31 Year NEET Previous Questions: Equilibrium - 2 Free MCQ Practice Test

Given s = 0.5 × 10^–4 moles/lit

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.

but pOH+ pH = 14  or pOH = 14 – pH

14 – 9.25 – 0 = pK_b pK_b = 4.75

For reaction to proceed from right to left

i.e the reaction will be fast in backward direction i.e r_b > r_f.

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.

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

K_sp 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

For 

For an acid-base indicator

Taking negative on both sides

or we can write  

IV^th group needs higher S^2– 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 II^nd group radicals.

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.

Given K_b = 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

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

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

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

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

Given,

We have to calculate

For this equation,

Now operate,

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

∴  pOH = 4

[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.

The equilibrium constant for the system can be written as K = [Fe³⁺][OH^-]³ (activity of the solid is taken as unity).

Let the initial concentration of OH^- be c. If OH^- is reduced to c/4, substitute into the expression for K to get K = [Fe³⁺]_new × (c/4)³.

Equating initial and final K: [Fe³⁺]_initial × c³ = [Fe³⁺]_new × (c³/4³). Cancelling c³ gives [Fe³⁺]_new = 4³ × [Fe³⁺]_initial.

Thus [Fe³⁺] increases by a factor of 4³ = 64, so the correct choice is option C

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.

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

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.

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.

Given 

K_a1 , = 1.5 × 10^{– 5 .}...(i)

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

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

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

pH = 12 or pOH = 2 

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

Δn = 2 – 1 = + 1

∴ K_c and K_p are not equal.