NEET Chemistry · NCERT Class 11

Equilibrium notes

Physical Equilibrium, Chemical Equilibrium and Law of Mass Action, Equilibrium Constant Kc, Kp and Reaction Quotient, Le Chatelier Principle and Factors Affecting Equilibrium, Ionic Equilibrium, Electrolytes and Ostwald Dilution Law, Acids, Bases, pH, pKa and pKb, Hydrolysis of Salts and pH of Salt Solutions, Buffer Solutions and Henderson Equation, Solubility Product, Precipitation and Common Ion Effect and Formula Sheet, Quick Revision and NEET Mind Map — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 10 topics
  • 83 min read
  • 33 formulas
  • 53 PYQs
1

Physical Equilibrium

7 min

Physical equilibrium occurs when two physical states or processes coexist with no observable macroscopic change, although microscopic changes continue. In a closed system, evaporation and condensation, melting and freezing, or sublimation and deposition can occur at equal rates. This dynamic nature is central to NCERT and NEET questions. Examples include ice and water at 273 K, saturated salt solution, water vapour in a closed container, and iodine crystals with iodine vapour. Important measurable ideas are vapour pressure, saturation, phase boundaries, and the effect of temperature and pressure. Physical equilibrium helps students understand why equilibrium does not mean stopping, but equal opposing rates.

Key points6

  • Physical equilibrium involves no chemical composition change.
  • Saturated solution represents solute-solid equilibrium with dissolved solute.
  • Boiling occurs when vapour pressure equals external pressure.
  • Phase diagrams show stable regions of solid, liquid, and gas.
  • Dynamic equilibrium is established faster when surface area or temperature is higher.
  • For pure substances, equilibrium conditions are fixed at a given pressure and temperature.

Memory tricks2

Dynamic Means Doing: Remember: equilibrium is not dead still. It is a busy balance where both opposite processes keep doing work at equal rates.

Closed Container Rule: For liquid-vapour equilibrium, think 'cap it to trap it' because escaping vapour prevents equilibrium in an open vessel.

Examples2

Water Bottle Example

In a closed water bottle, some water evaporates and some vapour condenses. After some time, the level appears unchanged because both rates become equal.

Saturated Sugar Solution

Undissolved sugar and dissolved sugar remain in equilibrium when dissolution and crystallization occur at equal rates.

Common mistakes2

Thinking Equilibrium Means No Motion

At equilibrium, molecules still move and processes continue. Only the rates and macroscopic properties become constant.

Confusing Evaporation with Boiling

Evaporation can occur at any temperature, while boiling requires vapour pressure to equal external pressure.

Reference tables1

Quick revision7

Formulas2

Boiling Condition
Pvapour=PexternalP_{\text{vapour}} = P_{\text{external}}

A liquid boils when its vapour pressure becomes equal to the external pressure.

P_vapour
Vapour pressure of the liquid
P_external
External atmospheric or applied pressure
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2

Chemical Equilibrium and Law of Mass Action

8 min

Chemical equilibrium is reached in a reversible reaction when the rate of forward reaction becomes equal to the rate of backward reaction. Reactants and products continue to interconvert, but their concentrations become constant with time. According to the law of mass action, the rate of a chemical reaction is proportional to the product of active masses of reactants raised to their stoichiometric powers. For a general reaction, this leads to the equilibrium constant expression. NEET frequently tests writing Kc correctly, excluding pure solids and liquids, interpreting equilibrium composition, and understanding that equilibrium can be reached from either direction in a closed system.

Key points6

  • Equilibrium is dynamic and molecular collisions continue.
  • Law of mass action connects concentration and reaction rate.
  • Stoichiometric coefficients become powers in Kc expression.
  • The value of K depends only on temperature for a given reaction.
  • Large K means product-favoured equilibrium; small K means reactant-favoured equilibrium.
  • Chemical equilibrium requires a closed system for gases and volatile substances.

Memory tricks2

Products on Top: For Kc, always place products in numerator and reactants in denominator: P over R.

Solid-Liquid Silent: Pure solids and pure liquids are silent in K expressions because their active masses are constant.

Examples2

Haber Process

N2 + 3H2 ⇌ 2NH3 is a classic reversible reaction where ammonia formation and decomposition occur simultaneously at equilibrium.

Dissociation of PCl5

PCl5 ⇌ PCl3 + Cl2 is frequently used in NEET to calculate Kc, Kp, and degree of dissociation.

Common mistakes2

Using Initial Concentrations in Kc

Kc must be calculated using equilibrium concentrations, not initial concentrations.

Assuming Equal Concentrations

Equilibrium means equal rates, not equal amounts of reactants and products.

Reference tables1

Quick revision7

Formulas2

Equilibrium Constant Kc
For aA +bB⇌cC+dD\text{For aA }+ bB ⇌ cC + dD
Kc=[C]c[D]d/[A]a[B]bKc = [C]^c[D]^d / [A]^a[B]^b

Gives concentration-based equilibrium constant for a homogeneous reaction.

[A], [B], [C], [D]
Equilibrium molar concentrations
a, b, c, d
Stoichiometric coefficients
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3

Equilibrium Constant Kc, Kp and Reaction Quotient

9 min

The equilibrium constant measures how far a reversible reaction proceeds before equilibrium is established. Kc uses molar concentrations, while Kp uses partial pressures for gaseous reactions. Their relation depends on the change in moles of gaseous species, Δn. The reaction quotient Q has the same expression as K but uses concentrations or pressures at any instant, not necessarily equilibrium. Comparing Q with K predicts the direction of reaction. This topic is very important for NEET numerical problems involving ICE tables, partial pressure, degree of dissociation, and deciding whether a reaction moves forward or backward.

Key points6

  • Kc and Kp are dimensionless in modern thermodynamics, though units may appear in older numerical treatment.
  • Only gaseous species enter Kp expression.
  • ICE tables help organize initial, change, and equilibrium amounts.
  • For reverse reaction, K_reverse = 1 / K_forward.
  • If a reaction is multiplied by n, new K = K^n.
  • If reactions are added, total K is product of individual K values.

Memory tricks2

Q is Question, K is Known Answer: Compare the current question Q with the equilibrium answer K to decide which direction the reaction must move.

Delta n is Gas Only: For Kp = Kc(RT)^Δn, count only gaseous moles. Ignore solids, liquids, and aqueous species.

Examples2

PCl5 Dissociation

For PCl5(g) ⇌ PCl3(g) + Cl2(g), Δn = 2 - 1 = 1, so Kp = KcRT.

No Change Case

For H2(g) + I2(g) ⇌ 2HI(g), Δn = 0, so Kp = Kc.

Common mistakes2

Counting All Species in Δn

Δn includes only gaseous species. Including solids or liquids gives wrong Kp-Kc relation.

Reversing Q Direction

If Q is less than K, products are deficient, so the reaction goes forward.

Reference tables2

Quick revision8

Formulas3

Kp-Kc Relation
Kp=Kc(RT)ΔnKp = Kc(RT)^Δn

Relates pressure equilibrium constant to concentration equilibrium constant for gaseous reactions.

Kp
Equilibrium constant in terms of partial pressure
Kc
Equilibrium constant in terms of molar concentration
R
Gas constant
T
Absolute temperature in kelvin
Δn
Moles of gaseous products minus gaseous reactants
Reaction Quotient
Qc=[C]c[D]d/[A]a[B]bQc = [C]^c[D]^d / [A]^a[B]^b

Same form as Kc but calculated at any instant to predict reaction direction.

Qc
Reaction quotient using concentration
[A], [B], [C], [D]
Instantaneous molar concentrations
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4

Le Chatelier Principle and Factors Affecting Equilibrium

9 min

Le Chatelier principle states that when a system at equilibrium is disturbed by changing concentration, pressure, volume, or temperature, the system shifts in a direction that reduces the disturbance. It is a prediction tool for equilibrium shifts and industrial optimization. Adding reactant usually shifts equilibrium forward; increasing pressure favours the side with fewer gaseous moles; increasing temperature favours the endothermic direction. A catalyst does not shift equilibrium; it only helps equilibrium reach faster by lowering activation energy of both directions equally. NEET commonly asks conceptual questions on Haber process, Contact process, colour changes, and pressure-volume effects.

Key points6

  • Le Chatelier principle predicts direction, not exact equilibrium composition.
  • Pressure affects only gaseous equilibria with unequal gaseous moles.
  • If Δn_gas = 0, pressure or volume change has no shift effect.
  • Temperature changes can change the numerical value of K.
  • Catalyst lowers activation energy for both forward and reverse reactions equally.
  • Industrial processes use compromise conditions for yield and rate.

Memory tricks2

Equilibrium Pushes Back: Le Chatelier is like a spring: push the system, and it pushes back to reduce the effect.

Heat as a Species: Write heat on product side for exothermic and reactant side for endothermic reactions to predict temperature shifts.

Examples2

Haber Process

N2 + 3H2 ⇌ 2NH3 has fewer gas moles on product side, so high pressure favours ammonia formation.

Brown NO2 Equilibrium

2NO2 ⇌ N2O4 is exothermic in forward direction. Cooling favours colourless N2O4, while heating favours brown NO2.

Common mistakes2

Saying Catalyst Increases Yield

Catalyst increases rate of reaching equilibrium, not equilibrium yield.

Applying Pressure Rule to Liquids

Pressure-volume shifts are significant mainly for gaseous equilibria.

Reference tables2

Quick revision8

Formulas2

Temperature Dependence of K
For exothermic reaction\text{For exothermic reaction}
increase in T decreases K\text{increase in }T\text{ decreases }K

Heat behaves like a product in exothermic reactions, so adding heat shifts equilibrium backward.

T
Absolute temperature
K
Equilibrium constant
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5

Ionic Equilibrium, Electrolytes and Ostwald Dilution Law

9 min

Ionic equilibrium deals with reversible ionization of weak electrolytes in aqueous solution. Strong electrolytes almost completely ionize, while weak electrolytes partially ionize and establish equilibrium between unionized molecules and ions. Degree of ionization, α, represents the fraction ionized and depends on concentration, dilution, temperature, and common ions. Ostwald dilution law relates the dissociation constant of a weak electrolyte with concentration and α. Dilution increases ionization of weak electrolytes, while addition of a common ion suppresses ionization. These concepts form the base for pH, buffers, salt hydrolysis, and solubility product problems in NEET.

Key points6

  • Ionic equilibrium applies to aqueous weak electrolytes.
  • For weak acid HA, Ka = Cα²/(1-α).
  • If α is very small, Ka ≈ Cα².
  • Common ion effect is a direct application of Le Chatelier principle.
  • Ionization increases as concentration decreases.
  • The dissociation constant depends on temperature.

Memory tricks2

Dilution Dissociates: For weak electrolytes, remember D-D: Dilution increases Dissociation.

Common Ion Compresses Ionization: A common ion is like crowding the product side, so the equilibrium shifts back.

Examples2

Acetic Acid Dilution

When CH3COOH is diluted, its percent ionization increases because the equilibrium shifts toward more ions.

Adding Sodium Acetate

Adding CH3COONa to CH3COOH supplies CH3COO- and suppresses acetic acid ionization.

Common mistakes2

Applying Ostwald Law to Strong Electrolytes

Ostwald dilution law is meant for weak electrolytes, not strong acids or salts that ionize almost completely.

Forgetting Approximation Condition

Use 1 - α ≈ 1 only when α is very small, generally less than about 0.05.

Reference tables2

Quick revision7

Formulas3

Degree of Ionization
α= number of molecules ionized / total number of molecules initially presentα =\text{ number of molecules ionized }/\text{ total number of molecules initially present}

Fraction of electrolyte molecules that dissociate into ions.

α
Degree of ionization or dissociation
Ostwald Dilution Law for Weak Acid
Ka=Cα2/(1−α)Ka = Cα² / (1 - α)

Relates acid dissociation constant to concentration and degree of ionization.

Ka
Acid dissociation constant
C
Initial molar concentration
α
Degree of ionization
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6

Acids, Bases, pH, pKa and pKb

10 min

Acids and bases can be understood through Arrhenius, Bronsted-Lowry, and Lewis concepts. Arrhenius acids produce H+ in water and bases produce OH-. Bronsted acids donate protons, while Bronsted bases accept protons. Lewis acids accept electron pairs and Lewis bases donate electron pairs, making this the broadest concept. pH measures hydrogen ion concentration and is crucial for NEET numericals. Strong acids and bases ionize almost completely, whereas weak acids and bases require Ka or Kb. pKa and pKb are logarithmic strength indicators: smaller pKa means stronger acid, and smaller pKb means stronger base.

Key points6

  • Lewis theory includes reactions without proton transfer.
  • Water is amphoteric because it can donate or accept H+.
  • Strong acid pH is calculated directly from concentration if fully ionized.
  • Weak acid pH requires Ka and approximation or quadratic solution.
  • pH scale is logarithmic: one unit change means tenfold change in [H+].
  • At 298 K, neutral solution has pH = 7 because [H+] = [OH-] = 10^-7 M.

Memory tricks3

BL: Base Likes H+: In Bronsted-Lowry theory, a base likes and accepts H+, while acid donates H+.

Lewis Lends Pair: Lewis base lends an electron pair; Lewis acid accepts it.

pH Power: Each pH unit is a power of 10 change in hydrogen ion concentration.

Examples2

Strong Acid pH

For 0.01 M HCl, [H+] = 10^-2 M, so pH = 2.

Lewis Acid Example

BF3 acts as a Lewis acid because boron has an incomplete octet and accepts an electron pair from NH3.

Common mistakes2

Forgetting Logarithmic Scale

pH 3 is ten times more acidic than pH 4, not just one unit more acidic.

Using pH + pOH = 14 at All Temperatures

pH + pOH = 14 strictly applies at 298 K because Kw changes with temperature.

Reference tables2

Quick revision8

Formulas5

pH
pH=−log[H+]pH = -log[H+]

Measures acidity of a solution using hydrogen ion concentration.

pH
Negative logarithm of hydrogen ion concentration
[H+]
Molar concentration of hydrogen ions
pOH
pOH=−log[OH−]pOH = -log[OH-]

Measures basicity using hydroxide ion concentration.

pOH
Negative logarithm of hydroxide ion concentration
[OH-]
Molar concentration of hydroxide ions
Ionic Product of Water
Kw=[H+][OH−]Kw = [H+][OH-]

At 298 K, Kw = 1.0 × 10^-14.

Kw
Ionic product of water
[H+]
Hydrogen ion concentration
[OH-]
Hydroxide ion concentration
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7

Hydrolysis of Salts and pH of Salt Solutions

8 min

Salt hydrolysis occurs when ions of a salt react with water to produce H+ or OH-, making the solution acidic or basic. Salts of strong acid and strong base do not hydrolyse appreciably and give neutral solution. Salts of strong acid and weak base produce acidic solutions because the cation hydrolyses. Salts of weak acid and strong base produce basic solutions because the anion hydrolyses. Salts of weak acid and weak base require comparison of Ka and Kb. NEET often asks direct pH nature, hydrolysis constant, and relation between hydrolysis and conjugate acid-base strength.

Key points6

  • Hydrolysis is reverse of neutralization for salt ions.
  • Conjugate base of weak acid is basic and hydrolyses.
  • Conjugate acid of weak base is acidic and hydrolyses.
  • NaCl solution is neutral because Na+ and Cl- do not hydrolyse significantly.
  • NH4Cl solution is acidic due to NH4+ hydrolysis.
  • CH3COONa solution is basic due to acetate ion hydrolysis.

Memory tricks2

Strong Parents, Neutral Child: A salt from strong acid and strong base is neutral because both parent ions are too weak to hydrolyse.

Weak Acid Salt is Basic: If the acid parent is weak, its conjugate base is strong enough to take H+ from water, giving OH-.

Examples2

NH4Cl is Acidic

NH4+ donates H+ to water to form H3O+, so ammonium chloride solution has pH less than 7.

Sodium Acetate is Basic

CH3COO- accepts H+ from water, leaving OH-, so sodium acetate solution has pH greater than 7.

Common mistakes2

Calling Every Salt Neutral

Only salts of strong acid and strong base are neutral. Many salts hydrolyse and change pH.

Ignoring Ka versus Kb for Weak-Weak Salts

For weak acid-weak base salts, compare Ka and Kb to decide acidic, basic, or neutral nature.

Reference tables1

Quick revision7

Formulas4

Hydrolysis Constant for Salt of Weak Acid and Strong Base
Kh=Kw/KaKh = Kw / Ka

Used for anion hydrolysis such as CH3COO- in sodium acetate.

Kh
Hydrolysis constant
Kw
Ionic product of water
Ka
Dissociation constant of weak acid
Hydrolysis Constant for Salt of Strong Acid and Weak Base
Kh=Kw/KbKh = Kw / Kb

Used for cation hydrolysis such as NH4+ in ammonium chloride.

Kh
Hydrolysis constant
Kw
Ionic product of water
Kb
Dissociation constant of weak base
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8

Buffer Solutions and Henderson Equation

8 min

A buffer solution resists change in pH when small amounts of acid or base are added. Acidic buffers contain a weak acid and its salt with a strong base, such as CH3COOH and CH3COONa. Basic buffers contain a weak base and its salt with a strong acid, such as NH4OH and NH4Cl. Buffer action occurs due to the common ion effect: added H+ or OH- is consumed by buffer components. Henderson-Hasselbalch equations allow quick pH calculation. Buffers are important in biological systems like blood, in analytical chemistry, and in NEET problems involving pKa, pKb, and concentration ratios.

Key points6

  • Buffers resist pH change but do not keep pH absolutely constant under unlimited acid or base.
  • The salt supplies conjugate ion needed to neutralize added acid or base.
  • Dilution changes buffer capacity but pH changes only slightly if ratio remains same.
  • Blood buffer mainly involves carbonic acid and bicarbonate ions.
  • Henderson equation uses concentration ratio, so units cancel.
  • A good buffer has pKa close to desired pH.

Memory tricks2

Buffer Has Both Shield and Sponge: The weak acid/base pair acts like a shield: one part absorbs H+, the other absorbs OH-.

Salt Over Acid: For acidic buffer, remember pH has salt over acid: pH = pKa + log S/A.

Examples2

Acetate Buffer

A mixture of CH3COOH and CH3COONa maintains pH near the pKa of acetic acid.

Blood Buffer

The H2CO3/HCO3- system helps maintain blood pH around 7.4.

Common mistakes2

Using Strong Acid and Strong Base as Buffer

A buffer needs a weak acid/base and its conjugate salt. Strong acid-base mixtures do not provide buffer action.

Forgetting pOH in Basic Buffer

For basic buffer, first calculate pOH, then convert using pH = 14 - pOH at 298 K.

Reference tables1

Quick revision7

Formulas2

Henderson Equation for Acidic Buffer
pH=pKa+log([salt]/[acid])pH = pKa + log([\text{salt}] / [\text{acid}])

Calculates pH of weak acid and conjugate base buffer.

pH
Acidity measure of buffer
pKa
Negative logarithm of acid dissociation constant
[salt]
Concentration of conjugate base salt
[acid]
Concentration of weak acid
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9

Solubility Product, Precipitation and Common Ion Effect

9 min

Solubility product, Ksp, is the equilibrium constant for dissolution of a sparingly soluble salt in water. When a salt such as AgCl dissolves, a dynamic equilibrium exists between solid salt and its ions in solution. The ionic product is calculated using actual ion concentrations at any instant. If ionic product is less than Ksp, no precipitate forms; if equal, the solution is saturated; if greater, precipitation occurs. Common ion effect decreases solubility by shifting dissolution equilibrium backward. This topic is highly important for NEET problems on precipitation, salt solubility, group separation, and comparing Ksp values.

Key points6

  • Ksp is a constant at a given temperature.
  • Molar solubility must be converted using stoichiometry before writing Ksp.
  • For AB salt, Ksp = s² if solubility is s.
  • For AB2 salt, Ksp = 4s³.
  • Selective precipitation depends on different Ksp values.
  • Adding a common ion shifts equilibrium toward undissolved solid.

Memory tricks2

Qsp Decides Precipitate: Think of Qsp as current ion crowding. If crowding exceeds Ksp capacity, the salt precipitates.

Common Ion Crashes Solubility: Adding an ion already present pushes the dissolved salt back into solid form.

Examples2

AgCl Precipitation

Mixing AgNO3 and NaCl gives Ag+ and Cl-. If [Ag+][Cl-] exceeds Ksp of AgCl, a white precipitate forms.

Group Analysis

Selective precipitation of metal ions is based on differences in Ksp values of their salts.

Common mistakes2

Comparing Ksp Without Stoichiometry

You can directly compare solubility using Ksp only for salts with the same ion ratio.

Forgetting Powers in Ksp

For AB2, [B-] = 2s, so Ksp = s(2s)^2 = 4s^3, not s^3.

Reference tables2

Quick revision8

Formulas3

Ksp for AB Type Salt
AB(s)⇌A+(aq)+B−(aq)AB(s) ⇌ A+(aq) + B-(aq)
Ksp=[A+][B−]=s2Ksp = [A+][B-] = s²

Used when one mole of salt gives one cation and one anion.

Ksp
Solubility product constant
s
Molar solubility of AB
Ksp for AB2 Type Salt
AB2(s)⇌A2+(aq)+2B−(aq)AB2(s) ⇌ A2+(aq) + 2B-(aq)
Ksp=[A2+][B−]2=4s3Ksp = [A2+][B-]^2 = 4s³

Stoichiometry must be included when ions are produced in unequal numbers.

Ksp
Solubility product constant
s
Molar solubility of AB2
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10

Formula Sheet, Quick Revision and NEET Mind Map

6 min

This topic collects the most exam-oriented formulas, decision rules, and NCERT concepts from the complete Equilibrium chapter. For NEET, students must rapidly connect physical equilibrium, chemical equilibrium, Le Chatelier principle, ionic equilibrium, pH calculations, buffer equations, hydrolysis, and solubility product. Most errors occur when students use initial concentration instead of equilibrium concentration, include solids in K expressions, forget Δn is only for gases, or apply pH formulas outside their conditions. The mind map helps revise the chapter as one connected system: equilibrium constants describe position, Q predicts direction, Le Chatelier predicts shift, and ionic equilibria explain pH, buffers, and precipitation.

Key points6

  • Pure solids and liquids are excluded from K, Q, and Ksp expressions.
  • K changes only with temperature, not with catalyst or concentration.
  • Strong electrolytes ionize almost completely; weak electrolytes use equilibrium constants.
  • Common ion effect suppresses ionization and solubility.
  • For salt hydrolysis, identify parent acid and base first.
  • For buffer problems, identify whether pH or pOH equation is needed.

Memory tricks2

K-Q-L-pH-Ksp Chain: Remember the chapter flow as K tells position, Q tells direction, Le Chatelier tells shift, pH tells acidity, and Ksp tells precipitation.

Products Over Reactants: Most equilibrium expressions begin with the same habit: products on top, reactants below, powers from coefficients.

Examples3

Fast Direction Check

If for a reaction K = 10 and Q = 2, products are deficient, so the reaction proceeds forward.

Fast Buffer Check

If [salt] = [acid] in an acidic buffer, pH = pKa because log 1 = 0.

Fast Ksp Check

If Ksp of AgCl is 1.8 × 10^-10 and [Ag+][Cl-] is 10^-8, precipitation occurs because Qsp > Ksp.

Common mistakes3

Mixing K and Q

K is only at equilibrium; Q is at any instant. Use Q for direction prediction.

Not Checking Approximation

Weak acid/base approximations should be verified when ionization is not very small.

Forgetting Temperature Dependence

Only temperature changes the equilibrium constant. Concentration and pressure change position but not K.

Reference tables2

Quick revision8

Formulas7

General Kc
Kc=[products](coefficients)/[reactants](coefficients)Kc = [\text{products}]^(\text{coefficients}) / [\text{reactants}]^(\text{coefficients})

Universal structure for concentration equilibrium constant.

Kc
Concentration equilibrium constant
Kp-Kc
Kp=Kc(RT)ΔnKp = Kc(RT)^Δn

Use only for gaseous reactions.

Δn
Gaseous product moles minus gaseous reactant moles
Water Ion Product
Kw=[H+][OH−]=10−14at298KKw = [H+][OH-] = 10^-14 at 298 K

Central relation for pH and pOH calculations.

Kw
Ionic product of water
pH Relations
pH=−log[H+]pH = -log[H+]
pOH=−log[OH−]pOH = -log[OH-]
pH+pOH=14at298KpH + pOH = 14 at 298 K

Most used acid-base formulas in NEET.

[H+]
Hydrogen ion concentration
[OH-]
Hydroxide ion concentration
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NEET PYQs from Equilibrium

The latest 3 of 53 previous-year questions from this chapter.

  1. NEET 2026Set 11MediumQ1

    At 298 K, a certain buffer solution contains equal concentrations of X⁻ and HX. K$_b$ for X⁻ is 10⁻¹⁰. What is the pH of this buffer solution?

    • A2
    • B4
    • C6
    • D10
  2. NEET 2026Set 50MediumQ2

    The correct order of solubility of the given salts in water at 298 K is Salt | $K_{sp}$ at 298 K $\mathrm{AgBr}$ | $5.0 \times 10^{-13}$ $\mathrm{Zn(OH)_2}$ | $1.0 \times 10^{-15}$ $\mathrm{Hg_2Cl_2}$ | $1.3 \times 10^{-18}$

    • A$\mathrm{Hg_2Cl_2 > Zn(OH)_2 > AgBr}$
    • B$\mathrm{AgBr > Zn(OH)_2 > Hg_2Cl_2}$
    • C$\mathrm{Hg_2Cl_2 > AgBr > Zn(OH)_2}$
    • D$\mathrm{Zn(OH)_2 > AgBr > Hg_2Cl_2}$
  3. NEET 2026Set 11EasyQ3

    Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change observed at alkaline pH close to the equivalence point is:

    • APinkish red to yellow
    • BYellow to pinkish red
    • CPink to colourless
    • DColourless to pink

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