Solved Example: Ionic Equilibrium | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Solved Example on Ionic Equilibrium 
JEE Mains 
Q1:  The conjugate acid of ????
?? -
is 
(A) ????
?? 
(B) ????
?? ???? 
(C) ????
?? +
 
(D) ?? ?? ?? ?? 
Ans: (A) The conjugate acid has one proton (H
+
) more. Hence, for NH
2
-
the conjugate acid would be 
NH
3
 (the positive charge of H
+
and the negative charge of NH
2
 
-
cancel each other out). 
Q2:  Out of the following, amphiprotic species are 
I. ??????
?? ?? -
 
II. ????
-
 
III. ?? ?? ????
?? -
 
IV. ?????? ?? -
 
(A) I, III, IV 
(B) I and III 
(C) III and IV 
(D) All 
Ans: (C) H
2
PO
4
-
And HCO
3
-
are amphoteric in nature. 
HCl+ H
2
PO
4
-
? H
3
PO
4
+ Cl
-1
NaOH+ H
2
PO
4
-
? HPO
4
-2
+ H
2
O
 
 
Q?? : ???? of an aqueous solution of ???????? at ????
°
?? should be 
(A) 7 
(B) > ?? 
(C) < ?? 
(D) 0 
Ans: (C) Halides and alkaline metals dissociate and do not affect the H
+
as the cation does not alter 
the H
+
and the anion does not attract the H
+
from water. This is why NaCl is a neutral salt. But pH of 
water decreases as the temperature increases. So option C is correct 
 
 
 
 
Page 2


JEE Solved Example on Ionic Equilibrium 
JEE Mains 
Q1:  The conjugate acid of ????
?? -
is 
(A) ????
?? 
(B) ????
?? ???? 
(C) ????
?? +
 
(D) ?? ?? ?? ?? 
Ans: (A) The conjugate acid has one proton (H
+
) more. Hence, for NH
2
-
the conjugate acid would be 
NH
3
 (the positive charge of H
+
and the negative charge of NH
2
 
-
cancel each other out). 
Q2:  Out of the following, amphiprotic species are 
I. ??????
?? ?? -
 
II. ????
-
 
III. ?? ?? ????
?? -
 
IV. ?????? ?? -
 
(A) I, III, IV 
(B) I and III 
(C) III and IV 
(D) All 
Ans: (C) H
2
PO
4
-
And HCO
3
-
are amphoteric in nature. 
HCl+ H
2
PO
4
-
? H
3
PO
4
+ Cl
-1
NaOH+ H
2
PO
4
-
? HPO
4
-2
+ H
2
O
 
 
Q?? : ???? of an aqueous solution of ???????? at ????
°
?? should be 
(A) 7 
(B) > ?? 
(C) < ?? 
(D) 0 
Ans: (C) Halides and alkaline metals dissociate and do not affect the H
+
as the cation does not alter 
the H
+
and the anion does not attract the H
+
from water. This is why NaCl is a neutral salt. But pH of 
water decreases as the temperature increases. So option C is correct 
 
 
 
 
Q4:  1 cc of ?? . ?? ???????? is added to 99 cc solution of ???????? . The ???? of the resulting solution will be 
(A) 7 
(B) 3 
(C) 4 
(D) 1 
Ans: (C) No' moles HCl =
(1×0.1)
1000
= 10
-4
 
Volume = 1dm
3
 Concentration = moles / volume =
10
-4
1
= 10
-4
 This gives a pH of 4 so option (C) is 
correct. 
 
Q5: ???????? of 
?? ??????
?? ?? ????
?? is mixed with ???????? of 
?? ??????
 ?? ?? ????
?? . The ???? of the resulting solution is 
(A) 1 
(B) 2 
(C) 2.3 
(D) None of these 
Ans: (B) 
We have 50 mL of 
M
200
H
2
SO
4
 
M
200
= 0.005M solution 
H
2
SO
4
? 2H
+
+ SO
4
2-
 Then [H
+
] = 2 × 0.005M= 0.01M
pH = -log [H
+
]
pH = -log 0.01M
pH = 2.00
 
 
Q6:  If ????
?? for fluoride ion at ????
°
?? is 10.83 , the ionisation constant of hydrofluoric acid in water 
at this temperature is: 
(A) ?? . ???? × ????
-?? 
(B) ?? . ???? × ????
-?? 
(C) ?? . ???? × ????
-?? 
(D) ?? . ???? × ????
-?? 
Ans: (C) The equation for the dissociation of HF is as follow: 
HF + H
2
O ? H
3
O
+
+ F
-
 
Here pK
b
= 10.83 
? -log K
b
= 10.83 
Hence, K
b
= 1.48 × 10
-11
 
Thus lonization constant of acid K
a
= K
W
/K
b
 
Page 3


JEE Solved Example on Ionic Equilibrium 
JEE Mains 
Q1:  The conjugate acid of ????
?? -
is 
(A) ????
?? 
(B) ????
?? ???? 
(C) ????
?? +
 
(D) ?? ?? ?? ?? 
Ans: (A) The conjugate acid has one proton (H
+
) more. Hence, for NH
2
-
the conjugate acid would be 
NH
3
 (the positive charge of H
+
and the negative charge of NH
2
 
-
cancel each other out). 
Q2:  Out of the following, amphiprotic species are 
I. ??????
?? ?? -
 
II. ????
-
 
III. ?? ?? ????
?? -
 
IV. ?????? ?? -
 
(A) I, III, IV 
(B) I and III 
(C) III and IV 
(D) All 
Ans: (C) H
2
PO
4
-
And HCO
3
-
are amphoteric in nature. 
HCl+ H
2
PO
4
-
? H
3
PO
4
+ Cl
-1
NaOH+ H
2
PO
4
-
? HPO
4
-2
+ H
2
O
 
 
Q?? : ???? of an aqueous solution of ???????? at ????
°
?? should be 
(A) 7 
(B) > ?? 
(C) < ?? 
(D) 0 
Ans: (C) Halides and alkaline metals dissociate and do not affect the H
+
as the cation does not alter 
the H
+
and the anion does not attract the H
+
from water. This is why NaCl is a neutral salt. But pH of 
water decreases as the temperature increases. So option C is correct 
 
 
 
 
Q4:  1 cc of ?? . ?? ???????? is added to 99 cc solution of ???????? . The ???? of the resulting solution will be 
(A) 7 
(B) 3 
(C) 4 
(D) 1 
Ans: (C) No' moles HCl =
(1×0.1)
1000
= 10
-4
 
Volume = 1dm
3
 Concentration = moles / volume =
10
-4
1
= 10
-4
 This gives a pH of 4 so option (C) is 
correct. 
 
Q5: ???????? of 
?? ??????
?? ?? ????
?? is mixed with ???????? of 
?? ??????
 ?? ?? ????
?? . The ???? of the resulting solution is 
(A) 1 
(B) 2 
(C) 2.3 
(D) None of these 
Ans: (B) 
We have 50 mL of 
M
200
H
2
SO
4
 
M
200
= 0.005M solution 
H
2
SO
4
? 2H
+
+ SO
4
2-
 Then [H
+
] = 2 × 0.005M= 0.01M
pH = -log [H
+
]
pH = -log 0.01M
pH = 2.00
 
 
Q6:  If ????
?? for fluoride ion at ????
°
?? is 10.83 , the ionisation constant of hydrofluoric acid in water 
at this temperature is: 
(A) ?? . ???? × ????
-?? 
(B) ?? . ???? × ????
-?? 
(C) ?? . ???? × ????
-?? 
(D) ?? . ???? × ????
-?? 
Ans: (C) The equation for the dissociation of HF is as follow: 
HF + H
2
O ? H
3
O
+
+ F
-
 
Here pK
b
= 10.83 
? -log K
b
= 10.83 
Hence, K
b
= 1.48 × 10
-11
 
Thus lonization constant of acid K
a
= K
W
/K
b
 
?? ?? = 10
-14
?? ?? = 10
-14
/1.48 × 10
-11
?? ?? = 6.76 × 10
-4
 
Thus Ionization constant of HF is 6.76 × 10
-4
 
 
Q7:  The ???? of an aqueous solution of ?? . ???? solution of a weak monoprotic acid which is ?? % 
ionised is 
(A) 1 
(B) 2 
(C) 3 
(D) 11 
 Ans: (C) [?? ?? ?? +
] = ?? ?? = ?? . ?? ×
?? ??????
= ?? × ????
-?? pH = -log [H
3
O
+
] = -log 10
-3
= 3
 
 
Q8:  If ?? ?? and ?? ?? be first and second ionization constant of ?? ?? ????
?? and ?? ?? >> ?? ?? which is 
incorrect. 
(A) [?? +
] = [?? ?? ????
?? -
] 
(B) [?? +
] = ?? ?? [?? ?? ????
?? ] 
(C) ?? ?? = [??????
?? -
] 
(D) [?? +
] = ?? [????
?? ?? -
] 
Ans: (D) 
H
3
PO
4
= H
+
+ H
2
PO
4
-
… … … . K
1
H
2
PO
4
-
= H
+
+ HPO
4
2-
… … … . . K
2
HPO
4
2-
= H
+
+ PO
4
3-
… … … … . K
3
 K
3
= [HPO
4
2-
]
H
+
= [PO
4
3-
]
 
 
Q9: Which of the following solution will have ???? close to 1.0 ? 
(A) ?????????? of ?? /???????????? + ?????????? of ?? /???? 
(B) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(C) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(D) ???????? of ?? /???????? + ???????? of ?? /?????????? 
Ans: (D) Molarity of both acid and base is same. Amount of acid used is thrice the amount of base. 
Thus the pH of the solution will be highly acidic, 
 
 
 
Page 4


JEE Solved Example on Ionic Equilibrium 
JEE Mains 
Q1:  The conjugate acid of ????
?? -
is 
(A) ????
?? 
(B) ????
?? ???? 
(C) ????
?? +
 
(D) ?? ?? ?? ?? 
Ans: (A) The conjugate acid has one proton (H
+
) more. Hence, for NH
2
-
the conjugate acid would be 
NH
3
 (the positive charge of H
+
and the negative charge of NH
2
 
-
cancel each other out). 
Q2:  Out of the following, amphiprotic species are 
I. ??????
?? ?? -
 
II. ????
-
 
III. ?? ?? ????
?? -
 
IV. ?????? ?? -
 
(A) I, III, IV 
(B) I and III 
(C) III and IV 
(D) All 
Ans: (C) H
2
PO
4
-
And HCO
3
-
are amphoteric in nature. 
HCl+ H
2
PO
4
-
? H
3
PO
4
+ Cl
-1
NaOH+ H
2
PO
4
-
? HPO
4
-2
+ H
2
O
 
 
Q?? : ???? of an aqueous solution of ???????? at ????
°
?? should be 
(A) 7 
(B) > ?? 
(C) < ?? 
(D) 0 
Ans: (C) Halides and alkaline metals dissociate and do not affect the H
+
as the cation does not alter 
the H
+
and the anion does not attract the H
+
from water. This is why NaCl is a neutral salt. But pH of 
water decreases as the temperature increases. So option C is correct 
 
 
 
 
Q4:  1 cc of ?? . ?? ???????? is added to 99 cc solution of ???????? . The ???? of the resulting solution will be 
(A) 7 
(B) 3 
(C) 4 
(D) 1 
Ans: (C) No' moles HCl =
(1×0.1)
1000
= 10
-4
 
Volume = 1dm
3
 Concentration = moles / volume =
10
-4
1
= 10
-4
 This gives a pH of 4 so option (C) is 
correct. 
 
Q5: ???????? of 
?? ??????
?? ?? ????
?? is mixed with ???????? of 
?? ??????
 ?? ?? ????
?? . The ???? of the resulting solution is 
(A) 1 
(B) 2 
(C) 2.3 
(D) None of these 
Ans: (B) 
We have 50 mL of 
M
200
H
2
SO
4
 
M
200
= 0.005M solution 
H
2
SO
4
? 2H
+
+ SO
4
2-
 Then [H
+
] = 2 × 0.005M= 0.01M
pH = -log [H
+
]
pH = -log 0.01M
pH = 2.00
 
 
Q6:  If ????
?? for fluoride ion at ????
°
?? is 10.83 , the ionisation constant of hydrofluoric acid in water 
at this temperature is: 
(A) ?? . ???? × ????
-?? 
(B) ?? . ???? × ????
-?? 
(C) ?? . ???? × ????
-?? 
(D) ?? . ???? × ????
-?? 
Ans: (C) The equation for the dissociation of HF is as follow: 
HF + H
2
O ? H
3
O
+
+ F
-
 
Here pK
b
= 10.83 
? -log K
b
= 10.83 
Hence, K
b
= 1.48 × 10
-11
 
Thus lonization constant of acid K
a
= K
W
/K
b
 
?? ?? = 10
-14
?? ?? = 10
-14
/1.48 × 10
-11
?? ?? = 6.76 × 10
-4
 
Thus Ionization constant of HF is 6.76 × 10
-4
 
 
Q7:  The ???? of an aqueous solution of ?? . ???? solution of a weak monoprotic acid which is ?? % 
ionised is 
(A) 1 
(B) 2 
(C) 3 
(D) 11 
 Ans: (C) [?? ?? ?? +
] = ?? ?? = ?? . ?? ×
?? ??????
= ?? × ????
-?? pH = -log [H
3
O
+
] = -log 10
-3
= 3
 
 
Q8:  If ?? ?? and ?? ?? be first and second ionization constant of ?? ?? ????
?? and ?? ?? >> ?? ?? which is 
incorrect. 
(A) [?? +
] = [?? ?? ????
?? -
] 
(B) [?? +
] = ?? ?? [?? ?? ????
?? ] 
(C) ?? ?? = [??????
?? -
] 
(D) [?? +
] = ?? [????
?? ?? -
] 
Ans: (D) 
H
3
PO
4
= H
+
+ H
2
PO
4
-
… … … . K
1
H
2
PO
4
-
= H
+
+ HPO
4
2-
… … … . . K
2
HPO
4
2-
= H
+
+ PO
4
3-
… … … … . K
3
 K
3
= [HPO
4
2-
]
H
+
= [PO
4
3-
]
 
 
Q9: Which of the following solution will have ???? close to 1.0 ? 
(A) ?????????? of ?? /???????????? + ?????????? of ?? /???? 
(B) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(C) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(D) ???????? of ?? /???????? + ???????? of ?? /?????????? 
Ans: (D) Molarity of both acid and base is same. Amount of acid used is thrice the amount of base. 
Thus the pH of the solution will be highly acidic, 
 
 
 
Q10: What is the percentage hydrolysis of ???????? in ?? /???? solution when the dissociation constant 
for ?????? is ?? . ?? × ????
-?? and ?? ?? = ?? . ?? × ????
-????
 
(A) 2.48 
(B) 5.26 
(C) 8.2 
(D) 9.6 
Ans: (A) Since NaCN is the salt of a weak acid (HCN) and strong base (NaOH ), the degree of 
hydrolysis, 
?? = v
?? ?? ?? ?? × ?? =
1.0 × 10
-14
× 80
1.3 × 10
-9
 = v6.16 × 10
-4
 = 2.48 × 10
-2
 
? Percentage hydrolysis of NaCN in N/80 Solution is 2.48 
 
Q11:  The compound whose ?? . ???? solution is basic is 
(A) Ammonium acetate 
(B) Ammonium chloride 
(C) Ammonium sulphate 
(D) Sodium acetate 
Ans: (D) Sodium acetate undergoes anionic hydrolysis 
CH
3
COO
-
+ H
2
O ? CH
3
COOH+ OH
-
 
 
Q. 12 The ˜ ???? of the neutralisation point of ?? . ?? ?? ammonium hydroxide with ?? . ?? ???????? is 
(A) 1 
(B) 6 
(C) 7 
(D) 9 
Ans: (B) NH
4
OH+ HCl ? NH
4
Cl 
NH
4
Cl Is a salt of weak base and strong acid .so it give s acidic solution with pH > 7 
 
Q13:  If equilibrium constant of 
????
?? ???????? + ?? ?? ?? ? ????
?? ?????? -
+ ?? ?? ?? +
 
Is ?? . ?? × ????
-?? , equilibrium constant for ????
?? ???????? + ????
-
????
?? ?????? -
+ ?? ?? ?? is 
(A) ?? . ?? × ????
-?? 
(B) ?? . ?? × ????
-?? 
Page 5


JEE Solved Example on Ionic Equilibrium 
JEE Mains 
Q1:  The conjugate acid of ????
?? -
is 
(A) ????
?? 
(B) ????
?? ???? 
(C) ????
?? +
 
(D) ?? ?? ?? ?? 
Ans: (A) The conjugate acid has one proton (H
+
) more. Hence, for NH
2
-
the conjugate acid would be 
NH
3
 (the positive charge of H
+
and the negative charge of NH
2
 
-
cancel each other out). 
Q2:  Out of the following, amphiprotic species are 
I. ??????
?? ?? -
 
II. ????
-
 
III. ?? ?? ????
?? -
 
IV. ?????? ?? -
 
(A) I, III, IV 
(B) I and III 
(C) III and IV 
(D) All 
Ans: (C) H
2
PO
4
-
And HCO
3
-
are amphoteric in nature. 
HCl+ H
2
PO
4
-
? H
3
PO
4
+ Cl
-1
NaOH+ H
2
PO
4
-
? HPO
4
-2
+ H
2
O
 
 
Q?? : ???? of an aqueous solution of ???????? at ????
°
?? should be 
(A) 7 
(B) > ?? 
(C) < ?? 
(D) 0 
Ans: (C) Halides and alkaline metals dissociate and do not affect the H
+
as the cation does not alter 
the H
+
and the anion does not attract the H
+
from water. This is why NaCl is a neutral salt. But pH of 
water decreases as the temperature increases. So option C is correct 
 
 
 
 
Q4:  1 cc of ?? . ?? ???????? is added to 99 cc solution of ???????? . The ???? of the resulting solution will be 
(A) 7 
(B) 3 
(C) 4 
(D) 1 
Ans: (C) No' moles HCl =
(1×0.1)
1000
= 10
-4
 
Volume = 1dm
3
 Concentration = moles / volume =
10
-4
1
= 10
-4
 This gives a pH of 4 so option (C) is 
correct. 
 
Q5: ???????? of 
?? ??????
?? ?? ????
?? is mixed with ???????? of 
?? ??????
 ?? ?? ????
?? . The ???? of the resulting solution is 
(A) 1 
(B) 2 
(C) 2.3 
(D) None of these 
Ans: (B) 
We have 50 mL of 
M
200
H
2
SO
4
 
M
200
= 0.005M solution 
H
2
SO
4
? 2H
+
+ SO
4
2-
 Then [H
+
] = 2 × 0.005M= 0.01M
pH = -log [H
+
]
pH = -log 0.01M
pH = 2.00
 
 
Q6:  If ????
?? for fluoride ion at ????
°
?? is 10.83 , the ionisation constant of hydrofluoric acid in water 
at this temperature is: 
(A) ?? . ???? × ????
-?? 
(B) ?? . ???? × ????
-?? 
(C) ?? . ???? × ????
-?? 
(D) ?? . ???? × ????
-?? 
Ans: (C) The equation for the dissociation of HF is as follow: 
HF + H
2
O ? H
3
O
+
+ F
-
 
Here pK
b
= 10.83 
? -log K
b
= 10.83 
Hence, K
b
= 1.48 × 10
-11
 
Thus lonization constant of acid K
a
= K
W
/K
b
 
?? ?? = 10
-14
?? ?? = 10
-14
/1.48 × 10
-11
?? ?? = 6.76 × 10
-4
 
Thus Ionization constant of HF is 6.76 × 10
-4
 
 
Q7:  The ???? of an aqueous solution of ?? . ???? solution of a weak monoprotic acid which is ?? % 
ionised is 
(A) 1 
(B) 2 
(C) 3 
(D) 11 
 Ans: (C) [?? ?? ?? +
] = ?? ?? = ?? . ?? ×
?? ??????
= ?? × ????
-?? pH = -log [H
3
O
+
] = -log 10
-3
= 3
 
 
Q8:  If ?? ?? and ?? ?? be first and second ionization constant of ?? ?? ????
?? and ?? ?? >> ?? ?? which is 
incorrect. 
(A) [?? +
] = [?? ?? ????
?? -
] 
(B) [?? +
] = ?? ?? [?? ?? ????
?? ] 
(C) ?? ?? = [??????
?? -
] 
(D) [?? +
] = ?? [????
?? ?? -
] 
Ans: (D) 
H
3
PO
4
= H
+
+ H
2
PO
4
-
… … … . K
1
H
2
PO
4
-
= H
+
+ HPO
4
2-
… … … . . K
2
HPO
4
2-
= H
+
+ PO
4
3-
… … … … . K
3
 K
3
= [HPO
4
2-
]
H
+
= [PO
4
3-
]
 
 
Q9: Which of the following solution will have ???? close to 1.0 ? 
(A) ?????????? of ?? /???????????? + ?????????? of ?? /???? 
(B) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(C) ???????? of ?? /?????????? + ???????? of ?? /???????????? 
(D) ???????? of ?? /???????? + ???????? of ?? /?????????? 
Ans: (D) Molarity of both acid and base is same. Amount of acid used is thrice the amount of base. 
Thus the pH of the solution will be highly acidic, 
 
 
 
Q10: What is the percentage hydrolysis of ???????? in ?? /???? solution when the dissociation constant 
for ?????? is ?? . ?? × ????
-?? and ?? ?? = ?? . ?? × ????
-????
 
(A) 2.48 
(B) 5.26 
(C) 8.2 
(D) 9.6 
Ans: (A) Since NaCN is the salt of a weak acid (HCN) and strong base (NaOH ), the degree of 
hydrolysis, 
?? = v
?? ?? ?? ?? × ?? =
1.0 × 10
-14
× 80
1.3 × 10
-9
 = v6.16 × 10
-4
 = 2.48 × 10
-2
 
? Percentage hydrolysis of NaCN in N/80 Solution is 2.48 
 
Q11:  The compound whose ?? . ???? solution is basic is 
(A) Ammonium acetate 
(B) Ammonium chloride 
(C) Ammonium sulphate 
(D) Sodium acetate 
Ans: (D) Sodium acetate undergoes anionic hydrolysis 
CH
3
COO
-
+ H
2
O ? CH
3
COOH+ OH
-
 
 
Q. 12 The ˜ ???? of the neutralisation point of ?? . ?? ?? ammonium hydroxide with ?? . ?? ???????? is 
(A) 1 
(B) 6 
(C) 7 
(D) 9 
Ans: (B) NH
4
OH+ HCl ? NH
4
Cl 
NH
4
Cl Is a salt of weak base and strong acid .so it give s acidic solution with pH > 7 
 
Q13:  If equilibrium constant of 
????
?? ???????? + ?? ?? ?? ? ????
?? ?????? -
+ ?? ?? ?? +
 
Is ?? . ?? × ????
-?? , equilibrium constant for ????
?? ???????? + ????
-
????
?? ?????? -
+ ?? ?? ?? is 
(A) ?? . ?? × ????
-?? 
(B) ?? . ?? × ????
-?? 
(C) ?? . ???? × -????
-?? 
(D) ?? . ???? × ????
????
 
Ans: (B) The pH of the solution at the equivalence point will be greater than 7 due to salt hydrolysis. 
So an indicator giving colour on the basic side will be suitable. 
 
Q14:  The ????
?? of a weak acid, ???? , is 4. 80. The ????
?? of a weak base, ?????? , is 4. 78. The ???? of an 
aqueous solution of the corresponding salt, BA, will be: 
(A) 8.58 
(B) 4.79 
(C) 7.01 
(D) 9.22 
Ans: (C) It is a salt of weak acid and weak base. 
[?? +
] = v
?? ?? × ?? ?? ?? ?? 
On solving we get 
pH = 7.01 
 
Q15:  The range of most suitable indicator which should be used for titration of 
?? -
????
+
(?? . ???? , ???????? ) with 0.1 ???????? should be (Given: ?? ?? (?? -
)
= ????
-?? ) 
(A) ?? - ?? 
(B) 3-5 
(C) 6-8 
(D) ?? - ???? 
Ans: (B) CH
3
COONa+ HCl ? CH
3
COOH+ NaCl 
It is an example of titration of weak base with strong acid. Observed pH range for the end point is 
3.00 to 6.00 . 
 
Q16:  The solubility of ?? ?? ?? ?? is ?? mol dm-3. Its solubility product is 
(A) ?? ?? ?? 
(B) ???? ?? ?? 
(C) ???? ?? ?? 
(D) ?????? ?? ?? 
Ans: (D) 
A
2
X
3
? 2 A
+3
+ 3X
-2
2y 3y
Ksp= [A
+3
]
2
[ B
-2
]
3
= (2y)
2
(3y)
3
= 108y
5