Equilibrium Class 11 MCQ Online Test With Answers Questions

Answer: (d) 108 y⁵
Explanation:

= (3y)³ × (2y)²
= 27 y³ × 4y²
= 108 y⁵

Answer: (b) Acetoacetic acid (pK_a = 3.58)
Explanation:
For small concentration of buffering agent and for maximum buffer capacity
[Salt]/ [Acid] ≈ 1
i.e., pH = pK_a

Answer: (c) Na₂CO₃
Explanation:
NaCl is salt of strong acid and strong base so it has pH of about 7.
NH₄Cl is salt of strong acid and weak base so it has pH is much less than 7.
CH₃​COONH₄ is salt of weak acid and weak base so it has pH of about 7.
Na₂​CO₃ is salt of weak acid and strong base so it has pH is much greater than 7.

Answer: (c) 13.30
Explanation:
pH = 13.30
Barium hydroxide is a strong base for both stages of dissociation:
Ba (OH)₂(s) → Ba²⁺ + 2OH^–
So the solution will have 0.20 M hydroxide ions. Now use the auto dissociation product for water:
[H⁺][OH^–] = 1.0 × 10^-14M
[OH^–] = 2.0 × 10^-1M
[H⁺] = 5.0 × 10^-14M
And then pH = −log₁₀ ([H+] = 5.0 × 10^-14)
= 13.30.

Answer: (a) CaO
Explanation:
CaO being a basic oxide does not react with NaOH, however SiO₂ (acidic oxide), BeO (amphoteric oxide) and Br₂O₃ (acidic oxide) react with NaOH.

Answer: (c) 4
Explanation:
pOH= −log [OH^–]
= −log 10^-10
= 10 log 10
= 10pH + pOH
= 14pH
= 14−pOH
= 14−10
= 4

Answer: (b) 0.01%
Explanation:
NaX: Salt of weak acid, strong base.
⇒ α = (√Kh/c)
= √ (Kw)/ (KaC)
= (10^-14)/(10^-5×0.1)
= 10^-4
⇒ % hydrolysis = 0.01%

Answer: (c) 12.65
Explanation:
When equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed, the hydroxide ion concentration is
(0.1 – 0.01)/(2)
​= 0.045 M
The pOH of the solution is pOH = −log [OH^–]
= −log 0.045
= 1.35
The pH of the solution is pH = 14 − pOH
= 14 − 1.35
= 12.65

Answer: (c) 5.3 × 10^-12
Explanation:
[Ag⁺] = 2.2 × 10^-4 mol L^-1
[C₂​O₄^2-] = 0.5[Ag⁺] = 0.5 × 2.2 × 10^-4 mol L^-1 = 1.1 × 10^-4 mol L^-1
K_sp ​=[Ag⁺]² [C₂O₄^2-​]K_sp
= (2.2 × 10^-4 mol L^-1)² × 1.1 × 10^-4 mol L^-1
K_sp = 5.3 × 10^-12

Answer: (c) Remain the Same
Explanation:
Equilibrium constants are not changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature.

The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chateliers Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.

According to Le Chateliers Principle, if you increase the concentration of Reactant, for example, the position of equilibrium will move to the right to decrease the concentration of reactant again.

Answer: (d) Increases as it loses electrons
Explanation:
The acronym “OILRIG” can be used to remember this: Oxidation is Losing electrons, Reduction is Gaining electrons. Therefore, when an atom undergoes oxidation, it loses electrons. This makes its net charge more positive, so its oxidation state increases.

Answer: (c) 4.5 × 10^-11
Explanation:
[Ag⁺] = 2.2 × 10^-4 mol L^-1
[C₂​O₄^2-] = 0.5[Ag⁺] = 0.5 × 2.2 × 10^-4 mol L^-1 =1.1×10^-4 mol L^-1
K_sp ​= [Ag⁺]² [C₂​O₄^2-​] K_sp
= (2.2 × 10^-4 mol L^-1)² × 1.1 × 10^-4 mol L^-1
K_sp = 5.3 × 10^-12

Answer: (d) I^–
Explanation:
According to this theory, an acid is a proton donor and a base is a proton acceptor. Every strong Bronsted acid has a weak conjugate base and every strong base has a weak conjugate acid. The acidity increases in halogen group atoms,
HF < HCl < HBr < HI.
So, HI is highly acidic and their conjugate bases decrease in order F^– > Cl^– > Br^– > I^–.

Answer: (b) 8
Explanation:
k_a k_b = k_w
k_b = 10^-8
k_a × 10^-8 = 10^-14
k_a = 10^-6 = [H⁺]
pH = −log [H⁺]
pH = −log 10^-6
= 6pH + pOH
= 14 pOH
= 14−6
= 8

Answer: (c) 13.30
Explanation:
pH =13.30
Barium hydroxide is a strong base for both stages of dissociation:
Ba (OH)₂(s) → Ba²⁺ + 2OH^–
So the solution will have 0.20 M hydroxide ions. Now use the auto dissociation product for water:
[H⁺][OH^–] = 1.0×10^-14M
[OH^–] = 2.0 × 10^-1M
[H⁺] = 5.0 × 10^-14M
And then pH = −log₁₀ ([H⁺] = 5.0 × 10^-14)
= 13.30.

Answer: (b) 1.7 × 10^-16
Explanation:
Let ”s“ be the solubility of salt M₂​S which undergoes dissociation as follows :-
M₂​S ⇌ 2M⁺ +S^-2
Initial concentration 1 0 0
Concentration at equilibrium 1-s 2s s
Hence, the solubility product (K_sp) = (s) × (2s)²
Therefore, K_sp = 4 × s³
Or K_sp​​ = 4× (3.5×10^-6)³
⇒ K_sp​
= 1.7 × 10^-16

Answer: (d) Be(OH)₂
Explanation:
Be(OH)₂ ​has the lowest value of K_sp at ordinary temperature because Be²⁺ ion is smaller than the other metal ions in the group, which results in a tighter bond with the 0H^– ions, thus much lower solubility.

The solubility of a hydroxide of group 2 elements increases down the group because as you go down the group size of metal increases thereby increasing the bond length and decreasing bond energy.

Answer: (c) NH₄OH(excess) + HCl
Explanation:
A mixture of ammonium hydroxide and HCl react to form ammonium chloride. This also contains unreacted ammonium hydroxide.
NH₄​ OH + HCl → NH₄Cl + H₂O
Thus, the resulting mixture of (NH₄OH+NH₄Cl) is a basic buffer solution. It contains a mixture of weak base ammonium hydroxide and its salt (ammonium chloride) with strong acid (HCl).
It is used in a qualitative analysis of group III radicals.
A mixture of HNO₃ and NaOH is a mixture of strong acid and strong base similarly a mixture of
H₂​SO₄ and KOH is also a mixture of strong acid and strong base, thus they do not form buffer solution.
Similarly, in option 4 mixture will form a strong base and salt of strong base weak acid hence it will not form a buffer solution.

Answer: (c) Na₂CO₃
Explanation:
NaCl is salt of strong acid and strong base so it has pH of about 7.
NH₄​Cl is salt of strong acid and weak base so it has pH is much less than 7.
CH₃COONH₄ is salt of weak acid and weak base so it has pH of about 7.
Na₂​CO₃ is salt of weak acid and strong base so it has pH is much greater than 7.

Answer: (b) Lewis Base
Explanation:
R−NH₂ (Amines) behaves as a Lewis base because it is capable of donating a lone pair of electron.