NEET Previous Year Questions (2014-2024): Equilibrium

NEET Previous Year Questions (2014-2024): Equilibrium | Chemistry Class 11 PDF Download

 Table of contents 2022 2021 2020 2019 2018 2017 2016 2015 2014

2022

Q1: The pH of the solution containing 50 mL each of 0.10 sodium acetate and 0.01 M acetic acid is
[Given pKa of CH3COOH = 4.57]       (NEET 2022 Phase 1)
(a) 4.57
(b) 2.57

(c) 5.57
(d) 3.57
Ans:
c
Given solution is acidic buffer solution.

Q2:for the above reaction at 298 K, KC is found to be 3.0 × 10–59. If the concentration of O2 at equilibrium is 0.040 M then concentration of O3 in M is       (NEET 2022 Phase 1)
(a) 2.4 × 1031
(b) 1.2 × 1021

(c) 4.38 × 10–32
(d) 1.9 × 10–63
Ans:
c

Q3:  0.01 M acetic acid solution is 1% ionised, then pH of this acetic acid solution is :      (NEET 2022 Phase 2)
(a) 1
(b) 3

(c) 2
(d) 4
Ans:
(d)

Q4: Kp for the following reaction is 3.0 at 1000 K.    (NEET 2022 Phase 2)
CO2(g) + C(s) $⇌$ 2CO(g)
What will be the value of Kc for the reaction at the same temperature?
(Given : R = 0.083 L bar K$−$1 mol$−$1)
(a) 3.6
(b) 0.36
(c) 3.6 $×$ 10$−$2
(d) 3.6 $×$ 10$−$3
Ans:
(c)

2021

Q1: The pKb of dimethylamine and pka of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate at solution is:      (NEET 2021)
(a) 7.75
(b) 6.25
(c) 8.50
(d) 5.50
Ans:
(a)

Calculate the pH of dimethylammonium acetate is as follows:

pH = 7.75
Hence, option 1 is the correct answer.

2020

Q1: Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2×10–15.     (NEET 2020)
(a) 1 × 10-13 M
(b) 1 × 108 M

(c) 2 × 10-13 M
(d) 2 × 10-8 M

Ans: (c)

2019

Q1:  pH of a saturated solution of Ca(OH)2 is 9. The solubility product (Ksp) of Ca(OH)2 is:    (NEET 2019)
(a) 0.5 × 10-15
(b) 0.25 × 10-10
(c) 0.125 × 10-15
(d) 0.5 × 10-10
Ans:
(a)

pH = 9 Hence pOH = 14 - 9 = 5
[OH-] = 10-5 M

Thus Ksp = [Ca2+][OH-]2

Q2: Conjugate base for Brönsted acids H2O and HF are :    (NEET 2019)
(a) OH- and H2F+, respectively
(b) H3O+ and F-, respectively
(c) OHand F-, respectively
(d) H3O+ and H2F+, respectively
Ans:
(c)

HF on loss of Hion becomes Fis the conjugate base of HF
Example :

Q3: Which will make basic buffer?    (NEET 2019)
(a) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH
(b) 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH
(c) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH
(d) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Ans:
c

This is basic solution due to NaOH.
This is not basic buffer.

Hydrolysis of salt takes place.
This is not basic buffer.

This is basic buffer

⇒ Neutral solution

2018

Q1: The solubility of BaSO4 in water 2.42×10gL-1 at 298 K. The value of solubility product (Ksp) will be(Given molar mass of BaSO4 = 233 g mol-1)    (NEET 2018)
(a) 1.08 × 10-10 mol2 L-2
(b) 1.08 × 10-12 mol2 L-2
(c) 1.08 × 10-14 mol2 L-2
(d) 1.08 × 10-8 mol2 L-2
Ans:
(a)

Q2: Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :

pH of which one of them will be equal to 1 ?    (NEET 2018)
(1) b
(2) a
(3) d
(4) c

Ans: (d)
(A) 60 mL M/$10$ HCl + 40 mL M/$10$ NaOH
Mili. moles of HCl = 60 $×$ 1/$10$ = 6
Mili. moles of NaOH = 40 $×$ 1/10 = 4
Mili. moles of HCl remaining = 6 - 4 = 2
Total volume will be 60 + 40 = 100 mL
Concentration of [H+] = $2$2/100 = 2 $×$ 10-2
$∴$ pH = 2 - log2 = 1.7
(B) 55 mL M/$10$ HCl + 45 mL M/$10$ NaOH
Mili. moles of HCl = 55 $×$ 1/10 = 5.5
Mili. moles of NaOH = 45 $×$ $1$/10 = 4.5
Mili. moles of HCl remaining = 5.5 - 4.5 = 1
Concentration of [H+] = $1100$ = 10-2
$∴$ pH = 2
(C) 75 mL M/$5$ HCl + 25 mL M/$5$ NaOH
Mili. moles of HCl = 75 $×$ $1$/5 = 15
Mili. moles of NaOH = 25 $×$1/5 = 5
Mili. moles of HCl remaining = 15 - 5 = 10
Total volume will be 75 + 25 = 100 mL
Concentration of [H+] = 10/100 = 10-1
$∴$ pH = 1
(D) 100 mL M/$10$ HCl + 100 mL M/ 1$0$ NaOH
Mili. moles of HCl = 100 $×$ $1$/10 = 10
Mili. moles of NaOH = 100 $×$ $1$/10 = 10
Mili. moles of HCl remaining = 10 - 10 = 0
So, it is neutral solution.
$∴$ pH = 7

Q3: Which one of the following conditions will favour maximum formation of the product in the reaction
A2(g) + B2(g) ⇌ X2(g) ΔrH = -X kJ ?    (NEET 2018)
(a) Low temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and high pressure
(d) High temperature and low pressure
Ans:
(a)
For reaction ΔH = - ve and Δng = - ve
∴ High P, Low T, favour product formation.

2017

Q1: Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2 × 10-4 mol L-1 Solubility product of Ag2C2O4 is :-    (NEET 2017)
(a) 2.66 × 10-12
(b) 4.5 × 10-11
(c) 5.3 × 10-12
(d) 2.42 × 10-8
Ans:
c

Q2: The equilibrium constant of the following are :    (NEET 2017)

The equilibrium constant (K) of the reaction:

(a)
(b)
(c)
(d)
Ans:
a

Q3: A 20 litre container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be :-    (NEET 2017)

(a) 10 litre
(b) 4 litre
(c) 2 litre
(d) 5 litre
Ans
:(d)

2016

Q1: MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 x 10-13 at room temperature, which statements would be true in regard to MY and NY3?    (NEET 2016 Phase 1)
(a) The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities.
(b) The molar solubilities of MY and NY3 in water are identical.
(c) The molar solubility of MY in water is less than that of NY3.
(d) The salts MY and NY3 are more soluble in 0.5 M KY than in pure water.
Ans:
(c)
For MY,

Q2: Consider the following liquid-vapour equilibrium.   (NEET 2016 Phase 1)
Liquid $⇌$ Vapour
Which of the following relations is correct ?
(a)

(b)

(c)

(d)

Ans: (b)
This is Clausius-Clapeyron equation.

Q3: The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) ina 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10−9) is
(a) 0.0060%
(b) 0.013%
(c) 0.77%
(d) 1.6%     (NEET 2016 Phase 2)
Ans:
(b)
C5H5N + H2O → C5H5N+H + OH
So, the amount of pyridine that forms pyridinium ion is α.

So, percentage of pyridine that forms pyridinium ion
= 1.30 × 10–4 × 100
= 1.30 × 10–2 = 0.013%

Q4: Which of the following fluro-compounds is most likely to behave as a Lewis base ?
(a) BF3
(b) PF3
(c) CF4
(d) SiF4     (NEET 2016 Phase 2)
Ans:
(b)
PF3 is lewis base because on P atom there is lone pair.

Q5: The solubility of AgCl(s) with solubility product 1.6 × 10−10 in 0.1 M NaCl solution would be
(a) 1.26 $×$ 10$−$5 M
(b) 1.6 $×$ 10$−$9 M
(c) 1.6 $×$ 10$−$11 M
(d) Zero     (NEET 2016 Phase 2)
Ans:
(b)

Concentration of Cl is (s + 0.1) mol L–1 because s mol L–1 from ionization of AgCl and 0.1 mol L–1 from ionization of 0.1 M NaCl.
Now, Ksp = [Ag+][Cl ]
$⇒$ 1.6 × 10–10 = s (s + 0.1)
$⇒$ 1.6 × 10–10 = s (0.1) {$∵$ s << 0.1}
$⇒$ s = 1.6 × 10–9 M

2015

Q1: The Ksp of Ag2CrO4, AgCl, AgBr and AgI are respectively, 1.1 x 10-12, 1.8 x 10-10, 5.0 x 10-13, 8.3 x 10-17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4 ?    (NEET / AIPMT Cancelled Paper 2015)
(a) Ag2CrO4
(b) AgI
(c) AgCl
(d) AgBr
Ans:
(a)

Q2: If the value of an equilibrium constant for a particular reaction is 1.6 x 1012, then at equilibrium the system will contain :
(a) similar amounts of reactants and products
(b) all reactants
(c) mostly reactants

(d) mostly products   (NEET / AIPMT Cancelled Paper 2015)
Ans:
(d)

Q3:What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
(a) 2.0
(b) 7.0
(c) 1.04
(d) 12.65                   (NEET / AIPMT 2015)
Ans:
(d)

As equal volumes of HCl and NaOH are added so the volume of resulting solution becomes double and the concentration of the solution becomes half.
$∴$ [OH-] = 0.09/2 = 0.045 M
$∴$ pOH = – log[OH ] = –log [0.045] = 1.35
$∴$ pH = 14 – pOH = 14 – 1.35 = 12.65

Q4: Which one of the following pairs of solution is not an acidic buffer ?
(a) CH3COOH and CH3COONa
(b) H2CO3 and Na2CO3
(c) H3PO4 and Na3PO4
(d) HClO4 and NaClO4         (NEET / AIPMT 2015)
Ans:
(d)
An acidic buffer is a mixture of a weak acid and its salt with a strong base. Among CH3COOH, H2CO3, H3PO4 and HClO4, the HClO4 is a strong acid while all other are weak acid thus, HClO4 and NaClO4 does not constitute to form an acidic buffer.

2014

Q1: For the reversible reaction :
N2(g) + 3H2(g) ⇔ 2NH3 (g) + heat
The equilibrium shifts in forward direction :    (NEET / AIPMT 2014)
(a) by decreasing the concentrations of N2(g) and H2(g)
(b) by increasing pressure and decreasing temperature
(c) by increasing the concentration of NH3(g)
(d) by decreasing the pressure

Ans: (b)
As the forward reaction is exothermic and leads to lowering of pressure (produces lesser number of gaseous moles). Hence, According to Le Chatelier's principle, at high pressure and low temperature, the given reversible reaction will shift in forward direction to form more product.

Q2: Using the Gibbs energy change, ΔG° = +63.3 kJ, for the following reaction,

(NEET / AIPMT 2014)
(a) 2.9 × 10-3
(b) 7.9 × 10-2
(c) 3.2 × 10-26
(d) 8.0 × 10-12
Ans:
(d)

Q3: For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2 , respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2 , it is readily observed that :  (NEET / AIPMT 2014)
(a)

(b)
(c)
(d)
Ans:
(c)
For exothermic reactions, on increasing the temperature the value of equilibrium constant decreases.

Q4: Which of the following salts will give highest pH in water ?    (NEET / AIPMT 2014)
(a) Na2CO3
(b) CuSO4
(c) KCl
(d) NaCl
Ans:
(a)
The highest pH refers to the basic solution containing OH- ions. Therefore, the basic salt releasing OH- ions on hydrolysis will give highest pH in water.
Only the salt of a strong base and weak acid would release OHion on hydrolysis. Among the given salts, Na2CO3 corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H2CO3 [weak acid]

The document NEET Previous Year Questions (2014-2024): Equilibrium | Chemistry Class 11 is a part of the NEET Course Chemistry Class 11.
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