The degree of dissociation of ${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$ is influenced by:

${\rm{C}}{{\rm{H}}_3}{\rm{COONa}}$

${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$

Pure ammonia is placed in a vessel at temperature where its dissociation constant (${\rm{\alpha }}$) is appreciable. At equilibrium,

Concentration of ${{\rm{H}}_2}$ is less than that of ${{\rm{N}}_2}$

${K_p}$ does not change significantly with pressure

${\rm{\alpha }}$ does not change with pressure

Concentration of ${\rm{N}}{{\rm{H}}_3}$ does not change with pressure

${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ is acidic because

On hydrolysis ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ gives weak base ${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$ and strong acid HCl

Nitrogen donates a pair of electrons

It is a salt of weak acid and strong base

On hydrolysis ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ gives strong base and weak acid

The conjugate base of ${{\rm{H}}_2}{\rm{PO}}_4^ - $ is:

${{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}$

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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 161 questions

1 Easy

1 of 161

Remembering +4 | -1

The indicator used in titrating oxalic acid with caustic soda solution is:

View answer and explanation

Correct option: D

Phenolphthalein

Option D (Phenolphthalein) is correct because phenolphthalein is a suitable indicator for titrating a weak acid (oxalic acid) with a strong base (caustic soda).

2 Easy

2 of 161

Understanding +4 | -1

\mathrm{M}{\mathrm{g}^{2 + }}

is … than

\mathrm{A}{\mathrm{l}^{3 + }}

View answer and explanation

Correct option: C

Weak Lewis acid

Option C (Weak Lewis acid) is correct because Mg²⁺ has a smaller charge density and weaker polarizing power compared to Al³⁺, making it a weaker Lewis acid.

3 Easy

3 of 161

Application +4 | -1

The solubility of

\mathrm{S}{\mathrm{b}_2}{\mathrm{S}_3}\;\mathrm{in }\;\mathrm{water}\;\mathrm{ is}\;1.0 \times {10^{ - 5}}\;mol/L

298\;K

. What will be its solubility product?

View answer and explanation

Correct option: A

108 \times {10^{ - 25}}

Option D (120 kJ) is correct because it is calculated using the Gibbs free energy equation: ∆G = ∆H - T∆S.

4 Easy

4 of 161

Analyzing +4 | -1

The degree of dissociation of

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is influenced by:

View answer and explanation

Correct option: D

Either of these

Option D (Either of these) is correct because the degree of dissociation of CH3COOH is influenced by the common ion effect from CH3COONa and HCl, and by acid-base reaction with NH4OH.

5 Easy

5 of 161

Understanding +4 | -1

Which of the following can act as a self-indicator?

View answer and explanation

Correct option: A

KMn{O_4}

Option A (

KMn{O_4}

) is correct because

KMn{O_4}

acts as a self-indicator due to its intense purple color, which changes upon reduction at the endpoint of the titration.

6 Easy

6 of 161

Analyzing +4 | -1

Assertion : The equilibrium constant for the reaction

\begin{aligned} & {CaSO}_4 \cdot 5 {H}_2 {O}({s})={CaSO}_4 \cdot 3 {H}_2 {O}({s})+2 {H}_2 {O}({g}) \\ & \text { is } {K}_{{C}}=\frac{\left[{CaSO}_4 \cdot 3 {H}_2 {O}\right]\left[{H}_2 {O}\right]^2}{\left[{CaSO}_4 \cdot 5 {H}_2 {O}\right]} \end{aligned}

Reason : Equilibrium constant is the ratio of the product of molar concentration of the substances produced to the product of the molar concentrations of reactants with each concentrations term raised to the power equal to the respective stoichiometric constant.

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D is correct because the equilibrium constant for a heterogeneous reaction only includes gaseous or aqueous species, and the concentration of solids is considered unity.

7 Easy

7 of 161

Analyzing +4 | -1

Pure ammonia is placed in a vessel at temperature where its dissociation constant (

\mathrm{\alpha }

) is appreciable. At equilibrium,

View answer and explanation

Correct option: D

Concentration of

{\mathrm{H}_2}

is less than that of

{\mathrm{N}_2}

Option D is correct because at equilibrium, the concentration of H2 will be less than N2 due to the stoichiometry of the ammonia dissociation reaction.

8 Easy

8 of 161

Remembering +4 | -1

2 \mathrm{SO_2(g})+\mathrm{O}_2(\mathrm{g}) \xrightarrow{\text { Catalyst }} 2 \mathrm{SO}_3(\mathrm{g})

Identify the catalyst from the options given below:

View answer and explanation

Correct option: A

\mathrm{V}_2 \mathrm{O}_5

Option A (

\mathrm{V}_2 \mathrm{O}_5

) is correct because Vanadium pentoxide (

\mathrm{V}_2 \mathrm{O}_5

) is the catalyst used in the Contact process for the production of

\mathrm{SO}_3

9 Easy

9 of 161

Understanding +4 | -1

The

{K_C}

for

{H_2}(g) + {I_2}(g)2HI(g)

is 64, if the volume of the container is reduced to one-half of its
original volume, the value of equilibrium constant will be

View answer and explanation

Correct option: B

Option B (64) is correct because the equilibrium constant, Kc, is independent of volume changes for the given reaction.

10 Easy

10 of 161

Understanding +4 | -1

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

is acidic because

View answer and explanation

Correct option: A

On hydrolysis

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

gives weak base

\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

and strong acid HCl

Option A is correct because ammonium chloride (

NH_4Cl

) is a salt of a weak base (

NH_4OH

) and a strong acid (HCl), making it acidic due to the hydrolysis of the ammonium ion.

11 Easy

11 of 161

Remembering +4 | -1

Phenolphthalein is a:

View answer and explanation

Correct option: A

Weak acid

Option A (Weak acid) is correct because phenolphthalein is a weak organic acid that exists in equilibrium between its protonated and deprotonated forms.

12 Easy

12 of 161

Application +4 | -1

For a system,

A+2 B \rightleftharpoons C

, the equilibrium concentration are

[A]=0.06

[B]=0.12

, and

[C]=0.216

. The

K_c

for the reaction is

View answer and explanation

Correct option: D

250

Option D (250) is correct because it is calculated by substituting the equilibrium concentrations into the Kc expression.

13 Easy

13 of 161

Remembering +4 | -1

Amines behave as

View answer and explanation

Correct option: B

Lewis base

Option B (Lewis base) is correct because amines have a lone pair of electrons on the nitrogen atom, making them electron-pair donors, which is the definition of a Lewis base.

14 Easy

14 of 161

Understanding +4 | -1

The salt that does not hydrolyse, is:

View answer and explanation

Correct option: D

\mathrm{CaC}{\mathrm{l}_2}

Option D (

\mathrm{CaC}{\mathrm{l}_2}

) is correct because it is a salt formed from a strong acid (HCl) and a strong base (Ca(OH)₂), and therefore does not hydrolyze.

15 Easy

15 of 161

Understanding +4 | -1

The conjugate base of

{\mathrm{H}_2}\mathrm{PO}_4^ -

is:

View answer and explanation

Correct option: A

\mathrm{HPO}_4^{2 - }

The correct answer is A (

\mathrm{HPO}_4^{2 - }

) because the conjugate base is formed by removing a proton (H+) from the acid.

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16 Easy

16 of 161

Understanding +4 | -1

s

and

S

are respectively solubility and solubility product of a sparingly soluble binary electrolyte then :

View answer and explanation

Correct option: C

{s}={S}^{1 / 2}

Option C (

{s}={S}^{1 / 2}

) is correct because for a binary electrolyte, the solubility is the square root of the solubility product.

17 Easy

17 of 161

Analyzing +4 | -1

Assertion : For the reaction,

2 {NH}_3({~g}) \rightleftharpoons {N}_2({~g})+3 {H}_2({~g})

, the unit of

K_p

will be atm.

Reason : Unit of

K_p

(a t m)^{\Delta n}

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D is correct. The assertion is false because the unit of Kp is atm², not atm, while the reason is true as the unit of Kp is (atm)^∆n.

18 Easy

18 of 161

Understanding +4 | -1

Consider a saturated solution. (i.e., a solution in equilibrium with solid solute) of sodium chloride. If the temperature is raised, then for

NaCl(s) + aq\, \rightleftharpoons \,NaCl(aq);\Delta H = + ve

View answer and explanation

Correct option: A

More solid will dissolve

Option A (More solid will dissolve) is correct because the dissolution of NaCl is endothermic, so increasing the temperature will shift the equilibrium towards dissolution according to Le Chatelier's principle.

19 Easy

19 of 161

Application +4 | -1

Degree of dissociation of

0.1\;\mathrm{ }N

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is (dissociation constant

= 1 \times {10^{ - 5}}

)

View answer and explanation

Correct option: D

{10^{ - 2}}

Option D (

{10^{ - 2}}

) is correct because the degree of dissociation is calculated using the formula

\alpha = \sqrt{\frac{K_a}{C}}

20 Easy

20 of 161

Remembering +4 | -1

In the titration of

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}\;\mathrm{and}\;\mathrm{ HCl}

, the indicator used is

View answer and explanation

Correct option: A

Methyl orange

Option A (Methyl orange) is correct because it is the appropriate indicator for titrations where the equivalence point is on the acidic side, such as with Na2CO3 and HCl.

21 Easy

21 of 161

Understanding +4 | -1

In which of the following equilibrium, change in volume of the system does not alter the number of moles?

View answer and explanation

Correct option: A

{\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{NO}\left( \mathrm{g} \right)

Option A (

{\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\\;\leftrightharpoons\\;2\mathrm{NO}\left( \mathrm{g} \right)

) is correct because the number of moles of gaseous reactants equals the number of moles of gaseous products, so a change in volume will not shift the equilibrium.

22 Easy

22 of 161

Application +4 | -1

A weak acid

HX

has dissociation constant

{10^{ - 5}}

. The

pH

0.1\;M

solution of this acid will be

View answer and explanation

Correct option: B

Option B (3) is correct because the pH of a weak acid solution is calculated using the formula pH = -log[H+], where [H+] = √(Ka * C).

23 Easy

23 of 161

Application +4 | -1

Which solution has pH equal to 10?

View answer and explanation

Correct option: A

{10^{ - 4}}\;\mathrm{M}\;\mathrm{ KOH}

Option A (

{10^{ - 4}}\;\mathrm{M}\;\mathrm{ KOH}

) is correct because a KOH concentration of 10^-4 M corresponds to a pOH of 4, and thus a pH of 10.

24 Easy

24 of 161

Remembering +4 | -1

When the

\mathrm{pH}

of a solution is 2, the hydrogen ion concentration is:

View answer and explanation

Correct option: B

1 \times {10^{ - 2}}\;M

Option B (

1 \times {10^{ - 2}}\;M

) is correct because pH is defined as the negative logarithm of hydrogen ion concentration, so if pH is 2, the hydrogen ion concentration is

10^{-2}

25 Easy

25 of 161

Application +4 | -1

A volume of

10 \mathrm{~ml}

of a strong acid solution of

\mathrm{pH}=2.0

are mixed with

990 \mathrm{~ml}

of a buffer solution of

\mathrm{pH}=4.0.

The

\mathrm{pH}

of the resulting solution will be (approximately)

View answer and explanation

Correct option: D

3.5

Option D (3.5) is correct because adding a strong acid to a buffer will decrease the pH, but the buffer resists large changes, resulting in a pH slightly lower than 4.

26 Easy

26 of 161

Understanding +4 | -1

Which aqueous solution will have pH less than 7?

View answer and explanation

Correct option: D

\mathrm{F}{\mathrm{e}^{3 + }}

Option D (

\mathrm{F}{\mathrm{e}^{3 + }}

) is correct because

\mathrm{F}{\mathrm{e}^{3 + }}

ions undergo hydrolysis to produce

H^+

ions, making the solution acidic (pH < 7).

27 Easy

27 of 161

Remembering +4 | -1

The indicators used in the titration of iodine against sodium thiosulphate is

View answer and explanation

Correct option: A

Starch

Option A (Starch) is correct because starch acts as an indicator in iodometric titrations, forming a blue complex with iodine, which disappears at the endpoint.

28 Easy

28 of 161

Remembering +4 | -1

Methyl orange is an indicator in acid-alkali titration. It gives:

View answer and explanation

Correct option: D

Yellow colour in alkaline medium and red colour in acid medium

Option D (Yellow colour in alkaline medium and red colour in acid medium) is correct because methyl orange exhibits these color changes in different pH environments.

29 Easy

29 of 161

Understanding +4 | -1

Which statement is correct?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is a weak acid

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is a weak acid) is correct because acetic acid is a weak acid that only partially dissociates in water.

30 Easy

30 of 161

Understanding +4 | -1

In an equilibrium reaction for which

\Delta G^{\circ}=0

, the equilibrium constant

\mathrm{K}

should be equal to:

View answer and explanation

Correct option: B

Option B (1) is correct because when

\Delta G^{\circ} = 0

, the equilibrium constant K is equal to 1, as derived from the equation

\Delta G^{\circ} = -RT \ln K

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 161 single-correct-answer MCQs on Equilibrium.
✓ Try each question before opening its answer. Read the explanation where one is available.
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