Question1
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Solution:
a. The vapour pressure will decrease due to increase in volume.
b. If the volume of the container is increased but the surface area remains the same then the rate of evaporation remains the same while the rate of condensation decreases.
c. When equilibrium is restored finally, the vapour pressure will have the same value as vapour pressure depends only on the temperature and not on the volume of the vessel.
Question2
2SO_{2}(g) + O_{2}(g) f 2SO_{3}(g)
Solution:
Kc = = = 12.23
âˆ´ Kc = 12.23.
Question3
p^{o} = 10^{5} Pa
pI_{2}(g) = 2.6 Ã— 10^{9} Pa (Experimentally)
K_{p} = Pa = 2.67 Ã— 10^{4} Pa.
Question4
Solution:
(i) K_{c} = (ii) K_{c} =
Question5
(i) 2NOCl(g) f 2NO(g) + Cl_{2}(g); Kp = 1.8 Ã— 10^{2} at 500 k
(ii) CaCO_{3}(S) f CaO(S) + CO_{2}(g); Kp = 167 at 1073 K
Solution:
(a) 2NOCl(g) 2NO(g) + Cl_{2}(g)
K_{p} = 1.8 Ã— 10^{2} at 500 K
For the reaction, Î”n = (2+1) â€“ 2 = 1
K_{p} = K_{c} (RT)^{n}
K_{C} = = = 4.4 Ã— 10^{4}
(b) CaCO_{3}(s) CaO(s) + CO_{2}(g)
K_{p} = 167 at 1073K
K_{c} =
The concentration of CaO(s) and CaCO_{3}(s) may be assumed to be constant i.e.
[CaO(s)] â€“ 1 and [CaCO_{3}(s)] = 1 and hence neglected so that K_{c} = [CO_{2}(g)]
Since the partial pressure of a gaseous component is proportional to its concentration
K_{p} = p[CO_{2}(g)]
K_{c} = = = 1.90.
Question6
K NO(g) + O_{3}(g) f NO_{2}(g) + O_{2}(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reactions?
Solution:
Kc for the reverse reaction is inverse of Kc for forward reaction.âˆ´ Kc (reverse) = = 1.6 Ã— 10^{15}.
Question7
Solution:
Concentration of solids and liquids are constant and taken as unity
[Solids] = 1; [liquids] = 1.
Therefore pure liquids and solids can be ignored while writing the equilibrium constant expression.
Question8
2N_{2}(g) + O_{2}(g) f 2N_{2}O(g)
If a mixture of 0.482 mol N_{2} and 0.933 mol of O_{2} is placed in a 10 L reaction vessel and allowed to form N_{2}O at a temperature for which Kc = 2.0 Ã— 10^{37}, determine the composition of equilibrium mixture.
Solution:
2N_{2}(g) + O_{2}(g) 2N_{2}O(g)
Initial moles: 0.482 + 0.933
Moles at equilibrium: (0.482 â€“ x) (0.933 â€“ x/2) x
Since the value of K is very small (2.0 Ã— 10^{37}) it indicates that the equilibrium is very much in favour of reactants and thus the value of x is very small. Therefore, the approximate concentrations at equilibrium may be written as
[N_{2}] = = 0.0482 mol L^{1 }
[O_{2}] = = 0.0933 mol L^{1 }
[N_{2}O] = or 0.1 x mol L^{1 }
Kc = = = 2.0 Ã— 10^{37 }
x = 6.6 Ã— 10^{20 }
âˆ´ [N_{2}O] = 0.1 Ã— 6.6 Ã— 10^{20} = 6.6 Ã— 10^{21} mol L^{1 }
[N_{2}] = 0.0482 mol L^{1 }
[O_{2}] = 0.0933 mol L^{1}.
Question9
When 0.087 mol of NO and 0.0437 mol of Br_{2} are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br_{2}.
Solution:
Amount of NO Br, formed at equilibrium = 0.0518 mol.
Initial amount of NO = 0.087 mol
Initial amount of Br_{2} = 0.0437 mol.
The reaction is
2NO + Br_{2(g)} 2NOBr 2NOBr 2NOBr
2 mols of NOBr is formed from 2 mols of NO
Therefore 0.0518 mol of NOBr is formed from 0.0518 mol of NO
So the amount of NO at the equilibrium = 0.087 â€“ 0.0518 = 0.0352 mol.
Amount of Br_{2} at the equilibrium
According to equation
Number of moles of Br_{2}, reacting with 0.0518 mol of NoBr = = 0.0259 mol.
Therefore Amount of Br_{2} at the equilibrium = 0.0437 â€“ 0.0259 = 0.0178 mol.
Question10
Solution:
Initial Pressure 0.2 atm.
At equilibrium 0.2 p p/2 p/2
(0.04 atm) 0.8 atm 0.8 atm
Kp = = = 400.
Question11
Solution:
[H_{2}] = M = 0.096 M
[N_{2}] = M = 0.079 M
[NH_{3}] = M = 0.407 M
Q_{c} = = = 2370
since Qc > Kc (equilibrium constant) the reaction is said to proceed in the reverse direction.
Question12
K_{e} = replace [H2O] by [H2O]^{6} replace [O2] by [O2]^{5}
Write the balanced chemical equation corresponding to this expression.
Solution:
K_{e} =
Hence chemical equation is
4NO + 6H_{2}O 4NH_{3} + 5O_{2}.
Question13
Solution:
H_{2}O (g) + CO (g) H_{2 }(g) + CO_{2} (g)
Initial moles of H_{2}O and CO taken = 1 mole each.
% dissociation of H_{2}O = 40%
Number of moles of H_{2}O dissociated =40/100 Ã—1= 0 .4 moles
Number of moles of CO at equilibrium = 1 0.4 = 0.6
Number of moles of H_{2}O at equilibrium = 1  0.4 = 0.6
No. of moles of H_{2} formed = 0.4
No. of moles of CO_{2} formed = 0.4
Therefore [H_{2}O] = 0.6/10 =0.06mol/L
[CO] = 0.6/10 = 0.06mol/L
[H_{2}] = 0.4/10 = 0.04mol/L
[CO_{2}] = 0.4/10 = 0.04mol/L
Hence,
= 4/9 = 0.44
Question14
H_{2}(g) + I_{2}(g) f 2HI(g) is 54.8. If 0.5 mol L^{1} of HI(g) is present at equilibrium at 700 K, what are the concentration of H_{2}(g) and I_{2}(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Solution:
For the equilibrium,
H_{2(g)} +I_{2(g)} 2HI_{(g)}
The value of equilibrium constant K is 54.8.
Hence, if we start with HI and allow the reaction to reach equilibrium then the equilibrium constant K' = 1/K = 1/54.8
K' = [H_{2}][I_{2}] /[HI]^{2} = 0.0182
Let the concentration of both H_{2} and I_{2} be x
Then K' = 0.0182 = x^{2}/ (0.5)^{2}
x^{2 }= 0.0182 Ã— 0.25 = 0.00456
x = 0.067mol/litre.
Question15
Solution:
2Icl(g) I_{2}(g) + Cl_{2}(g)
Initial Concentration 0.78M  
At equilibrium (0.78x)M x/2M x/2M
Kc =
0.14 = =
âˆ´ x = 0.33M
[ICl] = 0.78 â€“ 0.33 = 0.45M
[I_{2}] = = 0.165M
[Cl_{2}] = = 0.165M.
Question16
CH_{3}COOH + C_{2}H_{5}OH CH_{3}COOC_{2}H_{5} + H_{2}O
(i) Write the concentration ratio, Q, for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.0 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.0 mol of acetic acid and maintaining at 293 K, 0.214 mol of ethyl acetate is formed after some time. Has equilibrium been reached?
Solution:
(i)
(ii) At equilibrium, concentration of acetic acid is = 1.0  0.171 = 0.829 mol
and concentration of ethanol is = 0.1800.171
= 0.009
K = (0.171)/(0.829 Ã— 0.009) = 22.19
(iii) Initial conc. of ethanol = 0.5mol
Initial conc. of acetic acid = 1.0mol
After some time, conc. of ethanol = 0.5  0.214 = 0.286 mol
conc. of acetic acid = 1.0  0.214 = 0.786 mol
Hence, Q = (0.214)/(0.786 Ã— 0.286)
= 0.9520
Since the value of Q is lesser than that of K, the reaction must be proceeding in the forward direction readily.
Question17
Solution:
PCl_{5} PCl_{3} + Cl_{2}
Let the concentration of PCl_{5} and Cl_{2} be x at equilibrium.
K_{c} =
Or 8.3 Ã— 10^{3} =
Or 4.15 Ã— 10^{4} = x^{2}
Or x = 2 Ã— 10^{2} = 0.02 mol L^{1}
Hence the concentration of PCl_{5} and Cl_{2} each at equilibrium = 0.02 mol L^{1}.
Question18
Solution:
FeO(s) + CO(g) Fe(s) + CO_{2}(g)
K_{p} = 0.265 atm at 1050 K
FeO(s) + CO(g) Fe(s) + CO_{2}(g)
Initaial pressure 1.4 atm 0.8 atm
Equilibrium pressure (1.4 â€“ x) atm (0.8 + x) atm
K_{p} = = 0.265 atm
0.8 + x = 0.265 (1.4 â€“ x) = 0.371 â€“ 0.265 x
x + 0.265x = 0.371 â€“ 0.8
1.265x = 0.429
x = 0.34 atm
âˆ´ Equilibrium pressure of CO = 1.4 â€“ (0.34) = 1.74 atm
Equilibrium pressure of CO_{2} = 0.8 + (0.34) = 0.46 atm.
Question19
Solution:
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
K_{c} = == = = 0.01
Since Q_{c} < K_{c} the reaction is not at equilibrium. The reaction proceeds to form more product.
Question20
for which K_{c} = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.30 Ã— 10^{3} mol L^{1} what is its molar concentration in the mixture at equilibrium.
Solution:
2BrCl(g) Br_{2}(g) + Cl_{2}(g)
3.30 Ã— 10^{3 }0 0
K_{C} = 32 at 500K
From the equation 2 mols of BrCl on decomposition gives = 1 mol of Br and 1 mol of Cl_{2}
Therefore, 3.30 Ã— 10^{3} mol L^{1} of BrCl gives =
= 1.65 Ã— 10^{3} mol of Br_{2} and Cl_{2}
Let x be the molar concentration of BrCl at the equilibrium
Molar concentration of Br_{2} = 1.65 Ã— 10^{3} mol.
Molar concentration of I_{2} = 1.65 Ã— 10^{3} mol
Therefore K_{c} =
32 =
x^{2} =
x^{2} =
x = = 3 Ã— 10^{4} mol L^{1}
Therefore Molar concentration of BrCl = 3 Ã— 10^{4} mol L^{1}.
Question21
Solution:
Let total mass of the gaseous mixture be = 100g
Mass of CO = 90.55 g
Mass of CO_{2} = 9.45g
Moles of CO = = 3.234
Moles of CO_{2} = = 0.215
Total number of moles of gases = 3.234 + 0.215 = 3.449
P_{CO} = Ã— 1 = 0.938 atm
Ã— 1 = 0.0623 atm
K_{p} = = 14.12
K_{p} = K_{c}(RT)Î” ^{n} = K_{c}(RT) Q Î” n = (21) = 1
K_{c} = = 0.153.
Question22
Solution:
NO(g) + Â½ O_{2}(g) NO_{2}(g)
Î” _{f}G^{0} (NO_{2}) = 52.0 kJ/mol^{1 }
Î” _{f}G^{0} (NO) = 87.0 kJ/mol^{1 }
Î” _{f}G^{0} (O_{2}) = 0 kJ/mol^{1} T = 298 K
Î” GÂ° = Î” _{f}G^{0} (NO_{2}) â€“ [Î” _{f}G^{0} (NO) + Î” _{f}G^{0} (O_{2})]
= Î” _{f}G^{0} (NO_{2}) â€“ Î” _{f}G^{0} (NO)
= (5287) kJ mol^{1 }
We have, Î” GÂ° = 2.303 RT log K_{c }
35 kJ mol^{1} = 2.303 Ã— 8.314 JK^{1} Ã— 298 Ã— log K_{c }
log K_{c} =
=
=
log K_{c} = 6.134
K_{c} = 1.361 Ã— 10^{6}.
Question23
Solution:
a) No. of moles of the products increases.
b) No. of moles of the products decreases.
c) No. of moles of the products decreases.
Question24
(ii) CO_{2}(g) + C(s) 2CO(g)
(iii) 4NH_{3}(g) + 5O_{2}(g) 4NO(g) + 6H_{2}O(g)
Solution:
The following reactions will get effected by increase of pressure
(ii) CO_{2}(g) + C(s) 2CO(g)
The increase of pressure will shift the equilibrium to left hand direction.
(iii) 4NH_{3}(g) + 5O_{2}(g) 4NO(g) + 6H_{2}O(g)
The increase of pressure will shift the equilibrium to left hand direction.
Question25
Solution:
K_{p} = K_{c} =
Let (pH_{2})_{eq} = (pBr_{2})_{eq} = x bar
1.6 Ã— 10^{5} =
x^{2} =
x =
or x = 0.25 Ã— 10^{1}
or 2.5 Ã— 10^{2}
Therefore, (pH_{2})_{eq} = 2.5 Ã— 10^{2}
(pBr_{2})_{eq} = 2.5 Ã— 10^{2} bar
and (pHBr)_{eq} = 10.0 bar.
Question26
(a) Write an expression for K_{p} for the above reaction.
Solution:
(a) K_{p} =
(b)
(i) The value of K_{p} remains unchanged on increasing the pressure. When pressure is increased then according to LeChatelierâ€™s principle the equilibrium shifts in the direction where there is less number of moles of gases i. e., backward direction in case of the given reaction.
(ii) In case of endothermic reactions the value of K_{p} increases with increase in temperature. With increase in temperature, the equilibrium shifts in the endothermic directions i. e., forward direction in case of the given reaction.
(iii) K_{p} will remain undisturbed. Equilibrium composition will remain unchanged. However, in the presence of catalyst, the equilibrium would be attained quickly.
Question27
Solution:
2H_{2}(g) + CO(g) CH_{3}OH(g)
a) Effect of addition of H_{2}
This will favour forward reaction i. e., formation of CH_{3}OH because increase in the concentration of the reactants will result in the formation of more products.
b) Addition of CH_{3}OH
This will shift the equilibrium towards backward direction because when the product concentration increases, backward reaction is favoured.
c) Removal of CO
This favours the increase in formation of CO i. e., backward reaction is favoured.
d) Removal of CH_{3}OH
This will result in the increased production of CH_{3}OH i. e., forward reaction is favoured.
All these explanations are according to Le Chatelierâ€™s principles.
Question28
(b) What is the value of K_{c} for the reverse reaction at the same temperature?
(c) What would be the effect on K_{c} if (i) more PCl_{5} is added (ii) the pressure is increased (iii) the temperature is increased?
Solution:
PCl_{5}(g) PCl_{3}(g) + Cl_{2}(g)
(a) K_{c} =
(b) Reverse reaction
PCl_{3}(g) + Cl_{2}(g) PCl_{5}(g)
K_{c} = = =120.48
(c) (i) More PCl_{5} is added: The equilibrium will be shifted to the right. This will continue till the entire extra amount of PCl_{5} has been used up.
(ii) The pressure is increased:
PCl_{5}(g) PCl_{3}(g) + Cl_{2}(g)
1 mole 1 mole 1 mole
Since backward reaction takes place with decrease in number of moles, an increase in pressure will favour combination of PCl_{3} and Cl_{2} molecules to produce back PCl_{5}.
(iii) The temperature is increased: The reaction is endothermic. Therefore, an increase in temperature will shift the equilibrium to the right.
Question29
Solution:
a) Will have appreciable concentration of reactants.
b) Will have appreciable concentration of products.
c) Will have appreciable concentration of reactants and products.
Question30
Solution:
3O_{2}(g) 2O_{3}(g)
K_{c} = 2 Ã— 10^{5} T = 25Â° C = 298 K
K_{c} =
[O_{3}]^{2} = K_{c} Ã— [O_{2}]^{3 }
= 2 Ã— 10^{50} Ã— (1.6 Ã— 10^{2})^{3 }
= 2 Ã— 10^{50} Ã— 10^{6} Ã— 4.096
= 8.192 Ã— 10^{56}
[O_{3}] = âˆš 81.92 Ã— 10^{12} = 2.86 Ã— 10^{28}.M
Question31
Solution:
CO(g) + 3H_{2}(g) CH_{4}(g) + H_{2}O(g)
K_{c} =
3.9 =
[CH_{4}] = = 0.0585 mol.
Question32
Solution:
K_{sp} FeS ......................... 
= 6.3 x 10^{13} 
K_{sp} MnS ........................ 
= 2.5 x 10^{13} 
K_{sp }ZnS_{ ..............................} 
= 1.6 x 10^{24} 
K_{sp} CdS ...................... 
= 8.0 x 10^{27} 
FeSO_{4} Fe^{2+} + SO_{4}^{2 }  
H_{2}S .......................... 
= 2H^{+} = S^{2} 
Fe^{2+} + S^{2} FeS 

The total volume after mixing 
= (10 + 5) = 15 ml 
[Fe^{2+}] ........................ 
= 1.33 x 10^{2} 
[S^{2}] ......................... 
= 0.6 x 10^{19 } 
[Fe^{2+}] [S^{2}] .................  = 0.798 x 10^{21} 
Similarly [Zn^{2+}] [S^{2}] and [Cd^{2+}] [S^{2}] are also 0.798 x 10^{21}
Since in the case of ZnCl_{2} and CdCl_{2} Solutions, the [Zn^{2+}] [S^{2}] and [Cd^{2+}] [S^{2}] > K_{sp} of ZnS and Cds respectively.
Therefore ZnCl_{2} and CdCl_{2} give precipitation of their respective sulphides.
Question33
H_{2}O, BF_{3} , H^{+} and NH
Solution:
BF_{3}, H^{+}, NH_{4}^{+}.
Question34
Solution:
The conjugate base of HF is F^{} that of H_{2}SO_{4} is HSO_{4}^{} and that of HCO_{3}^{} is CO_{3}^{2}.
Question35
Solution:
Bronsted Bases 
Conjugate Acids 
NH_{2}^{} 
NH_{3} 
NH_{3} 
NH_{4}^{+} 
HCOO^{} 
HCOOH 
Question36
Solution:
Bronsted Acid/Base 
Conjugate Acids 
Conjugate Base 
H_{2}O 
H_{3}O^{+} 
OH^{} 
HCO_{3}^{} 
H_{2}CO_{3} 
CO_{3}^{2} 
HSO_{4}^{} 
H_{2}SO_{4} 
SO_{4}^{2} 
NH_{3} 
NH_{4}^{+} 
NH_{2}^{} 
Question37
Solution:
a) OH^{} ion is a Lewis base as it has lone pairs of electrons on oxygen which it can donate
b) F^{} ion has four lone pairs of electrons. Hence, it can act as a Lewis base by donating any one of these lone pair.
c) H^{+} acts as a Lewis acid as it has a vacant orbital and hence can accept a pair of electrons from Lewis bases.
d) BCl_{3} acts as a Lewis acid. In BCl_{3}, there are only six electrons in the valence shell of boron. It can accept a pair of electrons from bases such as NH_{3} to complete its octet.
Question38
Solution:
PH = log [H_{3}O^{+}]
=  log (3.8 Ã— 10^{3})
=  log 3.8 + (log 10^{3})
= log 3.8 + 3 log 10
pH = 2.42
Question39
Solution:
pH = 3.76
pH = log [H_{3}O^{+}]
3.76 = log [H_{3}O^{+}]
log [H_{3}O^{+}] = 3.76
log [H3O^{+}] = 4.2400
[H3O^{+}] = antilog (4.2400) = 1.7 x 10^{4}
Question40
Solution:
Given data: KÎ± of acetic acid = 1.74 Ã— 10^{5 }
Î± = ?
Concentration of acetic acid = 0.05 M
âˆ´ Î± = 0.0187
[CH_{3}COO^{}] = CÎ± = 0.05 Ã— 0.0187 mol L^{1} = 0.00094 mol L^{1}
[CH_{3}COO^{}] = 9.4 Ã— 10^{4} mol L^{1}.
[H_{3}O^{+}] = [CH_{3}COO] = 0.05M = 5 Ã— 10^{2} M
âˆ´ pH = log [H_{3}O^{+}]
= log (5 Ã— 10^{2})
=  log5 + 2 log 10
= 0.6990 + 2 = 1.301
âˆ´ pH = 1.301.
Question41
Solution:
We know that Ka x Kb = Kw = 1 x 10 ^{14} Kb = Kw/Ka.
i) Ionisation constant of HF (Ka) = 6.8 x 10^{4}Ionisation constant of F^{} (Kb) = Kw/Ka = 1.5 x 10^{11}
ii) Ionisation constant of HCOOH (Ka) = 1.8 x 10^{4}
Ionisation constant of HCOO^{} (Kb) = 5.6 x 10^{11}
iii) Ionisation constant of HCN (Ka) = 4.8 x 10^{9}
Ionisation constant of CN^{} (Kb) = 2.08 x 10^{6}.
Question42
Solution:
Ionisation constant of phenol 
= 1.0 Ã— 10^{10 } 
Degree of ionisation 
= Î± = 
= = 

[Phenolate ion] .................. 
= cÎ± = 
Degree of ionisation of the solution is also 0.01M

= 
Question43
Solution:
H_{2}S H^{+} + HS^{} Ka_{1} = 9.1 Ã— 10^{8}
HS^{} H^{+} + S^{2} Ka_{2} = 1.2 Ã— 10^{13}
HCl H^{+} + Cl^{}
In the case of HCl, the dissociation is complete.
So [H^{+}] = 0.1 M.
Let [H^{+}] from H_{2}S be x and also x be the [HS^{}].
In the solution of 0.1M HCl
[H^{+}] .......................... 
= 0.1 + x 
[HS] .......................... 
= X 
[H_{2}S] .......................... 
= [0.1 â€“ x] 
K_{a }.......................... 
= 
Ka_{1 }.......................... 
= 9.1 Ã— 10^{8 } 
Since x is very small 

9.1 Ã— 10^{18 }.......................... 
= 
x .......................... 
= = 9.1 Ã— 10^{8 } 
In 0.1 M HCl solution, concentration of [HS^{}] = 9.1 Ã— 10^{8} M / litre 

2^{nd} dissociation 

HS^{} H^{+} + S^{2 } 

Ka_{2 }.......................... 
= 1.2 Ã— 10^{13 } 
Ka_{2} .......................... 
= 
1.2 Ã— 10^{13} .................. 
= 
S^{2 }.......................... 
= 1.09 Ã— 10^{20} M / Litre 
Question44
a) 0.003 M HCl b) 0.005 M NaOH
c) 0.002 M HBr d) 0.002 M KOH
Solution:
a) 0.003 M HCl 

HCl + H_{2}O H_{3}O^{+} + Cl 

0.003 M 

pH .......................... 
= log [H_{3}O]^{+ } 

= log(3 Ã— 10^{3}) 

= (3log 3) 

= (30.4771) 

= 2.5229 = 2.52 


b) 0.005 M NaOH 

NaOH Na^{+}(aq) + OH^{}(aq) 

0.005M 

[OH^{}] .............................. 
= 0.005 M = 5 Ã— 10^{3} M 
[H_{3}O^{+}] .......................... 
= = 

= 0.2 Ã— 10^{11 }= 2 Ã— 10^{12 } 
PH .......................... 
=  log [H_{3}O^{+}] 

=  log (2 Ã— 10^{12}) 

= [12 â€“ log 2] 

= [12  0.3010 ] = 11.70 


c) 0.002 M HBr 

HBr + H_{2}O Br^{}(aq) + H_{3}O^{+}(aq) 

0.002 M 

[H_{3}O^{+}] .......................... 
= 0.002 M = 2 Ã— 10^{3} M 
pH ................................ 
=  log[H_{3}O^{+}] 

=  log(2 Ã— 10^{3}) 

= (3 â€“ log2) = (30.3010) 

= 2.699 = 2.70 


d) 0.002 KOH 

KOH â†’ K^{+}(aq) = OH^{}(aq) 

[OH^{}] .......................... 
= 0.002 M = 2 Ã— 10^{3} M 
[H_{3}O^{+}] ........................... 
= = = 0.5 Ã— 10^{11} M 
pH ............................ 
= log[H_{3}O^{+}] 

= log [5 Ã— 10^{12}] 

= (12 â€“ log 5) 

= 12 â€“ 0.699 

= 11.3 
Question45
Solution:
a) Molar mass of TlOH = 221.4
nTlOH = 2/221.4 = 9.03 Ã— 10^{âˆ’3}
^{ }^{[OH] = n TlOH_/ Vlit = 9.03 Ã— 103/ 2 = 4.51 Ã— 103 M}
^{ }^{pOH = âˆ’log (4.51 Ã— 103 ) = 3âˆ’log 4.51 = 3âˆ’0.6542 = 2.35}
^{ }^{pH + pOH = 14 ; pH = 14âˆ’ 2.35 = 11.65}
^{ }^{b) Molar mass of Ca(OH)2 = 74}
^{ }^{n Ca(OH)2 = 0.3 / 74 = 4.05 Ã— 10âˆ’3}
[OH^{âˆ’}] = 4.05 Ã— 10^{âˆ’3}/ 0.5 = 0.0081 M
2[OH] = 0.0081 Ã— 2 = 16.2 Ã— 10^{3} (Ca(OH)_{2} contains 2 OH groups.)
pOH = âˆ’log (16.2 Ã— 10^{3}) = 3 â€“ log 16.2 = 1.79
pH = 14 â€“ 1.79 = 12.21
(c) n NaOH = 0.3 / 40 = 7.5 Ã— 10^{3}M
[OH^{}] = 7.5Ã—10^{âˆ’3}/0.2 = 0.0375
pOH = âˆ’log(0.0375) = âˆ’log(3.75 Ã— 10^{âˆ’2}) = 2 âˆ’log3.75 = 1.43
pH = 14  1.43 = 12.57
(d) To get molarity(concentration) we can use V1M1 = V2M2
1 Ã— 13.6 = 1000 Ã— M2
M2 = 1Ã— 13.6/1000 = 0.0136 = [H^{+}]
pH =âˆ’log(1.36 Ã— 10^{1}) = 1 âˆ’ log 1.36 = 1.87
Question46
Solution:
Given Data:
C = 0.1 M
Î± = 0.132
[H_{3}O^{+}] = CÎ± = 0.1 Ã— 0.132 = 0.0132
pH =  log[H_{3}O^{+}]
=  log 0.0132
=  log (1.32 Ã— 10^{2})
=  log 1.32 + 2 log 10
= 0.1206 + 2
âˆ´ pH = 1.88
Î± =
Squaring both sides,
Î± ^{2} =
Ka = CÎ±^{2 }/ 1âˆ’Î± = 0.1 Ã— (0.132)^{2 }/ 1âˆ’0.132
âˆ´ Ka = 2.01 Ã— 10^{3}
pKa =  log Ka
pKa =  log (2.01 Ã— 10^{3}) = 3 âˆ’ log 2.01 = 2.70
pKa = 2.77
Question47
Solution:
pH = 9.95 ; [H^{+}] = antilog (9.95) = 1.12 X 10^{10}M
[OH] = 1X10^{14}/1.12 X 10^{10} = 8.91 X 10^{5}M
Kb = [M^{+}] [OH^{}] / [MOH] = (8.91 X 10^{5})^{2} /0.005 = 1.59 X 10^{6}
pKb = log Kb = log(1.59 X 10^{6}) = 5.80
Question48
Solution:
pH = 0.001M
From the table, the ionization constant of aniline is
K_{b} = 4.3 Ã— 10^{10 }
Î± _{b} = =
âˆ´ Î± _{b} = 0.002.
K_{b} Ã— KÎ± = K_{w }
KÎ± = = 2.33 Ã— 10^{5}.
âˆ´ KÎ± = 2.33 Ã— 10^{5}.
Question49
a) Human muscle â€“ fluid = 6.83
b) Human stomach fluid = 1.2
c) Human blood = 7.38
d) Human saliva = 6.4
Solution:
a) pH ................................ 
= 6.83 
log [H_{3}O^{+}] .......................... 
= pH = 6.83 
[H_{3}O^{+}] ............................... 
= antilog( 6.83) 

= 10^{7} Ã— 10^{0.17 } 
[H_{3}O^{+}] ............................... 
= 1.48 Ã— 10^{7} M 


b) pH ................................. 
= 1.2 
log [H_{3}O^{+}] ......................... 
= pH = 1.2 
[H_{3}O^{+}] ............................... 
= antilog(1.2) 

= 10^{2} Ã— 10^{0.8 } 

= 6.31 Ã— 10^{2} M 


c) pH ................................ 
= 7.38 
log [H_{3}O^{+}] .......................... 
= pH = 7.38 
[H_{3}O^{+}] ............................... 
= Antilog(7.38) 
[H_{3}O^{+}] ............................... 
= 10^{7.38 } 

= 10^{8} Ã— 10^{.62} 

= 4.17 Ã— 10^{8} 
d) pH ............................... 
= 6.4 
log [H_{3}O^{+}] ....................... 
= pH = 6.4 
[H_{3}O^{+}] ............................. 
= Antilog(6.4) 

= 3.98 Ã— 10^{7} M 
Question50
Solution:
i) pH 
= 6.8 
log[H_{3}O^{+}] 
= pH = 6.8 
[H_{3}O^{+}] 
= Antilog [6.8] 
= 1.5 Ã— 10^{7} M 

ii) pH = 5.0 

log[H_{3}O^{+}] 
= pH =5.0 
[H_{3}O^{+}] 
= Antilog(5.0) = 10^{5 }M 
iii) pH 
= 4.2 
log [H_{3}O^{+}] 
= pH = 4.2 
[H_{3}O^{+}] 
= Antilog(4.2) 
= 6.31 Ã— 10^{5} M 

iv) pH 
= 2.2 
log[H_{3}O^{+}] 
= pH = 2.2 
[H_{3}O^{+}] 
= Antilog(2.2) 
= 6.31 Ã— 10^{3 } 

v) pH 
= 7.8 
log[H_{3}O^{+}] 
= pH =  7.8 
[H_{3}O^{+}] 
= Antilog (7.8) 
Question51
Solution:
CaSO_{4} â†’ Ca^{2+} + SO_{4}^{2}
Let the solubility of CaSO_{4} be x moles/litre. Then the solution will contain x moles of Ca^{2+} and x moles SO_{4}^{2} ions respectively per litre. Hence, the solubility product, K_{sp}^{ }of CaSO_{4} is
K_{sp .......................} 
= [Ca^{2+}] [SO_{4}^{2}] 
= x Ã— (x) = x^{2 } 

We know K_{sp .................} 
= 9.1 x 10^{6 } 
Hence x^{2 .......................} 
= 9.1 Ã— 10^{6 } 
Or x ............................. 
= 3.01 Ã— 10^{3} mol L^{1 } 
1 mole of CaSO_{4 .............} 
= 40 + 32 + 64 = 136g 
3.01 Ã— 10^{3} of CaSO_{4 ......} 
= 136 Ã— 3.01 Ã— 10^{3} 
= 136 Ã— 0.00301 

= 0.409 g 
0.41 g needs 1 L of water for complete dissolution
1 g needs = = 2.38 L.
Question52
Solution:
Solubility of Sr(OH)_{2} at 298 K = 19.23 g/L
Sr(OH)_{2} Sr^{2+}(aq) + 2OH^{}
Mol. mass of Sr(OH)_{2} = 87 + 34 = 121
Molar concentration of Sr(OH)_{2} 
= 
Sr(OH)_{2 } ......................... _{ } 
= Sr^{2+}(aq) + 2OH^{ } 
0.16M 2(0.16) ...................... 
= 32 M 
[Sr^{2+}] ......................... 
= 0.1581 M 
[OH^{}] ......................... 
= 2 Ã— 0.158 
= 0.316 M 


= = = 
pH ........................._{ } 
=  log H_{3}O^{+ } 
=  log 10^{12} â€“ log 32 

= 12 + 1.51 

pH ......................... 
= 13.51 
Question53
Solution:
Ionisation constant of propionic acid 
= 1.32 Ã— 10^{5 } 
Concentration of solution 
= 0.05M 
Let the degree of ionisation is Î± 

Î±.................................................. 
= = 
= 1.63 Ã— 10^{2 } 

[H^{+}] .................................. 
= cÎ± 
= 0.05 Ã— 1.63 Ã— 10^{2 } 

= 0.0815 Ã— 10^{2 } 

= 815 Ã— 10^{6} 

pH ..................................... 
= log(815 Ã— 10^{6}) 
=  (log815 + log 10^{6}) 

= 6log 815 

= 62.9112 

= 3.088 

= 3.09 
In the case of molarity of HCl is 0.01 M, it is assumed that it is fully dissociated. Let [H^{+}] be x from the ionisation of propionic acid. This is (x) also the concentration of propionate ion.
[H^{+}] = 0.01 + X ; [Propionate] 
= x and [HP] = (Cx) 
K_{a }..................................... 
= [H^{+}] [P^{}] / [HP] 
= 

Since x is very small, Ka ....... 
= 
Or x .................................... 
= C. Ka / 0.01 
= 0.05 Ã— 1.32 Ã— 10^{5} / 0.01 

x .......................................... 
= 1.60 Ã— 10^{5 } 
Degree of ionisation in 0.1 M HCl = x/c 

= 1.60 Ã— 10^{5} / 0.05 

= 1.32 Ã— 10^{_3 } 
Question54
Solution:
PH ................................... 
= 2.34 
 log [H_{3}O^{+}] .................... 
= 2.34 
or [H_{3}O^{+}] ....................... 
= Antilog [2.34] 
= 4.57 Ã— 10^{3 } 

For weak acid [H^{+}] .......... 
= c Î± = 4.57 Ã— 10^{3 } 
C ................................ 
= 0.1M 
Î± ................................. 
= = 0.0457 
K_{a }................................. 
= Î± ^{2} / C 
= 

Ionisation constant K_{a} 
= 2.09 Ã— 10^{4 } 
Question55
Solution:
K_{a }................................. 
= 4.5 Ã— 10^{4 } 
Hydrolysis constant K_{h} 
= = = Ã— 10^{10 } 
Let the degree of hydrolysis 
= h 
K_{h }^{................................} 
= 
Since h is small 1h is negligible 

Concentration of NaNO_{3} 
= 0.04 M 
............................ 
= 0.04 Ã— h^{2 } 
âˆ´0.180h^{2 ......................} 
= 10^{10 } 
h^{2 }.............................. 
= 
h = ............. 
= 2.36 Ã— 10^{5 } 
PH.................................. 
= 7.0 + (pKa + log c) 
= 7.0 + (log 0.04 â€“ log(4.5 Ã— 10^{4} 

= 7.0 + [1.40 + 3.35] 

= 7.0 + [1.95] = 7.0 + 0.97 = 7.97 
Question56
Solution:
Since pyridinium hydrochloride is a salt of weak base and strong acid pH = 7 âˆ’ (log C + pkb), where kb is the dissociation constant of pyridine
3.44 = 7 âˆ’ (log 0.02 + pkb) 
âˆ’3.56 = âˆ’ (âˆ’1.70 + pkb) 
âˆ’7.12 = 1.70 âˆ’ pkb Pkb = 1.70 + 7.12 = 8.82 
Kb = antilog(âˆ’8.82) = 1.513 X 10^{9} 
Question57
Solution:
Nacl and NaNO_{2}^{ }solutions are neutral. KBr , NaCN and KF solutions are basic. NH_{4}NO_{3} solution is acidic.
Question58
Solution:
Sodium acetate is a salt of strong base and weak acid.
PH ............................... 
=  [logK_{w} + log K_{a} â€“ log c] 
K_{w} .............................. 
= 1.0 Ã— 10^{14}, K_{a} = 1.35 Ã— 10^{3} , c = 0.1 M 
PH ............................. 
=  [log^{14} + log (1.35 + 10^{3}) â€“ log(0.1) 
= 14[3+0.1303 + 1] 

=  [143+0.1303 + 1] 

=  [17 + 1.313] 

pH of its sat solution ........... 
= 7.94. 
Question59
Solution:
Ionic product of water 
= 2.7 Ã— 10^{14 } 
H_{2}O â†’ H^{+} + OH^{14 }x x 

[H^{+}] [OH^{}] ......................... 
= 2.7 Ã— 10^{14 } 
x^{2 ....................................} 
= 2.7 Ã— 10^{_14 } 
x .................................... 
= = 1.64 Ã— 10^{_7 } 
pH ................................... 
= log [H^{+}] 
= 7 â€“ log 1.64 

= 7 â€“ 0.2148 

= +6.7696 

pH .................................. 
= 6.7852. 
Question60
a) 10 ml of 0.2 M Ca(OH)_{2} + 25 ml of 0.1 M HCl
b) 10 ml of 0.01 M H_{2}SO_{4} + 10 ml of 0.01 M Ca(OH)_{2}
c) 10 ml of 0.1 M H_{2}SO_{4} + 10 ml of 0.1 M KOH
Solution:
a) n _{Ca(OH)2 }= 0.2 Ã—0.01 = 0.002
.n_{OH.} = 2_{.}.Ã—_{ }n _{Ca(OH)2}..= 2.Ã—_{ }0.002 = 0.004
n_{ H+ }= n_{HCl} = 0.025Ã— 0.1 = 0.0025
H+ is completely consumed.
âˆ´.n_{OH.}(left) = 0.004 â€“ 0.0025 = 0.0015
[OH^{}] = = 0.0429 M
p_{OH} = log(0.0429) = 20.6325 = 1.3675
pH = 14pOH = 141.3675 = 12.6325
...b) nH_{2}SO_{4}.= .01 Ã—..01 = 0.0001
n_{H+...}= 2_{..}Ã—_{ ..} ) nH_{2}SO_{4}.= 0.0002_{.....}
The base viz., Ca(OH)_{2} has same volume and concentration.
Hence n_{H+} = n_{OH} = 0.0002
pH of the mixture = 7
c) nH_{2}SO_{4} = .01 Ã—..0.1 = 0.001
n_{H+} = 2 nH_{2}SO_{4} =2 Ã—.0.001 = 0.002
n_{OH} = 0.01Ã— 0.1 = 0.001
n_{H+} (left) = 0.002 0.001 = 0.001
[H^{+}] = 0.001 /_{ }0.020 = 0.05 M
pH = log(5 Ã—10^{2}) = 2log 5 = 1.30
Question61
Solution:
i) Silver chromate Ag_{2} CrO_{4}, K_{sp} of Ag_{2}CrO_{4}
= 1.1 Ã— 10^{12}
Let the solubility of Ag_{2}CrO_{4} be equal to S,
Ag_{2}CrO_{4} 2Ag^{+} + CrO
1.1 Ã— 10^{12 .................} 
= (2s)^{2} (s) 
or 1.1 Ã— 10^{12 ..............} 
= 4s^{3 } 
or s^{3} .......................... 
= 
or s ............................ 
= 
âˆ´ Molarity of CrO ..... 
= 0.65 Ã— 10^{4} M 
Molarity of Ag^{+ .............} 
= 2 Ã— 0.65 Ã— 10^{4} M 

= 1.30 Ã— 10^{4} M 
(ii) Barium chromate : BaCrO_{4 } 

BaCrO_{4} Ba^{2+} + CrO 

K_{sp} of BaCrO_{4 .................} 
= 1.2 Ã— 10^{10 } 
1.2 Ã— 10^{10} ................... 
= s Ã— s 
or 1.2 Ã— 10^{10} ................ 
= s^{2 } 
or s ............................... 
= 

= 1.1 Ã— 10^{5} M 
Molarity of Ba^{2+} and CrO_{4}^{2} 
= 1.1 Ã— 10^{5} M each 
(iii) Ferric hydroxide: Fe(OH)_{3 } 

Fe(OH)_{3} Fe^{3+} + 3OH^{ } 

K_{sp} of Fe(OH)_{3} ................... 
= 1.0 Ã— 10^{_38 } 
Or 1.0 Ã— 10^{38 ...................} 
= s Ã— 3(s)^{3 } 
Or 1.0 Ã— 10^{38} = 3s^{4 } 
Or s^{4} = 
Or ................................. 
= 
= 1.39 Ã— 10^{10} M 

Molarity of Fe^{3+ ...............} 
= 1.39 Ã— 10^{10 }M 
Molarity of OH^{ ..............} 
= 3 Ã— 1.39 Ã— 10^{10 } 
............. 
= 4.17 Ã— 10^{10} M 
iv) Lead chloride : PbCl_{2 } 

PbCl_{2 }Pb^{2+} + 2Cl^{ } 

K_{sp} of PBCl_{2 ..................} 
= 1.6 Ã— 10^{_5 } 
1.6 Ã— 10^{5 .....................} 
= 2s^{3 } 
or s^{3 ...........................} 
= 0.8 Ã— 10^{5 } 
or s ........................... 
= (0.8 Ã— 10^{5})^{1/3 } 
= (8 Ã— 10^{6})^{1/3 } 

= 1.59 Ã— 10^{2} M 

Molarity of Pb^{2+} 
= 1.59 Ã— 10^{2 } 
Molarity of Cl 
= 2 Ã— 1.59 Ã— 10^{2 } 
= 3 .18 Ã— 10^{2} M 

v) Mercurous iodide : Hg_{2}I_{2 } 

Hg_{2}I_{2} 2Hg_{2}I_{2 } 

K_{sp} of Hg_{2}I_{2 .....................} 
= 4.5 Ã— 10^{29 } 
4.5 Ã— 10^{29 .....................} 
= 2s^{2} Ã— 2s^{2 } 
or 4.5 Ã— 10^{29} ................. 
= 4s^{4 } 
or s^{4} .............................. 
= Ã— 10^{29 } 
or s ............................... 
= 
= 2.24 Ã— 10^{10} M. 
Molarity of Hg_{2}^{2+} and I^{} each = 4.48 Ã— 10^{10} M.
Question62
Solution:
Ag_{2}CrO_{4} dissolves in water and the equilibrium in the saturated solution is
Ag_{2}CrO_{4}(s) = 2Ag^{+}(aq) + CrO_{4}^{2}(aq)
Let the solubility of Ag_{2}CrO_{4 }be S_{1}
[Ag^{+}(aq)] ....................... 
= 2s_{1 } 
and [CrO_{4}^{2}(aq)] ............... 
= s_{1} 
K_{sp ...........................} 
= [Ag^{+}(aq)]^{2} [CrO_{4}^{2}(aq) 
= (2s_{1})^{2} Ã— s = 4s_{1}^{3 } 

or s_{1 ........................} 
= = 
= 0.65 Ã— 10^{4 } 

Similalry AgBr 
= Ag(aq) + Br^{ }(aq) 
Let the solubility of AgBr be s_{2 } 

[Ag^{+}(aq)] ................... 
= s_{2} and [Be^{}(aq)] = s_{2 } 
K_{sp ....................} 
= [Ag(aq)] [Be^{}(aq)] 

= s_{2 Ã— }s_{2} = s_{1}^{2 } 
or s_{1 .........................} 
= (K_{sp})^{1/2} = (5.0 Ã— 10^{13})^{1/2 } 
= (0.5 Ã— 10^{12})^{1/2 } 

= 0.707 Ã— 10^{6} mol L^{1 } 

Raid of the molarities of silver chromate to silver bromide comes to = 0.65 Ã— 10^{4} ; 0.707 Ã— 10^{6} .
Question63
Solution:
NaIO_{3} Na^{+} + IO_{3}^{}
Cu(ClO_{4})_{2} Cu^{2+} + 2ClO_{4}^{}
Cu^{2+} + 2IO_{3}^{} Cu(IO_{3})_{2 }
Molarity of NaIO_{3} and Cu(ClO_{4}) each is 0.002 M.
They ionize completely
So [Cu^{2+}] [IO_{3}^{}] = 0.002 Ã— 0.002
= 4 Ã— 10^{_6 }
Since the products of [Cu^{2+}] [IO_{3}^{}] > KSP of Copper Iodate (7.4 Ã— 10^{8})
Copper iodate will be precipitated.
Question64
Solution:
FeSO_{4 ..............................} 
= Fe^{2+} + SO_{4}^{2 } 
Na_{2}S ............................. 
= 2Na^{+} + S^{2 } 
Fe^{2+} + S^{2} â†’ FeS 

Solubility product of iron sulphide 
= 6.3 Ã— 10^{18 } 
Let Fe^{2+ ........................} 
= [x] and S^{2} = [x] 
[x] [x] ...................... 
= 6.3 Ã— 10^{18 } 
x^{2} ............................... 
= 6.3 Ã— 10^{18 } 
x ................................ 
= (6.3 Ã— 10^{18})^{1/2} 
= 2.51 Ã— 10^{9 } 

Hence the highest molarity for the solution is 5.02 Ã— 10^{9} M. 