JEE Advanced Previous Year Questions (2021-2023): Ionic Equilibrium | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Advanced Previous Year Questions 
(2021-2023): Ionic Equilibrium 
2023 
Single Correct Type 
JEE Advanced 2023 Paper 1 Online 
Q1. On decreasing the ?? ?? from 7 to 2 , the solubility of a sparingly soluble salt (MX) of a weak acid 
(HX) increased from ????
-?? ?????? ?? -?? to ????
-?? ?????? ?? -?? . The ?? ?? ?? of ???? is 
A. 3 
B. 4 
C. 5 
D. 2 
Ans: (B) 
Explanation 
Relationship between solubility, H
+
and K
a
 is given by 
S = v
(K
SP
[H
+
] + K
a
)
K
a
 
If pH = 7 ? (H
+
) = 10
-7
 
S = 10
-4
 mol/L
? 10
-4
= v
K
SP
(10
-7
+ K
a
)
K
a
… (i)
 
10
-3
= v
K
SP
(10
-2
+ K
a
)
K
a
… (ii) 
Dividing and squaring equation (i) by equation (ii), 
(10
-4
)
2
(10
-3
)
2
=
?? ????
(10
-7
+ ?? ?? )
?? ?? ×
?? ?? ?? ????
(10
-2
+ ?? ?? )
 ? 10
-2
=
10
-7
+ K
a
10
-2
+ K
a
 ? 10
-4
+ 10
-2
· K
a
= 10
-7
+ K
a
 ? K
a
? 10
-4
 ? pK
a
= 4
 
 
 
Page 2


JEE Advanced Previous Year Questions 
(2021-2023): Ionic Equilibrium 
2023 
Single Correct Type 
JEE Advanced 2023 Paper 1 Online 
Q1. On decreasing the ?? ?? from 7 to 2 , the solubility of a sparingly soluble salt (MX) of a weak acid 
(HX) increased from ????
-?? ?????? ?? -?? to ????
-?? ?????? ?? -?? . The ?? ?? ?? of ???? is 
A. 3 
B. 4 
C. 5 
D. 2 
Ans: (B) 
Explanation 
Relationship between solubility, H
+
and K
a
 is given by 
S = v
(K
SP
[H
+
] + K
a
)
K
a
 
If pH = 7 ? (H
+
) = 10
-7
 
S = 10
-4
 mol/L
? 10
-4
= v
K
SP
(10
-7
+ K
a
)
K
a
… (i)
 
10
-3
= v
K
SP
(10
-2
+ K
a
)
K
a
… (ii) 
Dividing and squaring equation (i) by equation (ii), 
(10
-4
)
2
(10
-3
)
2
=
?? ????
(10
-7
+ ?? ?? )
?? ?? ×
?? ?? ?? ????
(10
-2
+ ?? ?? )
 ? 10
-2
=
10
-7
+ K
a
10
-2
+ K
a
 ? 10
-4
+ 10
-2
· K
a
= 10
-7
+ K
a
 ? K
a
? 10
-4
 ? pK
a
= 4
 
 
 
2022 
Numerical Type 
JEE Advanced 2022 Paper 2 Online  
Q1. Concentration of ?? ?? ????
?? and ????
?? ????
?? in a solution is ???? and ?? . ?? × ????
-?? ?? , respectively. Molar 
solubility of ???????? ?? in the same solution is ?? × ????
-?? ?? (expressed in scientific notation). The value of 
?? is 
[Given: Solubility product of ???????? ?? (?? ????
) = ?? . ?? × ????
-?? . For ?? ?? ????
?? , ?? ????
 is very large and ?? ?? ?? = ?? . ?? ×
????
-?? ] 
Ans: 6 
Given, 
Conentration of H
2
SO
4
= 1M 
Conentration of Na
2
SO
4
= 1.8 × 10
-2
M 
 H
2
SO
4
 ? HSO
4
-
+ H
+
  
At ?? = 0 1M - - 
At ?? = ?? - 1M 1M 
    
 Na
2
SO
4
 2Na
+
 +SO
4
 
2-
 
At ?? = 0 1.8 × 10
-2
 - - 
at ?? = ?? - 2(1.8 × 10
-2
) 1.8 × 10
-2
 
[H
+
] = 1M and [SO
4
2-
] = 1.8 × 10
-
 
 
 
?? ?? =
1.8×10
-2
×1
1
= 1.8 × 10
-2
 and it is given that K
?? 2
(Q
?? ) = 1.2 × 10
-2
M 
Since, K
?? 2
 (i.e, Q
?? ) < K
?? 
? Rather than dissociation of HSO
4
-
into H
+
and SO
4
2-
 ions, association between already present H
+
and SO
4
2-
 
will take place 
Assuming ' ?? ' mol/L of SO
4
2-
 and H
+
combines to form HSO
4
-
 
 
Page 3


JEE Advanced Previous Year Questions 
(2021-2023): Ionic Equilibrium 
2023 
Single Correct Type 
JEE Advanced 2023 Paper 1 Online 
Q1. On decreasing the ?? ?? from 7 to 2 , the solubility of a sparingly soluble salt (MX) of a weak acid 
(HX) increased from ????
-?? ?????? ?? -?? to ????
-?? ?????? ?? -?? . The ?? ?? ?? of ???? is 
A. 3 
B. 4 
C. 5 
D. 2 
Ans: (B) 
Explanation 
Relationship between solubility, H
+
and K
a
 is given by 
S = v
(K
SP
[H
+
] + K
a
)
K
a
 
If pH = 7 ? (H
+
) = 10
-7
 
S = 10
-4
 mol/L
? 10
-4
= v
K
SP
(10
-7
+ K
a
)
K
a
… (i)
 
10
-3
= v
K
SP
(10
-2
+ K
a
)
K
a
… (ii) 
Dividing and squaring equation (i) by equation (ii), 
(10
-4
)
2
(10
-3
)
2
=
?? ????
(10
-7
+ ?? ?? )
?? ?? ×
?? ?? ?? ????
(10
-2
+ ?? ?? )
 ? 10
-2
=
10
-7
+ K
a
10
-2
+ K
a
 ? 10
-4
+ 10
-2
· K
a
= 10
-7
+ K
a
 ? K
a
? 10
-4
 ? pK
a
= 4
 
 
 
2022 
Numerical Type 
JEE Advanced 2022 Paper 2 Online  
Q1. Concentration of ?? ?? ????
?? and ????
?? ????
?? in a solution is ???? and ?? . ?? × ????
-?? ?? , respectively. Molar 
solubility of ???????? ?? in the same solution is ?? × ????
-?? ?? (expressed in scientific notation). The value of 
?? is 
[Given: Solubility product of ???????? ?? (?? ????
) = ?? . ?? × ????
-?? . For ?? ?? ????
?? , ?? ????
 is very large and ?? ?? ?? = ?? . ?? ×
????
-?? ] 
Ans: 6 
Given, 
Conentration of H
2
SO
4
= 1M 
Conentration of Na
2
SO
4
= 1.8 × 10
-2
M 
 H
2
SO
4
 ? HSO
4
-
+ H
+
  
At ?? = 0 1M - - 
At ?? = ?? - 1M 1M 
    
 Na
2
SO
4
 2Na
+
 +SO
4
 
2-
 
At ?? = 0 1.8 × 10
-2
 - - 
at ?? = ?? - 2(1.8 × 10
-2
) 1.8 × 10
-2
 
[H
+
] = 1M and [SO
4
2-
] = 1.8 × 10
-
 
 
 
?? ?? =
1.8×10
-2
×1
1
= 1.8 × 10
-2
 and it is given that K
?? 2
(Q
?? ) = 1.2 × 10
-2
M 
Since, K
?? 2
 (i.e, Q
?? ) < K
?? 
? Rather than dissociation of HSO
4
-
into H
+
and SO
4
2-
 ions, association between already present H
+
and SO
4
2-
 
will take place 
Assuming ' ?? ' mol/L of SO
4
2-
 and H
+
combines to form HSO
4
-
 
 
K
?? 2
=
[H
+
][SO
4
2-
]
[HSO
4
-
]
1.2 × 10
-2
=
(1 - ?? )(1.8 × 10
-2
- ?? )
(1 + ?? )
 ? ?? << 1, so (1 + ?? ) ˜ 1 and (1 - ?? ) ˜ 1
1.2 × 10
-2
= 1.8 × 10
-2
- ?? ?? = (1.8 × 10
-2
) - (1.2 × 10
-2
)
?? = 0.6 × 10
-2
M
 
 So, [SO
4
2-
] = 1.8 × 10
-2
- ?? [SO
4
2-
] = (1.8 × 10
-2
) - (0.6 × 10
-2
)
[SO
4
2-
] = 1.2 × 10
-2
M
 
 
 
Given,  K
????
= 1.6 × 10
-8
 
? ?? (1.2 × 10
-2
+ ?? ) = 1.6 × 10
-8
 
Since, ?? << 1, So 1.2 × 10
-2
+ ?? ˜ 1.2 × 10
-2
 
So, ?? × 1.2 × 10
-2
= 1.6 × 10
-8
 
 ? ?? =
1.6 × 10
-8
1.2 × 10
-2
 ? ?? = 1.33 × 10
-6
 
?? × 10
-?? M = 1.33 × 10
-6
M 
So, Y = 6 
Hence, the value of ?? is 6 . 
JEE Advanced 2022 Paper 1 Online 
Q2. A solution is prepared by mixing 0.01 mol each of ?? ?? ????
?? , ?????????? ?? , ????
?? ????
?? , [Given: ?? ?? ????
 and 
?? ?? ????
 of ?? ?? ????
?? are 6.37 and 10.32, respectively; ?????? ?? = ?? . ???? 
Ans: 10.00 to 10.04 
Here, we have a buffer of NaHCO
3
 and Na
2
CO 
? ???? = ?? ?? ?? 2
+ log 
[ Salt ]
[ Acid ]
 
= 10.32 + log 
1
2
 
= 10.32 - log = 10.32 - 0.3 
? ???? = 10.02