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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 
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FAQs on JEE Advanced Previous Year Questions (2021 - 2024): Ionic Equilibrium - Chemistry for JEE Main & Advanced

1. What is ionic equilibrium and why is it important in JEE Advanced Chemistry?
Ans.Ionic equilibrium refers to the state in which the concentrations of ions in a solution remain constant over time, despite ongoing processes, such as dissociation and recombination of ions. It is crucial in JEE Advanced Chemistry because it forms the basis for understanding acid-base reactions, solubility, and buffer solutions, which are often tested in the exam.
2. How do you calculate the pH of a weak acid solution in ionic equilibrium?
Ans.To calculate the pH of a weak acid solution, you first need to determine the concentration of the acid and its dissociation constant (Ka). You can set up an equilibrium expression based on the dissociation of the acid (HA ⇌ H⁺ + A⁻) and use the ICE (Initial, Change, Equilibrium) table to find the concentration of H⁺ ions. Finally, pH is calculated using the formula pH = -log[H⁺].
3. What is the difference between strong and weak electrolytes in ionic equilibrium?
Ans.Strong electrolytes completely dissociate into ions in solution, leading to a high concentration of ions, while weak electrolytes only partially dissociate, resulting in a lower concentration of ions. This distinction affects properties like conductivity and pH, and it is essential for solving problems related to ionic equilibrium in JEE Advanced.
4. How can Le Chatelier's principle be applied to ionic equilibrium problems?
Ans.Le Chatelier's principle states that if a system at equilibrium is disturbed, the system will shift in a direction that counteracts the disturbance. In ionic equilibrium problems, this principle can be used to predict how changes in concentration, temperature, or pressure will affect the position of equilibrium, which is a common type of question in JEE Advanced.
5. What role do buffers play in maintaining ionic equilibrium?
Ans.Buffers are solutions that resist changes in pH when small amounts of acid or base are added. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid. In ionic equilibrium, buffers help maintain the pH of a solution, which is critical in biological and chemical systems. Understanding buffers is often essential for JEE Advanced exam questions on ionic equilibrium.
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