Important Questions Class 11 Chemistry Chapter 6 Equilibrium

Equilibrium is the state where forward and reverse processes occur at equal rates in a closed system. At equilibrium, measurable properties stay constant, but molecular activity continues.

Equilibrium explains why reactions do not always go to completion. Important Questions Class 11 Chemistry Chapter 6 help students understand physical equilibrium, chemical equilibrium, ionic equilibrium, pH, buffers, and solubility product. CBSE 2026 questions from this chapter usually test formulas, reaction direction, Le Chatelier’s principle, acid-base concepts, and numerical accuracy. Students should practise Kc, Kp, Qc, pH, Ka, Kb, Kw, and Ksp step by step.

Key Takeaways

• Dynamic Equilibrium: Forward and reverse processes continue at equal rates.
• Equilibrium Constant: Kc and Kp show the extent of a reversible reaction.
• Le Chatelier’s Principle: Equilibrium shifts to reduce concentration, pressure, or temperature stress.
• Ionic Equilibrium: pH, Ka, Kb, buffers, hydrolysis, and Ksp control aqueous solution behaviour.

Important Questions Class 11 Chemistry Chapter 6 Structure 2026

Important Questions Class 11 Chemistry Chapter 6 Overview

Important Questions Class 11 Chemistry Chapter 6 cover both reversible reactions and ionic behaviour in water. Students must understand the concept first, then apply formulas carefully.

What does equilibrium mean in Chemistry?

Equilibrium means a state where forward and reverse processes occur at equal rates. The concentration of reactants and products remains constant.

Equilibrium is dynamic, not static. Molecules continue reacting, evaporating, condensing, dissolving, or crystallising at the microscopic level.

Why is equilibrium important in CBSE 2026?

Equilibrium is important because it connects reaction direction, reaction yield, pH, salt behaviour, and solubility. CBSE 2026 can ask both theory and numericals.

The chapter also supports later chemistry topics. Acid-base chemistry, electrochemistry, metallurgy, and environmental chemistry use equilibrium ideas.

Equilibrium Class 11 Important Questions on Physical Processes

Equilibrium Class 11 physical process questions are usually concept-based. Students should mention closed system, constant temperature, and equal opposing rates.

What is dynamic equilibrium?

Dynamic equilibrium is a state where forward and reverse processes continue at equal rates. There is no net change in measurable properties.

Example:

H2O(l) ⇌ H2O(vap)

At equilibrium, water evaporates and vapour condenses at equal rates. The vapour pressure remains constant.

What are the main characteristics of physical equilibrium?

Physical equilibrium has constant measurable properties at a fixed temperature. The opposing processes continue at equal rates.

Key features:

1. It occurs only in a closed system.
2. Opposing processes occur together.
3. Both processes have equal rates at equilibrium.
4. Measurable properties remain constant.
5. The equilibrium is dynamic.

What is solid-liquid equilibrium?

Solid-liquid equilibrium occurs when solid and liquid phases coexist at a fixed temperature and pressure. Ice and water at 273 K show this equilibrium.

Example:

H2O(s) ⇌ H2O(l)

At 273 K and 1 atm, ice melts and water freezes at equal rates. The mass of ice and water remains constant.

What is liquid-vapour equilibrium?

Liquid-vapour equilibrium occurs when evaporation and condensation occur at equal rates. It happens in a closed vessel.

Example:

H2O(l) ⇌ H2O(g)

At equilibrium, vapour pressure becomes constant. This constant pressure is called equilibrium vapour pressure.

Why is equilibrium not possible in an open vessel?

Equilibrium is not possible in an open vessel because vapour molecules escape into the surroundings. Condensation cannot match evaporation.

In an open watch glass, acetone or water eventually disappears. Molecules spread into the room and do not return at the same rate.

What is solid-vapour equilibrium?

Solid-vapour equilibrium occurs when sublimation and deposition occur at equal rates. Iodine in a closed vessel shows this behaviour.

Example:

I2(s) ⇌ I2(vapour)

The violet colour of iodine vapour becomes constant at equilibrium. This shows no net change in vapour concentration.

What is Henry’s law?

Henry’s law states that the mass of a gas dissolved in a liquid is proportional to gas pressure above the liquid. Temperature must remain constant.

Copy-friendly equation:

Gas(g) ⇌ Gas(aq)

For carbon dioxide in soda water:

CO2(g) ⇌ CO2(aq)

When a soda bottle opens, pressure decreases. Dissolved carbon dioxide escapes as bubbles.

Chemical Equilibrium Class 11 Important Questions With Answers

Chemical equilibrium Class 11 questions require balanced equations and correct equilibrium expressions. Students should always use equilibrium concentrations.

What is chemical equilibrium?

Chemical equilibrium is the stage where forward and reverse reactions occur at equal rates. Reactant and product concentrations remain constant.

Example:

H2(g) + I2(g) ⇌ 2HI(g)

HI forms in the forward reaction. It also decomposes in the reverse reaction at equilibrium.

Why is chemical equilibrium called dynamic?

Chemical equilibrium is called dynamic because reactions continue in both directions. The forward and reverse rates become equal.

There is no visible change in concentration. Still, reactant molecules form products and product molecules form reactants continuously.

What is the law of chemical equilibrium?

The law of chemical equilibrium states that a fixed ratio of product and reactant concentrations becomes constant at a given temperature. Each concentration is raised to its stoichiometric coefficient.

For a general reaction:

aA + bB ⇌ cC + dD

Copy-friendly expression:

Kc = [C]^c [D]^d / [A]^a [B]^b

How do you write Kc for a reaction?

Kc is written by placing product concentrations in the numerator and reactant concentrations in the denominator. Each term uses its balanced coefficient as power.

For the reaction:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Copy-friendly expression:

Kc = [NH3]^2 / [N2][H2]^3

How do you write Kp for a gaseous reaction?

Kp is written using partial pressures instead of molar concentrations. It applies to gaseous equilibria.

For the reaction:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Copy-friendly expression:

Kp = (pNH3)^2 / (pN2)(pH2)^3

What is the relation between Kp and Kc?

The relation between Kp and Kc is Kp = Kc(RT)^Δn. Here Δn is gaseous product moles minus gaseous reactant moles.

Copy-friendly formula:

Kp = Kc(RT)^Δn

Where:

Δn = moles of gaseous products - moles of gaseous reactants

For:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Δn = 2 - 4
Δn = -2

So:

Kp = Kc(RT)^-2

What is homogeneous equilibrium?

Homogeneous equilibrium has all reactants and products in the same phase. The phase may be gas or solution.

Examples:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Fe3+(aq) + SCN-(aq) ⇌ [Fe(SCN)]2+(aq)

What is heterogeneous equilibrium?

Heterogeneous equilibrium has reactants and products in different phases. Pure solids and pure liquids are not included in K expression.

Example:

CaCO3(s) ⇌ CaO(s) + CO2(g)

Copy-friendly expression:

Kc = [CO2]

Kp = pCO2

CaCO3 and CaO are pure solids. Their concentrations stay constant.

Equilibrium Constant Class 11 Numericals

Equilibrium constant Class 11 numericals test formula selection and substitution. Write balanced reaction first, then place values correctly.

How do you calculate Kc for ammonia formation?

Kc is calculated by substituting equilibrium concentrations in the equilibrium expression.

Given:

[N2] = 1.5 × 10^-2 M
[H2] = 3.0 × 10^-2 M
[NH3] = 1.2 × 10^-2 M

Reaction:

N2(g) + 3H2(g) ⇌ 2NH3(g)

1. Formula Used:
   Kc = [NH3]^2 / [N2][H2]^3
2. Substitution:
   Kc = (1.2 × 10^-2)^2 / (1.5 × 10^-2)(3.0 × 10^-2)^3
3. Calculation:
   Kc = 1.44 × 10^-4 / (1.5 × 10^-2)(2.7 × 10^-5)
   Kc = 1.44 × 10^-4 / 4.05 × 10^-7
   Kc = 3.55 × 10^2
4. Final Result:
   Kc = 3.55 × 10^2

How do you calculate Kc for NO formation?

Use equilibrium concentrations and raise each term to its balanced coefficient.

Given:

[N2] = 3.0 × 10^-3 M
[O2] = 4.2 × 10^-3 M
[NO] = 2.8 × 10^-3 M

Reaction:

N2(g) + O2(g) ⇌ 2NO(g)

1. Formula Used:
   Kc = [NO]^2 / [N2][O2]
2. Substitution:
   Kc = (2.8 × 10^-3)^2 / (3.0 × 10^-3)(4.2 × 10^-3)
3. Calculation:
   Kc = 7.84 × 10^-6 / 12.6 × 10^-6
   Kc = 0.622
4. Final Result:
   Kc = 0.622

How do you calculate Kp from Kc?

Kp is calculated using Kp = Kc(RT)^Δn. First find Δn from the balanced gaseous equation.

Given:

Kc = 3.75 × 10^-6
T = 1069 K
R = 0.0831 bar L mol^-1 K^-1

Reaction:

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

1. Find Δn:
   Δn = gaseous product moles - gaseous reactant moles
   Δn = (2 + 1) - 2
   Δn = 1
2. Formula Used:
   Kp = Kc(RT)^Δn
3. Substitution:
   Kp = 3.75 × 10^-6 × (0.0831 × 1069)^1
4. Calculation:
   Kp = 3.75 × 10^-6 × 88.83
   Kp = 0.033
5. Final Result:
   Kp = 0.033

How do you predict reaction direction using Qc?

Reaction quotient Qc predicts whether a reaction moves forward, backward, or stays at equilibrium.

Rules:

1. If Qc < Kc, reaction moves forward.
2. If Qc > Kc, reaction moves backward.
3. If Qc = Kc, reaction is at equilibrium.

Example:

For 2A ⇌ B + C, Kc = 2 × 10^-3

Given:

[A] = [B] = [C] = 3 × 10^-4 M

1. Formula Used:
   Qc = [B][C] / [A]^2
2. Substitution:
   Qc = (3 × 10^-4)(3 × 10^-4) / (3 × 10^-4)^2
3. Calculation:
   Qc = 1
4. Comparison:
   Qc = 1
   Kc = 2 × 10^-3
   Qc > Kc
5. Final Result:
   The reaction proceeds in the reverse direction.

How do you calculate equilibrium concentration for PCl5 decomposition?

Use an ICE-style table and solve for the amount dissociated.

Given:

Initial [PCl5] = 3.00 M
Kc = 1.80

Reaction:

PCl5(g) ⇌ PCl3(g) + Cl2(g)

1. Initial Concentrations:
   [PCl5] = 3.00
   [PCl3] = 0
   [Cl2] = 0
2. Change:
   PCl5 decreases by x.
   PCl3 increases by x.
   Cl2 increases by x.
3. Equilibrium Concentrations:
   [PCl5] = 3 - x
   [PCl3] = x
   [Cl2] = x
4. Formula Used:
   Kc = [PCl3][Cl2] / [PCl5]
5. Substitution:
   1.80 = x^2 / (3 - x)
6. Calculation:
   x^2 = 1.80(3 - x)
   x^2 = 5.40 - 1.80x
   x^2 + 1.80x - 5.40 = 0
   x = 1.59
7. Final Concentrations:
   [PCl5] = 3 - 1.59 = 1.41 M
   [PCl3] = 1.59 M
   [Cl2] = 1.59 M
8. Final Result:
   [PCl5] = 1.41 M, [PCl3] = 1.59 M, [Cl2] = 1.59 M

Le Chatelier Principle Class 11 Important Questions

Le Chatelier principle Class 11 questions ask how equilibrium shifts after stress. Always state the stress first, then the shift direction.

What is Le Chatelier’s principle?

Le Chatelier’s principle states that a system at equilibrium shifts to reduce the effect of an applied change. The change may involve concentration, pressure, or temperature.

The principle applies to physical and chemical equilibrium. It helps predict product yield in industrial reactions.

How does concentration affect equilibrium?

Adding a reactant shifts equilibrium towards products, while removing a product also shifts it forward. The system tries to consume added species or replace removed species.

Example:

H2(g) + I2(g) ⇌ 2HI(g)

If H2 is added, equilibrium shifts right. More HI forms.

How does pressure affect equilibrium?

Increasing pressure shifts gaseous equilibrium towards the side with fewer gas moles. Decreasing pressure shifts it towards more gas moles.

Example:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Left side has 4 gas moles. Right side has 2 gas moles.

Increasing pressure shifts equilibrium right. Ammonia yield increases.

How does temperature affect equilibrium?

Increasing temperature favours the endothermic direction, while decreasing temperature favours the exothermic direction. Temperature changes the value of K.

Example:

N2(g) + 3H2(g) ⇌ 2NH3(g); ΔH = -92.38 kJ mol^-1

The forward reaction is exothermic. Lower temperature favours ammonia formation.

How does a catalyst affect equilibrium?

A catalyst does not change equilibrium composition or equilibrium constant. It only helps the system reach equilibrium faster.

A catalyst lowers activation energy for both forward and reverse reactions. It does not shift the equilibrium position.

What happens when inert gas is added at constant volume?

Adding inert gas at constant volume does not affect equilibrium. Partial pressures of reacting gases remain unchanged.

The reaction quotient does not change. Therefore, no equilibrium shift occurs.

Ionic Equilibrium Class 11 Important Questions

Ionic equilibrium Class 11 questions deal with ions in aqueous solution. Students should separate strong electrolytes from weak electrolytes.

What are electrolytes?

Electrolytes are substances that conduct electricity in aqueous solution due to ions. Acids, bases, and salts are common electrolytes.

Examples:

NaCl(aq) gives Na+ and Cl- ions.
HCl(aq) gives H+ and Cl- ions.

What is the difference between strong and weak electrolytes?

Strong electrolytes ionise almost completely, while weak electrolytes ionise partially. Weak electrolytes form equilibrium between ions and unionised molecules.

Examples:

Strong electrolyte: NaCl, HCl, NaOH
Weak electrolyte: CH3COOH, NH4OH

What are Arrhenius acids and bases?

Arrhenius acids produce H+ ions in water, while Arrhenius bases produce OH- ions in water.

Examples:

HCl(aq) → H+(aq) + Cl-(aq)

NaOH(aq) → Na+(aq) + OH-(aq)

Limitation:

Arrhenius theory applies mainly to aqueous solutions. It does not explain bases like NH3 fully.

What are Bronsted-Lowry acids and bases?

Bronsted-Lowry acids donate protons, while Bronsted-Lowry bases accept protons.

Example:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

Role:

1. H2O donates H+ and acts as an acid.
2. NH3 accepts H+ and acts as a base.
3. NH4+ is conjugate acid of NH3.
4. OH- is conjugate base of H2O.

What are Lewis acids and bases?

Lewis acids accept an electron pair, while Lewis bases donate an electron pair.

Example:

BF3 + NH3 → BF3:NH3

Role:

1. BF3 accepts an electron pair.
2. NH3 donates a lone pair.
3. BF3 is a Lewis acid.
4. NH3 is a Lewis base.

What are conjugate acid-base pairs?

A conjugate acid-base pair differs by one proton. The acid has one extra proton compared to its conjugate base.

Examples:

HF and F-
H2SO4 and HSO4-
NH4+ and NH3
H2O and OH-

Give conjugate bases of HF, H2SO4, and HCO3-.

The conjugate base has one proton less than the acid.

1. HF loses H+ to form F-.
2. H2SO4 loses H+ to form HSO4-.
3. HCO3- loses H+ to form CO3^2-.

Final Result: F-, HSO4-, CO3^2-

Give conjugate acids of NH2-, NH3, and HCOO-.

The conjugate acid has one proton more than the base.

1. NH2- gains H+ to form NH3.
2. NH3 gains H+ to form NH4+.
3. HCOO- gains H+ to form HCOOH.

Final Result: NH3, NH4+, HCOOH

pH and Ionization Constant Class 11 Questions

pH and ionization constant Class 11 questions need careful logarithm use. Equations should be copied clearly before substitution.

What is ionic product of water?

Ionic product of water is the product of H+ and OH- concentrations in water. It is represented by Kw.

Copy-friendly equation:

Kw = [H+][OH-]

At 298 K:

[H+] = 1.0 × 10^-7 M
[OH-] = 1.0 × 10^-7 M

So:

Kw = 1.0 × 10^-14

What is pH?

pH is the negative logarithm of hydrogen ion concentration. It shows whether a solution is acidic, basic, or neutral.

Copy-friendly formula:

pH = -log[H+]

At 298 K:

1. Acidic solution has pH < 7.
2. Neutral solution has pH = 7.
3. Basic solution has pH > 7.

What is the relation between pH and pOH?

The relation between pH and pOH at 298 K is pH + pOH = 14.

Copy-friendly formula:

pH + pOH = 14

Also:

pOH = -log[OH-]

How do you calculate pH of a soft drink?

pH is calculated by taking the negative logarithm of hydrogen ion concentration.

Given:

[H+] = 3.8 × 10^-3 M

1. Formula Used:
   pH = -log[H+]
2. Substitution:
   pH = -log(3.8 × 10^-3)
3. Calculation:
   pH = -[log 3.8 + log 10^-3]
   pH = -[0.58 - 3]
   pH = 2.42
4. Final Result:
   pH = 2.42

How do you calculate hydrogen ion concentration from pH?

Hydrogen ion concentration is calculated using [H+] = 10^-pH.

Given:

pH of vinegar = 3.76

1. Formula Used:
   [H+] = 10^-pH
2. Substitution:
   [H+] = 10^-3.76
3. Calculation:
   [H+] = 1.74 × 10^-4 M
4. Final Result:
   [H+] = 1.74 × 10^-4 M

What are Ka and Kb?

Ka is the ionization constant of a weak acid, while Kb is the ionization constant of a weak base. Larger value means stronger ionization.

For weak acid HA:

HA(aq) ⇌ H+(aq) + A-(aq)

Copy-friendly formula:

Ka = [H+][A-] / [HA]

For weak base BOH:

BOH(aq) ⇌ B+(aq) + OH-(aq)

Copy-friendly formula:

Kb = [B+][OH-] / [BOH]

What is the relation between Ka and Kb?

For a conjugate acid-base pair, Ka × Kb = Kw.

Copy-friendly formula:

Ka × Kb = Kw

At 298 K:

Kw = 1.0 × 10^-14

Also:

pKa + pKb = 14

How do you calculate pH of weak acid HOCl?

Use Ka and initial concentration to calculate [H+], then find pH.

Given:

[HOCl] = 0.08 M
Ka = 2.5 × 10^-5

Reaction:

HOCl(aq) ⇌ H+(aq) + OCl-(aq)

1. Let ionized concentration be x:
   [H+] = x
   [OCl-] = x
   [HOCl] = 0.08 - x
2. Formula Used:
   Ka = [H+][OCl-] / [HOCl]
3. Substitution:
   2.5 × 10^-5 = x^2 / (0.08 - x)
4. Approximation:
   Since x is small, 0.08 - x ≈ 0.08
5. Calculation:
   x^2 = 2.5 × 10^-5 × 0.08
   x^2 = 2.0 × 10^-6
   x = 1.41 × 10^-3
6. pH Calculation:
   pH = -log(1.41 × 10^-3)
   pH = 2.85
7. Final Result:
   pH = 2.85

Buffer Solution and Salt Hydrolysis Class 11 Questions

Buffer solution Class 11 questions connect weak acids, weak bases, and their salts. Salt hydrolysis questions test whether the final solution is acidic, basic, or neutral.

What is common ion effect?

Common ion effect is the suppression of ionization due to addition of an ion already present in equilibrium. It follows Le Chatelier’s principle.

Example:

CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)

Adding CH3COONa increases CH3COO-. Equilibrium shifts left and ionization of acetic acid decreases.

What is salt hydrolysis?

Salt hydrolysis is the reaction of salt ions with water to form acidic or basic solutions. It depends on the parent acid and base.

Rules:

1. Strong acid + strong base gives neutral solution.
2. Weak acid + strong base gives basic solution.
3. Strong acid + weak base gives acidic solution.
4. Weak acid + weak base depends on Ka and Kb.

Why is sodium acetate solution basic?

Sodium acetate solution is basic because acetate ion hydrolyses water to produce OH- ions.

Reaction:

CH3COO-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)

The solution contains extra OH- ions. Therefore, pH is greater than 7.

Why is ammonium chloride solution acidic?

Ammonium chloride solution is acidic because NH4+ hydrolyses water to produce H+ ions.

Reaction:

NH4+(aq) + H2O(l) ⇌ NH4OH(aq) + H+(aq)

The solution contains extra H+ ions. Therefore, pH is less than 7.

What is a buffer solution?

A buffer solution resists change in pH when small amounts of acid or base are added. It also resists pH change on dilution.

Examples:

1. CH3COOH + CH3COONa forms acidic buffer.
2. NH4OH + NH4Cl forms basic buffer.

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation relates buffer pH with pKa and salt-acid ratio.

For acidic buffer:

pH = pKa + log([Salt] / [Acid])

For basic buffer:

pOH = pKb + log([Salt] / [Base])

Use these equations when weak acid or weak base is mixed with its salt.

How do you calculate pH of ammonium acetate solution?

For a salt of weak acid and weak base, pH depends on pKa and pKb.

Given:

pKa of acetic acid = 4.76
pKb of ammonium hydroxide = 4.75

1. Formula Used:
   pH = 7 + 1/2(pKa - pKb)
2. Substitution:
   pH = 7 + 1/2(4.76 - 4.75)
3. Calculation:
   pH = 7 + 1/2(0.01)
   pH = 7 + 0.005
   pH = 7.005
4. Final Result:
   pH = 7.005

Solubility Product Class 11 Important Questions

Solubility product Class 11 questions are formula-based. Students should write ion concentrations in terms of molar solubility S.

What is solubility product?

Solubility product is the equilibrium constant for a sparingly soluble salt in saturated solution. It is represented by Ksp.

Example:

BaSO4(s) ⇌ Ba2+(aq) + SO4^2-(aq)

Copy-friendly expression:

Ksp = [Ba2+][SO4^2-]

How do you calculate molar solubility of BaSO4 from Ksp?

For BaSO4, molar solubility equals the concentration of Ba2+ and SO4^2-.

Given:

Ksp = 1.1 × 10^-10

1. Let molar solubility be S:
   [Ba2+] = S
   [SO4^2-] = S
2. Formula Used:
   Ksp = [Ba2+][SO4^2-]
3. Substitution:
   1.1 × 10^-10 = S × S
4. Calculation:
   S^2 = 1.1 × 10^-10
   S = 1.05 × 10^-5 mol L^-1
5. Final Result:
   Solubility of BaSO4 = 1.05 × 10^-5 mol L^-1

How do you calculate solubility of A2X3?

Use stoichiometry to express ion concentration in terms of S.

Given:

Ksp of A2X3 = 1.1 × 10^-23

Reaction:

A2X3(s) ⇌ 2A3+(aq) + 3X2-(aq)

1. Let solubility be S:
   [A3+] = 2S
   [X2-] = 3S
2. Formula Used:
   Ksp = [A3+]^2 [X2-]^3
3. Substitution:
   1.1 × 10^-23 = (2S)^2(3S)^3
4. Calculation:
   1.1 × 10^-23 = 4S^2 × 27S^3
   1.1 × 10^-23 = 108S^5
   S^5 = 1.0 × 10^-25
   S = 1.0 × 10^-5 mol L^-1
5. Final Result:
   Solubility of A2X3 = 1.0 × 10^-5 mol L^-1

How does common ion affect solubility?

Common ion decreases the solubility of a sparingly soluble salt. The equilibrium shifts towards the undissolved solid.

Example:

Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)

Adding NaOH increases OH- concentration. This shifts equilibrium left and reduces Ni(OH)2 solubility.

How do you calculate molar solubility of Ni(OH)2 in NaOH?

The OH- from NaOH acts as a common ion and reduces solubility.

Given:

Ksp of Ni(OH)2 = 2.0 × 10^-15
[NaOH] = 0.10 M

Reaction:

Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)

1. Let solubility be S:
   [Ni2+] = S
   [OH-] = 0.10 + 2S
2. Approximation:
   Since Ksp is very small, 2S is negligible.
   [OH-] ≈ 0.10
3. Formula Used:
   Ksp = [Ni2+][OH-]^2
4. Substitution:
   2.0 × 10^-15 = S(0.10)^2
5. Calculation:
   S = 2.0 × 10^-15 / 0.01
   S = 2.0 × 10^-13 M
6. Final Result:
   Molar solubility of Ni(OH)2 = 2.0 × 10^-13 M

Most Repeated Variations of Important Questions Class 11 Chemistry Chapter 6

Class 11 Chemistry Chapter 6 important questions often repeat the same formula logic. Students should identify the formula type before solving.

Which equilibrium questions are repeatedly asked?

Repeated questions usually come from Kc, Kp, Qc, Le Chatelier’s principle, pH, Ka, Kb, and Ksp. These areas combine concepts and calculations.

High-frequency patterns:

1. Write Kc expression.
2. Convert Kc to Kp.
3. Predict reaction direction using Qc.
4. Calculate pH from [H+].
5. Find conjugate acid-base pairs.
6. Use Le Chatelier’s principle.
7. Calculate solubility from Ksp.
8. Identify acidic, basic, or neutral salts.

Where do students lose marks in equilibrium numericals?

Students lose marks when they skip balanced equations or use initial concentration instead of equilibrium concentration. Most formulas need equilibrium values.

Common errors:

1. Including pure solids in Kc.
2. Forgetting powers in Kc expression.
3. Using wrong Δn in Kp = Kc(RT)^Δn.
4. Treating catalyst as an equilibrium shifter.
5. Confusing pH and pOH.
6. Ignoring common ion effect in Ksp questions.

Class 11 Chemistry Important Links