
pH = 3.44
As we know,
pH=log[H+] ∴[H+]=3.63×10−4
Now, Kh=3.63×10−40.02; (Given that concentration = 0.02M)
⇒Kh=6.6×10−6
As we know that,
Kh=KwKa Ka=KwKh=10−146.6×10−6
= 1.51×10−9
Q.63. Few salts are given below;
KBr
NH4NO3
KF
NaNO2
NaCN
NaCl
Determine the nature of solution of these salts i.e. Is it acidic or basic or neutral?
Ans.)
KBr
KBr + H2O ↔ KOH (Strong base)+ HBr (Strong acid)
Thus, it is a neutral solution.
NH4NO3
NH4NO3 + H2O ↔ NH4OH(Weak base) + HNO2 (Strong acid)
Thus, it is an acidic solution.
KF
KF + H2O ↔ KOH (Strong base) + HF (weak acid)
Thus, it is a basic solution.
NaNO2
NaNO2 + H2O ↔ NH4OH(Strong base) + HNO2(Weak acid)
Thus, it is a basic solution.
NaCN
NaCN + H2O ↔ HCN (Weak acid) + NaOH (Strong base)
Thus, it is a basic solution.
NaCl
NaCl + H2O ↔ NaOH (Strong base) + HCL (Strong acid)
Thus, it is a neutral solution.
Q.64. Find the pH of 0.1M acid and its 0.1M NaCl solution. The Ka for chloroacetic acid is 1.35×10−3.
Ans.)
The Ka for chloroacetic acid (ClCH2COOH) is 1.35×10−3.
⇒Ka=cα2 ∴α=Kac−−−√
= 1.35×10−30.1−−−−−−−√; (given concentration = 0.1M)
α=1.35×10−2−−−−−−−−−√
= 0.116
∴[H+]=cα=0.1∗0.116=0.0116
pH = −log[H+] = 1.94
ClCH2COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH2COOH
ClCH2COO−+H2O↔ClCH2COOH+OH− Kh=[ClCH2COO][OH−][ClCH2COO−]
Now, Kh=KwKa Kh=10−141.35×10−3 =0.740×10−11
Also, Kh=y20.1 ⇒0.740×10−11=y20.1 ⇒0.0740×10−11=y2 y=0.86×10−6 [OH−]=0.86×10−6 ∴[H+]=Kw0.86×10−6 =10−140.86×10−6 [H+]=1.162×10−3
pH = −log[H+]
= 7.94
Q.65. Determine the pH of neutral water at 310K temperature. Ionic product of H2O is 2.7×10−14.
Ans.)
Ionic Product,
Kw=[H+][OH−]
Assuming, [H+] = y
As, [H+]=[OH−], Kw = y2.
Kw at 310K is 2.7×10−14.
∴2.7×10−14=y2
y = 1.64×10−7
[H+]=1.64×10−7
pH = −log[H+]
= −log[1.64×10−7]
= 6.78
Thus, the pH of neutral water at 310K temperature is 6.78.
Q.66. Find out the pH of resultant mixture;
i) n10 ml of 0.02M H2SO4 + 10 ml of 0.02M Ca(OH)2
ii) 10 ml of 0.1M H2SO4 + 10 ml of 0.1M KOH
iii) 10 ml of 0.2M Ca(OH)2 + 25 ml of 0.1M HCl
Ans.)
i) Moles of OH−
= 2∗10∗0.021000=0.0004mol
Moles of H3O+
= 2∗10∗0.021000=0.0004mol
ii) Moles of OH−
= 2∗10∗0.11000=0.002mol
Moles of H3O+
= 2∗10∗0.11000=0.001mol
Here, the H3O+ is in excess is 0.01 mol
So, [H3O+]=0.00120×10−3=10−320×10−3=0.5
Thus, pH = -log(0.05)
= 1.3
As the solution id neutral pH = 7.
iii) Moles of OH−
= 2∗10∗0.21000=0.004mol
Moles of H3O+
= 25∗0.11000=0.0025mol
Here, the OH− is in excess is 0.0015 mol
So, [OH−]=0.001535×10−3=0.0428
Thus, pH = -log(OH)
= 1.36
pH = 14 – 1.36 = 12.63
As the solution id neutral pH = 12.63.
Q.67. Calculate the solubilities of
a) barium chromate
b) ferric hydroxide
c) lead chloride
d) mercurous iodide
e) silver chromate
At 300K from their solubility product constant. Also calculate the molarities of the individual ions.
Ans.)
a) Barium Chromate
BaCrO4 Ã Ba2+ + CrO2−4
Now, Ksp = [Ba2+][CrO2−4]
Aumming the solubility of BaCrO4 is ‘x’.
Thus,
[Ba2+] = x and CrO2−4 = x
Ksp = x2
1.2×10−10=x2
x = 1.09×10−10M
Molarity of Ba2+ = Molarity of CrO2−4 = x = 1.09×10−10M
b) Ferric Hydroxide
Fe(OH)3 Ã Fe3+ + OH−
Now, Ksp = [Fe3+][OH−]3
Aumming the solubility of Fe(OH)3 is ‘x’.
Thus,
[Fe3+] = x and OH− = 3x
Ksp = x(3x)3
= x*27x3
Ksp = 27x4
1.0×10−38=27x4
x = 0.00037×10−36M
Molarity of Fe3+ = x = 1.39×10−10M
Molarity of OH− = 3x = 4.17×10−10M
c) Lead Chloride
PbCl2 Ã Pb2+ + 2Cl−
Now, Ksp = [Pb2+][Cl−]
Aumming the solubility of PbCl2 is ‘x’.
Thus,
[Pb2+] = x and Cl− = 2x
Ksp = x(2x)2
= x*4x2
Ksp = 4x3
1.6×10−5=4x3
x = 1.58×10−2M
Molarity of Pb2+ = x = 1.58×10−2M
Molarity of Cl− = 2x = 3.16×10−2M
d) Mercurous iodide
Hg2I2 Ã Hg2+ + 2I−
Now, Ksp = [Hg2+]2[I−]2
Aumming the solubility of Hg2I2 is ‘x’.
Thus,
[Hg2+] = x and I− = 2x
Ksp = x(2x)2
= x*4x2
Ksp = 4x3
4.5×10−29=4x3
x = 2.24×10−10M
Molarity of Hg2+ = x = 2.24×10−10M
Molarity of I− = 2x = 4.48×10−10M
e) Silver Chromate
Ag2CrO4 Ã 2Ag2+ + CrO2−4
Now, Ksp = [Ag2+]2[CrO2−4]
Aumming the solubility of Ag2CrO4 is ‘x’.
Thus,
[Ag2+] = 2x and CrO2−4 = x
Ksp = (2x)2*x
1.1×10−12=4x3
x = 0.65×10−4M
Molarity of Ag2+ = 2x = 1.3×10−4M
Molarity of CrO2−4 = x = 0.65×10−4M
Q-68. Determine the ratio of molarities to their saturated solutions for the following:
Ag2CrO4 and AgBr
The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and5.0×10−13 respectively.
Ans.)
Ag2CrO4 Ã 2Ag2+ + CrO−4
Now, Ksp = [Ag2+]2[CrO−4]
Asuming the solubility of Ag2CrO4 is ‘x’.
Thus,
[Ag2+] = 2x and CrO−4 = x
Ksp = (2x)2*x
1.1×10−12=4x3
x = 0.65×10−4M
Assuming the solubility of AgBr is y.
AgBr(s) Ã Ag2+ + 2CrO−4
Ksp = (y)2
5.0×10−13=y2
y = 7.07×10−7M
The ratio of molarities to their saturated solution is:
xy=0.65×10−4M7.07×10−7M=91.9
Q.69. Cupric chlorate and sodium iodate having equal volume of 0.002M. Will the precipitation of copper iodate will occur or not?
Ans.)
Cupric chlorate and sodium iodate having equal volume are mixed together, then molar concentration of cupric chlorate and sodium iodate will reduce to half.
So, molar concentration of cupric chlorate and sodium iodate in mixture is 0.001M.
Na(IO3)2 Ã Na+ + IO−3
0.0001M 0.001M
Cu(ClO3)2 Ã Cu2+ + 2CIO−3
0.001M 0.001M
The Solubility for Cu(IO3)2 ⇒ Cu2+ (aq) + 2IO3– (aq)
Now, the ionic product of the copper iodate is:
= [Cu2+] [IO−3]2
= (0.001)(0.001)2
= 1.0×10−9M
As the value of Ksp is more than Ionic product.
Thus, the precipitation will not occur.
Q.70. For benzoic acid the ionization constant is 6.46×10−5M and for silver benzoate Ksp is 2.5×10−5M. Give relation between the solubility of silver benzoate in buffer of pH = 3.19 and its solubility in water.
Ans.)
Here, pH = 3.19
[H3O+] = 6.46×10−5M
C6H5COOH + H2O Ã C6H5COO− + H3O
Ka[C6H5COO−][H3O+]C6H5COOH Ka[C6H5COOH]C6H5COO−=[H3O+]Ka=6.46×10−46.46×10−5=10
Assuming the solubility of silver benzoate (C6H5COOAg) is y mol/L.
Now, [Ag+] = y
[C6H5COOH] = [C6H5COO−] = y
10[C6H5COO−] + [C6H5COO−] = y
[C6H5COO−] = y/11
Ksp[Ag+][C6H5COO−] = y
2.5×1013=yy11
y = 1.66×10−6 mol/L
Hence, solubility of C6H5COOAg in buffer of pH = 3.19 is 1.66×10−6 mol/L.
For, water:
Assuming the solubility of silver benzoate (C6H5COOAg) is x mol/L.
Now, [Ag+] = x M
Ksp = [Ag+][C6H5COO−]
Ksp = (y)2
y = Ksp−−−√=2.5×10−13−−−−−−−−−√=5×10−7mol/L ∴yx=1.66×10−65×10−7=3.32
Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.
Q.71. Calculate the maximum concentration of equimolar solutions of FeSO4 and Na2SO4 so that when they are mixed in equal volume than there is no precipitation of FeS? (Ksp for Fes is 6.3×10−18)
Ans.)
Assuming the maximum concentration of each solution is y mol/L
On mixing the solutions the volume of the concentration of each solution is reduced to half.
After mixing the maximum concentration of each solution is y/2 mol/L.
Thus, [FeSO4] = [Na2S] = y/2 M
So, [Fe2+] = [FeSO4] = y/2 M
FeS(s) ↔ Fe2+(aq)+S2−(aq)
Ksp = [Fe2+][S2-] 6.3×10−18=(y2)(y2) y24=6.3×10−18
Thus, y = 5.02×10−9
Thus, if the concentration of FeSO4 and Na2SO4 are equal to or less than that of 5.02×10−9M, then there won’t be precipitation of FeS.
Q.72. Find the minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K?
Ksp for CaSO4 is 9.1×10−6
Ans.)
CaSO4(s)↔Ca2+(aq)+SO2−4(aq)
Ksp = [Ca2+][SO2−4]
Assuming the solubility of calcium sulphate is x.
So, Ksp = x2
∴9.1×10−6=x2 ∴x=3.02×10−3mol/L
Now, molecular mass os calcium sulphate is 136g/mol.
Solubility in calcium sulphate in g/mol is
=3.02×10−3×136
= 0.41 g/L
i.e. 1 litre H2O will be required to dissolve 0.41g of calcium sulphate.
Thus, minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K is
= 10.41L=2.44L
Q.73. The concentration of S2- in 0.1M HCl solution saturated with H2S is 1.0×1019M. If 10mL of this added to 5mL of 0.04M solution given below:
MnCl2
ZnCl2
CdCl2
FeSO4
In which of the above solution the precipitation takes place?
For MnS, Ksp = 2.5×10−13
For ZnS, Ksp = 1.6×10−24
For CdS, Ksp = 8.0×10−27
For FeS, Ksp = 6.3×10−18
Ans.)
If the ionic product exceeds the Ksp value, then only precipitation can take place.
Before mixing:
[S2-] = Ksp = 1.0×10−19M [M2+] = 0.04M
Volume = 10mL Volume = 5mL
After mixing:
[S2-] = ? and [M2+] = ?
Total volume = (10 + 5) = 15mL Volume = 15mL
[S2-] = 1.0×10−19×1015=6.67×10−20M [M2+] = 0.04×515=1.33×10−2M
Now, the ionic product = [M2+][S2-]
= (1.33×10−2)(6.67×10−20)
= 8.87×10−22
Here, the ionic product of CdS and ZnS exceeds its corresponding Ksp value.
Thus, precipitation will occur in ZnCl2 and CdCl2 solutions.
I)What will be the final vapour pressure and what will happen when equilibrium is restored finally?
II)Write down, how initially the rates of evaporation and condensation got changed?
III)Write down the effect observed when there was a change in vapour pressure.
Ans.
(I)Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.
(II)On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(III)On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.
Q.2.Find out Kc for the given reaction in equilibrium state
: [SO2]= 0.6 M, [O2] = 0.82 M and [SO3] = 1.9 M ?
2SO2(g)+O2(g)↔2SO3(g)
Ans.
As per the question,
2SO2(g)+O2(g)↔2SO3(g) (Given)
Kc=[SO3]2[SO2]2[O2]=(1.9)2M2(0.6)2(0.82)M3=12.229M−1(approximately)
Hence, K for the equilibrium is 12.229 M–1.
Q.3. At a definite temperature and a total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms
I2(g)↔2I(g)
Find Kp for the equilibrium.
Ans.
Partial pressure of Iodine atoms (I)
pI=40100×ptotal=40100×105=4×104Pa
Partial pressure of I2 molecules,
pI=60100×ptotal=60100×105=6×104Pa
Now, for the given reaction,
Kp=(pI)2pI2=(4×104)2Pa26×104Pa=2.67×104Pa
Q.4. For the given reaction , find expression for the equilibrium constant
(i)2NOCl(g)↔2NO(g)+Cl2(g)
(ii)2Cu(NO3)2(s)↔2CuO(s)+4NO2(g)+O2(g)
(iii)CH3COOC2H5(aq)+H2O(1)↔CH3COOH(aq)+C2H5OH(aq)
(iv)Fe3+(aq)+3OH−(aq)↔Fe(OH)3(s)
(v)I2(s)+5F2↔2IF5
Ans.
KC=[NOg]2[Cl2(g)][NOCl(g)]2
(ii)KC=[CuO(s)]2[NO2(g)]4[O2(g)][Cu(NO3)2(g)]2=[NO2(g)]4[O2(g)]
(iii)KC=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)][H2O(l)]=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)]
(iv)KC=Fe(OH)3(s)[Fe3+(aq)][OH−(aq)]3=1[Fe3+(aq)][OH−(aq)]3
(v)KC=[IF5]2[I2(s)][F2]5=[IF5]2[F2]5
Q.5.Find the value of , Kc for each of the following equilibria from the given value of Kp:
(i)2NOCl(g)↔2NO(g)+Cl2(g);Kp=1.8×10−2at500K
(ii)CaCO3(s)↔CaO(s)+CO2(g);Kp=167at1073K
Ans.
The relation between Kp and Kc is given as:
Kp=Kc(RT)Δn
(a) Given,
R = 0.0831 barLmol–1K–1
Δn=3−2=1
T = 500 K
Kp=1.8×10−2
Now,
Kp = Kc (RT) ∆n
⇒1.8×10−2=Kc(0.0831×500)1⇒Kc=1.8×10−20.0831×500=4.33×10−4(approximately)
(b) Here,
∆n =2 – 1 = 1
R = 0.0831 barLmol–1K–1
T = 1073 K
Kp= 167
Now,
Kp = Kc (RT) ∆n
⇒167=Kc(0.0831×1073)Δn⇒Kc=1670.0831×1073=1.87(approximately)
Q.6. For the following equilibrium, Kc=6.3×1014at1000KNO(g)+O3(g)↔NO2(g)+O2(g)
Both the reverse and forward reactions in the equilibrium are elementary bimolecular reactions. Calculate Kc, for the reverse reaction?
Ans.
For the reverse reaction, Kc=1Kc=16.3×1014=1.59×10−15
Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression?
Ans.
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Mole concentration= NumberofmolesVolumeMass/molecularmassVolume=MassVolume×Molecularmass=DensityMolecularmass
Though density of solid and pure liquid is fixed and molar mass is also fixed .
∴Molar concentration are constatnt.
Q.8. When oxygen and nitrogen react with each other, then the following reaction takes place:
2N2(g)+O2↔2N2O(g)
If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc =2.0×10−37, determine the composition of equilibrium solution.
Ans.
Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g) + O2(g) 2N2O(g)
Initial conc. 0.482 mol 0.933 mol 0
At equilibrium(0.482-x)mol (1.933-x)mol x mol
[N2]=0.482−x10⋅[O2]=0.933−x210,[N2O]=x10
The value of equilibrium constant is extremely small. This means that only small amounts . Then,
[N2]=0.48210=0.0482molL−1and[O2]=0.93310=0.0933molL−1
Now,
Kc=[N2O(g)]2[N2(g)][O2(g)]⇒2.0×10−37=(x10)2(0.0482)2(0.0933)⇒x2100=2.0×10−37×(0.0482)2×(0.0933)⇒x2=43.35×10−40⇒x=6.6×10−20 [N2O]=x10=6.6×10−2010=6.6×10−21
Q.9. Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction is given below:
2NO(g)+Br2(g) ⇒ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Ans.
The given reaction is:
2NO(g)+Br2(g) 2NOBr(g)
2mol 1mol 2mol
Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.
Again, 2 mol of NOBr are formed from 1 mol of Br.
Therefore, 0.0518 mol of NOBr are formed from 0.05182 mol or Br, or 0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br2] = 0.0437 mol
Therefore, the amount of NO present at equilibrium is:
[NO] = 0.087 – 0.0518 = 0.0352 mol
And, the amount of Br present at equilibrium is:
[Br2] = 0.0437 – 0.0259 = 0.0178 mol
Q.10. At 450 K, Kp= 2.0×1010/bar for the given reaction at equilibrium.
2SO2(g)+O2(g) ⇒ 2SO3(g)
What is Kc at this temperature?
Ans.)
For the given reaction,
∆n = 2 – 3 = – 1
T = 450 K
R = 0.0831 bar L bar K–1 mol–1
Kp=2.0×1010bar−1
We know that,
Kp=Kc(RT)Δn⇒2.0×1010bar−1=Kc(0.0831LbarK−1mol−1×450K)−1⇒Kc=2.0×1010bar−1(0.0831LbarK−1mol−1×450K)−1=(2.0×1010bar−1)(0.0831LbarK−1mol−1×450K)=74.79×1010Lmol−1=7.48×1011Lmol−1=7.48×1011M−1
Q.11. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI(g) ⇒ H2(g)+I2(g)
Ans.
The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.
Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:
2HI(g) H2(g) + I2(g)
Initial conc. 0.2 atm 0 0
At equilibrium 0.4 atm 0.16 2.15
2 2
=0.08atm =0.08atm
Therefore,
Kp=pH2×pI2p2HI=0.08×0.08(0.04)2=0.00640.0016=4.0
Hence, the value of Kp for the given equilibrium is 4.0.
Q.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction
N2(g)+3H2(g) ⇒ 2NH3(g) is 1.7×102
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Ans.
The given reaction is:
N2(g)+3H2(g) ⇒ 2NH3(g)
The given concentration of various species is
[N2]= 1.5720molL−1 [H2]= 1.9220molL−1 [NH3]= 8.3120molL−1
Now, reaction quotient Qc is:
Q=[NH3]2[N2][H2]3=((8.13)20)2(1.5720)(1.9220)3=2.4×103
Since, ?? ≠ ??, the reaction mixture is not at equilibrium.
Again, ?? > ??. Hence, the reaction will proceed in the reverse direction.
Q.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:
4NO(g)+6H2o(g) ⇒ 4NH3(g)+5O2(g)
Q.14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 60% of water (by mass) reacts with CO according to the equation,
H2O(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)
Initial conc. 1 M 1 M 0 0
10 10
At equilibrium 1-0.6 M 1-0.6 M 0.6 M 0.6 M
10 10 10 10
=0.04 M =0.04M =0.06 M =0.06M
Therefore, the equilibrium constant for the reaction,
Kc=Kc=[H2][CO2][H2O][CO]=0.06×0.060.04×0.04=0.00360.0016=2.25(approximately)
Q.15. At 700 K, equilibrium constant for the reaction
H2(g)+I2(g) ⇒ 2HI(g)
is 54.8. If 0.5 molL–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 ?? for the reaction
H2(g)+I2(g) 2HI(g) is 54.8.
Therefore, at equilibrium, the equilibrium constant K’c for the reaction
H2(g)+I2(g) 2HI(g)
[HI]=0.5 molL-1 will be 1/54.8.
Let the concentrations of hydrogen and iodine at equilibrium be x molL–1
[H2]=[I2]=x mol L-1
Therefore, [H2][I2][HI]2=K‘c⇒x×x(0.5)2=154.8⇒x2=0.2554.8⇒x=0.06754x=0.068molL−1(approximately)
Hence, at equilibrium, [H2]=[I2]=0.068 mol L-1.
Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl(g) ⇒ I2(g)+Cl2(g) ; Kc =0.14
Ans.
The given reaction is:
2ICl(g) I2(g) + Cl2(g)
Initial conc. 0.78 M 0 0
At equilibrium (0.78-2x) M x M x M
Now, we can write, [I2][Cl2][IC]2=Kc⇒x×x(0.78−2x)2=0.14⇒x2(0.78−2x)2=0.14⇒x0.78−2x=0.374⇒x=0.292−0.748x⇒1.748x=0.292⇒x=0.167
Hence, at equilibrium,
[H2]=[I2]=0.167 M
[HI]= (0.78−2×0.167)M=0.446M
Q.17. Kp = 0.04 atm 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?
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)
Initial conc. 4.0 M 0 0
At equilibrium (4.0-p) p p
We can write,
pc2H4×pH2pc2H6=KP⇒p×p40−p=0.04⇒p2=0.16−0.04p⇒p2+0.04p−0.16=0
Now,
p=−0.04±(0.04)2−4×1×(−0.16)√2×1=−0.04±0.802=0.762(Takingpositivevalue)=0.38
Hence, at equilibrium,
[C2H6] – 4 – p = 4 – 0.38
= 3.62 atm
Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)
(i)Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii)At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Ans.
(i)Reaction quotient,
Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]
(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.
The given reaction is:
CH3COOH(l)+c2H5OH(l) CH3COOC2H5(l)+H2O(l)
Initial conc. 0 0
At equilibrium
= =
Therefore, equilibrium constant for the given reaction is:
Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.171V×0.171V0.829V×0.009V=3.919=3.92(approximately)
(iii)Let the volume of the reaction mixture be V.
CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)
Initial conc. 0 0
At equilibrium
= =
Therefore, the reaction quotient is,
Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.214V×0.214V0.786V×0.286V=0.2037=0.204(approximately)
Since Qc<Kc , equilibrium has not been reached.
Q.19. 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 0.5×10−1mol L–1. If value of Kc is 8.3×10−3, what are the concentrations of PCl3 and Cl2 at equilibrium?
PCl5(g) ⇒ PCl3(g) + Cl2(g)
Ans.
Consider the conc. Of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:
PCl5(g) ⇒ PCl3(g + Cl2(g)
At equilibrium 0.5×10−10molL−1 x mol L-1 x mol L-1
It is given that the value of equilibrium constant , Kc is 8.3×10−10molL−3
Now we can write the expression for equilibrium as:
[PCl2][Cl2][PCl3]=Kc⇒x×x0.5×10−10=8.3×10−3⇒x2=4.15×10−4⇒x=2.04×10−2=0.0204=0.02(approximately)
Therefore, at equilibrium,
[PCl3]=[Cl2]=0.02mol L-1
Q.20. 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= 0.265 at 1050 K.
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and ???2= 0.80 atm?
Ans.
For the given reaction,
FeO(g) + CO(g) Fe(s) + CO2(g)
Initialy, 1.4 atm 0.80 atm
Qp=pCO2pCO=0.801.4=0.571
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 = decrease in pressure of CO2 be p. Then, we can write,
Kp=pCO2pCO⇒0.265=0.80−p1.4+p⇒0.371+0.265p=0.80−p⇒1.265p=0.429⇒p=0.339atm
Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm
And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm
Q.21. A reaction is given:
N2(g)+3H2(g) ⇒ 2NH3
For the above equation, Equilibrium constant = 0.061 at 500 K
At a specific time, from the analysis we can conclude that composition of the reaction mixture is, 2.0 mol L–1 H2 , 3.0 mol L –1 N2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.
Ans.
N2(g) + 3H2(g) 2NH3
At a particular time: 3.0 mol L-1 2.0mol L-1 0.5 mol L-1
So,
Qc=[NH3]2[N2][H2]3=(0.5)2(3.0)(2.0)3=0.0104
It is given that Kc=0.061
∵ Qc≠Kc, the reaction is not at equilibrium.
∵ Qc<Kc, the reaction preceeds in the forward direction to reach at equilibrium.
Q.22.Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇒ Br2(g) + Cl2(g)
For which Kc= 42 at 600 K.
If initially pure BrCl is present at a concentration of 5.5×10−5molL-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)
Initial conc. 5.5×10−5 0 0
At equilibrium 5.5×10−5−2x x x
Now, we can write,
[Br2][Cl2][BrCl]2=Kc⇒x×x(5.5×10−5−2x)2=42⇒x5.5×10−5−2x=6.48⇒x=35.64×10−5−12.96x⇒13.96x=35.64×10−5⇒x=35.6413.96×10−5=2.55×10−5
So,at equilibrium
[BrCl]=5.5×10−5−(2×2.55×10−5)=5.5×10−5−5.1×10−5=0.4×10−5=4.0×10−6molL−1
Q.23. Find out Kc for the given reaction at temperature 1127K where the pressure is 1 atm. A solution of CO and CO2 is in equilibrium with carbon(solid). It has 93.55% CO by mass.
C(s)+CO2 (g) ⇒ 2CO (g)
Ans.)
Let us assume that the solution is of 100g in total.
Given, mass of CO = 93.55 g
Now, the mass of CO2 =(100 – 93.55)=6.45 g
Now, number of moles of CO, nCO=93.528=3.34mol
Number of moles of CO2, nCO2=6.4544=0.146mol
Partial pressure of CO,
PCO= nCOnCO+nCO2×ptotal=3.343.34+0.146×1=0.958atm
Partial pressure of CO2, PCO2=nCO2nCO+nCO2×ptotal=0.1463.34+0.146×1=atm
Therefore, Kp= [CO]2[CO2]=(0.938)20.062=14.19
For the given reaction,
∆n = 2 – 1 = 1
We know that,
Kp=Kc(RT) Δn
⇒14.19=Kc(0.082×1127)1⇒Kc=14.190.082×1127=0.154(approximately)
Q.24.Find out
(I)The equilibrium constant for the formation of NO2 from NO and O2 at 298 K and
(II) ∆G°
NO(g)+12O2(g)↔NO2(g)
Where;
∆fG° (NO2) = 52.0 kJ/mol
∆fG° (NO) = 87.0 kJ/mol
∆fG° (O2) = 0 kJ/mol
Ans.)
(I)We know that,
∆G° = RT log Kc
∆G° = 2.303 RT log Kc
Kc=−35.0×10−3−2.303×8.314×298=6.134∴Kc=antilog(6.134)=1.36×106
Therefore, the equilibrium constant for the given reaction Kc is 1.36×106
Q.25. When each of the following equilibria is subjected to a decrease in pressure by increasing the volume, does the number of moles of reaction products increase, decrease or remain same?
(I) PCl5(g) ⇒ PCl3 +Cl2 (g)
(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)
(III) 3Fe (s) + 4H2O (g) ⇒ Fe3O4 (s) +4H2 (g)
Ans.
(I) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(II) The number of moles of reaction products will decrease.
(III) The number of moles of reaction products remains the same.
Q.26. 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.)
When pressure is increased:
The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.
Since, the number of moles of gaseous reactants is more than that of gaseous products; the reaction given in (iv) will proceed in the forward direction
Since, the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction
Q.27. The equilibrium constant for the following reaction is 1.6×105at 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, ?? 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.
2HBr (g) H2 (g) + Br2 (g)
Initial conc. 10 0 0
At equilibrium 10-2p p p
Now, we can write,
pHBr×p2p2HBr=K‘pp×p(10−2p)2=6.25×10−6p10−2p=2.5×10−3p=2.5×10−2−(5.0×10−3)pp+(5.0×10−3)p=2.5×10−2(1005×10−3)=2.5×10−2p=2.49×10−2bar=2.5×10−2bar(approximately)
Therefore, at equilibrium,
[H2]=[Br2]= 2.49×10−2bar [HBr]= 10−2×(2.49×10−2)bar=9.95bar=10bar(approximately)
Q.28.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)
(I)Write as expression for Kp for the above reaction.
(II) How will the values of Kp and composition of equilibrium mixture be affected by
(i)Increasing the pressure
(ii)Increasing the temperature
(iii)Using a catalyst?
Ans.)
(I)For the given reaction,
Kp=pCO×p3H2pCH4×pH2O
(II) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.
Q.29. Describe the effect of:
I) Removal of CO
II) Addition of H2
III) Removal of CH3OH on the equilibrium of the reaction:
IV) Addition of CH3OH
2H2 (g)+CO (g) ⇒ CH3OH (g)
Ans.)
(I) On removing CO, the equilibrium will shift in the backward direction.
(II) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.
(III) On removing CH3OH, the equilibrium will shift in the forward direction.
(IV) On addition of CH3OH, the equilibrium will shift in the backward direction.
Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3×10−3. If decomposition is depicted as,
PCl5(g)↔PCl3(g)+Cl2(g)
∆rH° = 124.0 kJmol–1
a) Write an expression for Kc for the reaction.
b) What is the value of Kc for the reverse reaction at the same temperature?
c) What would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased?
(iii) The temperature is increased?
Ans.)
(a)Kc=[PCl3(g)][Cl2(g)][PCl3(g)]
(b)Value of Kc for the reverse reaction at the same temperature is:
K‘c=1Kc=18.3×10−3=1.2048×102=120.48
(c)(i)Kc would remain the same because in this case, the temperature remains the same.
(ii)Kc is constant at constant temperature. Thus, in this case, Kc would not change. (iii)In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.
Q.31. 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 ⇒ CO2(g) + H2(g)
Initial conc. 4.0 bar 4.0 bar 0 0
At equilibrium 4.0-p 4.0-p p p
Given Kp = 10.1
PCO2×PH2PCO×PH2O=KP⇒p×p(4.0−p)(4.0−p)=10.1⇒p4.0−p=3.178⇒p=12.712−3.178p4.178p=12.712p=12.7124.178p=3.04
So, partial pressure of H2 is 3.04 bar at equilibrium.
Q.32. Predict which of the following reaction will have appreciable concentration of reactants and products:
(a)Cl2(g)↔2Cl(g);Kc=5×10−39
(b)Cl2(g)+2NO(g)↔2NOCl(g);Kc=3.7×108
(c)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.
Q.33. 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.)
Given,
3O2 (g) ⇒ 2O3 (g)
Then, Kc = [O3(g)]2[O2(g)]3
Given that Kc = 2.0×10−50 and [O2(g)] = 1.6×10−2
Then,
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=8.192×10−56⇒[O3(g)]2=2.86×10−28M
So, the conc. of O3 is 2.86×10−28M.
Q.34. The reaction, CO(g)+3H2(g)→CH4(g)+H2O(g) at 1300K is at equilibrium in a 1L container. It has 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and y amount of CH4 in the container. Find the concentration of CH4 in the mixture.
The equilibrium constant, Kc is 3.90 at the given temp.
Ans.)
Let the concentration of CH4 at equilibrium be y.
\(CO_{(g)} \; + \; 3H_{2(g)} \; \rightarrow \; CH_{4(g)} \; + \; H_{2}O_{(g)}\)
At equilibrium,
For CO – 0.31=0.3M
For H2 – 0.11=0.1M
For H2O – 0.021=0.02M Kc = 3.90
Therefore,
[CH4(g)][H2O(g)][CO(g)][H2(g)]3=Kc y×0.020.3×(0.1)3=3.9 y=3.9×0.3×(0.1)30.02 y=0.001170.02 y=0.0585M y=5.85×10−2M
Therefore, the concentration of CH4 at equilibrium is 5.85×10−2M
Q.35. What is conjugate acid-base pair? Find the conjugate acid/base of the given species:
(i) HNO2
(ii) CN−
(iii) HClO4
(iv) F−
(v) OH−
(vi) CO2−3
(vii) S−
Ans.)
A conjugate acid-base pair is a pair that has a difference of only one proton.
The conjugate acid-base pair of the following are as follows:
(i) HNO2 – NO−2 (Base)
(ii) CN− – HCN (Acid)
(iii) HClO4 – ClO−4 (Base)
(iv) F− – HF (Acid)
(v) OH− – H2O (Acid)/ O2− (Base)
(vi) CO2−3 – HCO−3 (Acid)
(vii) S− – HS− (Acid)
Q.36. From the compounds given below which are Lewis acids?
(i) H2O
(ii) BF3
(iii) H+
(iv) NH+4
Ans.)
Lewis acids are the acids which can accept a pair of electrons.
(i) H2O – Not Lewis acid
(ii) BF3 – Lewis acid
(iii) H+ – Lewis acid
(iv) NH+4 – Lewis acid
Q.37. From the compounds given below which will be the conjugate base for the Bronsted acids?
(i) HF
(ii) H2SO4
(iii) HCO3
Ans.)
The following shows the conjugate bases for the Bronsted acids:
(i) HF – F−
(ii) H2SO4 – HSO−4
(iii) HCO3 – CO2−3
Q.38. For the Brönsted bases given below find their conjugate acids.
NH3
HCOO−
NH−2
Ans.)
Brönsted base Conjugate acid
1 NH3 NH+4
2 HCOO− HCOOH
3 NH−2 NH3
Q.39. The species given below can act as both Brönsted bases as well as Brönsted acids. For each of them give their conjugate acid and base.
HCO−3
HSO−4
NH3
H2O
Ans.)
Species Conjugate base Conjugate acid
1 HCO−3 CO2−3 H2CO3
2 HSO−4 SO2−4 H2SO4
3 NH3 NH−2 NH+4
4 H2O OH− H3O+
Q.40. Classify the species given below into bases and acids and also show that these species act as base/acid:
BCl3
H+
OH−
F−
Ans.)
BCl3:
It is a Lewis acid as it has tendency to accept a pair of electrons.
H+
It is a Lewis acid as it has tendency to accept a pair of electrons.
OH−
It is a Lewis base as it has tendency to lose a pair of electrons.
F−
It is a Lewis base as it has tendency to lose its lone pair of electrons.
Q.41.A sample soft drink is taken, whose hydrogen ion concentration is 2.5×10−4M.Find out pH.
Ans.)
pH=−log[H+]=−log(2.5×10−4)=−log2.5−log10−4=−log2.5+4=−0.398+4=3.602
Q.42.A sample of white vinegar is taken , whose pH is 2.36. Find out the hydrogen ion concentration in the sample.
Ans.)
pH=−log[H+]⇒log[H+]=−pH⇒[H+]=antilog(−pH)=antilog(−2.36)=0.004365=4.37×10−3 ∴ 4.37×10−3 is the concentration of white vinegar sample.
Q.43. Ionization constant for the following acids are given:
HF = 5.7×10−5 at 298K
HCOOH = 1.7×10−3 at 298K
HCN = 3.7×10−8 at 298K
Find out the conjugate bases for the above acids.
Ans.)
For F–, Kb=KwKa=10−14(5.7×10−5)=1.75×10−9
For HCOO–, Kb=10−14(1.7×10−3)=5.88×10−11
For CN– = Kb=10−14(3.7×10−8)=×10−11=2.70×10−6
Q.44.Phenol has ionization constant of 1.0×10−8. In a 0.06M of phenol solution calculate the presence of phenolate ion. Find out the degree of ionization if 0.02M of sodium phenolate is given.
Ans.)
C6H5OH C6H5O– + H+
Initial 0.06M
After dissociation 0.06-x x x
∴Ka=x×x0.06−x=1.0×10−8⇒x20.06=1.0×10−8⇒x2=6×10−10⇒x=2.4×105M
In presence of 0.02 sodium phenolate(C6H5Na), suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.06 – y ≃ 0.06,
[C6H5O–] = 0.02 + y ≃0.01M,
[H+]=y M
∴Ka=(0.02)(y)0.06=1.0×10−8⇒y=1.0×0.06(0.02)×10−8⇒y=6×10−8 ∴ degree of ionization = α=yc=6×10−86×10−2(Herec=0.06=6×10−2)=10−6
So, α=10−6
Q.45 Given, 9.1×10−8 is the initial(first) ionization constant of the gas H2S.Find out concentration of the ion HS– in 0.1M solution of H2S . Find the changes in concentration if the concentration is 0.1M in HCl. Find the concentration of S2- under both conditions, if 1.2×10−13 is the second dissociation constant of H2S.
Ans.)
To calculate [HS–]
H2S H+ + HS–
Intial 0.1 M
After dissociation 0.1-x x x
≃0.1 Ka=x×x0.1=9.1×10−8⇒x2=9.1×10−9⇒x=9.54×10−5
In the presence of 0.1 M HCl, suppose H2S dissociated is y. Then at equilibrium, [H2S] =0.1-y ≃0.1,
[H+]=0.1 + y ≃0.1,
[HS–] = y M
Ka=0.1×y0.1=9.1×10−8y=9.1×0.10.1×10−8y=9.1×10−8
Ka2
Ka1
To calculate [S2-]
H2S H+ + HS– , HS– H+ + S2-
For the overall reaction,
H2S 2H+ + S2-
Ka=Ka1×Ka2=9.1×10−8×1.2×10−13=1.092×10−20 Ka=[H+]2[S2−][H2S]
In the absence of 0.1M HCl,
[H+]= 2[S2-]
Hence, if [S2-] = x , [H+] =2x
∴(2x)20.1=1.092×10−20⇒4x3=1.092×10−21⇒x3=1.0924×10−21=273×10−24⇒logx3=log273−log10−24=2.4362−24⇒3logx=2.4362−24⇒logx=2.43623−243⇒x=0.8127−8=−7.1873⇒x=Antilog−7.1873=6.497×10−8=6.5×10−8
In presence of 0.1M HCl,
Suppose [S2-]=y, then
[H2S]=0.1-y≃0.1M,
[H+]=0.1+y≃0.1M
Ka=(0.1)2×y0.1=1.09×10−20y=1.09×10−19M
Q.46.Given,the ionization constant of acetic acid is 1.74×10−5. Find the degree of dissociation of acetic acid in its 0.05 M solution. Find the concentration of acetate ion in the solution and its pH.
Ans.)
CH3COOH ⇒ CH3COO– + H+
Ka=[CH3COO−][H+][CH3COOH]=[H+]2[CH3COOH]⇒[H+]=Ka[CH3COOH]−−−−−−−−−−−−−√=(1.74×10−5)(5×10−2)−−−−−−−−−−−−−−−−−−−√=9.33×10−4M[CH3COO−]=[H+]=9.33×10−4MpH=−log(9.33×1.0−4)=4−0.9699=4−0.97=3.03
Q.47. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Ans.)
HA H+ + A–
pH=-log[H+]
log[H+]=-4.15
[H+] = 7.08×10−5M [A–] = [H+] = 7.08×10−5M Ka=[H+][A−][HA]=(7.08×10−5)(7.08×10−5)10−2=5.0×10−7pKa=−logKa=−log(5.0×10−7)=7−0.699=6.301
Q.48.Consider complete dissociation, find out the pH of the following :
(I)0.004 M HCl
(II)0.003 M NaOH
(III)0.002 M HBr
(IV)0.002 M KOH
Ans.)
(I)HCl + aq H+ + Cl–
∴[H+]=[HCl]=4×10−3MpH=−log(4×10−3)=2.398
(II)NaOH + aq Na+ + OH–
∴[OH−]=3×10−3M[H+]=10−14(3×10−3)=3×10−12MpH=−log(3×10−12)=11.52
(III)HBr + aq H+ + Br–
∴[H+]=2×10−3MpH=−log(2×10−3M)=2.70
(IV)KOH + aq K+ + OH–
∴[OH+]=2×10−3M[H+]=10−14(2×10−3)=5×10−12pH=−log(5×10−12)=11.30
Q.49.Find out the pH of the following solution:
(I)2g of TIOH dissolved in water to give 2 litre of the solution
(II)0.3g of Ca(OH)2 dissolved in water to given 500mL of the solution
(III)0.3g of NaOH dissolved in water to give 200mL of the solution
(IV)1 mL of 13.6 M HCl is diluted with water to given 1 litre of the solution
Ans.)
(I)Molar conc. Of TlOH = 2g(204+16+1)gmol−1×12L=4.52×10−3M[OH−]=[TlOH]=4.52×10−3M[H+]=10−14(4.52×10−3)=2.21×10−12M∴pH=−log(2.21×10−12)=12−(0.3424)=11.66
(II)Molar conc. Of Ca(OH)2=0.3g(40+34)gmol−1×10.5L=8.11×10−3M[OH−]=2[Ca(OH)2]=2×(8.11×10−3)M=16.22×10−3MpOH=−log(16.22×10−3)=3−1.2101=1.79pH=14−1.79=12.21
(III)Molar conc. of NaOH = 0.3g(40+34)gmol−1×10.2L=3.75×10−2M[OH−]=3.75×10−2MpOH=−log(3.75×10−2)=2−0.0574=1.43pH=14−1.43=12.57
(IV)M1V1 = M2V2
∴13.6M××1mL=M2×1000mL∴M2=1.36×10−2M[H+]=[HCl]=1.36×10−2MpH=−log(1.36×10−2)=2−0.1335≃1.87
Q.50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans.)
CH2(Br)COOH CH2(Br)COO– + H+
Initial conc. C 0 0
Conc. at eqm. C – Cα Cα Cα
Ka=Cα⋅CαC(1−α)=Cα21−α≃Cα2=0.1×(0.132)2=1.74×10−3pKa=−log(1.74×10−3)=3−0.2405=2.76[H+]=Cα=0.1×0.132=1.32×10−2MpH=−log(1.32×10−2)=2−0.1206=1.88
Q.51. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans.)
CH2(Br)COOH CH2(Br)COO– + H+
Initial conc. C 0 0
Conc. at eqm. C – Cα Cα Cα
Ka=Cα⋅CαC(1−α)=Cα21−α≃Cα2=0.1×(0.132)2=1.74×10−3pKa=−log(1.74×10−3)=3−0.2405=2.76[H+]=Cα=0.1×0.132=1.32×10−2MpH=−log(1.32×10−2)=2−0.1206=1.88
Q.52. 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 = 4.27×10−10
c = 0.001M
pH =?
α =?
Kb=cα24.27×10−10=0.001×α24270×10−10=α265.34×10−5=α=6.53×10−4Then,[anion]=cα=0.001×65.34×10−5=0.065×10−5 pOH=−log(0.065×10−5)=6.187pH=7.813Now,Ka×Kb=KwKa=10−144.27×10−10=2.34×10−5 ∴2.34×10−5 is the ionization constant.
Q.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains
(I)0.01 M
(II)0.1 M in HCl?
Ans.)
c=0.05M
pKa=4.74
pKa=-log(Ka)
Ka=1.82×10−5Ka=cα2α=Kac−−−√α=1.82×10−55×10−2−−−−−−−√=1.908×10−2
When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case 1: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
CH3COOH H+ + CH3COO–
Initial conc. 0.05M 0 0
After dissociation 0.05-x 0.01+x x
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.
Ka=[CH3COO−][H+][CH3COOH]∴=(0.01)x0.05x=1.82×10−5×0.050.01x=1.82×10−3×0.05M
Now,
α=AmounofaciddissociationAmountofacidtaken=1.82×10−3×0.050.05=1.82×10−3
Case 2: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
[CH3COOH]=0.05 – X; 0.05 M
[CH3COO–]=X
[H+]=0.1+X ; 0.1M
Ka=[CH3COO−][H+][CH3COOH]∴Ka=(0.1)X0.05x=1.82×10−5×0.050.1x=1.82×10−4×0.05M
Now,
α=AmounofaciddissociationAmountofacidtaken=1.82×10−4×0.050.05=1.82×10−4
Q.54. The ionization constant of dimethylamine is5.4××10−4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?
Ans.)
Kb=5.4×10−4c=0.02MThen,α=Kbc−−−√=5.4×10−40.02−−−−−−√=0.1643
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
NaOH(aq) Na+(aq) + OH–(aq)
0.1M 0.1M
And,
(CH3)2 NH + H2O (CH3)2 NH2+ + OH
(0.02-x) x x
;0.02M ;0.1M
Then,[ (CH3)2 NH+2]=x
[OH– ]=x+0.1;0.1
⇒Kb=[(CH3)2NH+2][OH−][(CH3)2NH]5.4×10−4=x×0.10.02x=0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.
Q.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(I)Human saliva, 6.4
(II)Human stomach fluid, 1.2
(III)Human muscle-fluid, 6.83
(IV)Human blood, 7.38
Ans.)
(I)Human saliva, 6.4:
pH = 6.4
6.4 = – log [H+] [H+] = 3.98×10−7
(II)Human stomach fluid, 1.2:
pH =1.2
1.2 = – log [H+] ∴ [H+] = 0.063
(III)Human muscle fluid 6.83:
pH = 6.83
pH = – log [H+]
6.83 = – log [H+] [H+] =1.48×10−7M
(IV) Human blood, 7.38:
pH = 7.38 = – log [H+] [H+] = 4.17×10−8M
Q.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 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+]
(I)pH of milk = 6.8
Since, pH = –log [H+]
6.8 = –log [H+] log
[H+] = –6.8
[H+] = anitlog(–6.8)
= 1.5×10−7M
(II)pH ofblack coffee = 5.0
Since, pH = –log [H+]
5.0 = –log [H+] log
[H+] = –5.0
[H+] = anitlog(–5.0)
= 10−5M
(III)pH of tomato= 4.2
Since, pH = –log [H+]
4.2 = –log [H+] log
[H+] = –4.2
[H+] = anitlog(–4.2)
= 6.31×10−5M
(IV)pH of lemon juice= 2.2
Since, pH = –log [H+]
2.2 = –log [H+] log
[H+] = –2.2
[H+] = anitlog(–2.2)
= 6.31×10−3M
(V)pH of egg white= 7.8
Since, pH = –log [H+]
7.8 = –log [H+] log
[H+] = –7.8
[H+] = anitlog(–7.8)
= 1.58×10−8M
Q.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Ans.)
[KOHaq]= 0.56115g/L=2.805g/L=2.805×156.11M=0.05M
KOH(aq) K+(aq) + OH(aq)–
[OH–]=0.05M=[K+] [H+][H–]=Kw
[H+]=Kw[OH−] =10−140.05=2×10−13M∴pH=12.70
Q.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Ans.)
Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2
=19.23121.63M=0.1581M
Sr(OH)2(aq) Sr2+(aq) + 2(OH–)(aq)
∴[ Sr2+]=0.1581M
[OH–]= 2×0.1581M=0.3126
Now,
Kw=[OH–][H+] 10−140.3126=[H+]⇒[H+]=3.2×10−14∴pH=13.495;13.50
Q.59. The ionization constant of propanoic acid is 1.32××10−15. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Ans.)
Let the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
HA + H2O H3O+ + A–
(0.05-0.0α) ≈0.05 0.05α 0.05α
Kα=[H3O+][A−][HA]=(0.05α)(0.05α)0.05=0.05α2 α=Kα0.05−−−√=1.63×10−2
Then,[ H3O+ ]= 0.05α =0.05×1.63×10−2=Kb⋅15×10−4M∴pH=3.09
In the presence of 0.1M of HCl, let 뱫 be the degree of ionization.
Then,[ H3O+ ]=0.01
[A–]=0.05 α´
[HA]=0.05
Kα=0.01×0.05α′0.051.32×10−5=0.01×α′α′=1.32×10−3
Q.60. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Ans.)
c = 0.1 M
pH = 2.34
-log [H+] = pH
-log [H+] = 2.34
[H+]=4.5×10−3
Also,
[H+]=cα
4.5×10−3=0.1×α4.5×10−30.1=αα=4.5×10−3=0.045
Then,
Ka=cα2=0.1×(45×10−3)2=202.5×10−6=2.02×10−4
Q.61. for nitrous acid Ka = 4.5×10−4. Calculate degree of hydrolysis and pH for 0.04M of sodium nitrite.
Ans.)
Sodium nitrite is a salt of NaOH (strong base) and HNO2 (weak acid).
NO−2+H2O↔HNO2+OH− Kh=[HNO2][OH−][NO−2] ⇒KwKa=10−144.5×10−4=22×10−10
Let, y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:
[NO−2]=0.04–y;0.04 [HNO2]=y [OH−]=y Kh=y20.04=0.22×10−10 y2=0.0088×10−10 y=0.093×10−5 ∴[OH−]=0.093×10−5M [H3O+]=10−100.093×10−5=10.75×10−9M
Thus, pH=−log(10.75×10−9)
= 7.96
Thus, the degree of hydrolysis is
=y0.04=0.093×10−50.04 =2.325×10−5
Q.62. 0.02M solution of pyridinium hydrochloride (C5H6ClN) is having pH = 3.44. Determine the ionization constant of C5H5N (pyridine).
Ans.)
pH = 3.44
As we know,
pH=log[H+] ∴[H+]=3.63×10−4
Now, Kh=3.63×10−40.02; (Given that concentration = 0.02M)
⇒Kh=6.6×10−6
As we know that,
Kh=KwKa Ka=KwKh=10−146.6×10−6
= 1.51×10−9
Q.63. Few salts are given below;
KBr
NH4NO3
KF
NaNO2
NaCN
NaCl
Determine the nature of solution of these salts i.e. Is it acidic or basic or neutral?
Ans.)
KBr
KBr + H2O ↔ KOH (Strong base)+ HBr (Strong acid)
Thus, it is a neutral solution.
NH4NO3
NH4NO3 + H2O ↔ NH4OH(Weak base) + HNO2 (Strong acid)
Thus, it is an acidic solution.
KF
KF + H2O ↔ KOH (Strong base) + HF (weak acid)
Thus, it is a basic solution.
NaNO2
NaNO2 + H2O ↔ NH4OH(Strong base) + HNO2(Weak acid)
Thus, it is a basic solution.
NaCN
NaCN + H2O ↔ HCN (Weak acid) + NaOH (Strong base)
Thus, it is a basic solution.
NaCl
NaCl + H2O ↔ NaOH (Strong base) + HCL (Strong acid)
Thus, it is a neutral solution.
Q.64. Find the pH of 0.1M acid and its 0.1M NaCl solution. The Ka for chloroacetic acid is 1.35×10−3.
Ans.)
The Ka for chloroacetic acid (ClCH2COOH) is 1.35×10−3.
⇒Ka=cα2 ∴α=Kac−−−√
= 1.35×10−30.1−−−−−−−√; (given concentration = 0.1M)
α=1.35×10−2−−−−−−−−−√
= 0.116
∴[H+]=cα=0.1∗0.116=0.0116
pH = −log[H+] = 1.94
ClCH2COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH2COOH
ClCH2COO−+H2O↔ClCH2COOH+OH− Kh=[ClCH2COO][OH−][ClCH2COO−]
Now, Kh=KwKa Kh=10−141.35×10−3 =0.740×10−11
Also, Kh=y20.1 ⇒0.740×10−11=y20.1 ⇒0.0740×10−11=y2 y=0.86×10−6 [OH−]=0.86×10−6 ∴[H+]=Kw0.86×10−6 =10−140.86×10−6 [H+]=1.162×10−3
pH = −log[H+]
= 7.94
Q.65. Determine the pH of neutral water at 310K temperature. Ionic product of H2O is 2.7×10−14.
Ans.)
Ionic Product,
Kw=[H+][OH−]
Assuming, [H+] = y
As, [H+]=[OH−], Kw = y2.
Kw at 310K is 2.7×10−14.
∴2.7×10−14=y2
y = 1.64×10−7
[H+]=1.64×10−7
pH = −log[H+]
= −log[1.64×10−7]
= 6.78
Thus, the pH of neutral water at 310K temperature is 6.78.
Q.66. Find out the pH of resultant mixture;
i) n10 ml of 0.02M H2SO4 + 10 ml of 0.02M Ca(OH)2
ii) 10 ml of 0.1M H2SO4 + 10 ml of 0.1M KOH
iii) 10 ml of 0.2M Ca(OH)2 + 25 ml of 0.1M HCl
Ans.)
i) Moles of OH−
= 2∗10∗0.021000=0.0004mol
Moles of H3O+
= 2∗10∗0.021000=0.0004mol
ii) Moles of OH−
= 2∗10∗0.11000=0.002mol
Moles of H3O+
= 2∗10∗0.11000=0.001mol
Here, the H3O+ is in excess is 0.01 mol
So, [H3O+]=0.00120×10−3=10−320×10−3=0.5
Thus, pH = -log(0.05)
= 1.3
As the solution id neutral pH = 7.
iii) Moles of OH−
= 2∗10∗0.21000=0.004mol
Moles of H3O+
= 25∗0.11000=0.0025mol
Here, the OH− is in excess is 0.0015 mol
So, [OH−]=0.001535×10−3=0.0428
Thus, pH = -log(OH)
= 1.36
pH = 14 – 1.36 = 12.63
As the solution id neutral pH = 12.63.
Q.67. Calculate the solubilities of
a) barium chromate
b) ferric hydroxide
c) lead chloride
d) mercurous iodide
e) silver chromate
At 300K from their solubility product constant. Also calculate the molarities of the individual ions.
Ans.)
a) Barium Chromate
BaCrO4 Ã Ba2+ + CrO2−4
Now, Ksp = [Ba2+][CrO2−4]
Aumming the solubility of BaCrO4 is ‘x’.
Thus,
[Ba2+] = x and CrO2−4 = x
Ksp = x2
1.2×10−10=x2
x = 1.09×10−10M
Molarity of Ba2+ = Molarity of CrO2−4 = x = 1.09×10−10M
b) Ferric Hydroxide
Fe(OH)3 Ã Fe3+ + OH−
Now, Ksp = [Fe3+][OH−]3
Aumming the solubility of Fe(OH)3 is ‘x’.
Thus,
[Fe3+] = x and OH− = 3x
Ksp = x(3x)3
= x*27x3
Ksp = 27x4
1.0×10−38=27x4
x = 0.00037×10−36M
Molarity of Fe3+ = x = 1.39×10−10M
Molarity of OH− = 3x = 4.17×10−10M
c) Lead Chloride
PbCl2 Ã Pb2+ + 2Cl−
Now, Ksp = [Pb2+][Cl−]
Aumming the solubility of PbCl2 is ‘x’.
Thus,
[Pb2+] = x and Cl− = 2x
Ksp = x(2x)2
= x*4x2
Ksp = 4x3
1.6×10−5=4x3
x = 1.58×10−2M
Molarity of Pb2+ = x = 1.58×10−2M
Molarity of Cl− = 2x = 3.16×10−2M
d) Mercurous iodide
Hg2I2 Ã Hg2+ + 2I−
Now, Ksp = [Hg2+]2[I−]2
Aumming the solubility of Hg2I2 is ‘x’.
Thus,
[Hg2+] = x and I− = 2x
Ksp = x(2x)2
= x*4x2
Ksp = 4x3
4.5×10−29=4x3
x = 2.24×10−10M
Molarity of Hg2+ = x = 2.24×10−10M
Molarity of I− = 2x = 4.48×10−10M
e) Silver Chromate
Ag2CrO4 Ã 2Ag2+ + CrO2−4
Now, Ksp = [Ag2+]2[CrO2−4]
Aumming the solubility of Ag2CrO4 is ‘x’.
Thus,
[Ag2+] = 2x and CrO2−4 = x
Ksp = (2x)2*x
1.1×10−12=4x3
x = 0.65×10−4M
Molarity of Ag2+ = 2x = 1.3×10−4M
Molarity of CrO2−4 = x = 0.65×10−4M
Q-68. Determine the ratio of molarities to their saturated solutions for the following:
Ag2CrO4 and AgBr
The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and5.0×10−13 respectively.
Ans.)
Ag2CrO4 Ã 2Ag2+ + CrO−4
Now, Ksp = [Ag2+]2[CrO−4]
Asuming the solubility of Ag2CrO4 is ‘x’.
Thus,
[Ag2+] = 2x and CrO−4 = x
Ksp = (2x)2*x
1.1×10−12=4x3
x = 0.65×10−4M
Assuming the solubility of AgBr is y.
AgBr(s) Ã Ag2+ + 2CrO−4
Ksp = (y)2
5.0×10−13=y2
y = 7.07×10−7M
The ratio of molarities to their saturated solution is:
xy=0.65×10−4M7.07×10−7M=91.9
Q.69. Cupric chlorate and sodium iodate having equal volume of 0.002M. Will the precipitation of copper iodate will occur or not?
Ans.)
Cupric chlorate and sodium iodate having equal volume are mixed together, then molar concentration of cupric chlorate and sodium iodate will reduce to half.
So, molar concentration of cupric chlorate and sodium iodate in mixture is 0.001M.
Na(IO3)2 Ã Na+ + IO−3
0.0001M 0.001M
Cu(ClO3)2 Ã Cu2+ + 2CIO−3
0.001M 0.001M
The Solubility for Cu(IO3)2 ⇒ Cu2+ (aq) + 2IO3– (aq)
Now, the ionic product of the copper iodate is:
= [Cu2+] [IO−3]2
= (0.001)(0.001)2
= 1.0×10−9M
As the value of Ksp is more than Ionic product.
Thus, the precipitation will not occur.
Q.70. For benzoic acid the ionization constant is 6.46×10−5M and for silver benzoate Ksp is 2.5×10−5M. Give relation between the solubility of silver benzoate in buffer of pH = 3.19 and its solubility in water.
Ans.)
Here, pH = 3.19
[H3O+] = 6.46×10−5M
C6H5COOH + H2O Ã C6H5COO− + H3O
Ka[C6H5COO−][H3O+]C6H5COOH Ka[C6H5COOH]C6H5COO−=[H3O+]Ka=6.46×10−46.46×10−5=10
Assuming the solubility of silver benzoate (C6H5COOAg) is y mol/L.
Now, [Ag+] = y
[C6H5COOH] = [C6H5COO−] = y
10[C6H5COO−] + [C6H5COO−] = y
[C6H5COO−] = y/11
Ksp[Ag+][C6H5COO−] = y
2.5×1013=yy11
y = 1.66×10−6 mol/L
Hence, solubility of C6H5COOAg in buffer of pH = 3.19 is 1.66×10−6 mol/L.
For, water:
Assuming the solubility of silver benzoate (C6H5COOAg) is x mol/L.
Now, [Ag+] = x M
Ksp = [Ag+][C6H5COO−]
Ksp = (y)2
y = Ksp−−−√=2.5×10−13−−−−−−−−−√=5×10−7mol/L ∴yx=1.66×10−65×10−7=3.32
Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.
Q.71. Calculate the maximum concentration of equimolar solutions of FeSO4 and Na2SO4 so that when they are mixed in equal volume than there is no precipitation of FeS? (Ksp for Fes is 6.3×10−18)
Ans.)
Assuming the maximum concentration of each solution is y mol/L
On mixing the solutions the volume of the concentration of each solution is reduced to half.
After mixing the maximum concentration of each solution is y/2 mol/L.
Thus, [FeSO4] = [Na2S] = y/2 M
So, [Fe2+] = [FeSO4] = y/2 M
FeS(s) ↔ Fe2+(aq)+S2−(aq)
Ksp = [Fe2+][S2-] 6.3×10−18=(y2)(y2) y24=6.3×10−18
Thus, y = 5.02×10−9
Thus, if the concentration of FeSO4 and Na2SO4 are equal to or less than that of 5.02×10−9M, then there won’t be precipitation of FeS.
Q.72. Find the minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K?
Ksp for CaSO4 is 9.1×10−6
Ans.)
CaSO4(s)↔Ca2+(aq)+SO2−4(aq)
Ksp = [Ca2+][SO2−4]
Assuming the solubility of calcium sulphate is x.
So, Ksp = x2
∴9.1×10−6=x2 ∴x=3.02×10−3mol/L
Now, molecular mass os calcium sulphate is 136g/mol.
Solubility in calcium sulphate in g/mol is
=3.02×10−3×136
= 0.41 g/L
i.e. 1 litre H2O will be required to dissolve 0.41g of calcium sulphate.
Thus, minimum volume of H2O required to dissolve 1 gram of CaSO4 at 298K is
= 10.41L=2.44L
Q.73. The concentration of S2- in 0.1M HCl solution saturated with H2S is 1.0×1019M. If 10mL of this added to 5mL of 0.04M solution given below:
MnCl2
ZnCl2
CdCl2
FeSO4
In which of the above solution the precipitation takes place?
For MnS, Ksp = 2.5×10−13
For ZnS, Ksp = 1.6×10−24
For CdS, Ksp = 8.0×10−27
For FeS, Ksp = 6.3×10−18
Ans.)
If the ionic product exceeds the Ksp value, then only precipitation can take place.
Before mixing:
[S2-] = Ksp = 1.0×10−19M [M2+] = 0.04M
Volume = 10mL Volume = 5mL
After mixing:
[S2-] = ? and [M2+] = ?
Total volume = (10 + 5) = 15mL Volume = 15mL
[S2-] = 1.0×10−19×1015=6.67×10−20M [M2+] = 0.04×515=1.33×10−2M
Now, the ionic product = [M2+][S2-]
= (1.33×10−2)(6.67×10−20)
= 8.87×10−22
Here, the ionic product of CdS and ZnS exceeds its corresponding Ksp value.
Thus, precipitation will occur in ZnCl2 and CdCl2 solutions.
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