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31 Year NEET Previous Year Questions: Equilibrium - 2


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30 Questions MCQ Test Chemistry 31 Years NEET Chapterwise Solved Papers | 31 Year NEET Previous Year Questions: Equilibrium - 2

31 Year NEET Previous Year Questions: Equilibrium - 2 for NEET 2022 is part of Chemistry 31 Years NEET Chapterwise Solved Papers preparation. The 31 Year NEET Previous Year Questions: Equilibrium - 2 questions and answers have been prepared according to the NEET exam syllabus.The 31 Year NEET Previous Year Questions: Equilibrium - 2 MCQs are made for NEET 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for 31 Year NEET Previous Year Questions: Equilibrium - 2 below.
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31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 1

Solubility of MX2-type eletrolytes is 0.5 × 10–4 mole/lit, then find out Ksp of electrolytes [2002]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 1

Given s = 0.5 × 10–4 moles/lit

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 2

Which has the highest value of pH? [2002]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 2

Na2CO3 is a salt of weak acid H2CO3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 3

Solution of 0.1 N NH4OH and 0.1 N NH4Cl has pH 9.25. Then find out pKb of NH4OH [2002]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 3

but pOH+ pH = 14  or pOH = 14 – pH

14 – 9.25 – 0 = pKb pKb = 4.75

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 4

 The reaction quotient (Q) for the reaction

is given by

The reaction will proceed from right to left if [2003]

where Kc is the equilibrium constant

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 4

For reaction to proceed from right to left

i.e the reaction will be fast in backward direction i.e rb > rf.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 5

Which one of the following orders of acid strength is correct? [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 5

The higher is the tendency to donate proton, stronger is the acid. Thus the correct order is R – COOH > HOH > R – OH > CH ≡ CH depending upon the rate of donation of proton.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 6

Which one of the following compounds is not a protonic acid? [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 6

B(OH)3 does not provide H+ ions in water instead it accepts OH ion and hence it is Lewis acid

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 7

The solubility product of AgI at 25ºC is 1.0 × 10–16 mol2 L–2. The solubiliy of AgI in  10–4 N solution of KI at 25ºC is approximately(in mol L–1) [2003]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 7

Ksp for AgI = 1 × 10–16 In solution of KI, I would be due to the both AgI and KI, 10–4 solution KI would provide = 10–4 I
AgI would provide, say = x I (x is solubility of AgI)

as x is very small ∴ x2 can be ignored

∴      10–4 x = 10–16

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 8

The solubility product of a sparingly soluble salt AX2 is 3.2 x 10-11 . Its solubility ( in moles/litre) is
[2004] 

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 8

For 

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 9

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In) forms of the indicator by the expression [2004]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 9

For an acid-base indicator

Taking negative on both sides

or we can write  

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 10

H2S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is be cau se [2005]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 10

IVth group needs higher S2– i on concentration. In presence of HCl, the dissociation of H2S decreases hence produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H2S which is sufficient to precipitate IInd group radicals.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 11

What is the correct relation ship between the pHs of isomolar solutions of sodium oxide (pH1), sodium sulphide (pH2), sodium selenide (pH3) and sodium telluride (pH4)? [2 00 5]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 11

The solution formed from isomolar solutions of sodium oxide, sodium sulphide, sodium selenide H2O, H2S, H2Se & H2Te respectively.
As the acidic strengh increases from H2O to H2Te thus pH decreases and hence the correct of pHs is pH1 > pH2 > pH3 > pH4.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 12

At 25°C, the dissociation con stant of a base, BOH, is 1. 0 x 10-12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be [2005]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 12

Given Kb = 1.0 × 10–12

On calculation, we get, x = 1.0 × 10–5
Now [OH] =cx                      
= 0.01 × 10–5
= 1 × 10–7mol L–1

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 13

For the reaction [2006]

Which of the following statements is not true ?

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 13

First option is incorrect as the value of KP given is wrong. It should have been

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 14

The hydrogen ion concentration of a 10–8 M HCl aqueous solution at 298 K (Kw = 10–14) is  [2006]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 14

For a solution of 10–8 M HCl [H+] = 10–8 [H+] of water = 10–7
Total [H+] = 10–7 + 10–8 = 10 × 10–8 + 10–8
10–8 (10 + 1) = 11 × 10–8

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 15

Which of the following pairs constitutes a buffer?

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 15

HNO2 is a weak acid and NaNO2 is salt of that weak acid and strong base (NaOH).

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 16

A weak acid, HA, has a Ka of 1.00 × 10–5. If 0.100 mole of this acid dissolved in one litre of water, the percentage of acid dissociated at equilbrium is closest to [2007]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 16

Given Ka = 1.00×10–5,  C= 0.100 mol for a weak electrolyte, degree of dissociation

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 17

The following equilibrium constants are given:



The equilibrium constant for the oxidation of NH3 by oxygen to give NO is

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 17

Given,

We have to calculate

For this equation,

Now operate,

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 18

Calculate the pOH of a solution at 25°C that contains 1× 10–10 M of hydronium ions, i.e. H3O+.

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 18

Given [H3O+] = 1 × 10–10 M at 25º  [H3O+] [OH] = 10–14

∴  pOH = 4

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 19

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+ ion concentration in the mixture ? [2008]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 19

[H3O]+ for a solution having pH = 3 is given by [H3O]+ = 1×10–3 moles/litre  

Similarly for solution having pH = 4, [H3O]+ = 1 × 10–4 moles/ litre and for pH=5 [H3O+] = 1×10–5 moles/ litre Let the volume of each solution in mixture be IL, then total volume of mixture solution  L = (1 + 1 + 1) L =3L
Total [H3O]+ ion present in mixture solution = (10–3 + 10–4 + 10–5) moles Then [H3O]+ ion concentration of mixture solution

=  0.00037 M = 3.7 ×10–4 M.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 20

If the concentration of OH ions in the reaction decreased by   times, then equilibriumconcentration of Fe3+ will increase by : [2008]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 20

For this reaction Keq. is given by

If (OH) is decreased by times then forreaction equilibrium constant to remain constant, we have to increase the concentration of [Fe3+] by a factor of 43 i.e 4× 4 × = 64. Thus option (c) is correct answer.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 21

Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH ? [2008]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 21

The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals increase as we move from Mg to Ba and thus the solution of BaCl2 in water will be most basic and so it will have  highest pH.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 22

The dissociation equilibrium of a gas AB2 can be represented as : [2008]

The degree of dissociation is ‘x’ and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is :

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 22

For the reaction

=  x3 [(1–x) can be neglected in denominator

The partial pressure at equilibrium are calculated on the basis of total number of moles at equilibrium.
Total number of moles

= 2 (1–x) + 2x + x  = (2 + x)

where P is the total pressure.

Since x is very small so can be neglected in denominator Thus, we get

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 23

The values of Kp1 and Kp2 for the reactions

are in the ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio :

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 23

Given reaction are

Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then

After dissociation,  

At equilibrium (1–α) α α
[Let 1 mole of X dissociate with α as degree of dissociation ]
Total number of moles =   1– α + α + α = (1+α)

Thus  

We have,

Dividing (i) by (ii), we get

i.e. Option (c) is correct answer.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 24

The value of equilibrium constant of the reaction
[2008]
The equilibrium constant of the reaction

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 24

Given : Equilibrium constant (K1) for the reaction

  To find equilibrium constant for the following reaction

For this multiply (i) by 2, we get

[Note: When the equation for an equilibrium is multiplied by a factor, the equilibrium constant must be raised to the power equal to the factor]

Now reverse equation (iii), we get

[Note: For a reversible reaction, the equilibrium constant of the backward reaction is inverse of the equilibrium constant for the forward reaction.] Equation (iv) is the same as the required equation (ii), thus K2 for equation (ii) is i.e.
option (b) is correct.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 25

The dissociation constants for acetic acid and HCN at 25°C are 1.5 × 10–5 and 4.5 × 10–10 respectively. The equilibrium constant for the equilibrium [2009] would be:

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 25

Given 

Ka1 , = 1.5 × 10– 5 ....(i)

∴  From (i) and (ii), we find that the equilibrium constant (Ka) for the reaction ,

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 26

Which of the following molecules acts as a Lewis acid ?[2009]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 26

(CH3)3 B - is an electron deficient, th us behave as a lewis acid.

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 27

The ionization constant of ammonium hydroxide is 1.77 × 10–5 at 298 K. Hydrolysis constant of ammonium chloride is: [2009]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 27

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant Kh can be calculated as

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 28

If pH of a saturated solution of Ba (OH)2 is 12, the value of its K(sp) is : [2010]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 28

pH = 12 or pOH = 2 

[∴ Concentration of Ba 2+ is half of OH-]

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 29

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH?Ka for CH3COOH = 1.8 × 10-5 . [2010]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 29

31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 30

In which of the following equilibrium Kc and Kp are not equal? [2010]

Detailed Solution for 31 Year NEET Previous Year Questions: Equilibrium - 2 - Question 30

Δn = 2 – 1 = + 1

∴ Kc and Kp are not equal.

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