Equilibrium Numericals Sem II Echem 2

7/27/2019 Equilibrium Numericals Sem II Echem 2

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e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 300044331

4 EQUILIBRIUM

Type I : Chemical Equilibrium

Remember :

(1) For any hypothetical reaction

a A + b B c C + d D

Ke = f

r

K

K =

c d

a b

[C] [D]

[A] [B]

If reaction will occur in homogeneous solution Ke = Kc =c d

C D

a bA B

C C

C C

If reactants and products are in gaseous phase Ke = Kp =

c dC D

a bA B

p p

p p

Similarly

If reactants & products are expressed in mole fraction Kx =c d

C D

a bA B

X X

X X

(2) Kp= Kc(RT)n

If n = 0 then Kp= KcIf n > 0 then Kp> KcIf n < 0 then Kp< Kc

(3) Kp= Kx(P)n

(4) Kc = Kx(V)

n(5) If equilibrium constant of forward reaction is Kc, then equilibrium constant of

reverse reaction of the original reactionKc = (1/Kc).(6) If equilibrium constant of a reaction is Kc, and if the whole chemical reaction is

multiplied by n then the equilibrium constant of new chemical reaction Kc = (Kc)n.

(7) If equilibrium constant of two chemical equation is Kc1and Kc2respectively, and ifboth these chemical equations are added then equilibrium constant of resultantequation Kc = Kc1 Kc2

(8) If equilibrium constant of two chemical equation is Kc1and Kc2respectively, and ifone of the equation is subtracted from other then equilibrium constant of resultantequation Kc = Kc1/Kc2

(9) Direction in which chemical reaction will occur at any stage will be calculatedusing equilibrium quotient (Qc). Equilibrium quotient is calculated in a similar wayas equilibrium constant.

If Qc< Kc Chemical reaction will occur in forward direction.

If Qc> Kc Chemical reaction will not occur in forward direction; but it willoccur in reverse direction.

If Qc= Kc Chemical reaction will be in equilibrium state.


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1. The value of Kc for the reaction H2(g)+ I2(g) 2 HI(g)in a closed vessel at 400K is 4.0 102. If at equilibrium, the concentration of hydrogen is 0.6 mole/litand that of iodine is 0.8 mole/lit., calculate the concentration of hydrogen iodide.(Text. Book Illus. - 1 Page 98)

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

Equilibrium constant Kc = ]][I[H

[HI]

22

2

4.0 102 =[0.6][0.8]

[HI]2

[HI]2 = 4.0 102 0.6 0.8

= 192 104

[HI] = 13.86 102

[HI] = 1.386 101

2. When a mixture of 1 mole of water vapour and 1 mole of CO gas is heated at 725K in a 10 lit container, 40% (w/w%) water undergoes the following reaction withCO, when the reaction reaches equilibrium. Calculate the chemical equilibriumconstant. (Text. Book Illus. - 2 Page 99)

H2O(g) + CO(g) H2(g) + CO2(g) Initial mole 1 1 0 0Equili. mole 10.4 10.4 0.4 0.4

Total mole 10.4 + 10.4 + 0.4 + 0.4 = 2Conc. mole/lit 0.6/10 0.6/10 0.4/10 0.4/10

Kc =O][CO][H

]][CO[H

2

22

=]10/6.0][10/6.0[

]10/4.0][10/4.0[

=]6.0][6.0[

]4.0][4.0[

Kc = 0.44

3. The equilibrium constant Kp of the following reaction at 700 K is 1.80 103atmosphere. Calculate its Kc in mole/lit at the same temperature.

2 SO3(g) 2 SO2(g) + O2(g) Kp = Kc (RT)n

1.80 103 = Kc (0.082 700)1

Kc =700082.0

101.80 3

Kc = 3.136 105


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4. Solid ammonium carbamate (NH4COONH2) dissociates into CO2(g) and NH3(g).The total pressure of the system in the state of equilibrium is 27 atmosphere.Calculate chemical equilibrium constant of the system.

NH4COONH2(s) 2 NH3(g) + CO2(g) Initial mole 1 0 0

Euili. Mole 1x 2 x xHere ammonium carbamate is solid, hence its mole is neglected in calculation ofequilibrium constant.Total mole 2x + x = 3x

Mole fraction -3x

2x

3x

x

Partial pressure of NH3= Mole fraction Total Pressure

=3x

2x27 = 18 atm

Partial pressure of CO2= Mole fraction Total Pressure

=3x

x 27 = 9 atm

Equilibrium constant Ke =]COONH[NH

][CO][NH

24

2

2

3

Kp = (pNH3)2pCO2

= (18)29= 2916 (atm)3

5. 5 moles of NH3 gas was heated in a closed vessel of 200 ml capacity at atemperature of 600 K. At equilibrium, 30% of NH3remained unreacted. Calculate

the value of Kc for the reaction.N2 (g) + 3 H2 (g)2 NH3 (g)

Initial mole : 5 0 0

At equilibrium : 5 3.5 1.75 5.25Conc. Mole lit1 : 1.5/0.2 1.75/0.2 5.25/0.2(Because, 30% NH3 is unreacted at equilibrium, means 70% NH3 is reacted.Therefore 70% of 5 mole is 3.5 mole)

Thus, equilibrium constant Kc =2

3

3

22

][NH

]][H[N

=3

2[1.75/0.2][5.25/0.2]

[1.5/0.2]

=3

2

[8.75][26.25]

[7.5]

Kc = 2813.67 (mole/lit)2


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6. At a temperature of 327C, if the value of Kp is 3.2 102 atm for the reactionNH2COONH4(s) 2 NH3(g)+ CO2(g), calculate the value of Kc.

According to the relationship between Kp and Kc

Kp = Kc (RT)n

Kc = Kp (RT)n

= 3.2 102 (0.082 600) (30)

= 320 (49.2)

3

=3

[49.2]

320

Kc = 2.686 103

7. On heating solid ammonium carbamate at 400 K the value of Kp was found to be600 atm3 at equilibrium. Calculate the total pressure in the system atequilibrium.

NH4COONH2(s) 2 NH3(g) + CO2(g) Initial mole 1 0 0

Euili. Mole 1x 2 x xHere ammonium carbamate is solid, hence its mole is neglected in calculation ofequilibrium constant.Total mole 2x + x = 3x

Mole fraction -3x

2x

3x

x

Partial pressure of NH3= Mole fraction Total Pressure

=3x

2xP (Total pressure is assumed to be P)

Partial pressure of CO2= Mole fraction Total Pressure

= 3x

x

P (Total pressure is assumed to be P)

Equilibrium constant Ke =]COONH[NH

][CO][NH

24

2

2

3

Kp = (pNH3)2pCO2

600 = (3x

2xP)2

3x

xP

= 2P9

4 P

3

1

P3 =4

27600

P3 = 150 27 P = 15.9

8. On heating solid ammonium carbamate at 400 K the value of Kp was found to be600 atm3 at equilibrium. Calculate the total pressure in the system at

equilibrium. Reaction : (NH4)2CO3(s) 2 NH3(g) + CO2(g)+ H2O(g)


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9. The equilibrium constant of the reaction ( ) ( ) ( )2 g 2 g gH + I 2HI at 700 Ktemperature is 54.8. The concentration of HI at equilibrium is 0.5 mol lit 1.Suppose reaction is carried by taking HI(g) find concentrations of H2(g)and I2(g)at equilibrium. (Text. Book Illus. - 3 Page 100)

10. PCl5(g) can be obtained by chemical reaction between PCl3(g)and Cl2(g)in closedvessel. The concentrations of PCl3, Cl2and PCl5in this reaction at 500 K are 1.59

m, 1.59 m and 1.41 m respectively, calculate equilibrium constant of this reaction.(Text. Book Illus. - 4 Page 100)

11. Ethyl acetate is obtained by reaction of ethanol with acetic acid in presence of H+ions. Suppose 1 mole acetic acid and 0.18 mole ethanol are taken in this reaction.At equilibrium 0.171 mole Ethyl acetate is obtained. Calculate equilibrium

constant of this esterification reaction. (Text. Book Illus. - 5 Page 101)

12. The value of equilibrium constant for the reaction 2NOCl(g) 2NO(g)+ Cl2(g)isobtained as 0.033 atm at 1060 K temperature, then what will be the value of Kcforthis reaction ? (Text. Book Illus. - 6 Page 101)

13.

The value of Kc was found to be 1.6 103

at 127C at equilibrium for thefollowing reaction. If the reaction is carried out in a closed vessel of capacity 5

liter taking 50 mole of each reactants, calculate the concentration of CO and CO 2at the equilibrium.

H2O2(g) + CO(g) CO2(g)+ H2O(g)

H2O

2(g) + CO

(g) CO2(g) + H2O(g) Initial mole : 50 50 0 0

Equili. Mole : 50x 50x x x

Conc. Mole lit1: (50x)/5 (50x)/5 x/5

x/5

Equilibrium constant Kc for the above reaction is

Kc =][CO]O[H

O]][H[CO

22

22

Kc =x)/5]-x)5][(50-[(50

[x/5][x/5]

=2

2

)50(

x

x

40 =)50(

x

x

40 (50x) = x

2000 = x + 40x

x =2000

41

x = 48.78 mole

x/5 = 48.78/5 = 9.756 mole/lit


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50 x = 50 48.78 = 1.22 mole

(50x)/5 = 1.22/5 = 0.244 mole/lit

14. The value of Kc was found to be 1.6 103 at 127C at equilibrium for thefollowing reaction. If the reaction is carried out in a closed vessel of capacity 5liter taking 50 mole of each components (i.e. reactants and products), calculatethe concentration of CO and CO2at the equilibrium.

H2O2(g) + CO(g) CO2(g)+ H2O(g)

15. The equilibrium constant of the reaction CO(g)+ H2O(g) CO2(g)+ H2(g), Kcis4.25 at 810 K. Suppose, if this reaction begins with reactants havingconcentrations 0.1 mol lit1 or M, then calculate the concentrations of CO(g),H2O(g), CO2(g)and H2(g)at equilibrium at 810 K. (Text. Book Illus. - 7 Page 102)

16. The value of Kcfor the reaction H2(g)+ I2(g)2 HI(g) is found to be 57.0 at 700K temperature. Suppose in reaction mixture the concentrations of reactantsand products at some definite time are as follows : [H 2] = 0.05M. [I2] = 0.10Mand [HI] = 0.20 M. Determine in which direction the reaction will proceed ?

(Text. Book Illus. - 8 Page 103)

17. 6.9 gram N2O4is taken in a closed vessel of 0.5 litre volume. Temperature is 400K. If N2O4(g) undergo following reaction N2O4(g) 2 NO2(g) and attain

equilibrium state. At equilibrium total pressure is 9.15 bar, calculate Kc, Kp

and partial pressures at equilibrium. R = 0.082 lit atm mol1deg1. (Text. BookIllus. - 9 Page 104)

18. The value of equilibrium constant in reaction of phosphorylation of glucoseduring glycolysis was obtained 3.8 103at 298 K temperature. Calculate the

value of G for this reaction and mention your opinion about this reaction.

G

=

2.303 R T log K= 2.303 8.314 103 298 log (3.8 103)

= 2.303 8.314 103 298 [ log 3.8 + log 103]

= 2.303 8.314 103 298 [ 0.5798 +(3)]

= 2.303 8.314 103 298 [ 2.4202]

G = +13.809 KJoule/mol

Since value of G is positive, the reaction will not be spontaneous.

19. 2.0 moles of PCl5 were introduced in a 2 lit container and heated at 625 K toestablish equilibrium when 60% of PCl5was dissociated into PCl3and Cl2. Findthe value of equilibrium constant. ( Equilibrium constant K = 0.9)

20.

At a certain temperature and total pressure of 105

Pa, iodine vapour contains 40%by volume of I atoms. CalculateKpfor the following equilibrium.

I2(g) 2 I(g)21. Reaction between N2and O2takes place as follows.

2N2(g) + O2(g)2N2O (g)If a mixture of 0.482 mol N2and 0.933 mol of O2is placed in a 10 L reaction vesseland allowed to form N2O at a temperature for which Kc= 2.0 1037, determinethe composition of equilibrium mixture.


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22. A mixture of 1.57 mol of N2, 1.92 mol of H2and 8.13 mol of NH3 is introducedinto 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 1.7102. Is the reaction mixture atequilibrium ? If not, what is the direction of the net reaction ?

23. At 700 K, equilibrium constant for the reaction H2(g)+ I2(g)2HI(g) is 54.8. If 0.5mol L1of HI(g)is present at equilibrium at 700 K, what are the concentration ofH2(g)and I2(g)assuming that we initially started with HI (g)and allowed it to reachequilibrium at 700 K ?

24. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. Afterequilibrium was attained, concentration of PCl5 was found to be 0.05 mol L1. Ifvalue ofKcis 8.3 103, what are the concentrations of PCl3and Cl2at equilibrium ?

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

25. One of the reaction that takes place in producing steel from iron ore is thereduction of iron(II) oxide by carbon monoxide to give iron metal and CO2

FeO(s)+ CO(g) Fe(s) + CO2(g) ; Kp = 0.265 atm at 1050 KWhat are the equilibrium partial pressures of CO and CO2at 1050 K if the initialpartial pressures are :pCO= 1.4 atm andpCO2 = 0.80 atm ?

26. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2in equilibriumwith solid carbon has 90.55% CO by mass. Calculate Kc for this reaction at theabove temperature.

C(s)+ CO2(g) 2 CO(g)

27. Calculate a) Gand b) the equilibrium constant for the formation of NO2fromNO and O2at 298 K for the reactionNO(g)+ O2(g)NO2(g). Where Gf(NO2)= 52.0 KJ/mol, Gf(NO) = 87.0 KJ/mol.

28. The equilibrium constant for the following reaction is 1.6105at 1024 KH2(g) + Br2(g) 2 HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into asealed container at 1024K.


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Type II : Strong acid and Strong base

Remember :

(1) Strong acid : HCl, HNO3, H2SO4

(2) Strong base : NaOH, KOH

(3) Strong acid and strong base are completely dissociate.

(4) (i) [H3O+][OH] = Kw= 1014

(ii) pH = log [H3O+] = log [H+]

(iii) pOH = log [OH]

(iv) pH + pOH = 14

(v) Molarity (M) =1000

V in ml

Wt in gm

Mol. wt

Where M = Molarity of solution in mole/lit

V = Volume of solution in ml

W = Weight of solute in gmm = Molar mass of solute in gm/mole

[A] Molarity pH / pOH

29. 1 gm NaOH is dissolved in 100 ml solution. Calculate(i) [OH ] , (ii) [H+](iii) pH and (iv) pOH of solution.

Ans : Molarity of NaOH solution M =1000

V

Wt in gm

Mol. wt.

=1000

100

1

40

Molarity (M) = 0.25 MNaOH is strong base, therefore completely ionized

NaOH Na+ + OH

0.25 M 0.25 M 0.25 M

[OH] = 0.25 M

pOH = log [OH]

= log [2.5101]

= [0.3979 + (1)]

pOH = 0.6021

Now [H3O+] [OH] = 1014

[H+] =

1410

OH

=[ ]

1410

0 25.

=100

25 1014


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[H+] = 4 1014M

Now pH = log [H3O+]

= log [4.0 1014]

= [0.6021 14]

pH = 13.3979

30. Calculate [H+], [OH]pH & pOH of 0.06 M H2SO4solution.Ans : H2SO4is strong acid, therefore it is completely ionized

H2SO4 2H+ + SO42

0.06 M 2(0.06) 0.06 M

[H+] = 0.12 M

Now, [H3O+] [OH] = 1014

[OH] =

+

14

3

10

H O = [ ]

1410

0 12.

= 8.33 1014

pH = log [H3O+]

= log [0.12]

= [1 .0792]

= 1.0 0.0792

pH = 0.9208

pOH = 14 0.9208

pOH = 13.0792

31. Calculate [H+], [OH], pH & pOH of 0.002 M HNO3solution.32. Calculate [H+], [OH], pH & pOH of an aq. solution of 0.08 M H2SO4solution.33. (3.65 102) gm Hydrochloric acid is dissolved in 500 ml of solution. Find out of

its pH & pOH.

34. In one drink, the concentration of hydrogen ion is found to be 4 103. Whatwill be the value of its pH ? Also calculate the value of pOH.(Text. Book Illus. -11 Page 112)

35. Calculate pH and pOH of solution containing 0.03 M NaOH. (Text. Book Illus. -12 Page 112)

36. Calculate pH and pOH of following solutions. (Text. Book Exe 4 (1) Page 126)(a)0.1 M HCl, 0.1 M H2SO4, 0.1 M HNO3(b)0.1 M NaOH, 0.1 M KOH, 0.1 M Ba(OH)2

37. Calculate pH of following solutions. (Text. Book Exe 4 (2) Page 126)(a)3.65 gram HCl in 250 ml solution(b)9.80 gram H2SO4in 500 ml solution(c)1.6 gram NaOH in 250 ml solution(d) 11.2 gram KOH in 500 ml solution


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[B] pH / pOH Molarity

38. Calculate weight of H2SO4required to prepare 100 ml aq. solution having 13.6pOH.

Ans : pOH = 13.6

pH = 0.4

log [H3O+] = 0.4 log [H3O+] = 0.0 1.0 + 1.0 0.4

= 1 .6 [H3O+] = antilog ( 1 .6)

[H3O+] = 3.981 101

[H3O+] = 0.3981 M

H2SO4is strong electrolyte, therefore it is completely ionized.

H2SO4 2 H+ + SO42

? 0.3981

0 39812

.=

[H2SO4] =0 3981

2

. = 0.19905 M

Now,

Molarity of H2SO4solution M =1000

V

Wt in gm

Mol. wt.

0 3981

2

. =

1000

100

Wt in gm

98

Wt in gm =0 3981

2

.

100

1000 98

= 0.19905 100

1000 98

Wt in gm = 1.9507 gm

39. Calculate quantity of NaOH required to prepare 100 ml aq. solution having pH = 10.(Ans : [OH] = 104M, [NaOH] = 104M , W = 4.0 104gm )

Type III : pH and pOH of diluted solution

Remember :

(1) When any acid solution is diluted by adding water, [H3O+] ion decreases but pH ofsolution increases.

(2) When any base solution is diluted by adding water, [OH] ion decreases hencepOH of solution increases; but pH decreases.


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40. 10 ml 0.1 M NaOH solution is diluted to 50 ml by adding water. (1) Calculate pH oforiginal solution (2) pH of diluted solution (3) Change in pH.

Ans. pH of original solution :

NaOH is strong electrolyte, therefore completely ionized.

NaOH Na+ + OH

0.1 0.1 0.1

pOH = log [OH]

pOH = log [0.1]

= [1.0]

pOH = 1

pH = 13.0

pH of diluted solution :

M1V1 = M2V2

0.1 10 = M2 50

M2 =0 1 10

50

.

M2 = 0.02 M

Now, NaOH is strong electrolyte.

NaOH Na+ + OH

0.02 0.02 0.02

pOH = log [OH]

= log [0.02]

= log [2.0 102]

= [0.3010 2.0]

pH = 12 . 3010

Change in pH = 13.0 12.3010 = 0.6990

41. What would be the change in pH of 0.02 M H2SO4, which is diluted to 100 times.(Ans. Change in pH = 1.3979 3.3979 = 2)

42. Calculate pH of solution containing 3.65 gm HCl per lit. How can you increase pHof this solution by 1 unit using water.

43. Calculate pH of solution containing 4.9 gm H2SO4per lit. How much water shouldbe added to increase pH of this solution by 3 unit ?

Type IV : pH and pOH of Highly diluted solution

44. Calculate pH of solution prepared by diluting 1 ml 0.1 M HCl solution to 10000 lit.Ans : 1 ml = 103lit

10000 lit = 104lit


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Molarity of HCl solution after dilution :

M1V1 = M2V2

0.1 103 = M2 104

M2 =3

4

0 1 10

10

. = 108M

HCl is strong electrolyte, hence it will completely ionized.HCl H+(aq) + Cl(aq)

108M 108M

This concentration of H+ion (i.e. [H3O+]) is very less, hence [H3O+] ion produced dueto self ionization of water should be also considered.

2 H2O H3O+(aq) + OH(aq)

According to self ionization of water [H3O+][OH] = 1.0 1014

[H3O+] = [OH] = 1.0 107M

Total [H3O+] = (1.0 108) + (1.0 107)

= (1.0 108) + (10.0 108)

Total [H3O+] = 11.0 108M

= 1.1 107M

Therefore,

pH = log [H3O+]

= log [1.1 107]

= [0.0414 7.0000]

pH = 6.9586

45. Calculate pH of solution prepared by diluting 1ml 0.02 M NaOH solution to 10000lit.

46. 0.1 ml of 0.001 M HCl solution is diluted with water to make 10 litre solutions.Calculate pH of this dilute solution. (Text. Book Illus. - 13 Page 113)

Type V : pH and pOH of mixture of Strong acid and Strong base

47. Find out the pH of a mixture of 50 ml 0.03 M HCl and 60 ml of 0.02 M NaOH.Ans : Mole of H3O+before mixing :

1000 ml solution contains 0.03 mole HCl

50 ml solution contains ?

=50 0 03

1000

.

= 1.5 103mole HCl

Now, HCl is strong electrolyte

HCl H+ + Cl

1.5 103 1.5 103 1.5 103


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Mole of OHbefore mixing :

1000 ml solution contains 0.02 mole NaOH

60 ml solution contains ?

=60 0 02

1000

.

= 1.2 103mole HCl

Now, NaOH is strong electrolyte

NaOH Na+ + OH

1.2 103 1.2 103 1.2 103

Concentration of H3O+OR OHion in Mixture :

When acid and base are mixed, neutralization takes place according to followingreaction. According to given reaction, 1 mole of OHis neutralized by 1 mole H+ion.Hence 1.2 103mole OHion is neutralized by 1.2 103mole H+ion, when NaOH ismixed with HCl.

H+(aq) + OH(aq) H2O(1)

1 mole 1 mole 1 mole1.5 103 1.2 103 1.2 103

Un neutralized mole of H+ = (1.5 103) (1.2 103)

= (0.3 103)

= 3.0 104mole H+ion

Now,Total volume of solution = (50 + 60) ml = 110 ml

110 ml mixture of acid and base contains 3.0 104mole

1000 ml mixture of acid and base contains ?

=41000 3 0 10

110

.

= 2.727 103mole/lit

OR Molarity =Mole

lit =

43 10

0 110. = 2.727 103mole/lit

pH of Mixture :

pH = log [H3O+]

= log [2.727 103]

= [0.4357 3.0]

pH = 3 0.4357

pH = 2.5643

48. Calculate pH of a mixture of solution containing 24 ml of 0.1 M HCl and 25 ml 0.1M NaOH.

49. If 20 ml 0.15 M NaOH solution is added to a 10 ml 0.1 M H 2SO4solution, whatwill be the pH of the resulting mixture ? (Text. Book Exe 4 (13) Page 127)


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50. What would be pH of solution when following solution are added to 10.0 ml 0.1MHCl solution ?(a) 5.0 ml 0.1 M NaOH (pH = 1.4771)(b) 9.0 ml 0.1 M NaOH (pH = 2.2788)(c) 9.9 ml 0.1 M NaOH (pH = ? )(d) 10.0 ml 0.1 M NaOH (pH = 7.0)(e) 10.1 ml 0.1 M NaOH (pH = ? )

(f) 10.2 ml 0.1 M NaOH (pH = 10.9956)

51. If 20 ml 0.15 M HCl solution is added to 10 ml 0.1 M Ba(OH) 2solution what willbe the pH of the resulting mixture ? (Text. Book Exe 4 (14) Page 127)

52. Calculate the pH of the resultant mixtures.(a) 10 mL of 0.2 M Ca(OH)2+ 25 mL of 0.1 M HCl(b) 10 mL of 0.01 M H2SO4+ 10 mL of 0.01 M Ca(OH)2(c) 10 mL of 0.1 M H2SO4+ 10 mL of 0.1 M KOH

Type VI : Weak acid and Weak baseRemember :

(1) Weak acid : CH3COOH, HCOOH, C6H5COOH, C6H5OH etc.

(2) Weak base : NH3, NH4OH, CH3 NH2, C6H5NH2etc.

(3) For any weak acid HA

HA(aq)+ H2O(l) H3O+(aq) + A(aq)

Ka =+

3[H O ][A ]

[HA] =

+ 23[H O ]

C =

[ C][ ]

[1 ]C

C=

2[ C]

[1 ]

[H3O+] = C = aK C

= aK

C

Where C = M =1000

V in ml

Wt in gm

Mol. wt

C = Molarity of weak acid solution in mole/lit

= Degree of dissociation

Ka = Dissociation constant

(4) For any weak base MOH

MOH(aq) + H2O(l) M+(aq) + OH(aq)

Kb =+[M ][OH ]

[MOH] =

2[OH ]

C =

[ C][ ]

[1 ]C

C =

2[ C]

[1 ]

[OH] = C = bK C

= bK

C


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Where C = M =1000

V in ml

Wt in gm

Mol. wt

C = Molarity of weak base solution in mole/lit

= Degree of dissociation

Kb = Dissociation constant

(5) pH = log [H3O+] = log [H+]

pOH = log [OH]

53. If the value of ionization constant Ka = 1.28 10 10for an aq. solution of 0.05 MC6H5OH. Calculate the concn. of H3O+and pH of solution.

Ans : C6H5OH + H2O C6H5O + H3O+

[H3O+] = aK C

= 101 28 10 0 05. .

= 100 064 10.

= 126 4 10.

= 66 4 10.

[H3O+] = 2.5298 106M

pH = log [H3O+]

= log [2.5298 106]

= [0.4031 6.0]

= 6.0 0.4031

pH = 5.5969

54. Find out pH of 0.02M methyl amine solution Ionization constant of methylamine is 5.0 104.

CH3 NH2 + H2O CH3 NH3+ + OH

[OH] = bK C

= 45 0 10 0 02. .

= 40 1 10.

= 610 10

=3

10 10

[OH] = 3.1623 103M

pOH = log [OH]

= [0.5000 3.0000]

pOH = 2.5000

pH = 11.5


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55. At equilibrium, 0.02 M CH3NH2 ionizes to 1.5 % Calculate the value of itsionization constant.

Ans : CH3NH2 + H2O CH3 N+H3 + OH

Initial mole 0.02 0 0

At equili 0.02 0.0003

=0.0197

0.02 0.0003 0.0003

[OH] = 0.0003 M = 3.0 104M

Kb = 2[OH ]

C

=

243 0 10

0 02

.

.

Kb = 4.5 106

OR = 1.5 % =1 5

100

. = 0.015

Kb =

2[ C][1 ]

=

2[(0.015) 0.02]

[1 0 015]. But < < 1 11

= (0.015)2 0.02 = 4.5 106

56. If 0.02 M CH3COOH ionizes up to 2% at 25C. Calculate its ionization constant.57. 10 lit. of water contains 1.7 g NH3. Calculate its pOH Kb= 1.77 105.58. Calculate ionization constant of the acid when a solution of 0.043 M of HOOH has a

pH = 4.3759. NH3+ H2ONH4++ OH equilibrium exists in aqueous solution of ammonia.

The equilibrium constant Kbof NH3 is 1.77 105. Calculate the concn. of NH3,NH4+, OHand H3O+in 0.01 M solution of ammonia.

(Ans : [OH] = 4.207 104M, [NH4+] = 4.207 104 M, [NH3] = 9.5793 103M,[H3O+] = 2.378 1011M)

60. Calculate [H3O+], [CH3COOH], pH, pOH and pKw of 0.1 M CH3-COOH solution.Kaof acetic acid is 1.74 105. (Text. Book Illus. - Page 114)

61. pH of 0.1 M monobasic acid HA is 4.40. Calculate the value of [H3O+], [A], [HA],Kaand pKain the solution at equilibrium.

62. Calculate the values of [NH4+], [OH], pH and pOH of 0.1 M NH4OH solution atequilibrium. The ionization constant of NH4OH is 1.8 105. (Text. Book Illus. - 14Page 115)

63. The ionization constant of dimethyl amine [(CH3)2NH] is 5.4 104. If the initialconcentration of dimethyl amine is 0.02 M, calculate [OH], [(CH3)2NH], [H3O+],pH and pOH in solution at equilibrium. (Text. Book Illus. - 15 Page 116)

64. Calculate pH of 250 ml solution containing 6 gram acetic acid. Dissociationconstant of acetic acid is 1.8 105.(Text. Book Exe 4 (3) Page 126)


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65. Calculate pH of 500 ml ammonium hydroxide solution containing 7 gramammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 105. (Text. Book Exe 4 (4) Page 126)

66. The dissociation constant of benzoic acid is 6.5 105at 298 K. What will be thepH of its 0.15 M solution and its [H3+O] ion concentration ? (Text. Book Exe 4 (5)Page 126)

67. The dissociation constant of dimethyl amine is 5.4 105at 298 K. What will bethe pH of its 0.25 M solution and [H3O+] ion concentration ? (Text. Book Exe 4 (6) Page 127)

68. The ionization constant of acetic acid is 1.74105. Calculate the degree ofdissociation of acetic acid in its 0.05 M solution. Calculate the concentration ofacetate ion in the solution and its pH.

69. 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 andits pKa.

70. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculatethe pH of the solution and thepKaof bromoacetic acid.

71. The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionizationconstant and pKb.

Type VII : Relation between Kaand Kb

Remember :

(1) Ka Kb = Kw

Where Ka = Dissociation constant of acid OR conjugate acid

Kb = Dissociation constant of conjugate base OR base

(2) pKa+ pKb= pKw = 14

72. The ionization constant of acetic acid is 1.7 105. Calculate ionization constantof its conjugate base ?Also calculate pKaand pKbof acid and its conjugate base.

Ans : CH3-COOH + H2O CH3-COO + H3O+Acid Base con. base con. acid

Ka Kb = KwWhere Ka = Dissociation constant of acid

Kb = Dissociation constant of conjugate base of acid Kb = w

a

K

K =

14

5

1.0 10

1.7 10

= 5.882 1010

Now pKa = log Ka= log (1.7105)= [ log 1.7 + (5)]= [0.2304 5]= 5 0.2304= 4.7696


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Now, relation between pKaand pKbof acid and its conjugate base (vice versa) is pKa+ pKb = 14

pKb = 14 pKa = 9.2304

73. The ionization constant of formic acid (HCOOH) is 1.8 104. Calculateionization constant of its conjugate base format ion (HCOO) ?Also calculate pKa

and pKbof acid and its conjugate base. (Text. Book Illus. 16 Page 116)

74. The ionization constant of NH3is 1.8 105. Calculate ionization constant of itsconjugate acid ?Also calculate pKaand pKbof base and its conjugate acid. (Text.Book Illus. - Page 116)

75. If dissociation constant of aniline is 4.27 1010 at 298 K, what will be thedissociation constant of its conjugate acid ? Kw= 1.04 1014. (Text. Book Exe 4(15) Page 127)

76. If dissociation constant of acetic acid is 1.76 105 at 298 K, what will bedissociation constant of its conjugate base ? Kw= 1.04 1014. (Text. Book Exe

4 (16) Page 127)77. The ionization constant of HF, HCOOH and HCN at 298K are 6.8104, 1.8104

and 4.8109respectively. Calculate the ionization constants of the correspondingconjugate base.

78. What is the pH of 0.001 M aniline solution ? The ionization constant of aniline4.27 1010. Calculate the degree of ionization of aniline in the solution. Alsocalculate the ionization constant of the conjugate acid of aniline.

Type VIII : Common ion Effect on Dissociation constant of Weakacid and Weak base

79. What will be the change in pH of 0.1 M CH3COOH acid if 0.1 M CH3COONa isadded to its solution ? (pKaof CH3COOH = 4.74) (Text. Book Illus. 17 Page 118)

Ans : pH of solution in absence of common ion :pKa = 4.74

log Ka = 4.74log Ka = 4.74

= 4.0 0.74= 4.0 1.0 + 1.0 0.74

= 5.0 + 0.26

Ka = antilog (5 .26)= 1.82 105

Now, [H3O+] = aK C

= 51.82 10 0.1

= 1.82 103

= 1.349 103


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pH = log [H3O+]= log [1.349 103]= [ log 1.349 + (3)]= [0.1300 3]= 2.87 _____(1)

Now, if 0.1 M CH3COONa is added then it is strong electrolyte.

CH3COONa

CH3COO

+ Na+

0.1 0.1 0.1While CH3COOH is weak electrolyte and since value of Kais very small, degree ofdissociation is very less. So after addition of 0.1 M CH3COONa in aq. solution ofCH3COOH its equilibrium will be as follows.

CH3COOH + H2O CH3COO + H+

Initial mol 0.1 0.1 0Equi. mol 0.1 x 0.1+x x

0.1 0.1Therefore, Kaof CH3COOH will be

Ka =

+3 3

3

[CH COO ][H O ]

[CH COOH]

1.82 105 =+

3[0.1][H O ]

[0.1]

[H3O+] = 1.82 105

pH = log [1.82 105]= [log 1.82 + (5)]= [0.2601 5]= 5 0.2601= 4.74 _____(2)

From (1) and (2) it is clear that pH of solution is increased from 2.87 to 4.74; hence

concentration of [H3O+] decreases. Thus, in presence of common ion dissociation ofweak acid decreases.

80. What will be the change in pH of 0.1 M weak base NH4OH if 0.1 M NH4Cl isadded to the solution ? (Kbof NH4OH = 1.77 105) (Text. Book Illus. 18 Page118)

81. The ionization constant of phenol is 1.01010. What is the concentration ofphenolate ion in 0.05 M solution of phenol ? What will be its degree ofionization if the solution is also 0.01 M in sodium phenolate solution ?

82. The first ionization constant of H2S is 9.1108. Calculate the concentration ofHS ion in its 0.1M solution. How will this concentration be affected if thesolution is in 0.1 M HCl solution ? If the second dissociation constant of H2S is

1.21013, calculate the concentration of S2under both conditions.

83. 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(a) 0.01 M (b) 0.1 M HCl ?


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84. The ionization constant of dimethyl amine is 5.4 104. Calculate its degree ofionization in its 0.02 M solution. What percentage of dimethyl amine is ionizedif the solution is also 0.1 M in NaOH ?

85. The ionization constant of propanoic acid is 1.32 105. Calculate the degree ofionization of the acid in its 0.05M solution and also its pH. What will be itsdegree of ionization if the solution is in 0.01 M HCl solution also ?

Type IX : Hydrolysis of Salt

Remember :

(1) For a salt of Strong base and Weak acid : (e. g. CH3COONa. HCOOK, C6H5COONa,Na2CO3, K2CO3etc.

Kh = w

a

K

K =

2OH

C

(2) For a salt of Weak base and Strong acid : (e. g. NH4Cl, C6H5NH3+Cletc )

Kh = w

b

K

K =

2+H

Co

86. Find out pH of a solution of 0.025 M CH3COONa. Ionization constant ofCH3COOH is 1.75 105. Also calculate Hydrolysis constant of CH3COONa.

Ans : Kh = w

a

K

K =

2OH

Co

Kh =14

5

1 0 10

1 75 10

.

.

= 5.714 1010

2

OH0.025

=14

51 0 101 75 10

..

[OH]2 =14

5

1 0 10

1 75 10

.

.

0.025

= 0.571 109 0.025

= 0.0143 109

[OH]2 = 1.43 1011

[OH] = -111.43 10

= 1.43 106

[OH] = 3.7815 106M

pOH = log [OH]

= log [3.7815 106]

= [0.5777 6.000]

pOH = 6.0 0.5777

= 5.4223

pH = 8.5777


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87. Calculate pH of an aq. solution of 0.02 M CH3COONa. Ionization constant ofCH3COOH is 1.75 105.

88. Calculate pH of an aq. solution of 0.1 M Sodium acetate solution. Ionizationconstant of CH3COOH is 1.8 105. (Text. Book Illus. 19 Page 119)

89. Calculate hydrolysis constant and pH of a solution of 0.1 M sodium acetate.Dissociation constant of acetic acid is 1.8 105 at 298 K and ionic product ofwater is 1.04 1014. (Text. Book Exe 4 (11) Page 127)

90. Calculate hydrolysis constant and pH of 0.30 M NH4Cl solution. The dissociationconstant and ionic product of water are 1.8 105and 1.04 1014respectively at298 K. (Text. Book Exe 4 (12) Page 127)

91. Ionization constant of weak base NH4OHis 1.75 105.Calculate pH of 0.175 MNH4Cl solution. Also calculate Kh. (Ans : pH = 5, Kh= 5.714 1010)

92. The ionization constant of nitrous acid is 4.5104. Calculate the pH of 0.04 Msodium nitrite solution and also its degree of hydrolysis.

93.

A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate theionization constant of pyridine.

94. The ionization constant of chloro acetic acid is 1.35103. What will be the pH of0.1 M acid and its 0.1M sodium salt solution ?

Type X : Buffer solution

Remember :

(1) pH = pKa + log[ ]Salt[Acid]

(2) pOH = pKb + log [ ]Salt[Base]

95. To prepare one acidic buffer solution 0.125 M sodium acetate is added to 0.25 Msolution of acetic acid. If dissociation constant of acetic acid is 1.8 105, whatwill be the pH of this buffer solution ? (Text. Book Exe 4 (7) Page 127)

Ans : pH = pKa + log[ ]Salt[Acid]

= log Ka + log[ ]Salt[Acid]

= log (1.8 105

) + log

[ ]0.125

[0.25] = [log 1.8 + (5)] + log ( )= [0.2553 5] + (log 1 log 2)= 5 0.2553 + 0 0.3010

pH = 4.4437


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96. To prepare one basic buffer solution, 0.250 M NH4Cl is added to a solution of0.125 M ammonium hydroxide. If dissociation constant of ammonium hydroxideis 1.8 105, what will be the pH of this buffer solution ? (Text. Book Exe 4 (8)Page 127)

Type XI : Solubility product of Sparingly soluble saltsRemember :

(1)Type of Salt Solubility Product Some example

1:1 Ksp= S2(mol/lit)2 AgCl, AgI, HgS, ZnS, BaSO4,CaSO4, PbSO4

1:2 CaF2, BaF2, PbCl2

2:1Ksp= 4 S3(mole/lit)3

Ag2S, Ag2CrO4

1:3 Al(OH)3, Cr(OH)3

3:1Ksp= 27 S4(mole/lit)4

Ag3PO4

2:3 Bi2S3, As2S3, Sb2S33:2

Ksp= 108 S5(mole/lit)5Zr3(PO4)2

(2) Solubility (mol/lit) = S =1000

V in ml

Wt in gm

Mol. wt

Solubility in gm/lit = S Mol. Wt

Solubility in mg/lit = S Mol. Wt. 103

97. Find out solubility in water of PbSO4 in g/L whose solubility product is1.3 108. Mol. wt of PbSO4is 303 gm/mole. Also calculate (1) volume of solutioncontaining 25 mg PbSO4. (2) Wt of PbSO4dissolved in 234 ml.

Ans : PbSO4(s) Pb2+(aq) + SO42(aq)S S S

Ksp = S2

S = spK

= 81 3 10.

= 1 3. 104

S = 1.1402 104mole/L

Solubility in g/L = 1.1402 104 303

= 345.47 104

g/L= 3.4547 102g/L

OR S =1000

V in ml

Wt in gm

Mol. wt

1.1402 104 =1000

1000

Wt in gm

303

Wt in gm = 1.1402 104 303 = 3.4547 102g/L


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(1) S =1000

V in ml

Wt in gm

Mol. wt

1.1402 104 =1000

V in ml

0.025

303

V = 723.63 ml

(2) S =1000

V in ml

Wt in gm

Mol. wt

1.1402 104 =1000

234

Wt in gm

303

Wt in gm = 0.00808= 8.08 mg

98. The concentration of a saturated solution of Mg(OH)2 is 8.2 104 % (W/v).Calculate its Ksp(Molecular weight of Mg(OH)2is 58 g/mol)

Ans : Solubility of Mg(OH)2 = 8.2 104% (w/v)

= 8.2 104 gm/100 ml

Solubility in gm/lit = 8.2 103

gm/1000 ml

Solubility in mol/lit =38 2 10

58

. = 0.1414 103

S = 1.414 104mole/lit

OR S =1000

V in ml

Wt in gm

Mol. wt

S =1000

100

48.2 10

58

Now,

Mg(OH)2

Mg2+

(aq) + 2 OH

(aq) (1 : 2 type)S S 2 S

Ksp = 4 S3

= 4 (1.414 104)3

= 11.3086 1012

Ksp = 1.1309 1011(mole/lit)3

99. The concentration of Mg(OH)2in its saturated solution is found to be 8.2 104gm lit1at 298 K temperature. Calculate Kspof Mg(OH)2. (Mol. wt of Mg(OH)2is58 gm/mol) (Text. Book Illus. 20 Page 119)

100. Ksp of Ag2S is 1.0 1051. Molecular weight is 24.8 gm/mole. Calculate solubilityof Ag2S in mg/lit. Calculate solubility of Ag2S in mg in 400 ml solution.

101. A person drinks 2.5 lit waters every day, which is saturated with CaF2. Howmuch CaF2enters in the body every day ? (Mol. wt of CaF2is 78 gm/mole and Kspof CaF2is 1.7 1010) (Ans : S = 3.489 104M and 68.03 mg CaF2enters in bodyevery day)


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102. Solubility product of CaF2 is 1.7 1010. Calculate its solubility in water.Calculate vol. of saturated soln. containing 10 mg CaF2. Calculate wt of CaF2present in 786 ml.

103. Calculate Kspof Fe(OH)3whose solubility is 1.0 103M.104. 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.

105. The solubility product constant of Ag2CrO4and AgBr are 1.11012and 5.01013respectively. Calculate the ratio of the molarities of their saturated solutions.

106. What is the minimum volume of water required to dissolve 1 g of calciumsulphate at 298 K ? (For calcium sulphate,Kspis 9.110

6)

Type XII : Precipitations

Remember :

(1) IP = Ksp Solution is exactly saturated No pptn.

(2) IP > Ksp Solution is unsaturated No. pptn.

(3) IP > Ksp Solution concnis more than saturationpptn. will occur.

107. Will pptntakes place or not on mixing equal volumes of 2.0 104M BaCl2and5.0 103 M H2SO4? For BaSO4Ksp= 1.1 1010.

108. Whose precipitates will be obtained on addition of equal volumes of 2.0 102MH2SO4to water solutions containing 2.0 103M CaCl2and 5.0 103M BaCl2 ? Thesolubility product of CaSO4and BaSO4are respectively 2.4 105and 1.1 1010?

Ans : Calculation for CaSO4.

CaCl22 x 10-3

V ml

H2SO42 x 10-2

V ml

2 V ml

CaCl2 + H2SO4 CaSO4 + 2 HCl

[Ca2+

] after dilution :M1V1 = M2 V2

2.0 103 V = M2 2V

M2 = 1.0 103M

[CaCl2] = 1.0 103M

CaCl2 Ca2+(aq) + 2 Cl(aq)

1.0 103 1.0 103

[Ca2+] = 1.0 103

[SO42

] after dilution :M1V1 = M2 V2

2.0 102 V = M2 2V

M2 = 1.0 102M

[H2SO4] = 1.0 102M

H2SO4 2 H+(aq) + SO42(aq)

1.0 102 1.0 102

[SO42] = 1.0 102


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CaSO4 Ca2+

+ SO42

(1 : 1 type)

IP = [Ca2+] [SO42]

= [1.0 103] [1.0 102]

IP = 1.0 105

But, Kspof CaSO4is 2.4 105 IP < Ksp pptn will not occur.

BaCl25 x 10-3

V ml

H2SO42 x 10-2

V ml

2 V ml

BaCl2 + H2SO4 BaSO4 + 2 HCl

[Ba2+] after dilution :

M1V1 = M2 V2

5.0 103 V = M2 2V

M2 = 2.5 103M

[BaCl2] = 2.5 103M

BaCl2 Ba2+(aq) + 2 Cl(aq)

2.5 103 2.5 103

[Ba2+] = 2.5 103

[SO42] after dilution :

M1V1 = M2 V2

2.0 102 V = M2 2V

M2 = 1.0 102M

[H2SO4] = 1.0 102M

H2SO4 2 H+(aq) + SO42(aq)

1.0 102 1.0 102

[SO42] = 1.0 102

BaSO4 Ba2+

+ SO42

(1 : 1 type)

IP = [Ba2+] [SO42]

= [2.5 103] [1.0 102]

IP = 2.5 105

But, Kspof BaSO4is 1.1 1010 IP > Ksp pptn will occur.

BaSO4will be precipitate but CaSO4will not be precipitate.

109. 200 ml 1.0 103 M Pb(NO3)2 solution is mixed with 100 ml 3.0 103M NaClsolution. pptnof PbCl2will occur or not ? (Kspof PbCl2= 2.0 104)(Ans : IP = 6.667 1010, Therefore precipitation will not occur )

110.

Two solutions of (1) 100 ml 2.0 103

M Pb(NO3)2and (2) 100 ml 5.0 103

M H2SO4are mixed will there be precipitation of PbSO4? Kspof PbSO4is 1.3 108.

111. 20 ml 0.04 M NaCl solution is added to 10 ml 0.06 M AgNO3 solution. Ifsolubility product of AgCl is 1.8 1010at 298 K, predict whether precipitationwill occur or not ? (Text. Book Exe 4 (9) Page 127)

112. 20 ml 0.025 M K2CrO4 solution is added to 20 ml 0.05 M BaCl2 solution. Ifsolubility product of BaCrO4is 1.2 1010at 298 K, predict whether precipitationwill occur or not ? (Text. Book Exe 4 (10) Page 127)


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113. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate aremixed together. Will it lead to precipitation of copper iodate? (For cupric iodate

Ksp= 7.4108).

114. 250 ml solution contains 3.32 g Pb(NO3)2; will addition of 174 mg solid K2SO4to thissolution will precipitate PbSO4? (Kspof PbSO4is 1.3 108, Mol. wt of K2SO4is 174gm/mole and Pb(NO3)2is 332 gm/mol)

(Ans : IP = 1.6 104, IP > Ksp pptnwill occur)115. How many grams FeSO4should be added to 500 ml of 0.02 M NaOH solution to

precipitate Fe(OH)2? Ksp = 1.5 1015[Mol. wt of FeSO4= 152 gm/mol]

Ans : Fe(OH)2 Fe2+(aq) + 2OH(aq) (1 : 2 type)

S S S

Ksp = [Fe2+] [OH]2

[Fe2+] =sp

2

K

OH

= [ ]

15

2

1 5 10

0 02

.

.

=15

4

1 5 10

4 0 10

.

.

[Fe2+] = 3.75 1012

[FeSO4] = 3.75 1012 ( FeSO4Fe2+(aq)+ SO24(aq)) [Fe2+] = [FeSO4]

If [Fe2+] = [FeSO4] = 3.75 1012M, then IP = Ksp, hence pptnwill not occur. If [Fe2+] = [FeSO4] is slightly more than 3.75 1012M, then IP > Kspand pptnof

Fe(OH)2will occur.

S =1000

V in ml

Wt in gm

Mol. wt

3.75 1012=1000

500

Wt in gm

152

Wt in gm =123 75 10 500 152

1000

.

Wt in gm = 2.85 1010 gm

If we add FeSO4slightly more than 2.85 1010 gm, then IP > Kspand pptnofFe(OH)2will occur.

116. In a sample of water, the concn. of Fis 3.0 105M. How much solid CaCl2has tobe added to precipitate F? Kspof CaF2= 1.7 1010and Mol. wt of CaCl2is 111gm/mol.

117. Calculate minimum amount of NaOH required to precipitate Mg(OH)2from 500ml 0.1 M MgCl2Solution. (Ksp of Mg(OH)2is 2.8 1016& mol. wt of NaOH is40 g/mol).


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118. What is the maximum concentration of equimolar solutions of ferrous sulphateand sodium sulphide so that when mixed in equal volumes, there is no

precipitation of iron sulphide ? (For iron sulphide,Ksp= 6.31018)

119. The concentration of sulphide ion in 0.1 M HCl solution saturated withhydrogen sulphide is 1.0 1019M. If 10 mL of this is added to 5 mL of 0.04 Msolution of the following : FeSO4, MnCl2, ZnCl2 and CdCl2 in which of these

solutions precipitation will take place ?

Type XIII : Common Ion Effect on Solubility of Sparingly solublesalts

120. Find out the maximum amount in terms of gm of Zn(OH)2that can be dissolvedin 2 lit of 0.02 M NaOH solution. (Ksp for Zn(OH)2 is 4.5 1017). (Mol. wt. ofZn(OH)2is 99 gm/mol)

Ans : Solubility in water :

Zn(OH)2 Zn2+

(aq) + 2 OH

(aq) (1 : 2 type)S S 2 S

Ksp = 4S3

S = 3Ksp

4

=-17

34.5 10

4

= -173 1.125 10

=-183

11.125 10 S = -63 11.125 10

S = 2.2324 106mole/lit.

Solubility in 0.02 M NaOH solution :

NaOH Na+(aq) + OH(aq)

0.02 M 0.02 M

Suppose solubility of Zn(OH)2in 0.02 M NaOH solution is S mole/lit.

Therefore,

Zn(OH)2(s) Zn2+(aq) + 2 OH(aq)

S S 2 S + 0.02

Ksp = [Zn2+] [OH]2

Ksp = [S] [2S + 0.02]2

But 2S < < 0.02 (NaOH is strong electrolyte, while Zn (OH)2issparing soluble salt)

Ksp = [S] [0.02]2


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S =17

4

4 5 10

4 10.

= 1.125 1013mole/lit

Solubility in gm/lit = 1.125 1013 99

= 1.1137 1011gm/lit

Solubility in gm/2 lit = 2 1.1137 1011= 2.2275 1011gm/2 lit

OR S =1000

V in ml

Wt in gm

Mol. wt

1.125 1013 =1000

2000

Wt in gm

99

121. Calculate the maximum amount of AgCl (in terms of gm) that can be dissolved in2 lit solution of 0.02 M AgNO3 (Ksp of AgCl = 2.8 1010 and Mol. wt = 143.5

g/mol).122. The value of solubility product Ksp of sparingly soluble salt Mg(OH)2 is

1.8 1011M3at 298 K. If a solution of 0.1 M NaOH is added to it, what will bethe concentration of Mg(OH)2? Discuss result. (Text. Book Illus. 21 Page 122)

123. The ionization constant of benzoic acid is 6.46105andKspfor silver benzoate is2.51013. How many times is silver benzoate more soluble in a buffer of pH 3.19compared to its solubility in pure water ?

The best teachers will not give you somethingto drink, they will make you thirsty.

They will put you on a path to seek answers.

Equilibrium Numericals Sem II Echem 2

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